PHYS 305 - INTRODUCTION TO
MODERN PHYSICS - Wave functions
Question Bank - Set 1
Liberty University
Question 1
Question
Consider a particle in a one-dimensional box of length L. The wave function of
the particle is given by ψ(x) = Ax(L−x) for 0 < x < L and ψ(x) = 0 otherwise.
Determine the normalization constant A.
Solution
Step 1: Normalize the wave function by using the condition R∞
−∞ |ψ(x)|2dx = 1.
Step 2: Since the wave function is zero outside of the box, we only need to in-
tegrate over the range 0 < x < L. Step 3: The normalization integral becomes
RL
0|Ax(L−x)|2dx = 1. Step 4: Simplify the integral to RL
0A2x2(L−x)2dx = 1.
Step 5: Expand the expression under the integral to get RL
0A2(L2x2−2Lx3+
x4)dx = 1. Step 6: Integrate each term separately to get A21
3L2x3−1
2Lx4+1
5x5
L
0=
1. Step 7: Evaluate the definite integral at the limits to obtain A21
3L5−1
2L5+1
5L5=
1. Step 8: Simplify the expression to get A21
30 L5= 1. Step 9: Solve for Ato
obtain A=q30
L5.
Question 2
Question
Consider a particle confined to the region 0 ≤x≤ain one dimension. The
wave function of the particle is given by ψ(x) = Asin nπx
a, where Ais a
normalization constant and nis a positive integer. If the probability of finding
the particle in the range 0 ≤x≤a
4is 1
3, determine the value of n.
Solution
Step 1: Normalize the wave function. Given that the particle is confined to the
region 0 ≤x≤a, we can set up the normalization condition as follows:
Za
0|ψ(x)|2dx = 1
Substitute the given wave function:
Za
0|Asin nπx
a|2dx = 1
Simplify the integral:
A2Za
0|sin nπx
a|2dx = 1
A2Za
0
sin2nπx
adx = 1
A2Za
0
1−cos 2nπx
a
2dx = 1
Use the fact that Rcos(mx)dx =1
msin(mx):
A2x
2−a
2nπ sin 2nπx
aa
0
= 1
A2a
2−0= 1
A2·a
2= 1
A=r2
a
Step 2: Calculate the probability of finding the particle in the range 0 ≤x≤
a
4. Given that the probability of finding the particle in the range 0 ≤x≤a
4is
1
3, we can set up the probability condition as follows:
Za
4
0|ψ(x)|2dx =1
3
Substitute the wave function and the normalization constant:
Za
4
0r2
asin nπx
a
2
dx =1
3
Simplify the integral:
2
aZa
4
0
sin2nπx
adx =1
3
2
aZa
4
0
1−cos 2nπx
a
2dx =1
3
Continue with the integration to find n.
2
Question 3
Question
Let Ψ(x) = Asin(kx) be a wave function representing a particle in a 1D box
of length L. Given that the probability density function |Ψ(x)|2is normalized,
determine the normalization constant Ain terms of kand L.
Solution
To find the normalization constant A, we need to ensure that the probability of
finding the particle in the entire box is 1. This means we need to normalize the
wave function Ψ(x) such that
ZL
0|Ψ(x)|2dx = 1.
Step 1: Find the square of the wave function The square of the wave
function is given by |Ψ(x)|2=A2sin2(kx).
Step 2: Integrate the probability density function over the entire
box
ZL
0
A2sin2(kx)dx = 1.
Step 3: Solve the integral
ZL
0
A2sin2(kx)dx =A2ZL
0
1−cos(2kx)
2dx.
=A2
2x−sin(2kx)
2kL
0
.
=A2
2L−sin(2kL)
2k.
Step 4: Set the result equal to 1 and solve for A
A2
2L−sin(2kL)
2k= 1.
A2=2
L−sin(2kL)
2k
.
A=s2
L−sin(2kL)
2k
.
Therefore, the normalization constant Ain terms of kand Lis A=q2
L−sin(2kL)
2k
in order for the given wave function to be normalized.
3
Question 4
Question
Consider a particle in a one-dimensional infinite potential well of width L. The
wave function of the particle is given by Ψ(x) = Asin nπx
L, where Ais a
normalization constant and nis a positive integer.
Determine the probability density P(x) of finding the particle at a position
xwithin the well.
Solution
1. First, we need to normalize the wave function Ψ(x) by finding the nor-
malization constant A. The normalization condition is given by:
ZL
0|Ψ(x)|2dx = 1
2. Substituting the given wave function into the normalization condition:
ZL
0|Asin nπx
L|2dx = 1
3. Solving the integral to find A:
ZL
0
A2sin2nπx
Ldx = 1
4. Using the trigonometric identity sin2θ=1−cos(2θ)
2:
A2x
2−L
4nπ sin 2nπx
LL
0
= 1
5. Simplifying and solving for A:
AL
2= 1 ⇒A=2
L
6. With the normalized wave function Ψ(x) = 2
Lsin nπx
L, the probability
density P(x) is given by:
P(x) = |Ψ(x)|2=2
Lsin nπx
L2
=4
L2sin2nπx
L
7. Therefore, the probability density of finding the particle at a position x
within the well is P(x) = 4
L2sin2nπx
L.
4
Question 5
Question
Consider a particle in a one-dimensional infinite square well potential with
boundaries at x= 0 and x=a. The wave function for this particle is given by:
Ψ(x) = Asin nπx
a
where Ais a normalization constant and nis a positive integer.
Determine the probability of finding the particle in the interval 0 ≤x≤a
2.
Solution
To determine the probability of finding the particle in the interval 0 ≤x≤a
2,
we need to calculate the probability density function |Ψ(x)|2and integrate it
over this interval.
Step 1: Find the normalization constant A.The normalization condi-
tion for the wave function is:
Za
0|Ψ(x)|2dx = 1
Substitute the given wave function into the normalization condition:
Za
0|Asin nπx
a|2dx = 1
Solve for Aby evaluating the integral:
Za
0
A2sin2nπx
adx = 1
Za
0
A2 1−cos 2nπx
a
2!dx = 1
A2x
2−a
2nπ sin 2nπx
aa
0
= 1
A2a
2= 1
A=r2
a
Step 2: Calculate the probability density function |Ψ(x)|2.The prob-
ability density function is given by |Ψ(x)|2=|Asin nπx
a|2=A2sin2nπx
a.
Therefore, |Ψ(x)|2=2
asin2nπx
a.
5
Step 3: Determine the probability of finding the particle in the
interval 0≤x≤a2.The probability of finding the particle in the interval
0≤x≤a
2is given by the integral:
P=Za
2
0|Ψ(x)|2dx
Substitute the expression for |Ψ(x)|2:
P=Za
2
0
2
asin2nπx
adx
Solve this integral to find the probability.
Question 6
Question
Consider the wave function Ψ(x) = Asin(kx +ϕ) representing a particle in a
one-dimensional box of length L. If this wave function satisfies the boundary
conditions Ψ(0) = 0 and Ψ(L) = 0, determine the values of kthat are allowed.
Solution
Step 1: Apply the boundary condition Ψ(0) = 0.
Ψ(0) = Asin(0 + ϕ) = Asin(ϕ)=0
Since sin(ϕ) = 0 when ϕ=nπ for n∈Z, we have ϕ=nπ.
Step 2: Apply the boundary condition Ψ(L) = 0.
Ψ(L) = Asin(kL +nπ) = 0
For the sine function to be zero, we must have kL+nπ =mπ for m∈Z. Solving
for k, we get k=(m−n)
L.
Step 3: Determine the allowed values of k. Since the particle is in a box
of length L, we require the wave function to be periodic in this region. The
condition for periodicity is that the wave function must be an integer multiple
of 2πin that range. Thus, kshould be quantized:
k=(m−n)
L=2π
LN
where Nis an integer. This means that the allowed values of kare k=2π
LN
with N∈Z.
6
Question 7
Question
Let ψ(x) = Asin(kx) + Bcos(kx) be a wave function where A,B, and kare
constants. If ψ(0) = 0 and ψ(π
2) = A, determine the values of Aand B.
Solution
Step 1: Evaluate ψ(0) using the given wave function:
ψ(0) = Asin(0) + Bcos(0)
ψ(0) = A·0 + B·1
ψ(0) = B
Step 2: Since ψ(0) = 0, we have:
B= 0
Step 3: Evaluate ψ(π
2) using the wave function:
ψ(π
2) = Asinπ
2
Step 4: Since ψ(π
2) = A, we have:
A=Asinπ
2
Step 5: This implies that sinπ
2= 1, so we get:
A=A·1
Step 6: Therefore, A=A.
Step 7: Since Ais a constant, this equation holds for all values of A. Thus,
we cannot determine a unique value for A.
In summary, the value of Acan be any real number, but Bmust be equal
to 0 in order for the wave function to satisfy the given conditions.
Question 8
Question
Consider the following wave function for a particle in a one-dimensional box of
length L:
ψ(x) = Asin 2πx
L+Bsin 4πx
L
where Aand Bare normalization constants.
Determine the values of Aand Bthat normalize the wave function.
7
Solution
To normalize the wave function, we need to ensure that the integral of the abso-
lute square of the wave function over all space is equal to 1. This represents the
conservation of probability - the probability of finding the particle somewhere
in the box should be 100
Step 1: Set up the normalization integral
The normalization integral is given by:
ZL
0|ψ(x)|2dx = 1
Substitute the given wave function into the integral:
ZL
0
Asin 2πx
L+Bsin 4πx
L
2
dx = 1
Expand the square of the absolute value and simplify the expression.
Step 2: Solve the integral
The integral becomes:
ZL
0A2sin22πx
L+B2sin24πx
L+ 2AB sin 2πx
Lsin 4πx
Ldx = 1
Now, solve for Aand Bby integrating.
