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**
Contents
Preface 2
1 The Wave Function 3
2 Time-Independent Schrödinger Equation 14
3 Formalism 62
4 Quantum Mechanics in Three Dimensions 87
5 Identical Particles 132
6 Time-Independent Perturbation Theory 154
7 The Variational Principle 196
8 The WKB Approximation 219
9 Time-Dependent Perturbation Theory 236
10 The Adiabatic Approximation 254
11 Scattering 268
12 Afterword 282
Appendix Linear Algebra 283
2nd Edition – 1st Edition Problem Correlation Grid 299
**

2

**Preface
**

These are my own solutions to the problems in *Introduction to Quantum Mechanics, 2nd ed. *I have made every
effort to insure that they are clear and correct, but errors are bound to occur, and for this I apologize in advance.
I would like to thank the many people who pointed out mistakes in the solution manual for the first edition,
and encourage anyone who finds defects in this one to alert me (griffith@reed.edu). I’ll maintain a list of errata
on my web page (http://academic.reed.edu/physics/faculty/griffiths.html), and incorporate corrections in the
manual itself from time to time. I also thank my students at Reed and at Smith for many useful suggestions,
and above all Neelaksh Sadhoo, who did most of the typesetting.

At the end of the manual there is a grid that correlates the problem numbers in the second edition with those in the first edition.

David Griffiths

c*©*2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.

*CHAPTER 1. THE WAVE FUNCTION *3

**Chapter 1
**

**The Wave Function
**

**Problem 1.1
**

**(a)
**

*〈j〉*2 = 212 = 441.

*〈j*2*〉 *= 1
*N
*

∑
*j*2*N*(*j*) =

1 14

[ (142) + (152) + 3(162) + 2(222) + 2(242) + 5(252)

] =

1 14

(196 + 225 + 768 + 968 + 1152 + 3125) = 6434 14

= 459.571.

**(b)
**

*j *∆*j *= *j − 〈j〉
*14 14 *− *21 = *−*7
15 15 *− *21 = *−*6
16 16 *− *21 = *−*5
22 22 *− *21 = 1
24 24 *− *21 = 3
25 25 *− *21 = 4

*σ*2 =
1
*N
*

∑
(∆*j*)2*N*(*j*) =

1 14

[
(*−*7)2 + (*−*6)2 + (*−*5)2 *· *3 + (1)2 *· *2 + (3)2 *· *2 + (4)2 *· *5

] =

1 14

(49 + 36 + 75 + 2 + 18 + 80) = 260 14

= 18.571.

*σ *=
*√
*

18*.*571 = 4.309.

**(c)
**

*〈j*2*〉 − 〈j〉*2 = 459*.*571 *− *441 = 18*.*571*. *[Agrees with (b).]

c*©*2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.

4 *CHAPTER 1. THE WAVE FUNCTION
*

**Problem 1.2
**

**(a)
**

*〈x*2*〉 *=
∫ *h
*

0

*x*2
1

2
*√
hx
*

*dx *=
1

2
*√
h
*

(
2
5
*x*5*/*2

)∣∣∣∣*h
*0

=
*h*2

5
*.
*

*σ*2 = *〈x*2*〉 − 〈x〉*2 = *h
*2

5
*−
*

(
*h
*

3

)2 =

4 45

*h*2 *⇒ σ *= 2*h
*3
*√
*

5
= 0*.*2981*h.
*

**(b)
**

*P *= 1 *−
*∫ *x*+
*x−
*

1
2
*√
hx
*

*dx *= 1 *− *1
2
*√
h
*

(2
*√
x*)

∣∣∣∣*x*+
*x−
*

= 1 *− *1*√
h
*

(*√
x*+ *−
*

*√
x−
*

)
*.
*

*x*+ *≡ 〈x〉 *+ *σ *= 0*.*3333*h *+ 0*.*2981*h *= 0*.*6315*h*; *x− ≡ 〈x〉 − σ *= 0*.*3333*h− *0*.*2981*h *= 0*.*0352*h.
*

*P *= 1 *−
√
*

0*.*6315 +
*√
*

0*.*0352 = 0.393.

**Problem 1.3
**

**(a)
**

1 =
∫ *∞
−∞
*

*Ae−λ*(*x−a*)
2
*dx. *Let *u ≡ x− a*, *du *= *dx*, *u *: *−∞ → ∞*.

1 = *A
*∫ *∞
−∞
*

*e−λu
*2
*du *= *A
*

√
*π
*

*λ
⇒ A *=

√
*λ
*

*π
*.

**(b)
**

*〈x〉 *= *A
*∫ *∞
−∞
*

*xe−λ*(*x−a*)
2
*dx *= *A
*

∫ *∞
−∞
*

(*u *+ *a*)*e−λu
*2
*du
*

= *A
*[∫ *∞
*

*−∞
ue−λu
*

2
*du *+ *a
*

∫ *∞
−∞
*

*e−λu
*2
*du
*

]
= *A
*

(
0 + *a
*

√
*π
*

*λ
*

)
= *a*.

*〈x*2*〉 *= *A
*∫ *∞
−∞
*

*x*2*e−λ*(*x−a*)
2
*dx
*

= *A
*{∫ *∞
*

*−∞
u*2*e−λu
*

2
*du *+ 2*a
*

∫ *∞
−∞
*

*ue−λu
*2
*du *+ *a*2

∫ *∞
−∞
*

*e−λu
*2
*du
*

}
= *A
*

[
1
2*λ
*

√
*π
*

*λ
*+ 0 + *a*2

√
*π
*

*λ
*

]
= *a*2 +

1
2*λ
*

.

*σ*2 = *〈x*2*〉 − 〈x〉*2 = *a*2 + 1
2*λ
*

*− a*2 = 1
2*λ
*

; *σ *=
1*√
*2*λ
*

.

c*©*2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.

*CHAPTER 1. THE WAVE FUNCTION *5

**(c)
**

A

xa

ρ(x)

**Problem 1.4
**

**(a)
**

1 =
*|A|*2
*a*2

∫ *a
*0

*x*2*dx *+
*|A|*2

(*b− a*)2
∫ *b
a
*

(*b− x*)2*dx *= *|A|*2
{

1
*a*2

(
*x*3

3

)∣∣∣∣*a
*0

+ 1

(*b− a*)2
(
*− *(*b− x*)

3

3

)∣∣∣∣*b
a
*

}

= *|A|*2
[
*a
*

3 +

*b− a
*3

]
= *|A|*2 *b
*

3
*⇒ A *=

√
3
*b
*.

**(b)
**

xa

A

b

Ψ

**(c) **At *x *= *a*.

**(d)
**

*P *=
∫ *a
*

0

*|*Ψ*|*2*dx *= *|A|
*2

*a*2

∫ *a
*0

*x*2*dx *= *|A|*2 *a
*3

=
*a
*

*b
*.

{
*P *= 1 if *b *= *a, *
*P *= 1*/*2 if *b *= 2*a. *

**(e)
**

*〈x〉 *=
∫

*x|*Ψ*|*2*dx *= *|A|*2
{

1
*a*2

∫ *a
*0

*x*3*dx *+
1

(*b− a*)2
∫ *b
a
*

*x*(*b− x*)2*dx
*}

=
3
*b
*

{
1
*a*2

(
*x*4

4

)∣∣∣∣*a
*0

+ 1

(*b− a*)2
(
*b*2
*x*2

2
*− *2*bx
*

3

3 +

*x*4

4

)∣∣∣∣*b
a
*

}

= 3

4*b*(*b− a*)2
[
*a*2(*b− a*)2 + 2*b*4 *− *8*b*4*/*3 + *b*4 *− *2*a*2*b*2 + 8*a*3*b/*3 *− a*4

] =

3
4*b*(*b− a*)2

(
*b*4

3
*− a*2*b*2 + 2

3
*a*3*b
*

) =

1
4(*b− a*)2 (*b
*

3 *− *3*a*2*b *+ 2*a*3) = 2*a *+ *b
*4

.

*©*2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.

6 *CHAPTER 1. THE WAVE FUNCTION
*

**Problem 1.5
**

**(a)
**

1 = ∫

*|*Ψ*|*2*dx *= 2*|A|*2
∫ *∞
*

0

*e−*2*λxdx *= 2*|A|*2
(
*e−*2*λx
*

*−*2*λ
*

)∣∣∣∣*∞
*0

=
*|A|*2
*λ
*

; *A *=
*√
λ*.

**(b)
**

*〈x〉 *=
∫

*x|*Ψ*|*2*dx *= *|A|*2
∫ *∞
−∞
*

*xe−*2*λ|x|dx *= 0. [Odd integrand.]

*〈x*2*〉 *= 2*|A|*2
∫ *∞
*

0

*x*2*e−*2*λxdx *= 2*λ
*[

2
(2*λ*)3

] =

1
2*λ*2

.

**(c)
**

*σ*2 = *〈x*2*〉 − 〈x〉*2 = 1
2*λ*2

; *σ *=
1*√
*2*λ
*

. *|*Ψ(*±σ*)*|*2 = *|A|*2*e−*2*λσ *= *λe−*2*λ/
√
*

2*λ *= *λe−
√
*

2 = 0*.*2431*λ.
*

|Ψ|2 λ

σ−σ + x

.24λ

*Probability outside*:

2
∫ *∞
σ
*

*|*Ψ*|*2*dx *= 2*|A|*2
∫ *∞
σ
*

*e−*2*λxdx *= 2*λ
*(
*e−*2*λx
*

*−*2*λ
*

)∣∣∣∣*∞
σ
*

= *e−*2*λσ *= *e−
√
*

2 = 0*.*2431.

