# Griffiths D. J., Introduction to quantum mechanics Solutions (2nd ed., Pearson, 2005)(303s)

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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
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 (griﬃ[email protected]). 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
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.
1
1
j2 =
j2N (j) =
(142) + (152) + 3(162) + 2(222) + 2(242) + 5(252)
N
14
1
6434
=
(196 + 225 + 768 + 968 + 1152 + 3125) =
= 459.571.
14
14
j
∆j = j − j
14
14 − 21 = −7
15
15 − 21 = −6
(b)
16
16 − 21 = −5
22
22 − 21 = 1
24
24 − 21 = 3
25
25 − 21 = 4
1
1
σ2 =
(∆j)2N (j) =
(−7)2 + (−6)2 + (−5)2 · 3 + (1)2 · 2 + (3)2 · 2 + (4)2 · 5
N
14
1
260
=
(49 + 36 + 75 + 2 + 18 + 80) =
= 18.571.
14
14

σ =
18.571 = 4.309.
(c)
j2 − j 2 = 459.571 − 441 = 18.571.
[Agrees with (b).]
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publisher.

4
CHAPTER 1. THE WAVE FUNCTION
Problem 1.2
(a)
h
1
1
2
h
h2
x2 =
x2 √
dx =

x5/2
=
.
0
2 hx
2 h
5
5
0
h2
2
4
2h
σ2 = x2 − x 2 =
− h
=
h2 ⇒ σ =
√ = 0.2981h.
5
3
45
3 5
(b)
x+
1
√ x+

P = 1 −
√ dx = 1 − 1
√ (2 x)
= 1 − 1

x+ − x− .
x−
2 hx
2 h
x
h

x+ ≡ x + σ = 0.3333h + 0.2981h = 0.6315h;
x− ≡ x − σ = 0.3333h − 0.2981h = 0.0352h.

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 : −∞ → ∞.
−∞

π
λ
1 = A
e−λu2 du = A
⇒ A =
.
−∞
λ
π
(b)

x = A
xe−λ(x−a)2 dx = A
(u + a)e−λu2 du
−∞
−∞

π
= A
ue−λu2 du + a
e−λu2 du = A 0 + a
= a.
−∞
−∞
λ

x2 = A
x2e−λ(x−a)2 dx
−∞∞

= A
u2e−λu2 du + 2a
ue−λu2 du + a2
e−λu2 du
−∞
−∞
−∞
1
π
π
1
= A
+ 0 + a2
= a2 +
.

λ
λ

1
1
1
σ2 = x2 − x 2 = a2 +
− a2 =
;
σ = √
.

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publisher.

CHAPTER 1. THE WAVE FUNCTION
5
(c)
ρ(x)
A
a
x
Problem 1.4
(a)
|
a
b
A|2
a
|A|2
b
1
x3
1
1 =
x2dx +
(b − x)2dx = |A|2
+
−(b − x)3
a2
0
(b − a)2 a
a2
3
(b − a)2
3
0
a
a
b − a
b
3
= |A|2
+
= |A|2
⇒ A =
.
3
3
3
b
(b)
Ψ
A
a
b
x
(c) At x = a.
(d)
a
|A|2
a
a
a
P = 1
if b = a,
P =
|Ψ|2dx =
x2dx = |A|2
=
.
P = 1/2 if b = 2a.
0
a2
0
3
b
(e)
1
a
1
b
x =
x|Ψ|2dx = |A|2
x3dx +
x(b − x)2dx
a2 0
(b − a)2 a
a
b
3
1
x4
1
x2
x3
x4
=
+
b2
− 2b
+
b
a2
4
(b − a)2
2
3
4
0
a
3
=
a2(b − a)2 + 2b4 − 8b4/3 + b4 − 2a2b2 + 8a3b/3 − a4
4b(b − a)2
3
b4
2
1
2a + b
=
− a2b2 + a3b =
(b3 − 3a2b + 2a3) =
.
4b(b − a)2
3
3
4(b − a)2
4
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publisher.