Step 3: Set the integral equal to 1
After integrating, you should get an expression in terms of Aand B. Set
this expression equal to 1 and solve for Aand Bby imposing the normalization
condition.
Step 4: Find the normalization constants
By solving the integral and setting it equal to 1, you will obtain equations
involving Aand B. Solve these equations simultaneously to find the appropriate
values of Aand Bthat normalize the wave function.
Question 9
Question
Let ψ(x) = Asin kx+Bcos kx be a wave function describing a particle in a one-
dimensional box of length L. If the particle is in the ground state, determine
the normalization constant Ain terms of Land k.
8
Solution
Step 1: Normalize the wave function ψ(x) by requiring RL
0|ψ(x)|2dx = 1.
ZL
0|ψ(x)|2dx =ZL
0
(Asin kx +Bcos kx)2dx
=ZL
0
(A2sin2kx + 2AB sin kx cos kx +B2cos2kx)dx
Step 2: Use trigonometric identities sin2θ=1−cos 2θ
2and cos2θ=1+cos 2θ
2
to simplify the integral.
=ZL
0A21−cos 2kx
2+ 2AB sin kx cos kx +B21 + cos 2kx
2dx
=A2
2ZL
0
(1 −cos 2kx)dx +AB ZL
0
2 sin kx cos kxdx +B2
2ZL
0
(1 + cos 2kx)dx
Step 3: Evaluate the integrals.
=A2L
2−A2
2ksin 2kL +AB sin2kx
kL
0
+B2L
2+B2
2ksin 2kL
=A2L
2−A2
2ksin 2kL +AB sin2kL
k−0+B2L
2+B2
2ksin 2kL
=A2L
2−A2
2ksin 2kL +AB
k+B2L
2+B2
2ksin 2kL
Step 4: Set the integral equal to 1 and simplify to find Ain terms of B,L,
and k.A2L
2+B2L
2+AB
k= 1
A=s2
L1−B2L
2−B
k
Question 10
Question
For a particle in a one-dimensional box of length L, the wave function is given
by ψ(x) = Asin 2πx
L, where Ais a normalization constant.
Determine the probability that a measurement of the position of the particle
will yield a value between L
4and L
2.
9
Solution
Step 1: Normalize the wave function. The normalization condition for the wave
function is R∞
−∞ |ψ(x)|2dx = 1. In this case, since the particle is in a one-
dimensional box of length L, we need to consider the integral over that range:
ZL
0|Asin 2πx
L|2dx = 1
−A2ZL
0
sin22πx
Ldx = 1
−A2ZL
0
1−cos 4πx
L
2dx = 1
−A2
2x−L
4πsin 4πx
LL
0
= 1
−A2
2L−L
4πsin (4π)−0+0= 1
−A2
2L= 1
A=r2
L
Step 2: Find the probability that the measurement will yield a value between
L
4and L
2. The probability of finding the particle between x=aand x=bis
given by:
P(a<x<b) = Zb
a|ψ(x)|2dx
PL
4<x<L
2=ZL
2
L
4 r2
Lsin 2πx
L!2
dx
PL
4<x<L
2=ZL
2
L
4
2
Lsin22πx
Ldx
PL
4<x<L
2=2
LZL
2
L
4
1−cos 4πx
L
2dx
PL
4<x<L
2=1
LZL
2
L
4
1−cos 4πx
Ldx
PL
4<x<L
2=1
Lx−L
4πsin 4πx
LL
2
L
4
PL
4<x<L
2=1
LL
2−L
4πsin (2π)−L
4+L
4πsin (π)
10
PL
4<x<L
2=1
L[
*Question 11
Question
Let ψ(x) = Ax2+Bx +Cbe a wave function describing a particle in one
dimension. Determine the normalization constant Aassuming the particle is
confined to the interval −1≤x≤1.
Solution
Step 1: The normalization condition for a wave function ψ(x) over a region
−L≤x≤Lis given by:
ZL
−L|ψ(x)|2dx = 1
Step 2: Substituting ψ(x) = Ax2+Bx +Cinto the normalization condition
and solving for Agives:
Z1
−1|Ax2+Bx +C|2dx = 1
Step 3: Computing the integral in Step 2 gives:
Z1
−1|Ax2+Bx +C|2dx =Z1
−1|Ax2+Bx +C|2dx = 1
Step 4: Expand the square inside the absolute value and integrate term by
term:
Z1
−1
(A2x4+ 2ABx3+ (B2+ 2AC)x2+ 2BCx +C2)dx = 1
Step 5: Integrate each term separately:
A2Z1
−1
x4dx+2AB Z1
−1
x3dx+(B2+2AC)Z1
−1
x2dx+2BC Z1
−1
x dx+C2Z1
−1
dx = 1
Step 6: Evaluate the integrals and simplify:
A22
5+ 2AB (0) + (B2+ 2AC)2
3+ 2BC (0) + C2(2) = 1
Step 7: Since the integral is equal to 1, we have:
2
5A2+2
3(B2+ 2AC)+2C2= 1
Step 8: We can simplify this equation and solve for Ain terms of Band C:
2
5A2+2
3B2+4
3AC + 2C2= 1
11
Question 12
Question
Let ψ(x) = Ae−kx +Bekx be the wave function of a particle in a one-dimensional
box of length L. Determine the values of Aand Bsuch that the wave function
satisfies the boundary conditions ψ(0) = 0 and ψ(L) = 0.
Solution
Given wave function ψ(x) = Ae−kx +Bekx.
Step 1: Apply the boundary condition ψ(0) = 0
ψ(0) = A−B= 0
A=B
Step 2: Substitute A=Bback into the wave function
ψ(x) = Ae−kx +Aekx =A(e−kx +ekx)
Step 3: Apply the boundary condition ψ(L) = 0
ψ(L) = A(e−kL +ekL)=0
Step 4: Find the values of Aand Bthat satisfy the boundary
conditions Since e−kL +ekL = 0, the only way for ψ(L) = 0 is if A= 0.
Step 5: Conclusion The wave function that satisfies the boundary condi-
tions ψ(0) = 0 and ψ(L) = 0 is given by ψ(x) = 0.
Question 13
Question
Consider a particle confined to a 1-dimensional box of length L. The wave
function of the particle inside the box is given by:
ψ(x) = Asinnπx
L
where Ais a normalization constant and nis a positive integer. Determine the
probability of finding the particle in the region L
4<x<3L
4.
12
Solution
Step 1: Normalize the wave function ψ(x).
ZL
0|ψ(x)|2dx = 1
ZL
0|Asinnπx
L|2dx = 1
A2ZL
0
sin2(nπx
L)dx = 1
A2ZL
0
1−cos2nπx
L
2dx = 1
A2[1
2x−L
2nπ sin2nπx
L]L
0= 1
A2[1
2L] = 1
A2=2
L
A=r2
L
Step 2: Calculate the probability of finding the particle in the region L
4<
x < 3L
4.
P=Z3L
4
L
4|ψ(x)|2dx
=Z3L
4
L
4r2
Lsin nπx
L
2
dx
=Z3L
4
L
4
2
Lsin2nπx
Ldx
=2
LZ3L
4
L
4
1−cos 2nπx
L
2dx
=1
Lx−L
2nπ sin 2nπx
L3L
4
L
4
=1
LL
2−L
2nπ sin 3nπ
2−L
4+L
2nπ sin nπ
2
=1
LL
2+L
4
=3
4
13
Therefore, the probability of finding the particle in the region L
4<x<3L
4
is 3
4.
Question 14
Question
Let ψ(x) be a wave function given by ψ(x) = Asin(kx)+Bcos(kx), where A, B,
and kare constants. Determine the normalization condition for ψ(x).
Solution
To normalize the wave function ψ(x), we need to ensure that the integral of
|ψ(x)|2over all space is equal to 1. Mathematically, the normalization condition
can be expressed as:
Z∞
−∞ |ψ(x)|2dx = 1
Step 1: Calculate |ψ(x)|2
|ψ(x)|2= (ψ(x))∗·ψ(x)=(Asin(kx) + Bcos(kx))∗·(Asin(kx) + Bcos(kx))
= (Asin(kx) + Bcos(kx))(Asin(kx) + Bcos(kx))
=A2sin2(kx) + B2cos2(kx)+2AB sin(kx) cos(kx)
Step 2: Integrate |ψ(x)|2over all space
Z∞
−∞ |ψ(x)|2dx =Z∞
−∞
(A2sin2(kx) + B2cos2(kx)+2AB sin(kx) cos(kx))dx
=A2Z∞
−∞
sin2(kx)dx +B2Z∞
−∞
cos2(kx)dx + 2AB Z∞
−∞
sin(kx) cos(kx)dx
Step 3: Simplify the integrals and apply the normalization condi-
tion Since sin2(kx) and cos2(kx) have average values of 1
2over a full period, and
sin(kx) cos(kx) integrates to zero over a full period, the normalization condition
simplifies to:
A21
2+B21
2= 1
A2
2+B2
2= 1
A2+B2= 2
Therefore, the normalization condition for the wave function ψ(x) is A2+
B2= 2.
14
Question 15
Question
Consider a wave function given as ψ(x) = Ae−bx2, where Aand bare constants.
Find the normalization constant Afor this wave function.
Solution
Step 1: Determine the normalization condition for the wave function ψ(x). For
ψ(x) to be a valid wave function, it must satisfy the normalization condition:
Z∞
−∞ |ψ(x)|2dx = 1
Step 2: Substitute the given wave function into the normalization condition.
Z∞
−∞ |Ae−bx2|2dx = 1
Z∞
−∞
A2e−2bx2dx = 1
Step 3: Simplify the integral and solve for A.