**Problem 1.6
**

For integration by parts, the differentiation has to be with respect to the *integration *variable – in this case the
differentiation is with respect to *t*, but the integration variable is *x*. It’s true that

*∂
*

*∂t
*(*x|*Ψ*|*2) = *∂x
*

*∂t
|*Ψ*|*2 + *x ∂
*

*∂t
|*Ψ*|*2 = *x ∂
*

*∂t
|*Ψ*|*2*,
*

but this does *not *allow us to perform the integration:∫ *b
a
*

*x
∂
*

*∂t
|*Ψ*|*2*dx *=

∫ *b
a
*

*∂
*

*∂t
*(*x|*Ψ*|*2)*dx *= (*x|*Ψ*|*2)

∣∣*b
a
.
*

*©*2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.

*CHAPTER 1. THE WAVE FUNCTION *7

**Problem 1.7
**

From Eq. 1.33, *d〈p〉dt *= *−i*
∫

*∂
∂t
*

(
Ψ*∗ ∂*Ψ*∂x
*

)
*dx*. But, noting that *∂
*

2Ψ
*∂x∂t *=

*∂*2Ψ
*∂t∂x *and using Eqs. 1.23-1.24:

*∂
*

*∂t
*

(
Ψ*∗
*

*∂*Ψ
*∂x
*

) =

*∂*Ψ*∗
*

*∂t
*

*∂*Ψ
*∂x
*

+ Ψ*∗
∂
*

*∂x
*

(
*∂*Ψ
*∂t
*

) =

[
*− i*

2*m
∂*2Ψ*∗
*

*∂x*2
+

*i
*

*V *Ψ*∗
*

]
*∂*Ψ
*∂x
*

+ Ψ*∗
∂
*

*∂x
*

[
*i*

2*m
∂*2Ψ
*∂x*2

*− i
*
*V *Ψ

] =

*i*

2*m
*

[
Ψ*∗
*

*∂*3Ψ
*∂x*3

*− ∂
*2Ψ*∗
*

*∂x*2
*∂*Ψ
*∂x
*

] +

*i
*

[
*V *Ψ*∗
*

*∂*Ψ
*∂x
*

*− *Ψ*∗ ∂
∂x
*

(*V *Ψ)
]

The first term integrates to zero, using integration by parts twice, and the second term can be simplified to
*V *Ψ*∗ ∂*Ψ*∂x − *Ψ*∗V ∂*Ψ*∂x − *Ψ*∗ ∂V∂x *Ψ = *−|*Ψ*|*2 *∂V∂x . *So

*d〈p〉
dt
*

= *−i*
(
*i
*

) ∫
*−|*Ψ*|*2 *∂V
*

*∂x
dx *= *〈−∂V
*

*∂x
〉. *QED

**Problem 1.8
**

Suppose Ψ satisfies the Schrödinger equation *without V*0: *i**∂*Ψ*∂t *= *− *
2

2*m
∂*2Ψ
*∂x*2 + *V *Ψ. We want to find the solution

Ψ0 *with V*0: *i**∂*Ψ0*∂t *= *− *
2

2*m
∂*2Ψ0
*∂x*2 + (*V *+ *V*0)Ψ0.

*Claim*: Ψ0 = Ψ*e−iV*0*t/*.

*Proof*: *i**∂*Ψ0*∂t *= *i*
*∂*Ψ
*∂t e
*

*−iV*0*t/* + *i*Ψ
(
*− iV*0

)
*e−iV*0*t/* =

[
*− *22*m ∂
*

2Ψ
*∂x*2 + *V *Ψ

]
*e−iV*0*t/* + *V*0Ψ*e−iV*0*t/*

= *− *22*m
∂*2Ψ0
*∂x*2 + (*V *+ *V*0)Ψ0. QED

This has *no *effect on the expectation value of a dynamical variable, since the extra phase factor, being inde-
pendent of *x*, cancels out in Eq. 1.36.

**Problem 1.9
**

**(a)
**

1 = 2*|A|*2
∫ *∞
*

0

*e−*2*amx
*2*/**dx *= 2*|A|*2 1

2

√
*π
*

(2*am/*)
= *|A|*2

√
*π*

2*am
*; *A *=

(
2*am
π*

)1*/*4
.

**(b)
**

*∂*Ψ
*∂t
*

= *−ia*Ψ; *∂*Ψ
*∂x
*

= *−*2*amx
*

Ψ;
*∂*2Ψ
*∂x*2

= *−*2*am
*

(
Ψ + *x
*

*∂*Ψ
*∂x
*

)
= *−*2*am
*

(
1 *− *2*amx
*

2

)
Ψ*.
*

Plug these into the Schrödinger equation, *i**∂*Ψ*∂t *= *− *
2

2*m
∂*2Ψ
*∂x*2 + *V *Ψ:

*V *Ψ = *i*(*−ia*)Ψ +
2

2*m
*

(
*−*2*am
*

) (
1 *− *2*amx
*

2

) Ψ

=
[
*a− **a
*

(
1 *− *2*amx
*

2

)]
Ψ = 2*a*2*mx*2Ψ*, *so *V *(*x*) = 2*ma*2*x*2.

*©*2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.

8 *CHAPTER 1. THE WAVE FUNCTION
*

**(c)
**

*〈x〉 *=
∫ *∞
−∞
*

*x|*Ψ*|*2*dx *= 0. [Odd integrand.]

*〈x*2*〉 *= 2*|A|*2
∫ *∞
*

0

*x*2*e−*2*amx
*2*/**dx *= 2*|A|*2 1

22(2*am/*)

√
*π*

2*am
*=

4*am
.
*

*〈p〉 *= *md〈x〉
dt
*

= 0*.
*

*〈p*2*〉 *=
∫

Ψ*∗
*(

*i
*

*∂
*

*∂x
*

)2
Ψ*dx *= *−*2

∫
Ψ*∗
*

*∂*2Ψ
*∂x*2

*dx
*

= *−*2
∫

Ψ*∗
*[
*−*2*am
*

(
1 *− *2*amx
*

2

) Ψ

]
*dx *= 2*am*

{∫
*|*Ψ*|*2*dx− *2*am
*

∫
*x*2*|*Ψ*|*2*dx
*

}
= 2*am*

(
1 *− *2*am
*

*〈x*2*〉
*

)
= 2*am*

(
1 *− *2*am
*

4*am
*

)
= 2*am*

( 1 2

)
= *am*.

**(d)
**

*σ*2*x *= *〈x*2*〉 − 〈x〉*2 =

4*am
*=*⇒ σx *=

√

4*am
*; *σ*2*p *= *〈p*2*〉 − 〈p〉*2 = *am* =*⇒ σp *=

*√
am*.

*σxσp *=
√

4*am
*

*√
am* = 2 . This *is *(just barely) consistent with the uncertainty principle.

**Problem 1.10
**

From Math Tables: *π *= 3*.*141592653589793238462643 *· · ·
*

**(a)
***P *(0) = 0 *P *(1) = 2*/*25 *P *(2) = 3*/*25 *P *(3) = 5*/*25 *P *(4) = 3*/*25
*P *(5) = 3*/*25 *P *(6) = 3*/*25 *P *(7) = 1*/*25 *P *(8) = 2*/*25 *P *(9) = 3*/*25

In general, *P *(*j*) = *N*(*j*)*N *.

**(b) ***Most probable*: 3. *Median*: 13 are *≤ *4, 12 are *≥ *5, so median is 4.
*Average*: *〈j〉 *= 125 [0 *· *0 + 1 *· *2 + 2 *· *3 + 3 *· *5 + 4 *· *3 + 5 *· *3 + 6 *· *3 + 7 *· *1 + 8 *· *2 + 9 *· *3]
= 125 [0 + 2 + 6 + 15 + 12 + 15 + 18 + 7 + 16 + 27] =

118 25 = 4.72.

**(c) ***〈j*2*〉 *= 125 [0 + 12 *· *2 + 22 *· *3 + 32 *· *5 + 42 *· *3 + 52 *· *3 + 62 *· *3 + 72 *· *1 + 82 *· *2 + 92 *· *3]
= 125 [0 + 2 + 12 + 45 + 48 + 75 + 108 + 49 + 128 + 243] =

710 25 = 28.4.

*σ*2 = *〈j*2*〉 − 〈j〉*2 = 28*.*4 *− *4*.*722 = 28*.*4 *− *22*.*2784 = 6*.*1216; *σ *=
*√
*

6*.*1216 = 2.474.

*©*2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.

*CHAPTER 1. THE WAVE FUNCTION *9

**Problem 1.11
**

**(a) **Constant for 0 *≤ θ ≤ π*, otherwise zero. In view of Eq. 1.16, the constant is 1*/π*.

*ρ*(*θ*) =
{

1*/π, *if 0 *≤ θ ≤ π,
*0*, *otherwise*.
*

1/π

−π/2 0 π 3π/2

ρ(θ)

θ

**(b)
**

*〈θ〉 *=
∫

*θρ*(*θ*) *dθ *=
1
*π
*

∫ *π
*0

*θdθ *=
1
*π
*

(
*θ*2

2

)∣∣∣∣*π
*0

=
*π
*

2 [of course].