6
CHAPTER 1. THE WAVE FUNCTION
Problem 1.5
(a)

e−2λx
|A|2

1 =
|Ψ|2dx = 2|A|2
e−2λxdx = 2|A|2

=
;
A =
λ.
0

λ
0
(b)

x =
x|Ψ|2dx = |A|2
xe−2λ|x|dx = 0.
[Odd integrand.]
−∞

2
1
x2 = 2|A|2
x2e−2λxdx = 2λ
=
.
0
(2λ)3
2λ2
(c)
1
1

σ2 = x2 − x 2 =
;
σ = √
.
|Ψ(±σ)|2 = |A|2e−2λσ = λe−2λ/ 2λ = λe− 2 = 0.2431λ.
2λ2

|Ψ|2
λ
.24λ
−σ

x
Probability outside:

e−2λx

2
|Ψ|2dx = 2|A|2
e−2λxdx = 2λ
2

= e−2λσ = e−
= 0.2431.
σ
σ

σ
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

∂x

(x|Ψ|2) =
|Ψ|2 + x |Ψ|2 = x |Ψ|2,
∂t
∂t
∂t
∂t
but this does not allow us to perform the integration:
b

b ∂
b
x
|Ψ|2dx =
(x|Ψ|2)dx = (x|Ψ|2) .
a
a
∂t
a
∂t
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publisher.

CHAPTER 1. THE WAVE FUNCTION
7
Problem 1.7
From Eq. 1.33, d p = −i

Ψ∗ ∂Ψ dx. But, noting that ∂2Ψ = ∂2Ψ and using Eqs. 1.23-1.24:
dt
∂t
∂x
∂x∂t
∂t∂x

∂Ψ∗ ∂Ψ
∂Ψ
∂2Ψ∗
i
i
∂2Ψ
Ψ∗ ∂Ψ
=
+ Ψ∗ ∂
= − i
+
V Ψ∗ ∂Ψ + Ψ∗ ∂
− i V Ψ
∂t
∂x
∂t ∂x
∂x
∂t
2m ∂x2
∂x
∂x 2m ∂x2
i
∂Ψ
i
=
Ψ∗ ∂3Ψ − ∂2Ψ∗
+
V Ψ∗ ∂Ψ − Ψ∗ ∂ (V Ψ)
2m
∂x3
∂x2 ∂x
∂x
∂x
The ﬁrst term integrates to zero, using integration by parts twice, and the second term can be simpliﬁed to
V Ψ∗ ∂Ψ − Ψ∗V ∂Ψ − Ψ∗ ∂V Ψ = −|Ψ|2 ∂V . So
∂x
∂x
∂x
∂x
d p
i
∂V
= −i
−|Ψ|2
dx = − ∂V .
QED
dt
∂x
∂x
Problem 1.8
Suppose Ψ satisﬁes the Schr¨
odinger equation without V
∂2Ψ
0: i ∂Ψ = − 2
+ V Ψ. We want to ﬁnd the solution
∂t
2m ∂x2
Ψ
∂2Ψ0
0 with V0: i ∂Ψ0 = − 2
+ (V + V
∂t
2m ∂x2
0)Ψ0.
Claim: Ψ0 = Ψe−iV0t/ .
Proof: i ∂Ψ0 = i ∂Ψ e−iV0t/ + i Ψ − iV0 e−iV0t/ = − 2 ∂2Ψ + V Ψ e−iV0t/ + V
∂t
∂t
2m ∂x2
0Ψe−iV0t/
= − 2 ∂2Ψ0 + (V + V
2m ∂x2
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
π
π
2am 1/4
1 = 2|A|2
e−2amx2/ dx = 2|A|2
= |A|2
;
A =
.
0
2
(2am/ )
2am
π
(b)
∂Ψ
∂Ψ
∂2Ψ
∂Ψ
= −iaΨ;
= − 2amx Ψ;
= − 2am Ψ + x
= − 2am 1 − 2amx2
Ψ.
∂t
∂x
∂x2
∂x
Plug these into the Schr¨
odinger equation, i ∂Ψ = − 2 ∂2Ψ + V Ψ:
∂t
2m ∂x2
2
V Ψ = i (−ia)Ψ +
−2am
1 − 2amx2
Ψ
2m
=
a − a 1 − 2amx2
Ψ = 2a2mx2Ψ,
so
V (x) = 2ma2x2.
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publisher.

8
CHAPTER 1. THE WAVE FUNCTION
(c)

x =
x|Ψ|2dx = 0.
[Odd integrand.]
−∞

1
π
x2 = 2|A|2
x2e−2amx2/ dx = 2|A|2
=
.
0
22(2am/ )
2am
4am
d x
p = m
= 0.
dt

2
p2 =
Ψ∗
Ψdx = − 2
Ψ∗ ∂2Ψ dx
i ∂x
∂x2
= − 2
Ψ∗ − 2am 1 − 2amx2
Ψ dx = 2am
|Ψ|2dx − 2am
x2|Ψ|2dx
1
= 2am
1 − 2am x2
= 2am
1 − 2am
= 2am
= am .
4am
2
(d)

