Quantum Entanglement and Bell’s Theorem
Numerical Problems & Answers
1 PROBLEMS AND SOLUTIONS
1.1 PROBLEM 1
Two entangled particles are prepared in the singlet state. Alice measures the spin of her particle
along the z-axis, while Bob measures his particle’s spin along an axis that makes an angle of
30° with the z-axis. What is the probability that they obtain the same result?
Solution:
1. The probability of obtaining the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. In this case, 𝜃 = 30°.
3. 𝑃(same)=sin2(30°/2)=sin2(15°)
4. sin(15°)≈ 0.2588
5. 𝑃(same)≈(0.2588)2≈ 0.0670 or about 6.70%
1.2 PROBLEM 2
In a Bell test experiment, Alice and Bob each have three measurement settings (a, b, c). The
following correlation values are obtained:
𝐸(𝑎, 𝑏)= −0.5
𝐸(𝑎, 𝑐)= −0.7
𝐸(𝑏, 𝑐)= −0.6
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑐)| +|𝐸(𝑏, 𝑐)+ 𝐸(𝑏′,𝑐)|, where 𝑏′ is
the complement of b.
2. In this case, we can use 𝑏 as 𝑏′ since we only have three settings.
3. 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑐)| +|𝐸(𝑏, 𝑐)+ 𝐸(𝑏, 𝑐)|
4. 𝑆 = |−0.5 − (−0.7)| +|−0.6 + (−0.6)|
5. 𝑆 = |0.2|+|−1.2|= 0.2 + 1.2 = 1.4
1.3 PROBLEM 3
Alice and Bob share an ensemble of entangled particle pairs. Alice measures her particles along
directions 𝑎 and 𝑎′, while Bob measures along 𝑏
and 𝑏
′. The angles between these vectors are:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |−0.9239 − (−0.3827)| +|−0.3827 + 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.4 PROBLEM 4
In a Bell test experiment, the following probabilities are measured:
𝑃(+1, +1|𝑎, 𝑏)= 0.4
𝑃(−1, −1|𝑎, 𝑏)= 0.4
𝑃(+1, −1|𝑎, 𝑏)= 0.1
𝑃(−1, +1|𝑎, 𝑏)= 0.1
Calculate the correlation E(a,b).
Solution:
1. The correlation E(a,b) is given by:
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
𝐸(𝑎, 𝑏)= 𝑃(+1, +1|𝑎, 𝑏)+ 𝑃(−1, −1|𝑎, 𝑏)− 𝑃(+1, −1|𝑎, 𝑏)− 𝑃(−1, +1|𝑎, 𝑏)
2. Substituting the given probabilities:
𝐸(𝑎, 𝑏)= 0.4 + 0.4 − 0.1 − 0.1
3. 𝐸(𝑎, 𝑏)= 0.8 − 0.2 = 0.6
1.5 PROBLEM 5
Two entangled qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{3}}(\ket{00} + \ket{01} + \ket{10})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{11}$ component.
3. Therefore, the probability of measuring both qubits in state $\ket{1}$ is 0.
1.6 PROBLEM 6
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along three different axes: 𝑎, 𝑏
, and 𝑐, where:
∠(𝑎,𝑏
)=120°
∠(𝑏
,𝑐)=120°
∠(𝑐, 𝑎)=120°
Calculate the quantum mechanical prediction for the sum of correlations 𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+
𝐸(𝑐, 𝑎).
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. For each pair of directions:
𝐸(𝑎,𝑏
)= 𝐸(𝑏
, 𝑐)= 𝐸(𝑐, 𝑎)= −cos(120°)
3. cos(120°)= −0.5
4. So, 𝐸(𝑎,𝑏
)= 𝐸(𝑏
,𝑐)= 𝐸(𝑐, 𝑎)= −(−0.5)= 0.5
5. The sum of correlations is:
𝐸(𝑎,𝑏
)+ 𝐸(𝑏
,𝑐)+ 𝐸(𝑐, 𝑎)= 0.5 + 0.5 + 0.5 = 1.5
1.7 PROBLEM 7
In a Bell test experiment, Alice and Bob share 1000 entangled particle pairs. They measure their
particles’ spins along axes that are 60° apart. How many times do they expect to get the same
result (both +1 or both -1)?
Solution:
1. The probability of getting the same result is given by 𝑃(same)=sin2(𝜃/2), where 𝜃 is
the angle between the measurement axes.
2. 𝑃(same)=sin2(60°/2)=sin2(30°)= 0.25
3. The expected number of same results is:
𝑁(same)=1000 × 0.25 =250
1.8 PROBLEM 8
Two qubits are prepared in the Bell state:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$$
Alice measures her qubit in the basis $\{\ket{+}, \ket{-}\}$, where $\ket{+} =
\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$.
What is the probability that Bob’s qubit will be in the state $\ket{+}$ if he measures in the
same basis?
Solution:
1. First, rewrite the Bell state in the $\{\ket{+}, \ket{-}\}$ basis:
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{++} + \ket{--})$$
2. If Alice measures $\ket{+}$, Bob’s qubit will be in the state $\ket{+}$.
3. If Alice measures $\ket{-}$, Bob’s qubit will be in the state $\ket{-}$.
4. The probability of Alice measuring $\ket{+}$ is 0.5.
5. Therefore, the probability of Bob’s qubit being in the state $\ket{+}$ is also 0.5.
1.9 PROBLEM 9
In a Bell test experiment, Alice and Bob measure their particles along axes 𝑎 and 𝑏
,
respectively. The angle between 𝑎 and 𝑏
is 45°. What is the quantum mechanical prediction for
the correlation E(a,b)?