*〈θ*2*〉 *= 1
*π
*

∫ *π
*0

*θ*2 *dθ *=
1
*π
*

(
*θ*3

3

)∣∣∣∣*π
*0

=
*π*2

3 .

*σ*2 = *〈θ*2*〉 − 〈θ〉*2 = *π
*2

3
*− π
*

2

4 =

*π*2

12
; *σ *=

*π
*

2
*√
*

3 .

**(c)
**

*〈*sin *θ〉 *= 1
*π
*

∫ *π
*0

sin *θ dθ *=
1
*π
*

(*− *cos *θ*)*|π*0 =
1
*π
*

(1 *− *(*−*1)) = 2
*π
*

.

*〈*cos *θ〉 *= 1
*π
*

∫ *π
*0

cos *θ dθ *=
1
*π
*

(sin *θ*)*|π*0 = 0.

*〈*cos2 *θ〉 *= 1
*π
*

∫ *π
*0

cos2 *θ dθ *=
1
*π
*

∫ *π
*0

(1*/*2)*dθ *=
1
2
.

[Because sin2 *θ *+ cos2 *θ *= 1, and the integrals of sin2 and cos2 are equal (over suitable intervals), one can
replace them by 1/2 in such cases.]

**Problem 1.12
**

**(a) ***x *= *r *cos *θ ⇒ dx *= *−r *sin *θ dθ. *The probability that the needle lies in range *dθ *is *ρ*(*θ*)*dθ *= 1*πdθ*, so the
probability that it’s in the range *dx *is

*ρ*(*x*)*dx *=
1
*π
*

*dx
*

*r *sin *θ
*=

1
*π
*

*dx
*

*r
*√

1 *− *(*x/r*)2
=

*dx
*

*π
√
r*2 *− x*2

*.
*

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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.

10 *CHAPTER 1. THE WAVE FUNCTION
*

ρ(x)

xr 2r-r-2r

∴ *ρ*(*x*) =
{ 1

*π
√
r*2*−x*2 *, *if *− r < x < r,
*

0*, *otherwise*.
*[*Note: *We want the *magnitude *of *dx *here.]

Total:
∫ *r
−r
*

1
*π
√
r*2*−x*2 *dx *=

2
*π
*

∫ *r
*0

1*√
r*2*−x*2 *dx *=

2
*π *sin

*−*1 *x
r
*

∣∣*r
*0

= 2*π *sin
*−*1(1) = 2*π · π*2 = 1*.*

**(b)
**

*〈x〉 *= 1
*π
*

∫ *r
−r
*

*x
*1*√
*

*r*2 *− x*2
*dx *= 0 [odd integrand, even interval].

*〈x*2*〉 *= 2
*π
*

∫ *r
*0

*x*2*√
r*2 *− x*2

*dx *=
2
*π
*

[
*−x
*

2

√
*r*2 *− x*2 + *r
*

2

2
sin*−*1

(*x
r
*

)]∣∣∣∣*r
*0

=
2
*π
*

*r*2

2
sin*−*1(1) =

*r*2

2 .

*σ*2 = *〈x*2*〉 − 〈x〉*2 = *r*2*/*2 =*⇒ σ *= *r/
√
*

2.

To get *〈x〉 *and *〈x*2*〉 *from Problem 1.11(c), use *x *= *r *cos *θ*, so *〈x〉 *= *r〈*cos *θ〉 *= 0*, 〈x*2*〉 *= *r*2*〈*cos2 *θ〉 *= *r*2*/*2.

**Problem 1.13
**

Suppose the eye end lands a distance *y *up from a line (0 *≤ y < l*), and let *x *be the projection along that same
direction (*−l ≤ x < l*). The needle crosses the line above if *y *+ *x ≥ l *(i.e. *x ≥ l − y*), and it crosses the line
below if *y *+ *x < *0 (i.e. *x < −y*). So for a given value of *y*, the probability of crossing (using Problem 1.12) is

*P *(*y*) =
∫ *−y
−l
*

*ρ*(*x*)*dx *+
∫ *l
l−y
*

*ρ*(*x*)*dx *=
1
*π
*

{∫ *−y
−l
*

1*√
l*2 *− x*2

*dx *+
∫ *l
l−y
*

1*√
l*2 *− x*2

*dx
*

}

=
1
*π
*

{
sin*−*1

(*x
l
*

)∣∣∣*−y
−l
*

+ sin*−*1
(*x
l
*

)∣∣∣*l
l−y
*

} =

1
*π
*

[
*− *sin*−*1(*y/l*) + 2 sin*−*1(1) *− *sin*−*1(1 *− y/l*)

]
= 1 *− *sin

*−*1(*y/l*)
*π
*

*− *sin
*−*1(1 *− y/l*)

*π
.
*

Now, all values of *y *are equally likely, so *ρ*(*y*) = 1*/l*, and hence the probability of crossing is

*P *=
1
*πl
*

∫ *l
*0

[
*π − *sin*−*1

(*y
l
*

)
*− *sin*−*1

(
*l − y
l
*

)]
*dy *=

1
*πl
*

∫ *l
*0

[
*π − *2 sin*−*1(*y/l*)

]
*dy
*

=
1
*πl
*

[
*πl − *2

(
*y *sin*−*1(*y/l*) + *l
*

√
1 *− *(*y/l*)2

)∣∣∣*l
*0

]
= 1 *− *2

*πl
*[*l *sin*−*1(1) *− l*] = 1 *− *1 + 2

*π
*=

2
*π
*

.

*©*2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.

*CHAPTER 1. THE WAVE FUNCTION *11

**Problem 1.14
**

**(a) ***Pab*(*t*) =
∫ *b
a
|*Ψ(*x, t*)2*dx, *so *dPabdt *=

∫ *b
a
*

*∂
∂t |*Ψ*|*2*dx. *But (Eq. 1.25):

*∂|*Ψ*|*2
*∂t
*

=
*∂
*

*∂x
*

[
*i*

2*m
*

(
Ψ*∗
*

*∂*Ψ
*∂x
*

*− ∂*Ψ
*∗
*

*∂x
*Ψ

)]
= *− ∂
*

*∂t
J*(*x, t*)*.
*

∴ *dPab
dt
*

= *−
*∫ *b
a
*

*∂
*

*∂x
J*(*x, t*)*dx *= *− *[*J*(*x, t*)]*|ba *= *J*(*a, t*) *− J*(*b, t*)*. *QED

Probability is dimensionless, so *J *has the dimensions 1/time, and units seconds*−*1.

**(b) **Here Ψ(*x, t*) = *f*(*x*)*e−iat*, where *f*(*x*) *≡ Ae−amx*2*/*, so Ψ*∂*Ψ*∗∂x *= *fe−iat
df
dxe
*

*iat *= *f dfdx *,

and Ψ*∗ ∂*Ψ*∂x *= *f
df
dx *too, so *J*(*x, t*) = 0.

**Problem 1.15
**

**(a) **Eq. 1.24 now reads *∂*Ψ
*∗
*

*∂t *= *− i*2*m ∂
*2Ψ*∗
*

*∂x*2 +
*i
*
*V ∗*Ψ*∗*, and Eq. 1.25 picks up an extra term:

*∂
*

*∂t
|*Ψ*|*2 = *· · · *+ *i
*

*|*Ψ*|*2(*V ∗ − V *) = *· · · *+ *i
*

*|*Ψ*|*2(*V*0 + *i*Γ *− V*0 + *i*Γ) = *· · · −
*

2Γ
*|*Ψ*|*2*,
*

and Eq. 1.27 becomes *dPdt *= *− *2Γ
∫ *∞
−∞ |*Ψ*|*2*dx *= *−
*

2Γ
*P *. QED

**(b)
**

*dP
*

*P
*= *−*2Γ

*dt *=*⇒ *ln*P *= *−*2Γ

*t *+ constant =*⇒ P *(*t*) = *P *(0)*e−*2Γ*t/*, so *τ *=

2Γ
*.
*

**Problem 1.16
**

Use Eqs. [1.23] and [1.24], and integration by parts:

*d
*

*dt
*

∫ *∞
−∞
*

Ψ*∗*1Ψ2 *dx *=
∫ *∞
−∞
*

*∂
*

*∂t
*(Ψ*∗*1Ψ2) *dx *=

∫ *∞
−∞
*

(
*∂*Ψ*∗*1
*∂t
*

Ψ2 + Ψ*∗*1
*∂*Ψ2
*∂t
*

)
*dx
*

=
∫ *∞
−∞
*

[(*−i*
2*m
*

*∂*2Ψ*∗*1
*∂x*2

+
*i
*

*V *Ψ*∗*1

)
Ψ2 + Ψ*∗*1

(
*i*

2*m
∂*2Ψ2
*∂x*2

*− i
*
*V *Ψ2

)]
*dx
*

= *− i*
2*m
*

∫ *∞
−∞
*

(
*∂*2Ψ*∗*1
*∂x*2

Ψ2 *− *Ψ*∗*1
*∂*2Ψ2
*∂x*2

)
*dx
*

= *− i*
2*m
*

[
*∂*Ψ*∗*1
*∂x
*

Ψ2

∣∣∣∣*∞
−∞
*

*−
*∫ *∞
−∞
*

*∂*Ψ*∗*1
*∂x
*

*∂*Ψ2
*∂x
*

*dx− *Ψ*∗*1
*∂*Ψ2
*∂x
*

∣∣∣∣*∞
−∞
*

+
∫ *∞
−∞
*

*∂*Ψ*∗*1
*∂x
*

*∂*Ψ2
*∂x
*

*dx
*

]
= 0*. *QED

*©*2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.