A2Z∞
−∞
e−2bx2dx = 1
A2Z∞
−∞
e−(√2x)2dx = 1
A2rπ
2= 1
A2=1
pπ
2
=r2
π
Step 4: Determine the value of A.
A=r2
π
Therefore, the normalization constant for the wave function ψ(x) = Ae−bx2
is A=q2
π.
Question 16
Question
Given a wave function Ψ(x) = Asin(kx) + Bcos(kx), where A,B, and kare
constants, determine the normalization constant Nfor the wave function.
15
Solution
Step 1: Normalize the wave function by finding Nsuch that R∞
−∞ |Ψ(x)|2dx = 1.
Step 2: Calculate |Ψ(x)|2= Ψ(x)·Ψ(x).
|Ψ(x)|2= (Asin(kx) + Bcos(kx))(Asin(kx) + Bcos(kx))
Step 3: Expand the expression and simplify.
|Ψ(x)|2=A2sin2(kx) + B2cos2(kx)+2AB sin(kx) cos(kx)
Step 4: Use trigonometric identities to simplify further.
|Ψ(x)|2=A2sin2(kx) + B2cos2(kx) + AB sin(2kx)
Step 5: Integrate |Ψ(x)|2from −∞ to ∞.
Z∞
−∞ |Ψ(x)|2dx =Z∞
−∞
(A2sin2(kx) + B2cos2(kx) + AB sin(2kx))dx
Step 6: Evaluate the integral for each term to obtain an equation for N.
Step 7: Finally, solve for the normalization constant Nto complete the
normalization of the wave function.
Question 17
Question
Consider a wave function given by ψ(x) = Asin(kx)+Bcos(kx), where Aand B
are constants and kis the wave number. Determine the normalization constant
Ain terms of B.
Solution
Given wave function: ψ(x) = Asin(kx) + Bcos(kx)
To determine the normalization constant A, we need to normalize the wave
function by ensuring that
Z∞
−∞ |ψ(x)|2dx = 1
Step 1: Compute |ψ(x)|2
|ψ(x)|2= (Asin(kx) + Bcos(kx))2
=A2sin2(kx) + B2cos2(kx)+2AB sin(kx) cos(kx)
Step 2: Evaluate the integral
Z∞
−∞ |ψ(x)|2dx =Z∞
−∞
(A2sin2(kx) + B2cos2(kx)+2AB sin(kx) cos(kx))dx
16
Since the sine and cosine functions are orthogonal, the integral of their prod-
uct over one period is 0. Therefore, the only term that contributes to the integral
is the constant term.
Z∞
−∞ |ψ(x)|2dx =A2Z∞
−∞
sin2(kx)dx +B2Z∞
−∞
cos2(kx)dx
The integrals of sin2(kx) and cos2(kx) over one period are both equal to π,
so:
A2π+B2π= 1
Step 3: Solve for A
A2π+B2π= 1
A2+B2=1
π
A=r1
π−B2
Thus, the normalization constant Ain terms of Bis A=q1
π−B2.
Question 18
Question
Consider a particle in a one-dimensional box of length L. The wave function of
the particle is given by:
ψ(x) = Ax(L−x)
where Ais a normalization constant.
Determine the normalization constant Afor the wave function ψ(x).
Solution
Step 1: Normalize the wave function by integrating |ψ(x)|2over the entire range
of x.
ZL
0|ψ(x)|2dx =ZL
0|Ax(L−x)|2dx
Step 2: Simplify the integral by plugging in the given wave function ψ(x).
ZL
0
A2x2(L−x)2dx
Step 3: Expand and simplify the integrand.
ZL
0
A2(x4−2Lx3+L2x2)dx
17
Step 4: Integrate each term separately.
ZL
0
A2x4−2A2Lx3+A2L2x2dx
=A2x5
5−2Lx4
4+L2x3
3
L
0
Step 5: Evaluate the integral over the range 0 to L.
A2L5
5−2L5
4+L5
3
Step 6: Set the result equal to 1 (the normalization condition) and solve for
A.
A2L5
5−2L5
4+L5
3= 1
Step 7: Simplify and solve for A.
A2L5
5−2A2L5
4+A2L5
3= 1
3A2L5−10A2L5+ 15A2L5
60 = 1
8A2L5
60 = 1
A2=15
4L5
A=r15
4L5
Therefore, the normalization constant Afor the wave function ψ(x) is A=
q15
4L5.
Question 19
Question
Let ψ(x) = Ax2e−x
2abe a wave function representing a particle in a one-
dimensional potential well. Find the normalization constant A.
18
Solution
1. We need to normalize the wave function, which means we must have
R∞
−∞ |ψ(x)|2dx = 1.
2. First, let’s square the wave function: |ψ(x)|2=|Ax2e−x
2a|2=A2x4e−x
a.
3. Next, we can rewrite the integral as: R∞
−∞ A2x4e−x
adx = 1.
4. We can simplify the integral by factoring out the constant A2:A2R∞
0x4e−x
adx =
1.
5. We will now solve the integral R∞
0x4e−x
adx.
6. Let’s make the substitution u=x
a,dx =adu, which gives us: A2R∞
0(au)4e−uadu =
1.
7. Simplifying, we get A2a5R∞
0u4e−udu = 1.
8. The integral now becomes A2a5·4! = 1, where 4! represents the factorial
of 4.
9. Hence, A2a5·24 = 1, and solving for Agives us A=1
2√6a5/2.
Question 20
Question
Consider a wave function ψ(x) = A·e−αx2, where Aand αare constants.
Determine the normalization constant A.
Solution
To normalize the wave function ψ(x), we must ensure that the total probability
of finding the particle in all space is equal to 1.
Step 1: Calculate the normalization condition. The normalization condition
is given by:
Z∞
−∞ |ψ(x)|2dx = 1
Step 2: Substitute the given wave function into the normalization condition.
Z∞
−∞ |A·e−αx2|2dx = 1
Z∞
−∞ |A|2· |e−2αx2|dx = 1
Z∞
−∞ |A|2·e−2αx2dx = 1
19
Step 3: Solve the integral. To solve the integral, we use the fact that
R∞
−∞ e−ax2dx =pπ
afor a > 0. Therefore, we have:
|A|2·rπ
2α= 1
|A|2=1
pπ/(2α)
A=r2α
π
Therefore, the normalization constant Ais q2α
π.
Question 21
Question
Consider a particle in a 1-dimensional box of length L. The particle is in the
ground state with wave function ψ(x) = Asin(πx/L), where Ais a normalization
constant. Determine the probability of finding the particle in the interval 0 <
x < L/4.
Solution
Step 1: Normalize the wave function. Step 2: Find the probability of finding
the particle in the interval 0 < x < L/4 using the normalized wave function.
Step 1: Normalize the wave function.
The normalization condition states that the integral of the absolute square
of the wave function over all space is equal to 1:
Z∞
−∞ |ψ(x)|2dx = 1
For our wave function ψ(x) = Asin(πx/L), the normalization condition
becomes:
ZL
0
A2sin2(πx/L)dx = 1
A2ZL
0
sin2(πx/L)dx = 1
A2ZL
0
1−cos(2πx/L)
2dx = 1
A2x
2−L
2πsin(2πx/L)
L
0
= 1
20
A2L
2−L
2πsin(2π)= 1
A2L
2= 1
A2=2
L
A=r2
L
Therefore, the normalized wave function is ψ(x) = q2
Lsin(πx/L).
Step 2: Find the probability of finding the particle in the interval 0 < x <
L/4 using the normalized wave function.
The probability of finding the particle in the interval 0 < x < L/4 is given
by the integral of the absolute square of the wave function over that interval:
P=ZL/4
0|ψ(x)|2dx
P=ZL/4
0 r2
Lsin πx
L!2
dx
P=ZL/4
0
2
Lsin2πx
Ldx
Using the angle addition formula sin2(u) = 1−cos(2u)
2, we have:
P=ZL/4
0
1−cos 2πx
L
Ldx
P=x
L−L
2πsin 2πx
L
L/4
0
P=L
4L−L
2πsin π
2−[0 −0]
P=1
4−L
2π
Therefore, the probability of finding the particle in the interval 0 < x < L/4
is 1
4−L
2π.
21
Question 22
Question
Consider a wave function Ψ(x) = Asin(kx) + Bcos(kx), where A,B, and kare
constants. Determine the normalization constant Nsuch that R∞
−∞ |Ψ(x)|2dx =
1.
Solution
Step 1: Compute |Ψ(x)|2.
|Ψ(x)|2=|Asin(kx) + Bcos(kx)|2
= (Asin(kx) + Bcos(kx)) (Asin(kx) + Bcos(kx))
=A2sin2(kx)+2AB sin(kx) cos(kx) + B2cos2(kx)
=A2(1 −cos2(kx)) + 2AB sin(kx) cos(kx) + B2cos2(kx)
=A2+B2+ 2AB sin(kx) cos(kx)
=A2+B2+AB sin(2kx)
Step 2: Integrate |Ψ(x)|2from −∞ to ∞.
Z∞
−∞ |Ψ(x)|2dx =Z∞
−∞
(A2+B2+AB sin(2kx)) dx
=A2x+B2x−AB
2kcos(2kx)∞
−∞
= lim
a→∞ A2a+B2a−AB
2kcos(2ka)−(A2(−a) + B2(−a)−AB
2kcos(−2ka))
= lim
a→∞ A2a+B2a−AB
2k(cos(2ka)−cos(−2ka)) + A2a+B2a−AB
2k(cos(−2ka)−cos(2ka))
= lim
a→∞ 2(A2+B2)a
=∞
Since the integral diverges, there is no normalization constant Nthat can
make the wave function Ψ(x) normalized.