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑎,𝑏
)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. In this case, 𝜃 = 45°.
3. 𝐸(𝑎, 𝑏
)= −cos(45°)
4. cos(45°)=1
√2≈ 0.7071
5. 𝐸(𝑎, 𝑏
)= −0.7071
1.10 PROBLEM 10
Alice and Bob share 10,000 entangled particle pairs. They measure their particles’ spins along
axes that are 30° apart. In how many cases do they expect to get opposite results (one +1 and
one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(30°/2)= cos2(15°)≈ 0.9330
3. The expected number of opposite results is:
𝑁(opposite)=10,000 × 0.9330 = 9,330
1.11 PROBLEM 11
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{5}}(2\ket{00} + \ket{11})$$
What is the probability of measuring the first qubit in state $\ket{0}$ and the second qubit in
state $\ket{1}$?
Solution:
1. The probability of measuring $\ket{01}$ is given by the square of the amplitude of the
$\ket{01}$ component in the state $\ket{\psi}$.
2. In the given state, there is no $\ket{01}$ component.
3. Therefore, the probability of measuring the first qubit in state $\ket{0}$ and the second
qubit in state $\ket{1}$ is 0.
1.12 PROBLEM 12
In a Bell test experiment, the following correlations are measured:
𝐸(𝑎, 𝑏)= −0.6
𝐸(𝑎, 𝑏′)= −0.8
𝐸(𝑎′,𝑏)= −0.7
𝐸(𝑎′, 𝑏′)= 0.5
Calculate the value of the CHSH inequality parameter S.
Solution:
1. The CHSH inequality is given by 𝑆 = |𝐸(𝑎, 𝑏)− 𝐸(𝑎, 𝑏′)| +|𝐸(𝑎′, 𝑏)+ 𝐸(𝑎′,𝑏′)|
2. Substituting the given correlations:
𝑆 = |(−0.6)−(−0.8)| +|(−0.7)+ 0.5|
3. 𝑆 = |0.2|+|−0.2|= 0.2 + 0.2 = 0.4
1.13 PROBLEM 13
Alice and Bob share a large number of entangled particle pairs. They measure their particles’
spins along four different axes: 𝑎, 𝑎′, 𝑏
, and 𝑏
′, where:
∠(𝑎,𝑏
)=22.5°
∠(𝑎, 𝑏
′)=67.5°
∠(𝑎′,𝑏
)=67.5°
∠(𝑎′, 𝑏
′)=112.5°
Calculate the quantum mechanical prediction for the CHSH inequality parameter S.
Solution:
1. The quantum correlation for two particles is given by 𝐸(𝑥, 𝑦)= −cos(𝜃), where 𝜃 is the
angle between the measurement directions.
2. Calculate each correlation:
𝐸(𝑎,𝑏
)= −cos(22.5°)≈ −0.9239
𝐸(𝑎,𝑏
′)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
)= −cos(67.5°)≈ −0.3827
𝐸(𝑎′,𝑏
′)= −cos(112.5°)≈ 0.3827
3. The CHSH parameter is given by 𝑆 = |𝐸(𝑎, 𝑏
)− 𝐸(𝑎,𝑏
′)| +|𝐸(𝑎′, 𝑏
)+ 𝐸(𝑎′,𝑏
′)|
4. 𝑆 = |(−0.9239)−(−0.3827)| +|(−0.3827)+ 0.3827|
5. 𝑆 = |−0.5412|+|0|= 0.5412 + 0 = 0.5412
6. 𝑆 = 2√2 ≈ 2.8284
1.14 PROBLEM 14
Two qubits are prepared in the state:
$$\ket{\psi} = \frac{1}{\sqrt{10}}(2\ket{00} + \ket{01} + 2\ket{10} + \ket{11})$$
What is the probability of measuring both qubits in the state $\ket{1}$?
Solution:
1. The probability of measuring both qubits in state $\ket{1}$ is given by the square of the
amplitude of the $\ket{11}$ component in the state $\ket{\psi}$.
2. In the given state, the amplitude of $\ket{11}$ is 1
√10.
3. Therefore, the probability is:
$$P(\ket{11}) = \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} = 0.1$$
4. The probability of measuring both qubits in state $\ket{1}$ is 0.1 or 10%.
1.15 PROBLEM 15
In a Bell test experiment, Alice and Bob share 5000 entangled particle pairs. They measure their
particles’ spins along axes that are 45° apart. How many times do they expect to get opposite
results (one +1 and one -1)?
Solution:
1. The probability of getting opposite results is given by 𝑃(opposite)= cos2(𝜃/2), where 𝜃
is the angle between the measurement axes.
2. 𝑃(opposite)= cos2(45°/2)= cos2(22.5°)≈ 0.8536
3. The expected number of opposite results is:
𝑁(opposite)=5000 × 0.8536 =4268
4. They expect to get opposite results approximately 4268 times.
