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Contents
Preface
2
1
The
Wave
Function
3
2
TimeIndependent
Schrödinger
Equation
14
3 Formalism
62
4 Quantum Mechanics in Three Dimensions
87
5
Identical
Particles
132
6
TimeIndependent
Perturbation
Theory
154
7
The
Variational
Principle
196
8
The
WKB
Approximation
219
9
TimeDependent
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.
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).]
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)
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 +
.
2λ
λ
λ
2λ
1
1
1
σ2 = x2 − x 2 = a2 +
− a2 =
;
σ = √
.
2λ
2λ
2λ
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)
ρ(x)
A
a
x
Problem 1.4
(a)

a
b
A2
a
A2
b
1
x3
1
1 =
x2dx +
(b − x)2dx = A2
+
−(b − x)3
a2
0
(b − a)2 a
a2
3
(b − a)2
3
0
a
a
b − a
b
3
= A2
+
= A2
⇒ A =
.
3
3
3
b
(b)
Ψ
A
a
b
x
(c) At x = a.
(d)
a
A2
a
a
a
P = 1
if b = a,
P =
Ψ2dx =
x2dx = A2
=
.
P = 1/2 if b = 2a.
0
a2
0
3
b
(e)
1
a
1
b
x =
xΨ2dx = A2
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
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.
6
CHAPTER 1. THE WAVE FUNCTION
Problem 1.5
(a)
∞
∞
e−2λx
A2
√
1 =
Ψ2dx = 2A2
e−2λxdx = 2A2
−
=
;
A =
λ.
0
2λ
λ
0
(b)
∞
x =
xΨ2dx = A2
xe−2λxdx = 0.
[Odd integrand.]
−∞
∞
2
1
x2 = 2A2
x2e−2λxdx = 2λ
=
.
0
(2λ)3
2λ2
(c)
1
1
√
√
σ2 = x2 − x 2 =
;
σ = √
.
Ψ(±σ)2 = A2e−2λσ = λe−2λ/ 2λ = λe− 2 = 0.2431λ.
2λ2
2λ
Ψ2
λ
.24λ
−σ
+σ
x
Probability outside:
∞
∞
∞
e−2λx
√
2
Ψ2dx = 2A2
e−2λxdx = 2λ
2
−
= e−2λσ = e−
= 0.2431.
σ
σ
2λ
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
∂
∂
(xΨ2) =
Ψ2 + x Ψ2 = x Ψ2,
∂t
∂t
∂t
∂t
but this does not allow us to perform the integration:
b
∂
b ∂
b
x
Ψ2dx =
(xΨ2)dx = (xΨ2) .
a
a
∂t
a
∂t
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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.
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. 1.231.24:
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 = −Ψ2 ∂V . So
∂x
∂x
∂x
∂x
d p
i
∂V
= −i
−Ψ2
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 = 2A2
e−2amx2/ dx = 2A2
= A2
;
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.
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.
8
CHAPTER 1. THE WAVE FUNCTION
(c)
∞
x =
xΨ2dx = 0.
[Odd integrand.]
−∞
∞
1
π
x2 = 2A2
x2e−2amx2/ dx = 2A2
=
.
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
Ψ2dx − 2am
x2Ψ2dx
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.
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
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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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)
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
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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
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