Question 23
Question
Let f(x) be a wave function of a particle in one dimension. Given that the
wave function is normalized, i.e., R∞
−∞ |f(x)|2dx = 1, determine whether the
following functions are valid wave functions:
1. g(x) = e2ix
2. h(x) = sin(3x) + cos(4x)
22
Solution
To determine whether a function is a valid wave function, we need to check if
it satisfies the normalization condition:
Z∞
−∞ |f(x)|2dx = 1
1. For function g(x) = e2ix:
Z∞
−∞ |g(x)|2dx =Z∞
−∞ |e2ix|2dx
=Z∞
−∞
e2ixe−2ix dx
=Z∞
−∞
1dx
=∞ = 1
Since the integral diverges, g(x) = e2ix is not a valid wave function.
2. For function h(x) = sin(3x) + cos(4x):
Z∞
−∞ |h(x)|2dx =Z∞
−∞ |sin(3x) + cos(4x)|2dx
=Z∞
−∞
(sin2(3x) + 2 sin(3x) cos(4x) + cos2(4x)) dx
=Z∞
−∞
(1 + sin(3x+ 4x)) dx
=Z∞
−∞
1dx +Z∞
−∞
sin(7x)dx
=∞ = 1
Since the integral diverges, h(x) = sin(3x) + cos(4x) is not a valid wave
function.
Therefore, neither g(x) = e2ix nor h(x) = sin(3x) + cos(4x) are valid wave
functions.
Question 24
Question
Consider the following wave function for a particle in a one-dimensional box of
length L:
ψ(x) = Asin2nπx
L
23
where Ais a normalization constant.
What is the probability of finding the particle in the interval 0 ≤x≤L
4for
the ground state (n= 1) of the particle?
Solution
Step 1: Normalize the wave function.
ZL
0|ψ(x)|2dx = 1
A2ZL
0
sin4πx
Ldx = 1
A23L
8= 1
A=r8
3L
Step 2: Find the probability of finding the particle in the interval 0 ≤x≤L
4.
P=ZL
4
0|ψ(x)|2dx
P=ZL
4
0 r8
3Lsin2πx
L!2
dx
P=8
3LZL
4
0
sin4πx
Ldx
Step 3: Use the trigonometric identity sin2(θ) = 1−cos(2θ)
2.
P=8
3LZL
4
0
1−cos 2πx
L
2dx
P=8
6Lx
2−L
2πsin 2πx
L
L
4
0
P=1
3
Therefore, the probability of finding the particle in the interval 0 ≤x≤L
4
for the ground state of the particle is 1
3.
24
Question 25
Question
Consider a particle in a one-dimensional box of length L. The wave function
of the particle is given by Ψ(x) = Asin nπx
L, where Ais a normalization
constant. Determine the value of nfor which the probability of finding the
particle between 0.2Land 0.3Lis maximum.
Solution
Step 1: Normalize the wave function. The normalization condition for a wave
function Ψ(x) in a one-dimensional box is given by
ZL
0|Ψ(x)|2dx = 1
Given Ψ(x) = Asin nπx
L, we have
ZL
0|Asin nπx
L|2dx = 1
Solve this integral, set it equal to 1, and solve for A.
Step 2: Find the probability of finding the particle between 0.2Land 0.3L.
The probability of finding the particle between 0.2Land 0.3Lis given by
P=Z0.3L
0.2L|Ψ(x)|2dx
Substitute the normalized wave function in Pand simplify the integral.
Step 3: Maximize the probability. To maximize the probability, differentiate
Pwith respect to n, set it equal to zero, and solve for n.
Question 26
Question
Find the general solution to the time-independent Schr¨odinger equation for a
particle confined to move in a one-dimensional box of length L, with boundary
conditions Ψ(0) = Ψ(L) = 0.
Solution
To find the general solution to the Schr¨odinger equation, we start with the
time-independent Schr¨odinger equation given by:
−ℏ2
2m
d2Ψ
dx2+V(x)Ψ = EΨ,
25
where Eis the total energy of the particle, ℏis the reduced Planck constant, m
is the mass of the particle, V(x) is the potential energy function, and Ψ(x) is
the wave function.
The potential energy function inside the box is zero, so V(x) = 0 for 0 <
x<L. Therefore, the Schr¨odinger equation simplifies to:
−ℏ2
2m
d2Ψ
dx2=EΨ.
Step 1: Solving the differential equation Let’s solve the differential
equation above to find the general solution for Ψ(x). To do this, we assume
Ψ(x) = Asin(kx) + Bcos(kx), where Aand Bare constants to be determined,
and k=q2mE
ℏ2.
Taking the second derivative of Ψ(x) and substituting it into the Schr¨odinger
equation, we get:
−ℏ2
2m(−Ak2sin(kx)−Bk2cos(kx)) = E(Asin(kx) + Bcos(kx))
Step 2: Applying boundary conditions Since we have boundary condi-
tions Ψ(0) = Ψ(L) = 0 for a particle in a well, we can use these to determine
the constants Aand B.
Substituting x= 0 into Ψ(x) gives Ψ(0) = B= 0.
Substituting x=Linto Ψ(x) gives Ψ(L) = Asin(kL) = 0. This implies
either A= 0 or sin(kL) = 0. If A= 0, then the wave function is trivial.
Therefore, we must have sin(kL) = 0, which gives us kL =nπ, where nis a
positive integer.
Therefore, the general solution for the wave function Ψ(x) is given by:
Ψn(x) = Asin nπx
L,
where nis a positive integer.
Thus, we have found the general solution for the time-independent Schr¨odinger
equation for a particle in a one-dimensional box.
Question 27
Question
Find the normalized wave function for a particle in a one-dimensional box of
length L. The potential inside the box is zero while the potential outside is
infinite.
Solution
To find the normalized wave function for a particle in a one-dimensional box,
we will follow the steps below:
26
Step 1: Express the general solution of the Schr¨odinger equation
The general solution for the wave function inside the box is given by:
ψ(x) = Asin(kx) + Bcos(kx)
where k=q2mE
ℏ.
Step 2: Apply boundary conditions At x= 0, |ψ(0)|2must be finite.
Therefore, B= 0. So, ψ(x) = Asin(kx).
Step 3: Apply the second boundary condition at x=LAt x=L,
ψ(L) = 0 since the potential outside the box is infinite.
Asin(kL)=0
This implies that kL =nπ for n= 1,2,3, . . ..
Step 4: Determine the energy eigenvalues Since kL =nπ, we have
k=nπ
L. Substitute this into the expression for kto get the energy eigenvalues:
E=n2π2ℏ2
2mL2
Step 5: Normalize the wave function Since the particle must be found
somewhere in the box, the normalized wave function is:
ψ(x) = r2
Lsin nπx
L
Question 28
Question
Let f(x) = (cx if 0 ≤x≤1
0 otherwise be a wave function. Determine the value of c
that makes f(x) a valid wave function.
Solution
Step 1: To find the value of c, we need to ensure that the wave function f(x)
satisfies the normalization condition:
Z∞
−∞ |f(x)|2dx = 1
Step 2: First, let’s determine the interval over which f(x) is non-zero. Since
f(x) is only non-zero when 0 ≤x≤1, the integral simplifies to:
Z1
0|f(x)|2dx
27
Step 3: Substituting the expression for f(x) into the integral, we get:
Z1
0|cx|2dx =Z1
0
c2x2dx
Step 4: Solving the integral, we find:
c2Z1
0
x2dx =c2x3
31
0
=c2
3
Step 5: Set the integral equal to 1 and solve for c:
c2
3= 1
c2= 3
c=√3
Therefore, the value of cthat makes f(x) a valid wave function is √3.
Question 29
Question
Consider a wave function in one dimension given by Ψ(x) = Asin(kx)+Bcos(kx),
where Aand Bare constants. Determine the normalization constant Ain terms
of B.
Solution
To normalize the wave function, we require that the integral of the square of
the wave function over all space is equal to 1, i.e., R∞
−∞ |Ψ(x)|2dx = 1.
Step 1: Calculate |Ψ(x)|2using the given wave function.
|Ψ(x)|2= (Asin(kx)+Bcos(kx))2=A2sin2(kx)+B2cos2(kx)+2AB sin(kx) cos(kx)
Step 2: Integrate |Ψ(x)|2from −∞ to ∞to normalize the wave function.
Z∞
−∞ |Ψ(x)|2dx =Z∞
−∞
(A2sin2(kx) + B2cos2(kx)+2AB sin(kx) cos(kx))dx
Step 3: Recognize that the integral of the sine and cosine terms over a full
period is zero.
Z∞
−∞
B2cos2(kx)dx =B2Z∞
−∞
cos2(kx)dx =B21
2Z∞
−∞
(1+cos(2kx))dx =B21
2(∞+∞) = ∞
28
Step 4: Set up the integral for A2sin2(kx) term to evaluate.
Z∞
−∞
A2sin2(kx)dx =A2Z∞
−∞
sin2(kx)dx =A21
2Z∞
−∞
(1 −cos(2kx))dx
Step 5: Recognize that the integral of the cosine term over a full period is
zero, so we have
A21
2(∞−∞) = A21
2·0=0
Step 6: The integral of the cross term 2AB sin(kx) cos(kx) over a full period
is also zero.
Z∞
−∞
2AB sin(kx) cos(kx)dx = 2AB Z∞
−∞
sin(kx) cos(kx)dx = 0
Step 7: Combining all the terms and setting the result to 1 to normalize
the wave function:
∞+ 0 + 0 = 1
As ∞ = 1, the wave function cannot be normalized.
Question 30
Question
Let f(x) be a wave function defined on the interval [−π, π] such that f(x) is an
odd function and satisfies the property f(0) = 0. If f(x) can be expressed as
f(x) = ∞
X
n=1
bnsin(nx),
where bn=2
πRπ
0f(x) sin(nx)dx, find the constant b1.
