Solutions Manual for Introduction to Quantum Mechanics, 3rd Edition by Griffiths & Schroeter | All 12 Chapters Covered

Introduction to Quantum Mechanics, 3rd ed,
by Griffiths, Schroeter. Chapter 1-12




SOLUTION MANUAL

,2




Contents

1 The Wave Function 4

2 The Time-Independent Schrödinger Equation 16

3 Formalism 78

4 Quantum Mechanics in Three Dimensions 109

5 Identical Particles 168

6 Symmetries and Conservation Laws 197

7 Time-Independent Perturbation Theory 235

8 The Variational Principle 301

9 The WKB Approximation 333

10 Scattering 354

11 Quantum Dynamics 372

12 Afterword 420

A Linear Algebra 427

,4 CHAPTER 1. THE WAVE FUNCTION




Chapter 1

The Wave Function

Problem 1.1
(a)

hji2 = 212 = 441.

1 X 2 1  2 
hj 2 i = j N (j) = (14 ) + (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 − hji
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 X 1  
σ2 = (∆j)2 N (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)

hj 2 i − hji2 = 459.571 − 441 = 18.571. [Agrees with (b).]

,CHAPTER 1. THE WAVE FUNCTION 5


Problem 1.2
(a)
Z h   h
2 1
2 1 2 5/2 h2
hx i = x √ dx = √ x = .
0 2 hx 2 h 5 0 5

 2
2 2 h2 2 h 4 2 2h
σ = hx i − hxi = − = h ⇒ σ = √ = 0.2981h.
5 3 45 3 5

(b)
Z x+ x+
1 1 √ 1 √ √

P =1− √ dx = 1 − √ (2 x) =1− √ x+ − x− .
x− 2 hx 2 h x− h


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


√ √
P =1− 0.6315 + 0.0352 = 0.393.




Problem 1.3
(a)
Z ∞
2
1= Ae−λ(x−a) dx. Let u ≡ x − a, du = dx, u : −∞ → ∞.
−∞

Z ∞
r r
−λu2 π λ
1=A e du = A ⇒ A= .
−∞ λ π

(b)
Z ∞ Z ∞
2 2
hxi = A xe−λ(x−a) dx = A (u + a)e−λu du
−∞ −∞
Z ∞ Z ∞  r 

−λu2 −λu2 π
=A ue du + a e du = A 0 + a = a.
−∞ −∞ λ
Z ∞
2
hx2 i = A x2 e−λ(x−a) dx
−∞
Z ∞ Z ∞ Z ∞ 
2 2 2
=A u2 e−λu du + 2a ue−λu du + a2 e−λu du
−∞ −∞ −∞
 r r 
1 π π 1
=A + 0 + a2 = a2 + .
2λ λ λ 2λ

1 1 1
σ 2 = hx2 i − hxi2 = a2 + − a2 = ; σ=√ .
2λ 2λ 2λ

,6 CHAPTER 1. THE WAVE FUNCTION


(c)
l(x)
A




a x




Problem 1.4
(a)
Z Z (   a   b)
a b
|A|2 2 |A|2 2 2 1 x3 1 (b − x)3
1= 2 2 x dx +
(b − x) dx = |A| + −
a (b − a) a
0 a2 3 0 (b − a)2 3 a
  r
2 a b−a 2b 3
= |A| + = |A| ⇒ A= .
3 3 3 b


(b)
^
A




a b x


(c) At x = a.

(d)
Z a Z a 
|A|2
2 2 a 2a P = 1 if b = a, X
P = |Ψ| dx = 2 x dx = |A| = .
0 a 0 3 b P = 1/2 if b = 2a. X

(e)
Z  Z a Z b 
1 1
hxi = x|Ψ|2 dx = |A|2 2 x3 dx + x(b − x) 2
dx
a 0 (b − a)2 a
(  4 a   b)
2
3 1 x 1 2x x3 x4
= + b − 2b +
b a2 4 0 (b − a)2 2 3 4 a
3  
= 2
a2 (b − a)2 + 2b4 − 8b4 /3 + b4 − 2a2 b2 + 8a3 b/3 − a4
4b(b − a)
 4 
3 b 2 2 2 3 1 2a + b
= 2
−a b + a b = 2
(b3 − 3a2 b + 2a3 ) = .
4b(b − a) 3 3 4(b − a) 4

,CHAPTER 1. THE WAVE FUNCTION 7


Problem 1.5
(a)
Z Z ∞   ∞
2 2 −2λx 2 e−2λx |A|2 √
1= |Ψ| dx = 2|A| e dx = 2|A| = ; A= λ.
0 −2λ 0 λ