12 *CHAPTER 1. THE WAVE FUNCTION
*

**Problem 1.17
**

**(a)
**

1 = *|A|*2
∫ *a
−a
*

(
*a*2 *− x*2

)2
*dx *= 2*|A|*2

∫ *a
*0

(
*a*4 *− *2*a*2*x*2 + *x*4

)
*dx *= 2*|A|*2

[
*a*4*x− *2*a*2*x
*

3

3 +

*x*5

5

]∣∣∣∣*a
*0

= 2*|A|*2*a*5
(

1 *− *2
3

+ 1 5

) =

16 15

*a*5*|A|*2*, *so *A *=
√

15
16*a*5

*.
*

**(b)
**

*〈x〉 *=
∫ *a
−a
*

*x|*Ψ*|*2 *dx *= 0. (Odd integrand.)

**(c)
**

*〈p〉 *=
*i
A*2

∫ *a
−a
*

(
*a*2 *− x*2

) *d
dx
*

(
*a*2 *− x*2

) ︸ ︷︷ ︸

*−*2*x
*

*dx *= 0. (Odd integrand.)

Since we only know *〈x〉 *at *t *= 0 we cannot calculate *d〈x〉/dt *directly.

**(d)
**

*〈x*2*〉 *= *A*2
∫ *a
−a
*

*x*2
(
*a*2 *− x*2

)2
*dx *= 2*A*2

∫ *a
*0

(
*a*4*x*2 *− *2*a*2*x*4 + *x*6

)
*dx
*

= 2 15

16*a*5

[
*a*4

*x*3

3
*− *2*a*2*x
*

5

5 +

*x*7

7

]∣∣∣∣*a
*0

=
15
8*a*5

(
*a*7

)(1
3
*− *2

5 +

1 7

)

= ✚
15*a*2

8

(
35 *− *42 + 15

✁3 *· *✁5 *· *7

) =

*a*2

8
*· *8
7

=
*a*2

7
*.
*

**(e)
**

*〈p*2*〉 *= *−A*22
∫ *a
−a
*

(
*a*2 *− x*2

) *d*2
*dx*2

(
*a*2 *− x*2

) ︸ ︷︷ ︸

*−*2

*dx *= 2*A*222
∫ *a
*

0

(
*a*2 *− x*2

)
*dx
*

= 4 *· *15
16*a*5

2

(
*a*2*x− x
*

3

3

)∣∣∣∣*a
*0

= 152

4*a*5

(
*a*3 *− a
*

3

3

) =

152

4*a*2
*· *2
3

= 5 2

2

*a*2
*.
*

**(f)
**

*σx *=
√
*〈x*2*〉 − 〈x〉*2 =

√
1
7
*a*2 =

*a√
*7
*.
*

**(g)
**

*σp *=
√
*〈p*2*〉 − 〈p〉*2 =

√ 5 2 2

*a*2
=

√ 5 2

*a
.
*

*©*2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.

*CHAPTER 1. THE WAVE FUNCTION *13

**(h)
**

*σxσp *=
*a√
*7
*·
*√

5 2

*a
*=

√ 5 14 =

√ 10 7

2
*>
*

2
*. *

**Problem 1.18
**

*h√
*3*mkBT
*

*> d ⇒ T < h
*2

3*mkBd*2
*.
*

**(a) **Electrons (*m *= 9*.*1 *× *10*−*31 kg):

*T <
*(6*.*6 *× *10*−*34)2

3(9*.*1 *× *10*−*31)(1*.*4 *× *10*−*23)(3 *× *10*−*10)2 = 1*.*3 *× *10
5 K.

Sodium nuclei (*m *= 23*mp *= 23(1*.*7 *× *10*−*27) = 3*.*9 *× *10*−*26 kg):

*T <
*(6*.*6 *× *10*−*34)2

3(3*.*9 *× *10*−*26)(1*.*4 *× *10*−*23)(3 *× *10*−*10)2 = 3*.*0 K.

**(b) ***PV *= *NkBT *; volume occupied by one molecule (*N *= 1*, V *= *d*3) *⇒ d *= (*kBT/P *)1*/*3*.
*

*T <
h*2

2*mkB
*

(
*P
*

*kBT
*

)2*/*3
*⇒ T *5*/*3 *< h
*

2

3*m
P *2*/*3

*k
*5*/*3
*B
*

*⇒ T < *1
*kB
*

(
*h*2

3*m
*

)3*/*5
*P *2*/*5*.
*

For helium (*m *= 4*mp *= 6*.*8 *× *10*−*27 kg) at 1 atm = 1*.*0 *× *105 N/m2:

*T <
*1

(1*.*4 *× *10*−*23)

(
(6*.*6 *× *10*−*34)2
3(6*.*8 *× *10*−*27)

)3*/*5
(1*.*0 *× *105)2*/*5 = 2.8 K.

For hydrogen (*m *= 2*mp *= 3*.*4 *× *10*−*27 kg) with *d *= 0*.*01 m:

*T <
*(6*.*6 *× *10*−*34)2

3(3*.*4 *× *10*−*27)(1*.*4 *× *10*−*23)(10*−*2)2 = 3*.*1 *× *10
*−*14 K.

At 3 K it is definitely in the classical regime.

*©*2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.

14 *CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
*

**Chapter 2
**

**Time-Independent Schrödinger
Equation
**

**Problem 2.1
**

**(a)
**

Ψ(*x, t*) = *ψ*(*x*)*e−i*(*E*0+*i*Γ)*t/* = *ψ*(*x*)*e*Γ*t/**e−iE*0*t/* =*⇒ |*Ψ*|*2 = *|ψ|*2*e*2Γ*t/**.
*

∫ *∞
−∞
*

*|*Ψ(*x, t*)*|*2*dx *= *e*2Γ*t/*
∫ *∞
−∞
*

*|ψ|*2*dx.
*

The second term is independent of *t*, so if the product is to be 1 for all time, the first term (*e*2Γ*t/*) must
also be constant, and hence Γ = 0. QED

**(b) **If *ψ *satisfies Eq. 2.5, *− *22*m
∂*2*ψ
dx*2 + *V ψ *= *Eψ*, then (taking the complex conjugate and noting that *V *and

*E *are real): *− *22*m
∂*2*ψ∗
*

*dx*2 + *V ψ
∗ *= *Eψ∗*, so *ψ∗ also *satisfies Eq. 2.5. Now, if *ψ*1 and *ψ*2 satisfy Eq. 2.5, so

too does any linear combination of them (*ψ*3 *≡ c*1*ψ*1 + *c*2*ψ*2):

*− *
2

2*m
∂*2*ψ*3
*dx*2

+ *V ψ*3 = *−
*

2

2*m
*

(
*c*1

*∂*2*ψ*1
*dx*2

+ *c*2
*∂*2*ψ*2
*∂x*2

)
+ *V *(*c*1*ψ*1 + *c*2*ψ*2)

= *c*1

[
*− *

2

2*m
d*2*ψ*1
*dx*2

+ *V ψ*1

]
+ *c*2

[
*− *

2

2*m
d*2*ψ*2
*dx*2

+ *V ψ*2

]
= *c*1(*Eψ*1) + *c*2(*Eψ*2) = *E*(*c*1*ψ*1 + *c*2*ψ*2) = *Eψ*3*.
*

Thus, (*ψ *+ *ψ∗*) and *i*(*ψ − ψ∗*) – both of which are *real *– satisfy Eq. 2.5. *Conclusion: *From any complex
solution, we can always construct two *real *solutions (of course, if *ψ *is already real, the second one will be
zero). In particular, since *ψ *= 12 [(*ψ *+ *ψ
*

*∗*) *− i*(*i*(*ψ − ψ∗*))]*, ψ *can be expressed as a linear combination of
two real solutions. QED

**(c) **If *ψ*(*x*) satisfies Eq. 2.5, then, changing variables *x → −x *and noting that *∂*2*/∂*(*−x*)2 = *∂*2*/∂x*2,

*− *
2

2*m
∂*2*ψ*(*−x*)

*dx*2
+ *V *(*−x*)*ψ*(*−x*) = *Eψ*(*−x*);

so if *V *(*−x*) = *V *(*x*) then *ψ*(*−x*) *also *satisfies Eq. 2.5. It follows that *ψ*+(*x*) *≡ ψ*(*x*) + *ψ*(*−x*) (which is
*even*: *ψ*+(*−x*) = *ψ*+(*x*)) and *ψ−*(*x*) *≡ ψ*(*x*) *− ψ*(*−x*) (which is *odd*: *ψ−*(*−x*) = *−ψ−*(*x*)) both satisfy Eq.

*©*2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.

*CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION *15

2.5. But *ψ*(*x*) = 12 (*ψ*+(*x*) + *ψ−*(*x*)), so any solution can be expressed as a linear combination of even and
odd solutions. QED

**Problem 2.2
**

Given *d
*2*ψ
*

*dx*2 =
2*m
*2

[*V *(*x*)*−E*]*ψ*, if *E < V*min, then *ψ′′ *and *ψ *always have the same sign: If *ψ *is positive(negative),
then *ψ′′ *is also positive(negative). This means that *ψ *always curves away from the axis (see Figure). However,
it has got to go to zero as *x → −∞ *(else it would not be normalizable). At some point it’s got to *depart *from
zero (if it *doesn’t*, it’s going to be identically zero *everywhere*), in (say) the positive direction. At this point its
slope is positive, and *increasing*, so *ψ *gets bigger and bigger as *x *increases. It can’t ever “turn over” and head
back toward the axis, because that would requuire a negative second derivative—it always has to bend away
from the axis. By the same token, if it starts out heading negative, it just runs more and more negative. In
neither case is there any way for it to come back to zero, as it must (at *x → ∞*) in order to be normalizable.
QED

x

ψ

**Problem 2.3
**

Equation 2.20 says *d
*2*ψ
*

*dx*2 = *− *2*mE*2 *ψ*; Eq. 2.23 says *ψ*(0) = *ψ*(*a*) = 0. If *E *= 0, *d*2*ψ/dx*2 = 0, so *ψ*(*x*) = *A *+ *Bx*;
*ψ*(0) = *A *= 0 *⇒ ψ *= *Bx*; *ψ*(*a*) = *Ba *= 0 *⇒ B *= 0, so *ψ *= 0. If *E < *0, *d*2*ψ/dx*2 = *κ*2*ψ*, with *κ ≡
*

*√
−*2*mE/*

real, so *ψ*(*x*) = *Aeκx *+ *Be−κx*. This time *ψ*(0) = *A *+ *B *= 0 *⇒ B *= *−A*, so *ψ *= *A*(*eκx − e−κx*), while
*ψ*(*a*) = *A
*

(
*eκa − eiκa
*

)
= 0 *⇒ *either *A *= 0, so *ψ *= 0, or else *eκa *= *e−κa*, so *e*2*κa *= 1, so 2*κa *= ln(1) = 0,

so *κ *= 0, and again *ψ *= 0. In all cases, then, the boundary conditions force *ψ *= 0, which is unacceptable
(non-normalizable).

**Problem 2.4
**

*〈x〉 *=
∫

*x|ψ|*2*dx *= 2
*a
*

∫ *a
*0

*x *sin2
(*nπ
*

*a
x
*)
*dx. *Let *y ≡ nπ
*

*a
x, *so *dx *=

*a
*

*nπ
dy*; *y *: 0 *→ nπ.
*

=
2
*a
*

( *a
nπ
*

)2 ∫ *nπ
*0

*y *sin2 *y dy *=
2*a
*

*n*2*π*2

[
*y*2

4
*− y *sin 2*y
*

4
*− *cos 2*y
*

8

]∣∣∣∣*nπ
*0

=
2*a
*

*n*2*π*2

[
*n*2*π*2

4
*− *cos 2*nπ
*

8 +

1 8

] =

*a
*

2
. (Independent of *n.*)

*©*2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.

16 *CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
*

*〈x*2*〉 *= 2
*a
*

∫ *a
*0

*x*2 sin2
(*nπ
*

*a
x
*)
*dx *=

2
*a
*

( *a
nπ
*

)3 ∫ *nπ
*0

*y*2 sin2 *y dy
*

=
2*a*2

(*nπ*)3

[
*y*3

6
*−
*

(
*y*3

4
*− *1

8

)
sin 2*y − y *cos 2*y
*

4

]*nπ
*0

=
2*a*2

(*nπ*)3

[
(*nπ*)3

6
*− nπ *cos(2*nπ*)

4

]
= *a*2

[
1
3
*− *1

2(*nπ*)2

]
*.
*

*〈p〉 *= *md〈x〉
dt
*

= 0. (*Note *: Eq*. *1*.*33 is much faster than Eq*. *1*.*35*.*)

*〈p*2*〉 *=
∫

*ψ∗n
*

(

*i
*

*d
*

*dx
*

)2
*ψn dx *= *−*2

∫
*ψ∗n
*

(
*d*2*ψn
dx*2

)
*dx
*

= (*−*2)
(
*−*2*mEn
*

2

) ∫
*ψ∗nψn dx *= 2*mEn *=

(
*nπ*

*a
*

)2
*.
*

*σ*2*x *= *〈x*2*〉 − 〈x〉*2 = *a*2
(

1
3
*− *1

2(*nπ*)2
*− *1

4

) =

*a*2

4

(
1
3
*− *2

(*nπ*)2

)
; *σx *=

*a
*

2

√
1
3
*− *2

(*nπ*)2
.

*σ*2*p *= *〈p*2*〉 − 〈p〉*2 =
(
*nπ*

*a
*

)2
; *σp *=

*nπ*

*a
*. ∴ *σxσp *=

2

√
(*nπ*)2

3
*− *2.

The product *σxσp *is smallest for *n *= 1; in that case, *σxσp *= 2

√
*π*2

3 *− *2 = (1*.*136)*/*2 *> **/*2*. *

**Problem 2.5
**

**(a)
**

*|*Ψ*|*2 = Ψ2Ψ = *|A|*2(*ψ∗*1 + *ψ∗*2)(*ψ*1 + *ψ*2) = *|A|*2[*ψ∗*1*ψ*1 + *ψ∗*1*ψ*2 + *ψ∗*2*ψ*1 + *ψ∗*2*ψ*2]*.
*

1 = ∫

*|*Ψ*|*2*dx *= *|A|*2
∫

[*|ψ*1*|*2 + *ψ∗*1*ψ*2 + *ψ∗*2*ψ*1 + *|ψ*2*|*2]*dx *= 2*|A|*2 *⇒ A *= 1*/
√
*

2.

**(b)
**

Ψ(*x, t*) =
1*√
*2

[
*ψ*1*e
*

*−iE*1*t/* + *ψ*2*e−iE*2*t/*
]

(but
*En
*

= *n*2*ω*)

=
1*√
*2

√
2
*a
*

[ sin

(*π
a
x
*)
*e−iωt *+ sin

(
2*π
a
x
*

)
*e−i*4*ωt
*

] =

1*√
a
e−iωt
*

[ sin

(*π
a
x
*)

+ sin (

2*π
a
x
*

)
*e−*3*iωt
*

] .

*|*Ψ(*x, t*)*|*2 = 1
*a
*

[ sin2

(*π
a
x
*)

+ sin
(*π
a
x
*)

sin (

2*π
a
x
*

) (
*e−*3*iωt *+ *e*3*iωt
*

) + sin2

(
2*π
a
x
*

)] =

1
*a
*

[ sin2

(*π
a
x
*)

+ sin2 (

2*π
a
x
*

) + 2 sin

(*π
a
x
*)

sin (

2*π
a
x
*

)
cos(3*ωt*)

]
*.
*

*©*2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.

*CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION *17

**(c)
**

*〈x〉 *=
∫

*x|*Ψ(*x, t*)*|*2*dx
*

=
1
*a
*

∫ *a
*0

*x
*

[ sin2

(*π
a
x
*)

+ sin2 (

2*π
a
x
*

) + 2 sin

(*π
a
x
*)

sin (

2*π
a
x
*

)
cos(3*ωt*)

]
*dx
*

∫ *a
*0

*x *sin2
(*π
a
x
*)
*dx *=

[
*x*2

4
*− x *sin

(
2*π
a x
*

)
4*π/a
*

*− *cos
(

2*π
a x
*

)
8(*π/a*)2

]∣∣∣∣∣
*a
*

0

=
*a*2

4 =

∫ *a
*0

*x *sin2
(

2*π
a
x
*

)
*dx.
*

∫ *a
*0

*x *sin
(*π
a
x
*)

sin (

2*π
a
x
*

)
*dx *=

1 2

∫ *a
*0

*x
*

[ cos

(*π
a
x
*)
*− *cos

(
3*π
a
x
*

)]
*dx
*

= 1 2

[
*a*2

*π*2
cos

(*π
a
x
*)

+
*ax
*

*π
*sin

(*π
a
x
*)
*− a
*

2

9*π*2
cos

(
3*π
a
x
*

)
*− ax
*

3*π
*sin

(
3*π
a
x
*

)]*a
*0

= 1 2

[
*a*2

*π*2
(
cos(*π*) *− *cos(0)

)
*− a
*

2

9*π*2
(
cos(3*π*) *− *cos(0)

)]
= *−a
*

2

*π*2

(
1 *− *1

9

)
= *− *8*a
*

2

9*π*2
*.
*

∴ *〈x〉 *= 1
*a
*

[
*a*2

4 +

*a*2

4
*− *16*a
*

2

9*π*2
cos(3*ωt*)

] =

*a
*

2

[
1 *− *32

9*π*2
cos(3*ωt*)

] .

*Amplitude:
*32
9*π*2

(*a
*2

)
= 0*.*3603(*a/*2); *angular frequency: *3*ω *=

3*π*2
2*ma*2

*.
*

**(d)
**

*〈p〉 *= *md〈x〉
dt
*

= *m
*(*a
*

2

) (
*− *32

9*π*2

)
(*−*3*ω*) sin(3*ωt*) = 8

3*a
*sin(3*ωt*)*.
*

**(e) **You could get either *E*1 = *π*22*/*2*ma*2 or *E*2 = 2*π*22*/ma*2, with equal probability *P*1 = *P*2 = 1*/*2*.
*

So *〈H〉 *= 1
2
(*E*1 + *E*2) =

5*π*22

4*ma*2
; it’s the *average *of *E*1 and *E*2.