Solution
Step 1: Since f(x) is an odd function, f(x) = −f(−x) for all xin the interval
[−π, π]. Therefore, we have
f(x) = −f(−x).
Given that f(x) satisfies the property f(0) = 0, we can write
f(0) = −f(0)
0 = −f(0)
f(0) = 0.
29
Step 2: We now substitute f(x) into the formula for bnand simplify. We
have
bn=2
πZπ
0
f(x) sin(nx)dx =2
πZπ
0 ∞
X
m=1
bmsin(mx)!sin(nx)dx.
Using the orthogonality property of sine functions, we can simplify the integral
to
bn=2
πZπ
0
bnsin2(nx)dx.
Step 3: Further simplifying the integral, we have
bn=2
πbnZπ
0
1−cos(2nx)
2dx.
bn=1
πbnx−sin(2nx)
2n
π
0
.
Plugging in the limits of integration and simplifying, we get
bn=1
πbnπ−sin(2nπ)
2n−0.
bn=bn1−sin(2nπ)
2nπ .
Step 4: Since sin(2nπ) = 0 for all integer values of n, we have
bn=bn
1=1.
Thus, bncancels out from the equation.
Step 5: Finally, we find the expression for b1. Substituting n= 1 into the
expression derived in Step 3, we have
b1=b11−sin(2π)
2π.
Since sin(2π) = 0, we get
b1=b1·1
b1=b1.
Therefore, the constant b1is equal to itself, which means b1can take any
value.
30
Question 3
Question
Let Ψ(x) = Asin(kx) be a wave function representing a particle in a 1D box
of length L. Given that the probability density function |Ψ(x)|2is normalized,
determine the normalization constant Ain terms of kand L.
Solution
To find the normalization constant A, we need to ensure that the probability of
finding the particle in the entire box is 1. This means we need to normalize the
wave function Ψ(x) such that
ZL
0|Ψ(x)|2dx = 1.
Step 1: Find the square of the wave function The square of the wave
function is given by |Ψ(x)|2=A2sin2(kx).
Step 2: Integrate the probability density function over the entire
box
ZL
0
A2sin2(kx)dx = 1.
Step 3: Solve the integral
ZL
0
A2sin2(kx)dx =A2ZL
0
1−cos(2kx)
2dx.
=A2
2x−sin(2kx)
2kL
0
.
=A2
2L−sin(2kL)
2k.
Step 4: Set the result equal to 1 and solve for A
A2
2L−sin(2kL)
2k= 1.
A2=2
L−sin(2kL)
2k
.
A=s2
L−sin(2kL)
2k
.
Therefore, the normalization constant Ain terms of kand Lis A=q2
L−sin(2kL)
2k
in order for the given wave function to be normalized.
3
Question 4
Question
Consider a particle in a one-dimensional infinite potential well of width L. The
wave function of the particle is given by Ψ(x) = Asin nπx
L, where Ais a
normalization constant and nis a positive integer.
Determine the probability density P(x) of finding the particle at a position
xwithin the well.
Solution
1. First, we need to normalize the wave function Ψ(x) by finding the nor-
malization constant A. The normalization condition is given by:
ZL
0|Ψ(x)|2dx = 1
2. Substituting the given wave function into the normalization condition:
ZL
0|Asin nπx
L|2dx = 1
3. Solving the integral to find A:
ZL
0
A2sin2nπx
Ldx = 1
4. Using the trigonometric identity sin2θ=1−cos(2θ)
2:
A2x
2−L
4nπ sin 2nπx
LL
0
= 1
5. Simplifying and solving for A:
AL
2= 1 ⇒A=2
L
6. With the normalized wave function Ψ(x) = 2
Lsin nπx
L, the probability
density P(x) is given by:
P(x) = |Ψ(x)|2=2
Lsin nπx
L2
=4
L2sin2nπx
L
7. Therefore, the probability density of finding the particle at a position x
within the well is P(x) = 4
L2sin2nπx
L.
4
Question 5
Question
Consider a particle in a one-dimensional infinite square well potential with
boundaries at x= 0 and x=a. The wave function for this particle is given by:
Ψ(x) = Asin nπx
a
where Ais a normalization constant and nis a positive integer.
Determine the probability of finding the particle in the interval 0 ≤x≤a
2.
Solution
To determine the probability of finding the particle in the interval 0 ≤x≤a
2,
we need to calculate the probability density function |Ψ(x)|2and integrate it
over this interval.
Step 1: Find the normalization constant A.The normalization condi-
tion for the wave function is:
Za
0|Ψ(x)|2dx = 1
Substitute the given wave function into the normalization condition:
Za
0|Asin nπx
a|2dx = 1
Solve for Aby evaluating the integral:
Za
0
A2sin2nπx
adx = 1
Za
0
A2 1−cos 2nπx
a
2!dx = 1
A2x
2−a
2nπ sin 2nπx
aa
0
= 1
A2a
2= 1
A=r2
a
Step 2: Calculate the probability density function |Ψ(x)|2.The prob-
ability density function is given by |Ψ(x)|2=|Asin nπx
a|2=A2sin2nπx
a.
Therefore, |Ψ(x)|2=2
asin2nπx
a.
5
Step 3: Determine the probability of finding the particle in the
interval 0≤x≤a2.The probability of finding the particle in the interval
0≤x≤a
2is given by the integral:
P=Za
2
0|Ψ(x)|2dx
Substitute the expression for |Ψ(x)|2:
P=Za
2
0
2
asin2nπx
adx
Solve this integral to find the probability.
Question 6
Question
Consider the wave function Ψ(x) = Asin(kx +ϕ) representing a particle in a
one-dimensional box of length L. If this wave function satisfies the boundary
conditions Ψ(0) = 0 and Ψ(L) = 0, determine the values of kthat are allowed.
Solution
Step 1: Apply the boundary condition Ψ(0) = 0.
Ψ(0) = Asin(0 + ϕ) = Asin(ϕ)=0
Since sin(ϕ) = 0 when ϕ=nπ for n∈Z, we have ϕ=nπ.
Step 2: Apply the boundary condition Ψ(L) = 0.
Ψ(L) = Asin(kL +nπ) = 0
For the sine function to be zero, we must have kL+nπ =mπ for m∈Z. Solving
for k, we get k=(m−n)
L.
Step 3: Determine the allowed values of k. Since the particle is in a box
of length L, we require the wave function to be periodic in this region. The
condition for periodicity is that the wave function must be an integer multiple
of 2πin that range. Thus, kshould be quantized:
k=(m−n)
L=2π
LN
where Nis an integer. This means that the allowed values of kare k=2π
LN
with N∈Z.
6
Question 7
Question
Let ψ(x) = Asin(kx) + Bcos(kx) be a wave function where A,B, and kare
constants. If ψ(0) = 0 and ψ(π
2) = A, determine the values of Aand B.
Solution
Step 1: Evaluate ψ(0) using the given wave function:
ψ(0) = Asin(0) + Bcos(0)
ψ(0) = A·0 + B·1
ψ(0) = B
Step 2: Since ψ(0) = 0, we have:
B= 0
Step 3: Evaluate ψ(π
2) using the wave function:
ψ(π
2) = Asinπ
2
Step 4: Since ψ(π
2) = A, we have:
A=Asinπ
2
Step 5: This implies that sinπ
2= 1, so we get:
A=A·1
Step 6: Therefore, A=A.
Step 7: Since Ais a constant, this equation holds for all values of A. Thus,
we cannot determine a unique value for A.
In summary, the value of Acan be any real number, but Bmust be equal
to 0 in order for the wave function to satisfy the given conditions.
Question 8
Question
Consider the following wave function for a particle in a one-dimensional box of
length L:
ψ(x) = Asin 2πx
L+Bsin 4πx
L
where Aand Bare normalization constants.
Determine the values of Aand Bthat normalize the wave function.
7
Solution
To normalize the wave function, we need to ensure that the integral of the abso-
lute square of the wave function over all space is equal to 1. This represents the
conservation of probability - the probability of finding the particle somewhere
in the box should be 100
Step 1: Set up the normalization integral
The normalization integral is given by:
ZL
0|ψ(x)|2dx = 1
Substitute the given wave function into the integral:
ZL
0
Asin 2πx
L+Bsin 4πx
L
2
dx = 1
Expand the square of the absolute value and simplify the expression.
Step 2: Solve the integral
The integral becomes:
ZL
0A2sin22πx
L+B2sin24πx
L+ 2AB sin 2πx
Lsin 4πx
Ldx = 1
Now, solve for Aand Bby integrating.
Step 3: Set the integral equal to 1
After integrating, you should get an expression in terms of Aand B. Set
this expression equal to 1 and solve for Aand Bby imposing the normalization
condition.
Step 4: Find the normalization constants
By solving the integral and setting it equal to 1, you will obtain equations
involving Aand B. Solve these equations simultaneously to find the appropriate
values of Aand Bthat normalize the wave function.
Question 9
Question
Let ψ(x) = Asin kx+Bcos kx be a wave function describing a particle in a one-
dimensional box of length L. If the particle is in the ground state, determine
the normalization constant Ain terms of Land k.
8
Solution
Step 1: Normalize the wave function ψ(x) by requiring RL
0|ψ(x)|2dx = 1.
ZL
0|ψ(x)|2dx =ZL
0
(Asin kx +Bcos kx)2dx
=ZL
0
(A2sin2kx + 2AB sin kx cos kx +B2cos2kx)dx
Step 2: Use trigonometric identities sin2θ=1−cos 2θ
2and cos2θ=1+cos 2θ
2
to simplify the integral.
=ZL
0A21−cos 2kx
2+ 2AB sin kx cos kx +B21 + cos 2kx
2dx
=A2
2ZL
0
(1 −cos 2kx)dx +AB ZL
0
2 sin kx cos kxdx +B2
2ZL
0
(1 + cos 2kx)dx
Step 3: Evaluate the integrals.