(b)
Z Z ∞
2 2
hxi = x|Ψ| dx = |A| xe−2λ|x| dx = 0. [Odd integrand.]
−∞

Z ∞  
2 2 2 −2λx 2 1
hx i = 2|A| x e dx = 2λ = .
0 (2λ)3 2λ2

(c)

1 1 √ √
σ 2 = hx2 i − hxi2 = ; σ=√ . |Ψ(±σ)|2 = |A|2 e−2λσ = λe−2λ/ 2λ
= λe− 2
= 0.2431λ.
2λ2 2λ


|^| 2
h



.24h

x
<m +m
Probability outside:
Z ∞ Z ∞   ∞
e−2λx √
2 |Ψ|2 dx = 2|A|2 e−2λx dx = 2λ = e−2λσ = e− 2
= 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 2 ∂ ∂
(x|Ψ|2 ) = |Ψ| + x |Ψ|2 = x |Ψ|2 ,
∂t ∂t ∂t ∂t
but this does not allow us to perform the integration:
Z b Z b
∂ ∂ b
x |Ψ|2 dx = (x|Ψ|2 )dx 6= (x|Ψ|2 ) a .
a ∂t a ∂t

,8 CHAPTER 1. THE WAVE FUNCTION


Problem 1.7
dhpi R ∂

∂2Ψ ∂2Ψ
From Eq. 1.33, dt = −i~ ∂t Ψ∗ ∂Ψ
∂x dx. But, noting that ∂x∂t = ∂t∂x and using Eqs. 1.23-1.24:
       
∂ ∗ ∂Ψ ∂Ψ∗ ∂Ψ ∗ ∂ ∂Ψ i~ ∂ 2 Ψ∗ i ∗ ∂Ψ ∗ ∂ i~ ∂ 2 Ψ i
Ψ = +Ψ = − + VΨ +Ψ − VΨ
∂t ∂x ∂t ∂x ∂x ∂t 2m ∂x2 ~ ∂x ∂x 2m ∂x2 ~
 3 2 ∗
  
i~ ∂ Ψ ∂ Ψ ∂Ψ i ∂Ψ ∂
= Ψ∗ 3 − + V Ψ∗ − Ψ∗ (V Ψ)
2m ∂x ∂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 2 ∂V
∂x − Ψ V ∂x − Ψ ∂x Ψ = −|Ψ| ∂x . So
 Z
dhpi i ∂V ∂V
= −i~ −|Ψ|2 dx = h− i. QED
dt ~ ∂x ∂x



Problem 1.8
2 2
Suppose Ψ satisfies the Schrödinger equation without V0 : i~ ∂Ψ ~ ∂ Ψ
∂t = − 2m ∂x2 + V Ψ. We want to find the solution
2 2
∂Ψ0 ~ ∂ Ψ
Ψ0 with V0 : i~ ∂t = − 2m ∂x2 + (V + V0 )Ψ0 .
0



Claim: Ψ0 = Ψe−iV0 t/~ .

−iV t/~ h ~2 ∂ 2 Ψ i
Proof: i~ ∂Ψ
∂t
0
= i~ ∂Ψ −iV0 t/~
∂t e + i~Ψ − iV0
~ e 0
= − 2m ∂x 2 + V Ψ e−iV0 t/~ + V0 Ψe−iV0 t/~
2 2
~ ∂ Ψ0
= − 2m ∂x2 + (V + V0 )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)

Z ∞ r r  1/4
2 −2amx2 /~ 21 π π~ 2am
1 = 2|A| e dx = 2|A| = |A|2 ; A= .
0 2 (2am/~) 2am π~


(b)
   
∂Ψ ∂Ψ 2amx ∂2Ψ 2am ∂Ψ 2am 2amx2
= −iaΨ; =− Ψ; =− Ψ+x =− 1− Ψ.
∂t ∂x ~ ∂x2 ~ ∂x ~ ~
2 2
Plug these into the Schrödinger equation, i~ ∂Ψ ~ ∂ Ψ
∂t = − 2m ∂x2 + V Ψ:
  
~2 2am 2amx2
V Ψ = i~(−ia)Ψ + − 1− Ψ
2m ~ ~
  
2amx2
= ~a − ~a 1 − Ψ = 2a2 mx2 Ψ, so V (x) = 2ma2 x2 .
~

,CHAPTER 1. THE WAVE FUNCTION 9


(c)
Z ∞
hxi = x|Ψ|2 dx = 0. [Odd integrand.]
−∞

Z ∞
r
2 2 2 −2amx2 /~ 2 1 π~ ~
hx i = 2|A| x e dx = 2|A| 2 = .
0 2 (2am/~) 2am 4am

dhxi
hpi = m = 0.
dt

Z 2  Z
2 ~ ∂ ∗ ∂2Ψ
hp i = Ψ Ψdx = −~2 Ψ∗ 2 dx
i ∂x ∂x
Z   2
  Z Z 
2 ∗ 2am 2amx 2 2am 2 2
= −~ Ψ − 1− Ψ dx = 2am~ |Ψ| dx − x |Ψ| dx
~ ~ ~
     