**Problem 2.6
**

From Problem 2.5, we see that

Ψ(*x, t*) = 1*√
a
e−iωt
*

[ sin

(
*π
ax
*

) + sin

(
2*π
a x
*

)
*e−*3*iωteiφ
*

] ;

*|*Ψ(*x, t*)*|*2 = 1*a
*[
sin2

(
*π
ax
*

) + sin2

(
2*π
a x
*

) + 2 sin

(
*π
ax
*

) sin

(
2*π
a x
*

)
cos(3*ωt− φ*)

] ;

*©*2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.

18 *CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
*

and hence *〈x〉 *= *a*2
[
1 *− *329*π*2 cos(3*ωt− φ*)

] . This amounts physically to starting the clock at a different time

(i.e., shifting the *t *= 0 point).

If *φ *=
*π
*

2
*, *so Ψ(*x, *0) = *A*[*ψ*1(*x*) + *iψ*2(*x*)]*, *then cos(3*ωt− φ*) = sin(3*ωt*); *〈x〉 *starts at

*a
*

2
*.
*

If *φ *= *π, *so Ψ(*x, *0) = *A*[*ψ*1(*x*) *− ψ*2(*x*)]*, *then cos(3*ωt− φ*) = *− *cos(3*ωt*); *〈x〉 *starts at
*a
*

2

( 1 +

32
9*π*2

)
*.
*

**Problem 2.7
**

Ψ(x,0)

xaa/2

Aa/2

**(a)
**

1 = *A*2
∫ *a/*2

0

*x*2*dx *+ *A*2
∫ *a
a/*2

(*a− x*)2*dx *= *A*2
[
*x*3

3

∣∣∣∣*a/*2
0

*− *(*a− x*)
3

3

∣∣∣∣*a
a/*2

]

=
*A*2

3

(
*a*3

8 +

*a*3

8

) =

*A*2*a*3

12
*⇒ A *= 2

*√
*3*√
a*3

*.
*

**(b)
**

*cn *=

√
2
*a
*

2
*√
*

3
*a
√
a
*

[ ∫ *a/*2
0

*x *sin
(
*nπ
*

*a
x
*

)
*dx *+

∫ *a
a/*2

(*a− x*) sin
(
*nπ
*

*a
x
*

)
*dx
*

] =

2
*√
*

6
*a*2

{[(
*a
*

*nπ
*

)2 sin

(
*nπ
*

*a
x
*

)
*− xa
*

*nπ
*cos

(
*nπ
*

*a
x
*

)]∣∣∣∣*a/*2
0

+ *a
*[
*− a
*

*nπ
*cos

(
*nπ
*

*a
x
*

)]∣∣∣∣*a
a/*2

*−
*[(

*a
*

*nπ
*

)2 sin

(
*nπ
*

*a
x
*

)
*−
*

(
*ax
*

*nπ
*

) cos

(
*nπ
*

*a
x
*

)]∣∣∣∣*a
a/*2

}

=
2
*√
*

6
*a*2

[(
*a
*

*nπ
*

)2 sin

(
*nπ
*

2

)
*−*✘✘✘

✘✘✘
✘*a*2

2*nπ
*cos

(
*nπ
*

2

)
*−
*✟✟

✟✟
✟*a*2

*nπ
*cos*nπ *+

✟✟ ✟✟

✟✟*a*2

*nπ
*cos

(
*nπ
*

2

) +

(
*a
*

*nπ
*

)2 sin

(
*nπ
*

2

) + ✟✟

✟✟
✟*a*2

*nπ
*cos*nπ −*✘✘✘

✘✘✘
✘*a*2

2*nπ
*cos

(
*nπ
*

2

)] =

2
*√
*

6

*a*2
2

*a*2

(*nπ*)2
sin

(
*nπ
*

2

) =

4
*√
*

6
(*nπ*)2

sin
(
*nπ
*

2

) =

{
0*, n *even,
(*−*1)(*n−*1)*/*2 4

*√
*6

(*nπ*)2 *, n *odd*.
*

So Ψ(*x, t*) =
4
*√
*

6
*π*2

√
2
*a
*

∑
*n*=1*,*3*,*5*,...
*

(*−*1)(*n−*1)*/*2 1
*n*2

sin
(
*nπ
*

*a
x
*

)
*e−Ent/**, *where *En *=

*n*2*π*22

2*ma*2
*.
*

*©*2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.

*CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION *19

**(c)
**

*P*1 = *|c*1*|*2 =
16 *· *6
*π*4

= 0*.*9855*.
*

**(d)
**

*〈H〉 *=
∑

*|cn|*2*En *=
96
*π*4

*π*22

2*ma*2

( 1 1

+ 1 32

+ 1 52

+ 1 72

+ *· · ·*︸ ︷︷ ︸
*π*2*/*8

) =

482

*π*2*ma*2
*π*2

8 =

62

*ma*2
*.
*

**Problem 2.8
**

**(a)
**

Ψ(*x, *0) =

{
*A, *0 *< x < a/*2;
0*, *otherwise*.
*

1 = *A*2
∫ *a/*2

0

*dx *= *A*2(*a/*2) *⇒ A *=
√

2
*a
.
*

**(b) **From Eq. 2.37,

*c*1 = *A
*

√
2
*a
*

∫ *a/*2
0

sin
(*π
a
x
*)
*dx *=

2
*a
*

[
*−a
π
*

cos
(*π
a
x
*)] ∣∣∣∣*a/*2

0

= *− *2
*π
*

[ cos

(*π
*2

)
*− *cos 0

] =

2
*π
.
*

*P*1 = *|c*1*|*2 = (2*/π*)2 = 0*.*4053*.
*

**Problem 2.9
**

*ĤΨ(x, *0) = *− *
2

2*m
∂*2

*∂x*2
[*Ax*(*a− x*)] = *−A *

2

2*m
∂
*

*∂x
*(*a− *2*x*) = *A*

2

*m
.
*

∫
Ψ(*x, *0)∗ĤΨ(x, 0) *dx *= *A*2

2

*m
*

∫ *a
*0

*x*(*a− x*) *dx *= *A*2
2

*m
*

(
*a
x*2

2
*− x
*

3

3

) ∣∣∣∣*a
*0

= *A*2

2

*m
*

(
*a*3

2
*− a
*

3

3

) =

30
*a*5

2

*m
*

*a*3

6 =

52

*ma*2

(same as Example 2.3).

*©*2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.

20 *CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
*

**Problem 2.10
**

**(a) **Using Eqs. 2.47 and 2.59,

*a*+*ψ*0 =
1*√
*

2*mω
*

(
*−* *d
*

*dx
*+ *mωx
*

) (*mω
π*

)1*/*4
*e−
*

*mω
*2 *x
*

2

=
1*√
*

2*mω
*

(*mω
π*

)1*/*4 [
*−*

(
*−mω
*

2

)
2*x *+ *mωx
*

]
*e−
*

*mω
*2 *x
*

2 =

1*√
*2*mω
*

(*mω
π*

)1*/*4
2*mωxe−
*

*mω
*2 *x
*

2
*.
*

(*a*+)2*ψ*0 =
1

2*mω
*

(*mω
π*

)1*/*4
2*mω
*

(
*−* *d
*

*dx
*+ *mωx
*

)
*xe−
*

*mω
*2 *x
*

2

= 1

(*mω
π*

)1*/*4 [
*−*

(
1 *− xmω
*

2
2*x
*

)
+ *mωx*2

]
*e−
*

*mω
*2 *x
*

2 =

(*mω
π*

)1*/*4 (2*mω
*

*x*2 *− *1
)
*e−
*

*mω
*2 *x
*

2
*.
*

Therefore, from Eq. 2.67,

*ψ*2 =
1*√
*2
(*a*+)2*ψ*0 =

1*√
*2

(*mω
π*

)1*/*4 (2*mω
*

*x*2 *− *1
)
*e−
*

*mω
*2 *x
*

2
*.
*

**(b)
**ψ ψ ψ1 20

**(c) **Since *ψ*0 and *ψ*2 are even, whereas *ψ*1 is odd,
∫
*ψ∗*0*ψ*1*dx *and

∫
*ψ∗*2*ψ*1*dx *vanish automatically. The only one

we need to check is
∫
*ψ∗*2*ψ*0 *dx*:∫

*ψ∗*2*ψ*0 *dx *=
1*√
*2

√
*mω
*

*π*

∫ *∞
−∞
*

(
2*mω
*

*x*2 *− *1
)
*e−
*

*mω
*

*x*2*dx
*

= *−
*√

*mω
*

2*π*

( ∫ *∞
−∞
*

*e−
mω
*

*x*2*dx− *2*mω
*

∫ *∞
−∞
*

*x*2*e−
mω
*

*x*2*dx
*

)
= *−
*

√
*mω
*

2*π*

(√
*π*

*mω
− *2*mω
*

2*mω
*

√
*π*

*mω
*

)
= 0*. *

**Problem 2.11
**

**(a) **Note that *ψ*0 is even, and *ψ*1 is odd. In either case *|ψ|*2 is even, so *〈x〉 *=
∫
*x|ψ|*2*dx *= 0*. *Therefore

*〈p〉 *= *md〈x〉/dt *= 0*. *(These results hold for *any *stationary state of the harmonic oscillator.)
From Eqs. 2.59 and 2.62, *ψ*0 = *αe−ξ
*

2*/*2*, ψ*1 =
*√
*

2*αξe−ξ
*2*/*2. So

*n *= 0**:
**

*〈x*2*〉 *= *α*2
∫ *∞
−∞
*

*x*2*e−ξ
*2*/*2*dx *= *α*2

(

*mω
*

)3*/*2 ∫ *∞
−∞
*

*ξ*2*e−ξ
*2
*dξ *=

1*√
π
*

(

*mω
*

)*√
π
*

2 =

2*mω
.
*

*©*2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.

*CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION *21

*〈p*2*〉 *=
∫

*ψ*0

(

*i
*

*d
*

*dx
*

)2
*ψ*0 *dx *= *−*2*α*2

√
*mω
*

∫ *∞
−∞
*

*e−ξ
*2*/*2

(
*d*2

*dξ*2
*e−ξ
*

2*/*2

)
*dξ
*

= *−m**ω√
π
*

∫ *∞
−∞
*

(
*ξ*2 *− *1

)
*e−ξ
*

2*/*2*dξ *= *−m**ω√
π
*

(*√
π
*

2
*−
√
π
*

) =

*m**ω
*

2
*.
*

*n *= 1**:
**

*〈x*2*〉 *= 2*α*2
∫ *∞
−∞
*

*x*2*ξ*2*e−ξ
*2
*dx *= 2*α*2

(

*mω
*

)3*/*2 ∫ *∞
−∞
*

*ξ*4*e−ξ
*2
*dξ *=

2*√
πmω
*

3
*√
π
*

4 =

3
2*mω
*

*.
*

*〈p*2*〉 *= *−*22*α*2
√

*mω
*

∫ *∞
−∞
*

*ξe−ξ
*2*/*2

[
*d*2

*dξ*2
(
*ξe−ξ
*

2*/*2
)]
*dξ
*

= *−*2*mω**√
π
*

∫ *∞
−∞
*

(
*ξ*4 *− *3*ξ*2

)
*e−ξ
*

2
*dξ *= *−*2*mω**√
*

*π
*

(
3
4
*√
π − *3

*√
π
*

2

) =

3*m**ω
*2

*.
*

**(b) ***n *= 0**:
**

*σx *=
√
*〈x*2*〉 − 〈x〉*2 =

√

2*mω
*; *σp *=

√
*〈p*2*〉 − 〈p〉*2 =

√
*m**ω
*

2 ;

*σxσp *=

√

2*mω
*

√
*mω*

2 =

2
*. *(Right *at *the uncertainty limit.)

*n *= 1**:
**

*σx *=

√ 3

2*mω
*; *σp *=

√
3*m**ω
*

2
; *σxσp *= 3

2
*>
*

2
*. *

**(c)
**

*〈T 〉 *= 1
2*m
*

*〈p*2*〉 *=

1
4*ω *(*n *= 0)

3
4*ω *(*n *= 1)

; *〈V 〉 *= 12*mω*2*〈x*2*〉 *=

1
4*ω *(*n *= 0)

3
4*ω *(*n *= 1)

*.
*

*〈T 〉 *+ *〈V 〉 *= *〈H〉 *=

1
2*ω *(*n *= 0) = *E*0

3
2*ω *(*n *= 1) = *E*1

*, *as expected.

**Problem 2.12
**

From Eq. 2.69,

*x *=

√

2*mω
*(*a*+ + *a−*)*, p *= *i
*

√
*mω
*

2
(*a*+ *− a−*)*,
*

so

*〈x〉 *=
√

2*mω
*

∫
*ψ∗n*(*a*+ + *a−*)*ψn dx.
*

*©*2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.

22 *CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
*

But (Eq. 2.66)
*a*+*ψn *=

*√
n *+ 1*ψn*+1*, a−ψn *=

*√
nψn−*1*.
*

So

*〈x〉 *=
√

2*mω
*

[*√
n *+ 1

∫
*ψ∗nψn*+1 *dx *+

*√
n
*

∫
*ψ∗nψn−*1 *dx
*

]
= 0 (by orthogonality)*.
*

*〈p〉 *= *md〈x〉
dt
*

= 0*. x*2 =

2*mω
*(*a*+ + *a−*)2 =

2*mω
*(
*a*2+ + *a*+*a− *+ *a−a*+ + *a
*

2
*−
*

)
*.
*

*〈x*2*〉 *=
2*mω
*

∫
*ψ∗n
*

(
*a*2+ + *a*+*a− *+ *a−a*+ + *a
*

2
*−
*

)
*ψn. *But

*a*2+*ψn *= *a*+

(*√
n *+ 1*ψn*+1

) =

*√
n *+ 1

*√
n *+ 2*ψn*+2 =

√
(*n *+ 1)(*n *+ 2)*ψn*+2*.
*

*a*+*a−ψn *= *a*+
(*√
*

*nψn−*1
)

=
*√
n
√
nψn *= *nψn.
*

*a−a*+*ψn *= *a−
*(*√
*

*n *+ 1*ψn*+1
)

=
√
*n *+ 1)

*√
n *+ 1*ψn *= (*n *+ 1)*ψn.
*

*a*2*−ψn *= *a−
*(*√
*

*nψn−*1
)

=
*√
n
√
n− *1*ψn−*2 =

√
(*n− *1)*nψn−*2*.
*

So

*〈x*2*〉 *=
2*mω
*

[
0 + *n
*

∫
*|ψn|*2*dx *+ (*n *+ 1)

∫
*|ψn|*2 *dx *+ 0

] =

2*mω
*(2*n *+ 1) =

(
*n *+

1 2

)

*mω
.
*

*p*2 = *−**mω
*2

(*a*+ *− a−*)2 = *−
**mω
*

2
(
*a*2+ *− a*+*a− − a−a*+ + *a*2*−
*

)
*⇒
*

*〈p*2*〉 *= *−**mω
*2

[0 *− n− *(*n *+ 1) + 0] = *mω
*2

(2*n *+ 1) =
(
*n *+

1 2

)
*m**ω.
*

*〈T 〉 *= *〈p*2*/*2*m〉 *= 1
2

(
*n *+

1 2

)
*ω .
*

*σx *=
√
*〈x*2*〉 − 〈x〉*2 =

√
*n *+

1 2

√

*mω
*; *σp *=

√
*〈p*2*〉 − 〈p〉*2 =

√
*n *+

1 2

*√
m**ω*; *σxσp *=

(
*n *+

1 2

)
*≥ *

2
*. *

**Problem 2.13
**

**(a)
**

1 = ∫

*|*Ψ(*x, *0)*|*2*dx *= *|A|*2
∫ (

9*|ψ*0*|*2 + 12*ψ∗*0*ψ*1 + 12*ψ∗*1*ψ*0 + 16*|ψ*1*|*2
)
*dx
*

= *|A|*2(9 + 0 + 0 + 16) = 25*|A|*2 *⇒ A *= 1*/*5*.
*

*©*2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.

*CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION *23

**(b)
**

Ψ(*x, t*) =
1
5

[
3*ψ*0(*x*)*e−iE*0*t/* + 4*ψ*1(*x*)*e−iE*1*t/*

] =

1 5

[
3*ψ*0(*x*)*e−iωt/*2 + 4*ψ*1(*x*)*e−*3*iωt/*2

]
*.
*

(Here *ψ*0 and *ψ*1 are given by Eqs. 2.59 and 2.62; *E*1 and *E*2 by Eq. 2.61.)

*|*Ψ(*x, t*)*|*2 = 1
25

[
9*ψ*20 + 12*ψ*0*ψ*1*e
*

*iωt/*2*e−*3*iωt/*2 + 12*ψ*0*ψ*1*e−iωt/*2*e*3*iωt/*2 + 16*ψ*21
]

= 1 25

[
9*ψ*20 + 16*ψ
*

2
1 + 24*ψ*0*ψ*1 cos(*ωt*)

]
*.
*

**(c)
**

*〈x〉 *= 1
25

[ 9

∫
*xψ*20 *dx *+ 16

∫
*xψ*21 *dx *+ 24 cos(*ωt*)

∫
*xψ*0*ψ*1 *dx
*

]
*.
*

But
∫
*xψ*20 *dx *=

∫
*xψ*21 *dx *= 0 (see Problem 2.11 or 2.12), while∫

*xψ*0*ψ*1 *dx *=
√

*mω
*

*π*

√
2*mω
*

∫
*xe−
*

*mω
*2 *x
*

2
*xe−
*

*mω
*2 *x
*

2
*dx *=

√
2
*π
*

(*mω
*

) ∫ *∞
−∞
*

*x*2*e−
mω
*

*x*2*dx
*

=

√
2
*π
*

(*mω
*

)
2
*√
π*2

( 1 2

√

*mω
*

)3 =

√

2*mω
.
*

So

*〈x〉 *= 24
25

√

2*mω
*cos(*ωt*); *〈p〉 *= *m d
*

*dt
〈x〉 *= *−*24

25

√
*mω*

2
sin(*ωt*)*.
*

(With *ψ*2 in place of *ψ*1 the frequency would be (*E*2 *− E*0)*/* = [(5*/*2)*ω − *(1*/*2)*ω*]*/* = 2*ω.*)
Ehrenfest’s theorem says *d〈p〉/dt *= *−〈∂V/∂x〉*. Here

*d〈p〉
dt
*

= *−*24
25

√
*mω*

2
*ω *cos(*ωt*)*, V *=

1
2
*mω*2*x*2 *⇒ ∂V
*

*∂x
*= *mω*2*x,
*

so

*−
*〈*∂V
∂x
*

〉
= *−mω*2*〈x〉 *= *−mω*2 24

25

√

2*mω
*cos(*ωt*) = *−*24

25

√
*mω
*

2
*ω *cos(*ωt*)*,
*

so Ehrenfest’s theorem holds.