=A2L
2−A2
2ksin 2kL +AB sin2kx
kL
0
+B2L
2+B2
2ksin 2kL
=A2L
2−A2
2ksin 2kL +AB sin2kL
k−0+B2L
2+B2
2ksin 2kL
=A2L
2−A2
2ksin 2kL +AB
k+B2L
2+B2
2ksin 2kL
Step 4: Set the integral equal to 1 and simplify to find Ain terms of B,L,
and k.A2L
2+B2L
2+AB
k= 1
A=s2
L1−B2L
2−B
k
Question 10
Question
For a particle in a one-dimensional box of length L, the wave function is given
by ψ(x) = Asin 2πx
L, where Ais a normalization constant.
Determine the probability that a measurement of the position of the particle
will yield a value between L
4and L
2.
9
Solution
Step 1: Normalize the wave function. The normalization condition for the wave
function is R∞
−∞ |ψ(x)|2dx = 1. In this case, since the particle is in a one-
dimensional box of length L, we need to consider the integral over that range:
ZL
0|Asin 2πx
L|2dx = 1
−A2ZL
0
sin22πx
Ldx = 1
−A2ZL
0
1−cos 4πx
L
2dx = 1
−A2
2x−L
4πsin 4πx
LL
0
= 1
−A2
2L−L
4πsin (4π)−0+0= 1
−A2
2L= 1
A=r2
L
Step 2: Find the probability that the measurement will yield a value between
L
4and L
2. The probability of finding the particle between x=aand x=bis
given by:
P(a<x<b) = Zb
a|ψ(x)|2dx
PL
4<x<L
2=ZL
2
L
4 r2
Lsin 2πx
L!2
dx
PL
4<x<L
2=ZL
2
L
4
2
Lsin22πx
Ldx
PL
4<x<L
2=2
LZL
2
L
4
1−cos 4πx
L
2dx
PL
4<x<L
2=1
LZL
2
L
4
1−cos 4πx
Ldx
PL
4<x<L
2=1
Lx−L
4πsin 4πx
LL
2
L
4
PL
4<x<L
2=1
LL
2−L
4πsin (2π)−L
4+L
4πsin (π)
10
PL
4<x<L
2=1
L[
*Question 11
Question
Let ψ(x) = Ax2+Bx +Cbe a wave function describing a particle in one
dimension. Determine the normalization constant Aassuming the particle is
confined to the interval −1≤x≤1.
Solution
Step 1: The normalization condition for a wave function ψ(x) over a region
−L≤x≤Lis given by:
ZL
−L|ψ(x)|2dx = 1
Step 2: Substituting ψ(x) = Ax2+Bx +Cinto the normalization condition
and solving for Agives:
Z1
−1|Ax2+Bx +C|2dx = 1
Step 3: Computing the integral in Step 2 gives:
Z1
−1|Ax2+Bx +C|2dx =Z1
−1|Ax2+Bx +C|2dx = 1
Step 4: Expand the square inside the absolute value and integrate term by
term:
Z1
−1
(A2x4+ 2ABx3+ (B2+ 2AC)x2+ 2BCx +C2)dx = 1
Step 5: Integrate each term separately:
A2Z1
−1
x4dx+2AB Z1
−1
x3dx+(B2+2AC)Z1
−1
x2dx+2BC Z1
−1
x dx+C2Z1
−1
dx = 1
Step 6: Evaluate the integrals and simplify:
A22
5+ 2AB (0) + (B2+ 2AC)2
3+ 2BC (0) + C2(2) = 1
Step 7: Since the integral is equal to 1, we have:
2
5A2+2
3(B2+ 2AC)+2C2= 1
Step 8: We can simplify this equation and solve for Ain terms of Band C:
2
5A2+2
3B2+4
3AC + 2C2= 1
11
Question 12
Question
Let ψ(x) = Ae−kx +Bekx be the wave function of a particle in a one-dimensional
box of length L. Determine the values of Aand Bsuch that the wave function
satisfies the boundary conditions ψ(0) = 0 and ψ(L) = 0.
Solution
Given wave function ψ(x) = Ae−kx +Bekx.
Step 1: Apply the boundary condition ψ(0) = 0
ψ(0) = A−B= 0
A=B
Step 2: Substitute A=Bback into the wave function
ψ(x) = Ae−kx +Aekx =A(e−kx +ekx)
Step 3: Apply the boundary condition ψ(L) = 0
ψ(L) = A(e−kL +ekL)=0
Step 4: Find the values of Aand Bthat satisfy the boundary
conditions Since e−kL +ekL = 0, the only way for ψ(L) = 0 is if A= 0.
Step 5: Conclusion The wave function that satisfies the boundary condi-
tions ψ(0) = 0 and ψ(L) = 0 is given by ψ(x) = 0.
Question 13
Question
Consider a particle confined to a 1-dimensional box of length L. The wave
function of the particle inside the box is given by:
ψ(x) = Asinnπx
L
where Ais a normalization constant and nis a positive integer. Determine the
probability of finding the particle in the region L
4<x<3L
4.
12
Solution
Step 1: Normalize the wave function ψ(x).
ZL
0|ψ(x)|2dx = 1
ZL
0|Asinnπx
L|2dx = 1
A2ZL
0
sin2(nπx
L)dx = 1
A2ZL
0
1−cos2nπx
L
2dx = 1
A2[1
2x−L
2nπ sin2nπx
L]L
0= 1
A2[1
2L] = 1
A2=2
L
A=r2
L
Step 2: Calculate the probability of finding the particle in the region L
4<
x < 3L
4.
P=Z3L
4
L
4|ψ(x)|2dx
=Z3L
4
L
4r2
Lsin nπx
L
2
dx
=Z3L
4
L
4
2
Lsin2nπx
Ldx
=2
LZ3L
4
L
4
1−cos 2nπx
L
2dx
=1
Lx−L
2nπ sin 2nπx
L3L
4
L
4
=1
LL
2−L
2nπ sin 3nπ
2−L
4+L
2nπ sin nπ
2
=1
LL
2+L
4
=3
4
13
Therefore, the probability of finding the particle in the region L
4<x<3L
4
is 3
4.
Question 14
Question
Let ψ(x) be a wave function given by ψ(x) = Asin(kx)+Bcos(kx), where A, B,
and kare constants. Determine the normalization condition for ψ(x).
Solution
To normalize the wave function ψ(x), we need to ensure that the integral of
|ψ(x)|2over all space is equal to 1. Mathematically, the normalization condition
can be expressed as:
Z∞
−∞ |ψ(x)|2dx = 1
Step 1: Calculate |ψ(x)|2
|ψ(x)|2= (ψ(x))∗·ψ(x)=(Asin(kx) + Bcos(kx))∗·(Asin(kx) + Bcos(kx))
= (Asin(kx) + Bcos(kx))(Asin(kx) + Bcos(kx))
=A2sin2(kx) + B2cos2(kx)+2AB sin(kx) cos(kx)
Step 2: Integrate |ψ(x)|2over all space
Z∞
−∞ |ψ(x)|2dx =Z∞
−∞
(A2sin2(kx) + B2cos2(kx)+2AB sin(kx) cos(kx))dx
=A2Z∞
−∞
sin2(kx)dx +B2Z∞
−∞
cos2(kx)dx + 2AB Z∞
−∞
sin(kx) cos(kx)dx
Step 3: Simplify the integrals and apply the normalization condi-
tion Since sin2(kx) and cos2(kx) have average values of 1
2over a full period, and
sin(kx) cos(kx) integrates to zero over a full period, the normalization condition
simplifies to:
A21
2+B21
2= 1
A2
2+B2
2= 1
A2+B2= 2
Therefore, the normalization condition for the wave function ψ(x) is A2+
B2= 2.
14
Question 15
Question
Consider a wave function given as ψ(x) = Ae−bx2, where Aand bare constants.
Find the normalization constant Afor this wave function.
Solution
Step 1: Determine the normalization condition for the wave function ψ(x). For
ψ(x) to be a valid wave function, it must satisfy the normalization condition:
Z∞
−∞ |ψ(x)|2dx = 1
Step 2: Substitute the given wave function into the normalization condition.
Z∞
−∞ |Ae−bx2|2dx = 1
Z∞
−∞
A2e−2bx2dx = 1
Step 3: Simplify the integral and solve for A.
A2Z∞
−∞
e−2bx2dx = 1
A2Z∞
−∞
e−(√2x)2dx = 1
A2rπ
2= 1
A2=1
pπ
2
=r2
π
Step 4: Determine the value of A.
A=r2
π
Therefore, the normalization constant for the wave function ψ(x) = Ae−bx2
is A=q2
π.
Question 16
Question
Given a wave function Ψ(x) = Asin(kx) + Bcos(kx), where A,B, and kare
constants, determine the normalization constant Nfor the wave function.
15
Solution
Step 1: Normalize the wave function by finding Nsuch that R∞
−∞ |Ψ(x)|2dx = 1.
Step 2: Calculate |Ψ(x)|2= Ψ(x)·Ψ(x).
|Ψ(x)|2= (Asin(kx) + Bcos(kx))(Asin(kx) + Bcos(kx))
Step 3: Expand the expression and simplify.
|Ψ(x)|2=A2sin2(kx) + B2cos2(kx)+2AB sin(kx) cos(kx)
Step 4: Use trigonometric identities to simplify further.
|Ψ(x)|2=A2sin2(kx) + B2cos2(kx) + AB sin(2kx)
Step 5: Integrate |Ψ(x)|2from −∞ to ∞.
Z∞
−∞ |Ψ(x)|2dx =Z∞
−∞
(A2sin2(kx) + B2cos2(kx) + AB sin(2kx))dx
Step 6: Evaluate the integral for each term to obtain an equation for N.
