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QUANTUM MECHANICS PROBLEMS AND
DETAILED SOLUTIONS FOR PRACTICE
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
2. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
3. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
4. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
5. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
6. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
7. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
1. Evaluate Expectation Value of Position
⟨𝑥⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
⟨𝑥⟩=∫ 𝑥
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥⟩=𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑥⟩=0
2. Evaluate Expectation Value of Momentum
⟨𝑝⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
⟨𝑝⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−𝑖ℏ 𝜕
𝜕𝑥)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜓∗(𝑥)(−𝑖ℏ𝜕𝜓(𝑥)
𝜕𝑥 )=−𝑖ℏ(−2𝛼𝑥𝐴2𝑒−2𝛼𝑥2)
⟨𝑝⟩=2𝑖ℏ𝛼𝐴2∫ 𝑥
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
Since 𝑥𝑒−2𝛼𝑥2 is an odd function
⟨𝑝⟩=0
3. Evaluate Uncertainty in Position 𝛥𝑥 for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝛥𝑥=√⟨𝑥2⟩−⟨𝑥⟩2
⟨𝑥⟩=0
⟨𝑥2⟩=∫ 𝑥2
∞
−∞ |𝜓(𝑥)|2 𝑑𝑥
|𝜓(𝑥)|2=𝐴2𝑒−2𝛼𝑥2
⟨𝑥2⟩=𝐴2∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=𝐴2(√𝜋
4𝛼3/2)
𝛥𝑥=√√𝜋
4𝛼3/2
4. Evaluate Expectation Value of the Hamiltonian
⟨𝐻⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution
𝐻= 𝑝2
2𝑚+1
2𝑚𝜔2𝑥2
⟨𝐻⟩=∫ 𝜓∗
∞
−∞ (𝑥)𝐻𝜓(𝑥) 𝑑𝑥
⟨𝑝2
2𝑚⟩=∫ 𝜓∗
∞
−∞ (𝑥)(−ℏ2
2𝑚 𝜕2
𝜕𝑥2)𝜓(𝑥) 𝑑𝑥
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑝2
2𝑚⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑝2
2𝑚⟩=ℏ2𝛼
2𝑚
⟨1
2𝑚𝜔2𝑥2⟩=1
2𝑚𝜔2⟨𝑥2⟩
=1
2𝑚𝜔21
2𝛼
=𝑚𝜔2
4𝛼
⟨𝐻⟩=ℏ2𝛼
2𝑚+𝑚𝜔2
4𝛼
Problem 5: Evaluate Expectation Value of Kinetic Energy
⟨𝑇⟩ for 𝜓(𝑥)=𝐴𝑒−𝛼𝑥2
Solution:
𝑇= 𝑝2
2𝑚
⟨𝑇⟩=∫ 𝜓∗
∞
−∞ (𝑥)(𝑝2
2𝑚)𝜓(𝑥) 𝑑𝑥
𝑝2
2𝑚=−ℏ2
2𝑚 𝜕2
𝜕𝑥2
𝜕𝜓(𝑥)
𝜕𝑥 =−2𝛼𝑥𝐴𝑒−𝛼𝑥2
𝜕2𝜓(𝑥)
𝜕𝑥2=−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2
−ℏ2
2𝑚(−2𝛼𝐴𝑒−𝛼𝑥2+4𝛼2𝑥2𝐴𝑒−𝛼𝑥2)
=ℏ2𝛼𝐴
𝑚𝑒−𝛼𝑥2−2ℏ2𝛼2𝑥2𝐴
𝑚𝑒−𝛼𝑥2
⟨𝑇⟩=ℏ2𝛼𝐴2
𝑚∫ 𝑒−2𝛼𝑥2
∞
−∞  𝑑𝑥−2ℏ2𝛼2𝐴2
𝑚∫ 𝑥2
∞
−∞ 𝑒−2𝛼𝑥2 𝑑𝑥
=ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼 −ℏ2𝛼𝐴2√𝜋
𝑚√2𝛼
⟨𝑇⟩=ℏ2𝛼
2𝑚
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