**(d) **You could get *E*0 = 12*ω, *with probability *|c*0*|*2 = 9*/*25*, *or *E*1 = 32*ω, *with probability *|c*1*|*2 = 16*/*25*.
*

**Problem 2.14
**

The new allowed energies are *E′n *= (*n *+
1
2 )*ω
*

*′ *= 2(*n *+ 12 )*ω *= *ω, *3*ω, *5*ω, . . . *. So the probability of
getting 12*ω *is zero. The probability of getting *ω *(the new ground state energy) is *P*0 = *|c*0*|*2, where *c*0 =∫

Ψ(*x, *0)*ψ′*0 *dx*, with

Ψ(*x, *0) = *ψ*0(*x*) =
(*mω
π*

)1*/*4
*e−
*

*mω
*2 *x
*

2
*, ψ*0(*x*)*′ *=

(
*m*2*ω
π*

)1*/*4
*e−
*

*m*2*ω
*2 *x
*

2
*.
*

*©*2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.

24 *CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
*

So

*c*0 = 21*/*4
√

*mω
*

*π*

∫ *∞
−∞
*

*e−
*3*mω
*2 *x
*

2
*dx *= 21*/*4

√
*mω
*

*π*
2
*√
π
*

( 1 2

√ 2

3*mω
*

)
= 21*/*4

√
2
3
*.
*

Therefore

*P*0 =
2
3

*√
*2 = 0*.*9428*.
*

**Problem 2.15
**

*ψ*0 =
(*mω
π*

)1*/*4
*e−ξ
*

2*/*2*, *so *P *= 2
√

*mω
*

*π*

∫ *∞
x*0

*e−ξ
*2
*dx *= 2

√
*mω
*

*π*

√

*mω
*

∫ *∞
ξ*0

*e−ξ
*2
*dξ.
*

Classically allowed region extends out to: 12*mω
*2*x*20 = *E*0 =

1
2*ω, *or *x*0 =

√

*mω , *so *ξ*0 = 1*.
*

*P *=
2*√
π
*

∫ *∞
*1

*e−ξ
*2
*dξ *= 2(1 *− F *(

*√
*2)) (in notation of CRC Table) = 0*.*157*.
*

**Problem 2.16
**

*n *= 5: *j *= 1 *⇒ a*3 = *−*2(5*−*1)(1+1)(1+2)*a*1 = *− *43*a*1; *j *= 3 *⇒ a*5 =
*−*2(5*−*3)

(3+1)(3+2)*a*3 = *− *15*a*3 = 415*a*1; *j *= 5 *⇒ a*7 = 0*. *So
*H*5(*ξ*) = *a*1*ξ − *43*a*1*ξ*3 + 415*a*1*ξ*5 =

*a*1
15 (15*ξ − *20*ξ*3 + 4*ξ*5)*. *By convention the coefficient of *ξ*5 is 25, so *a*1 = 15 *· *8,

and *H*5(*ξ*) = 120*ξ − *160*ξ*3 + 32*ξ*5 (which agrees with Table 2.1).

*n *= 6: *j *= 0 *⇒ a*2 = *−*2(6*−*0)(0+1)(0+2)*a*0 = *−*6*a*0; *j *= 2 *⇒ a*4 =
*−*2(6*−*2)

(2+1)(2+2)*a*2 = *− *23*a*2 = 4*a*0; *j *= 4 *⇒ a*6 =
*−*2(6*−*4)

(4+1)(4+2)*a*4 = *− *215*a*4 = *− *815*a*0; *j *= 6 *⇒ a*8 = 0*. *So *H*6(*ξ*) = *a*0 *− *6*a*0*ξ*2 + 4*a*0*ξ*4 *− *815*ξ*6*a*0*. *The coefficient of *ξ*6

is 26, so 26 = *− *815*a*0 *⇒ a*0 = *−*15 *· *8 = *−*120. *H*6(*ξ*) = *−*120 + 720*ξ
*2 *− *480*ξ*4 + 64*ξ*6*.
*

**Problem 2.17
**

**(a)
**

*d
*

*dξ
*(*e−ξ
*

2
) = *−*2*ξe−ξ*2 ;

(
*d
*

*dξ
*

)2
*e−ξ
*

2 =

*d
*

*dξ
*(*−*2*ξe−ξ*2) = (*−*2 + 4*ξ*2)*e−ξ*2 ;

(
*d
*

*dξ
*

)3
*e−ξ
*

2 =

*d
*

*dξ
*

[
(*−*2 + 4*ξ*2)*e−ξ*2

] =

[
8*ξ *+ (*−*2 + 4*ξ*2)(*−*2*ξ*)

]
*e−ξ
*

2
= (12*ξ − *8*ξ*3)*e−ξ*2 ;

(
*d
*

*dξ
*

)4
*e−ξ
*

2 =

*d
*

*dξ
*

[
(12*ξ − *8*ξ*3)*e−ξ*2

] =

[
12 *− *24*ξ*2 + (12*ξ − *8*ξ*3)(*−*2*ξ*)

]
*e−ξ
*

2
= (12 *− *48*ξ*2 + 16*ξ*4)*e−ξ*2 *.
*

*H*3(*ξ*) = *−eξ
*2
(

*d
*

*dξ
*

)3
*e−ξ
*

2
= *−*12*ξ *+ 8*ξ*3; *H*4(*ξ*) = *eξ
*

2 (

*d
*

*dξ
*

)4
*e−ξ
*

2
= 12 *− *48*ξ*2 + 16*ξ*4*.
*

*©*2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.

*CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION *25

**(b)
**

*H*5 = 2*ξH*4 *− *8*H*3 = 2*ξ*(12 *− *48*ξ*2 + 16*ξ*4) *− *8(*−*12*ξ *+ 8*ξ*3) = 120*ξ − *160*ξ*3 + 32*ξ*5*.
*

*H*6 = 2*ξH*5 *− *10*H*4 = 2*ξ*(120*ξ − *160*ξ*3 + 32*ξ*5) *− *10(12 *− *48*ξ*2 + 16*ξ*4) = *−*120 + 720*ξ*2 *− *480*ξ*4 + 64*ξ*6*.
*

**(c)
**

*dH*5
*dξ
*

= 120 *− *480*ξ*2 + 160*ξ*4 = 10(12 *− *48*ξ*2 + 16*ξ*4) = (2)(5)*H*4*. *

*dH*6
*dξ
*

= 1440*ξ − *1920*ξ*3 + 384*ξ*5 = 12(120*ξ − *160*ξ*3 + 32*ξ*5) = (2)(6)*H*5*. *

**(d)
**

*d
*

*dz
*(*e−z
*

2+2*zξ*) = (*−*2*z *+ *ξ*)*e−z*2+2*zξ*; setting *z *= 0*, H*0(*ξ*) = 2*ξ.
*

(
*d
*

*dz
*

)2
(*e−z
*

2+2*zξ*) =
*d
*

*dz
*

[
(*−*2*z *+ 2*ξ*)*e−z*2+2*zξ
*

] =

[
*− *2 + (*−*2*z *+ 2*ξ*)2

]
*e−z
*

2+2*zξ*; setting *z *= 0*, H*1(*ξ*) = *−*2 + 4*ξ*2*.
*

(
*d
*

*dz
*

)3
(*e−z
*

2+2*zξ*) =
*d
*

*dz
*

{[
*− *2 + (*−*2*z *+ 2*ξ*)2

]
*e−z
*

2+2*zξ
*

} =

{
2(*−*2*z *+ 2*ξ*)(*−*2) +

[
*− *2 + (*−*2*z *+ 2*ξ*)2

]
(*−*2*z *+ 2*ξ*)

}
*e−z
*

2+2*zξ*;

setting *z *= 0*, H*2(*ξ*) = *−*8*ξ *+ (*−*2 + 4*ξ*2)(2*ξ*) = *−*12*ξ *+ 8*ξ*3*.
*

**Problem 2.18
**

*Aeikx *+ *Be−ikx *= *A*(cos *kx *+ *i *sin *kx*) + *B*(cos *kx− i *sin *kx*) = (*A *+ *B*) cos *kx *+ *i*(*A−B*) sin *kx
*= *C *cos *kx *+ *D *sin *kx, *with *C *= *A *+ *B*; *D *= *i*(*A−B*)*.
*

*C *cos *kx *+ *D *sin *kx *= *C
*(
*eikx *+ *e−ikx
*

2

)
+ *D
*

(
*eikx − e−ikx
*

2*i
*

) =

1
2
(*C − iD*)*eikx *+ 1

2
(*C *+ *iD*)*e−ikx
*

= *Aeikx *+ *Be−ikx, *with *A *=
1
2
(*C − iD*); *B *= 1

2
(*C *+ *iD*)*.
*

*©*2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.