Step 7: Finally, solve for the normalization constant Nto complete the
normalization of the wave function.
Question 17
Question
Consider a wave function given by ψ(x) = Asin(kx)+Bcos(kx), where Aand B
are constants and kis the wave number. Determine the normalization constant
Ain terms of B.
Solution
Given wave function: ψ(x) = Asin(kx) + Bcos(kx)
To determine the normalization constant A, we need to normalize the wave
function by ensuring that
Z∞
−∞ |ψ(x)|2dx = 1
Step 1: Compute |ψ(x)|2
|ψ(x)|2= (Asin(kx) + Bcos(kx))2
=A2sin2(kx) + B2cos2(kx)+2AB sin(kx) cos(kx)
Step 2: Evaluate the integral
Z∞
−∞ |ψ(x)|2dx =Z∞
−∞
(A2sin2(kx) + B2cos2(kx)+2AB sin(kx) cos(kx))dx
16
Since the sine and cosine functions are orthogonal, the integral of their prod-
uct over one period is 0. Therefore, the only term that contributes to the integral
is the constant term.
Z∞
−∞ |ψ(x)|2dx =A2Z∞
−∞
sin2(kx)dx +B2Z∞
−∞
cos2(kx)dx
The integrals of sin2(kx) and cos2(kx) over one period are both equal to π,
so:
A2π+B2π= 1
Step 3: Solve for A
A2π+B2π= 1
A2+B2=1
π
A=r1
π−B2
Thus, the normalization constant Ain terms of Bis A=q1
π−B2.
Question 18
Question
Consider a particle in a one-dimensional box of length L. The wave function of
the particle is given by:
ψ(x) = Ax(L−x)
where Ais a normalization constant.
Determine the normalization constant Afor the wave function ψ(x).
Solution
Step 1: Normalize the wave function by integrating |ψ(x)|2over the entire range
of x.
ZL
0|ψ(x)|2dx =ZL
0|Ax(L−x)|2dx
Step 2: Simplify the integral by plugging in the given wave function ψ(x).
ZL
0
A2x2(L−x)2dx
Step 3: Expand and simplify the integrand.
ZL
0
A2(x4−2Lx3+L2x2)dx
17
Step 4: Integrate each term separately.
ZL
0
A2x4−2A2Lx3+A2L2x2dx
=A2x5
5−2Lx4
4+L2x3
3
L
0
Step 5: Evaluate the integral over the range 0 to L.
A2L5
5−2L5
4+L5
3
Step 6: Set the result equal to 1 (the normalization condition) and solve for
A.
A2L5
5−2L5
4+L5
3= 1
Step 7: Simplify and solve for A.
A2L5
5−2A2L5
4+A2L5
3= 1
3A2L5−10A2L5+ 15A2L5
60 = 1
8A2L5
60 = 1
A2=15
4L5
A=r15
4L5
Therefore, the normalization constant Afor the wave function ψ(x) is A=
q15
4L5.
Question 19
Question
Let ψ(x) = Ax2e−x
2abe a wave function representing a particle in a one-
dimensional potential well. Find the normalization constant A.
18
Solution
1. We need to normalize the wave function, which means we must have
R∞
−∞ |ψ(x)|2dx = 1.
2. First, let’s square the wave function: |ψ(x)|2=|Ax2e−x
2a|2=A2x4e−x
a.
3. Next, we can rewrite the integral as: R∞
−∞ A2x4e−x
adx = 1.
4. We can simplify the integral by factoring out the constant A2:A2R∞
0x4e−x
adx =
1.
5. We will now solve the integral R∞
0x4e−x
adx.
6. Let’s make the substitution u=x
a,dx =adu, which gives us: A2R∞
0(au)4e−uadu =
1.
7. Simplifying, we get A2a5R∞
0u4e−udu = 1.
8. The integral now becomes A2a5·4! = 1, where 4! represents the factorial
of 4.
9. Hence, A2a5·24 = 1, and solving for Agives us A=1
2√6a5/2.
Question 20
Question
Consider a wave function ψ(x) = A·e−αx2, where Aand αare constants.
Determine the normalization constant A.
Solution
To normalize the wave function ψ(x), we must ensure that the total probability
of finding the particle in all space is equal to 1.
Step 1: Calculate the normalization condition. The normalization condition
is given by:
Z∞
−∞ |ψ(x)|2dx = 1
Step 2: Substitute the given wave function into the normalization condition.
Z∞
−∞ |A·e−αx2|2dx = 1
Z∞
−∞ |A|2· |e−2αx2|dx = 1
Z∞
−∞ |A|2·e−2αx2dx = 1
19
Step 3: Solve the integral. To solve the integral, we use the fact that
R∞
−∞ e−ax2dx =pπ
afor a > 0. Therefore, we have:
|A|2·rπ
2α= 1
|A|2=1
pπ/(2α)
A=r2α
π
Therefore, the normalization constant Ais q2α
π.
Question 21
Question
Consider a particle in a 1-dimensional box of length L. The particle is in the
ground state with wave function ψ(x) = Asin(πx/L), where Ais a normalization
constant. Determine the probability of finding the particle in the interval 0 <
x < L/4.
Solution
Step 1: Normalize the wave function. Step 2: Find the probability of finding
the particle in the interval 0 < x < L/4 using the normalized wave function.
Step 1: Normalize the wave function.
The normalization condition states that the integral of the absolute square
of the wave function over all space is equal to 1:
Z∞
−∞ |ψ(x)|2dx = 1
For our wave function ψ(x) = Asin(πx/L), the normalization condition
becomes:
ZL
0
A2sin2(πx/L)dx = 1
A2ZL
0
sin2(πx/L)dx = 1
A2ZL
0
1−cos(2πx/L)
2dx = 1
A2x
2−L
2πsin(2πx/L)
L
0
= 1
20
A2L
2−L
2πsin(2π)= 1
A2L
2= 1
A2=2
L
A=r2
L
Therefore, the normalized wave function is ψ(x) = q2
Lsin(πx/L).
Step 2: Find the probability of finding the particle in the interval 0 < x <
L/4 using the normalized wave function.
The probability of finding the particle in the interval 0 < x < L/4 is given
by the integral of the absolute square of the wave function over that interval:
P=ZL/4
0|ψ(x)|2dx
P=ZL/4
0 r2
Lsin πx
L!2
dx
P=ZL/4
0
2
Lsin2πx
Ldx
Using the angle addition formula sin2(u) = 1−cos(2u)
2, we have:
P=ZL/4
0
1−cos 2πx
L
Ldx
P=x
L−L
2πsin 2πx
L
L/4
0
P=L
4L−L
2πsin π
2−[0 −0]
P=1
4−L
2π
Therefore, the probability of finding the particle in the interval 0 < x < L/4
is 1
4−L
2π.
21
Question 22
Question
Consider a wave function Ψ(x) = Asin(kx) + Bcos(kx), where A,B, and kare
constants. Determine the normalization constant Nsuch that R∞
−∞ |Ψ(x)|2dx =
1.
Solution
Step 1: Compute |Ψ(x)|2.
|Ψ(x)|2=|Asin(kx) + Bcos(kx)|2
= (Asin(kx) + Bcos(kx)) (Asin(kx) + Bcos(kx))
=A2sin2(kx)+2AB sin(kx) cos(kx) + B2cos2(kx)
=A2(1 −cos2(kx)) + 2AB sin(kx) cos(kx) + B2cos2(kx)
=A2+B2+ 2AB sin(kx) cos(kx)
=A2+B2+AB sin(2kx)
Step 2: Integrate |Ψ(x)|2from −∞ to ∞.
Z∞
−∞ |Ψ(x)|2dx =Z∞
−∞
(A2+B2+AB sin(2kx)) dx
=A2x+B2x−AB
2kcos(2kx)∞
−∞
= lim
a→∞ A2a+B2a−AB
2kcos(2ka)−(A2(−a) + B2(−a)−AB
2kcos(−2ka))
= lim
a→∞ A2a+B2a−AB
2k(cos(2ka)−cos(−2ka)) + A2a+B2a−AB
2k(cos(−2ka)−cos(2ka))
= lim
a→∞ 2(A2+B2)a
=∞
Since the integral diverges, there is no normalization constant Nthat can
make the wave function Ψ(x) normalized.
Question 23
Question
Let f(x) be a wave function of a particle in one dimension. Given that the
wave function is normalized, i.e., R∞
−∞ |f(x)|2dx = 1, determine whether the
following functions are valid wave functions:
1. g(x) = e2ix
2. h(x) = sin(3x) + cos(4x)
22
Solution
To determine whether a function is a valid wave function, we need to check if
it satisfies the normalization condition:
Z∞
−∞ |f(x)|2dx = 1
1. For function g(x) = e2ix:
Z∞
−∞ |g(x)|2dx =Z∞
−∞ |e2ix|2dx
=Z∞
−∞
e2ixe−2ix dx
=Z∞
−∞
1dx
=∞ = 1
Since the integral diverges, g(x) = e2ix is not a valid wave function.
2. For function h(x) = sin(3x) + cos(4x):
Z∞
−∞ |h(x)|2dx =Z∞
−∞ |sin(3x) + cos(4x)|2dx
=Z∞
−∞
(sin2(3x) + 2 sin(3x) cos(4x) + cos2(4x)) dx
=Z∞
−∞
(1 + sin(3x+ 4x)) dx
=Z∞
−∞
1dx +Z∞
−∞
sin(7x)dx
=∞ = 1
Since the integral diverges, h(x) = sin(3x) + cos(4x) is not a valid wave
function.
Therefore, neither g(x) = e2ix nor h(x) = sin(3x) + cos(4x) are valid wave
functions.
Question 24
Question
Consider the following wave function for a particle in a one-dimensional box of
length L:
ψ(x) = Asin2nπx
L
23
where Ais a normalization constant.