2am 2 2am ~ 1
= 2am~ 1 − hx i = 2am~ 1 − = 2am~ = am~.
~ ~ 4am 2

(d)
r
~ ~ √
σx2 2
= hx i − hxi = 2
=⇒ σx = ; σp2 = hp2 i − hpi2 = am~ =⇒ σp = am~.
4am 4am

q √
~
σx σp = 4am am~ = ~2 . This is (just barely) consistent with the uncertainty principle.




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


(b) Most probable: 3. Median: 13 are ≤ 4, 12 are ≥ 5, so median is 4.
1
Average: hji = 25 [0 · 0 + 1 · 2 + 2 · 3 + 3 · 5 + 4 · 3 + 5 · 3 + 6 · 3 + 7 · 1 + 8 · 2 + 9 · 3]
1 118
= 25 [0 + 2 + 6 + 15 + 12 + 15 + 18 + 7 + 16 + 27] = 25 = 4.72.
1
(c) hj 2 i = 25 [0 + 12 · 2 + 22 · 3 + 32 · 5 + 42 · 3 + 52 · 3 + 62 · 3 + 72 · 1 + 82 · 2 + 92 · 3]
1 710
= 25 [0 + 2 + 12 + 45 + 48 + 75 + 108 + 49 + 128 + 243] = 25 = 28.4.
√
σ 2 = hj 2 i − hji2 = 28.4 − 4.722 = 28.4 − 22.2784 = 6.1216; σ = 6.1216 = 2.474.

,10 CHAPTER 1. THE WAVE FUNCTION


Problem 1.11
(a)
r
1 2
mv 2 + V = E → v(x) = (E − V (x)) .
2 m

(b) r
Z b Z b
1 m 1
T = q
dx = p dx.
2
E− 1 2 k (2E/k) − x2
2 kx
a a
m
p
Turning points: v = 0 ⇒ E = V = 12 kb2 ⇒ b = 2E/k; a = −b.
r Z b r   b r
m 1 m −1 x m
T =2 √ dx = 2 sin =2 sin−1 (1)
k 0 2
b −x 2 k b 0 k
r   r
m π m
=2 =π .
k 2 k

1 1 ρ(x)
ρ(x) = pmq 2 1 2

= π √ b2 − x 2 .
π k m E− 2 kx 1
Z Z
b
2 b
1 2 π
ρ(x) dx = √ dx = = 1. X
a π 0 b2 − x 2 π 2


x
-b b

(c) hxi = 0.
Z b Z
1 x2 2 b x2
hx2 i = √
dx = √ dx
π−b b2 − x 2 π 0 b2 − x 2
  x  b
2 xp 2 b2 b2 b2 π b2 E
= − b − x2 + sin−1 = sin−1 (1) = = = .
π 2 2 b 0 π π 2 2 k
r
p p b E
σx = hx2 i − hxi2 = hx2 i = √ = .
2 k



Problem 1.12
(a)
dt |dt/dp| dp
= ρ(p) dp =
T T
where dt is now the time it spends with momentum in the range dp (dt is intrinsically positive, but
dp/dt = F = −kx runs negative—hence the absolute value). Now
s  
p2 1 2 2 p2
+ kx = E ⇒ x = ± E− ,
2m 2 k 2m

, CHAPTER 1. THE WAVE FUNCTION 11


so


1 1 1
ρ(p) = r  = p = p ,
pm 2 p2 π 2mE − p2 π c2 − p2
π k k k E− 2m



√
where c ≡ 2mE. This is the same as ρ(x) (Problem 1.11(b)), with c in place of b (and, of course, p in
place of x).



c2 c √
(b) From Problem 1.11(c), hpi = 0, hp2 i = , σp = √ = mE .
2 2



r r
E√ m E 1 1
(c) σx σp = mE = E = . If E ≥ 2 ~ω, then σx σp ≥ 2 ~, which is precisely the Heisenberg
k k ω
uncertainty principle!




Problem 1.13

x@t_D := Cos@tD

snapshots = Table@x@ p RandomReal@jDD, 8j, 10 000<D



Histogram@snapshots, 100, "PDF", PlotRange Æ 80, 2<D
2.0




1.5


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2.0 1.0




1.5 0.5




1.0
-0.5 0.0 0.5 1.0




0.5



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0.5 0.0 2.0 0.5 1.0




1.5


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