σ2x = x2 − x 2 =
=⇒ σ
;
σ2
am .
4am
x =
4am
p =
p2 − p 2 = am =⇒ σp =

σxσp =
am =
. This is (just barely) consistent with the uncertainty principle.
4am
2
Problem 1.10
From Math Tables: π = 3.141592653589793238462643 · · ·
P (0) = 0
P (1) = 2/25
P (2) = 3/25
P (3) = 5/25
P (4) = 3/25
(a)
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 = 1 [0 · 0 + 1 · 2 + 2 · 3 + 3 · 5 + 4 · 3 + 5 · 3 + 6 · 3 + 7 · 1 + 8 · 2 + 9 · 3]
25
= 1 [0 + 2 + 6 + 15 + 12 + 15 + 18 + 7 + 16 + 27] = 118 = 4.72.
25
25
(c) j2 = 1 [0 + 12 · 2 + 22 · 3 + 32 · 5 + 42 · 3 + 52 · 3 + 62 · 3 + 72 · 1 + 82 · 2 + 92 · 3]
25
= 1 [0 + 2 + 12 + 45 + 48 + 75 + 108 + 49 + 128 + 243] = 710 = 28.4.
25
25

σ2 = j2 − j 2 = 28.4 − 4.722 = 28.4 − 22.2784 = 6.1216;
σ =
6.1216 = 2.474.
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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.
1/π, if 0 ≤ θ ≤ π,
ρ(θ) =
0,
otherwise.
ρ(θ)
1/π
θ
−π/2
0
π
3π/2
(b)
π
1
π
1
θ2
π
θ =
θρ(θ) dθ =
θdθ =
=
[of course].
π 0
π
2
2
0
π
1
π
1
θ3
π2
θ2 =
θ2 dθ =
=
.
π 0
π
3
3
0
π2
π2
π
σ2 = θ2 − θ 2 =
− π2 =
;
σ = √ .
3
4
12
2 3
(c)
1
π
1
1
2
sin θ =
sin θ dθ =
(− cos θ)|π =
(1 − (−1)) =
.
π
0
0
π
π
π
1
π
1
cos θ =
cos θ dθ =
(sin θ)|π = 0.
π
0
0
π
1
π
1
π
1
cos2 θ =
cos2 θ dθ =
(1/2)dθ =
.
π 0
π 0
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 dθ, so the
π
probability that it’s in the range dx is
1
dx
1
dx
dx
ρ(x)dx =
=
=

.
π r sin θ
π r 1 − (x/r)2
π r2 − x2
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10
CHAPTER 1. THE WAVE FUNCTION
ρ(x)
x
-2r
-r
r
2r
1

, if − r < x < r,
ρ(x) =
π
r2−x2
[Note: We want the magnitude of dx here.]
0,
otherwise.
r
r
r
Total:
1

1

dx = 2
dx = 2 sin−1 x
= 2 sin−1(1) = 2 · π = 1.
r π r2−x2
π
0
r2−x2
π
r 0
π
π
2
(b)
1
r
1
x =
x √
dx = 0
[odd integrand, even interval].
π −r
r2 − x2
r
2
r
x2
2
r2
2 r2
r2
x2 =

dx =
−x r2 − x2 +
sin−1 x
=
sin−1(1) =
.
π 0
r2 − x2
π
2
2
r
π 2
2
0

σ2 = x2 − x 2 = r2/2 =⇒ σ = r/ 2.
To get x and x2 from Problem 1.11(c), use x = r cos θ, so x = r cos θ = 0, x2 = r2 cos2 θ = r2/2.
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
−y
l
1
−y
1
l
1
P (y) =
ρ(x)dx +
ρ(x)dx =

dx +

dx
−l
l−y
π
−l
l2 − x2
l−y
l2 − x2
1
−y
l
1
=
sin−1 x
+ sin−1 x
=
− sin−1(y/l) + 2 sin−1(1) − sin−1(1 − y/l)
π
l
−l
l
l−y
π
= 1 − sin−1(y/l) − sin−1(1 − y/l) .
π
π
Now, all values of y are equally likely, so ρ(y) = 1/l, and hence the probability of crossing is
1
l
1
l
P =
π − sin−1 y − sin−1
l − y
dy =
π − 2 sin−1(y/l) dy
πl 0
l
l
πl 0
1
l
2
2
=
πl − 2 y sin−1(y/l) + l 1 − (y/l)2
= 1 − 2 [l sin−1(1) − l] = 1 − 1 +
=
.
πl
0
πl
π
π