What is the probability of finding the particle in the interval 0 ≤x≤L
4for
the ground state (n= 1) of the particle?
Solution
Step 1: Normalize the wave function.
ZL
0|ψ(x)|2dx = 1
A2ZL
0
sin4πx
Ldx = 1
A23L
8= 1
A=r8
3L
Step 2: Find the probability of finding the particle in the interval 0 ≤x≤L
4.
P=ZL
4
0|ψ(x)|2dx
P=ZL
4
0 r8
3Lsin2πx
L!2
dx
P=8
3LZL
4
0
sin4πx
Ldx
Step 3: Use the trigonometric identity sin2(θ) = 1−cos(2θ)
2.
P=8
3LZL
4
0
1−cos 2πx
L
2dx
P=8
6Lx
2−L
2πsin 2πx
L
L
4
0
P=1
3
Therefore, the probability of finding the particle in the interval 0 ≤x≤L
4
for the ground state of the particle is 1
3.
24
Question 25
Question
Consider a particle in a one-dimensional box of length L. The wave function
of the particle is given by Ψ(x) = Asin nπx
L, where Ais a normalization
constant. Determine the value of nfor which the probability of finding the
particle between 0.2Land 0.3Lis maximum.
Solution
Step 1: Normalize the wave function. The normalization condition for a wave
function Ψ(x) in a one-dimensional box is given by
ZL
0|Ψ(x)|2dx = 1
Given Ψ(x) = Asin nπx
L, we have
ZL
0|Asin nπx
L|2dx = 1
Solve this integral, set it equal to 1, and solve for A.
Step 2: Find the probability of finding the particle between 0.2Land 0.3L.
The probability of finding the particle between 0.2Land 0.3Lis given by
P=Z0.3L
0.2L|Ψ(x)|2dx
Substitute the normalized wave function in Pand simplify the integral.
Step 3: Maximize the probability. To maximize the probability, differentiate
Pwith respect to n, set it equal to zero, and solve for n.
Question 26
Question
Find the general solution to the time-independent Schr¨odinger equation for a
particle confined to move in a one-dimensional box of length L, with boundary
conditions Ψ(0) = Ψ(L) = 0.
Solution
To find the general solution to the Schr¨odinger equation, we start with the
time-independent Schr¨odinger equation given by:
−ℏ2
2m
d2Ψ
dx2+V(x)Ψ = EΨ,
25
where Eis the total energy of the particle, ℏis the reduced Planck constant, m
is the mass of the particle, V(x) is the potential energy function, and Ψ(x) is
the wave function.
The potential energy function inside the box is zero, so V(x) = 0 for 0 <
x<L. Therefore, the Schr¨odinger equation simplifies to:
−ℏ2
2m
d2Ψ
dx2=EΨ.
Step 1: Solving the differential equation Let’s solve the differential
equation above to find the general solution for Ψ(x). To do this, we assume
Ψ(x) = Asin(kx) + Bcos(kx), where Aand Bare constants to be determined,
and k=q2mE
ℏ2.
Taking the second derivative of Ψ(x) and substituting it into the Schr¨odinger
equation, we get:
−ℏ2
2m(−Ak2sin(kx)−Bk2cos(kx)) = E(Asin(kx) + Bcos(kx))
Step 2: Applying boundary conditions Since we have boundary condi-
tions Ψ(0) = Ψ(L) = 0 for a particle in a well, we can use these to determine
the constants Aand B.
Substituting x= 0 into Ψ(x) gives Ψ(0) = B= 0.
Substituting x=Linto Ψ(x) gives Ψ(L) = Asin(kL) = 0. This implies
either A= 0 or sin(kL) = 0. If A= 0, then the wave function is trivial.
Therefore, we must have sin(kL) = 0, which gives us kL =nπ, where nis a
positive integer.
Therefore, the general solution for the wave function Ψ(x) is given by:
Ψn(x) = Asin nπx
L,
where nis a positive integer.
Thus, we have found the general solution for the time-independent Schr¨odinger
equation for a particle in a one-dimensional box.
Question 27
Question
Find the normalized wave function for a particle in a one-dimensional box of
length L. The potential inside the box is zero while the potential outside is
infinite.
Solution
To find the normalized wave function for a particle in a one-dimensional box,
we will follow the steps below:
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Step 1: Express the general solution of the Schr¨odinger equation
The general solution for the wave function inside the box is given by:
ψ(x) = Asin(kx) + Bcos(kx)
where k=q2mE
ℏ.
Step 2: Apply boundary conditions At x= 0, |ψ(0)|2must be finite.
Therefore, B= 0. So, ψ(x) = Asin(kx).
Step 3: Apply the second boundary condition at x=LAt x=L,
ψ(L) = 0 since the potential outside the box is infinite.
Asin(kL)=0
This implies that kL =nπ for n= 1,2,3, . . ..
Step 4: Determine the energy eigenvalues Since kL =nπ, we have
k=nπ
L. Substitute this into the expression for kto get the energy eigenvalues:
E=n2π2ℏ2
2mL2
Step 5: Normalize the wave function Since the particle must be found
somewhere in the box, the normalized wave function is:
ψ(x) = r2
Lsin nπx
L
Question 28
Question
Let f(x) = (cx if 0 ≤x≤1
0 otherwise be a wave function. Determine the value of c
that makes f(x) a valid wave function.
Solution
Step 1: To find the value of c, we need to ensure that the wave function f(x)
satisfies the normalization condition:
Z∞
−∞ |f(x)|2dx = 1
Step 2: First, let’s determine the interval over which f(x) is non-zero. Since
f(x) is only non-zero when 0 ≤x≤1, the integral simplifies to:
Z1
0|f(x)|2dx
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Step 3: Substituting the expression for f(x) into the integral, we get:
Z1
0|cx|2dx =Z1
0
c2x2dx
Step 4: Solving the integral, we find:
c2Z1
0
x2dx =c2x3
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0
=c2
3
Step 5: Set the integral equal to 1 and solve for c:
c2
3= 1
c2= 3
c=√3
Therefore, the value of cthat makes f(x) a valid wave function is √3.
Question 29
Question
Consider a wave function in one dimension given by Ψ(x) = Asin(kx)+Bcos(kx),
where Aand Bare constants. Determine the normalization constant Ain terms
of B.
Solution
To normalize the wave function, we require that the integral of the square of
the wave function over all space is equal to 1, i.e., R∞
−∞ |Ψ(x)|2dx = 1.
Step 1: Calculate |Ψ(x)|2using the given wave function.
|Ψ(x)|2= (Asin(kx)+Bcos(kx))2=A2sin2(kx)+B2cos2(kx)+2AB sin(kx) cos(kx)
Step 2: Integrate |Ψ(x)|2from −∞ to ∞to normalize the wave function.
Z∞
−∞ |Ψ(x)|2dx =Z∞
−∞
(A2sin2(kx) + B2cos2(kx)+2AB sin(kx) cos(kx))dx
Step 3: Recognize that the integral of the sine and cosine terms over a full
period is zero.
Z∞
−∞
B2cos2(kx)dx =B2Z∞
−∞
cos2(kx)dx =B21
2Z∞
−∞
(1+cos(2kx))dx =B21
2(∞+∞) = ∞
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Step 4: Set up the integral for A2sin2(kx) term to evaluate.
Z∞
−∞
A2sin2(kx)dx =A2Z∞
−∞
sin2(kx)dx =A21
2Z∞
−∞
(1 −cos(2kx))dx
Step 5: Recognize that the integral of the cosine term over a full period is
zero, so we have
A21
2(∞−∞) = A21
2·0=0
Step 6: The integral of the cross term 2AB sin(kx) cos(kx) over a full period
is also zero.
Z∞
−∞
2AB sin(kx) cos(kx)dx = 2AB Z∞
−∞
sin(kx) cos(kx)dx = 0
Step 7: Combining all the terms and setting the result to 1 to normalize
the wave function:
∞+ 0 + 0 = 1
As ∞ = 1, the wave function cannot be normalized.
Question 30
Question
Let f(x) be a wave function defined on the interval [−π, π] such that f(x) is an
odd function and satisfies the property f(0) = 0. If f(x) can be expressed as
f(x) = ∞
X
n=1
bnsin(nx),
where bn=2
πRπ
0f(x) sin(nx)dx, find the constant b1.
Solution
Step 1: Since f(x) is an odd function, f(x) = −f(−x) for all xin the interval
[−π, π]. Therefore, we have
f(x) = −f(−x).
Given that f(x) satisfies the property f(0) = 0, we can write
f(0) = −f(0)
0 = −f(0)
f(0) = 0.
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Step 2: We now substitute f(x) into the formula for bnand simplify. We
have
bn=2
πZπ
0
f(x) sin(nx)dx =2
πZπ
0 ∞
X
m=1
bmsin(mx)!sin(nx)dx.
Using the orthogonality property of sine functions, we can simplify the integral
to
bn=2
πZπ
0
bnsin2(nx)dx.
Step 3: Further simplifying the integral, we have
bn=2
πbnZπ
0
1−cos(2nx)
2dx.
bn=1
πbnx−sin(2nx)
2n
π
0
.
Plugging in the limits of integration and simplifying, we get
bn=1
πbnπ−sin(2nπ)
2n−0.
bn=bn1−sin(2nπ)
2nπ .
Step 4: Since sin(2nπ) = 0 for all integer values of n, we have
bn=bn
1=1.
Thus, bncancels out from the equation.
Step 5: Finally, we find the expression for b1. Substituting n= 1 into the
expression derived in Step 3, we have
b1=b11−sin(2π)
2π.
Since sin(2π) = 0, we get
b1=b1·1
b1=b1.
Therefore, the constant b1is equal to itself, which means b1can take any
value.
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