Quantum Mechanics by D.G. Griffiths 3rd Edition, Other for Physics. Indian Institute of Technology (IIT)
shikhar-mittal
shikhar-mittal29 July 2017

Quantum Mechanics by D.G. Griffiths 3rd Edition, Other for Physics. Indian Institute of Technology (IIT)

PDF (4 MB)
303 pages
3Number of download
83Number of visits
1Number of comments
Description
This is the solution manual of the book
20 points
Download points needed to download
this document
Download the document
Preview3 pages / 303

This is only a preview

3 shown on 303 pages

Download the document

This is only a preview

3 shown on 303 pages

Download the document

This is only a preview

3 shown on 303 pages

Download the document

This is only a preview

3 shown on 303 pages

Download the document
ISMComplete

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.

〈j2= 1 N

j2N(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

σ2 = 1 N

∑ (∆j)2N(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)

〈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)

〈x2= ∫ h

0

x2 1

2 √ hx

dx = 1

2 √ h

( 2 5 x5/2

)∣∣∣∣h 0

= h2

5 .

σ2 = 〈x2〉 − 〈x〉2 = h 2

5

( h

3

)2 =

4 45

h2 ⇒ σ = 2h 3

5 = 0.2981h.

(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 = 0.3333h + 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 : −∞ → ∞.

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.

〈x2= A ∞ −∞

x2e−λ(x−a) 2 dx

= A {∫

−∞ u2e−λu

2 du + 2a

∞ −∞

ue−λu 2 du + a2

∞ −∞

e−λu 2 du

} = A

[ 1 2λ

π

λ + 0 + a2

π

λ

] = a2 +

1 2λ

.

σ2 = 〈x2〉 − 〈x〉2 = a2 + 1 2λ

− a2 = 1 2λ

; σ = 12λ

.

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 a2

a 0

x2dx + |A|2

(b− a)2 ∫ b a

(b− x)2dx = |A|2 {

1 a2

( x3

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|2dx = |A| 2

a2

a 0

x2dx = |A|2 a 3

= a

b .

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

(e)

〈x〉 = ∫

x|Ψ|2dx = |A|2 {

1 a2

a 0

x3dx + 1

(b− a)2 ∫ b a

x(b− x)2dx }

= 3 b

{ 1 a2

( x4

4

)∣∣∣∣a 0

+ 1

(b− a)2 ( b2 x2

2 2bx

3

3 +

x4

4

)∣∣∣∣b a

}

= 3

4b(b− a)2 [ a2(b− a)2 + 2b4 8b4/3 + b4 2a2b2 + 8a3b/3 − a4

] =

3 4b(b− a)2

( b4

3 − a2b2 + 2

3 a3b

) =

1 4(b− a)2 (b

3 3a2b + 2a3) = 2a + b 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)

1 = ∫

|Ψ|2dx = 2|A|2 ∫

0

e−2λxdx = 2|A|2 ( e−2λx

2λ

)∣∣∣∣0

= |A|2 λ

; A = √ λ.

(b)

〈x〉 = ∫

x|Ψ|2dx = |A|2 ∫ ∞ −∞

xe−2λ|x|dx = 0. [Odd integrand.]

〈x2= 2|A|2 ∫

0

x2e−2λxdx = 2λ [

2 (2λ)3

] =

1 2λ2

.

(c)

σ2 = 〈x2〉 − 〈x〉2 = 1 2λ2

; σ = 12λ

. |Ψ(±σ)|2 = |A|2e−2λσ = λe−2λ/ √

2λ = λe− √

2 = 0.2431λ.

|Ψ|2 λ

σ−σ + x

.24λ

Probability outside:

2 ∫ ∞ σ

|Ψ|2dx = 2|A|2 ∫ ∞ σ

e−2λxdx = 2λ ( e−2λx

2λ

)∣∣∣∣∞ σ

= e−2λσ = 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|2dx =

b a

∂t (x|Ψ|2)dx = (x|Ψ|2)

∣∣b a .

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 7

Problem 1.7

From Eq. 1.33, d〈p〉dt = −i ∫

∂ ∂t

( Ψ∗ ∂Ψ∂x

) dx. But, noting that

∂x∂t =

∂t∂x and using Eqs. 1.23-1.24:

∂t

( Ψ

Ψ ∂x

) =

Ψ

∂t

Ψ ∂x

+ Ψ∗ ∂

∂x

( Ψ ∂t

) =

[ − i

2m ∂

∂x2 +

i

 V Ψ

] Ψ ∂x

+ Ψ∗ ∂

∂x

[ i

2m ∂∂x2

− i  V Ψ

] =

i

2m

[ Ψ

∂x3

− ∂

∂x2 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 V0: iΨ∂t =  2

2m ∂∂x2 + V Ψ. We want to find the solution

Ψ0 with V0: iΨ0∂t =  2

2m ∂2Ψ0 ∂x2 + (V + V0)Ψ0.

Claim: Ψ0 = Ψe−iV0t/.

Proof: iΨ0∂t = i Ψ ∂t e

−iV0t/ + iΨ ( − iV0 

) e−iV0t/ =

[ 22m ∂

∂x2 + V Ψ

] e−iV0t/ + Ve−iV0t/

= 22m ∂2Ψ0 ∂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)

1 = 2|A|2 ∫

0

e−2amx 2/dx = 2|A|2 1

2

π

(2am/) = |A|2

π

2am ; A =

( 2am π

)1/4 .

(b)

Ψ ∂t

= −iaΨ; Ψ ∂x

= 2amx 

Ψ; ∂x2

= 2am 

( Ψ + x

Ψ ∂x

) = 2am



( 1 2amx

2



) Ψ.

Plug these into the Schrödinger equation, iΨ∂t =  2

2m ∂∂x2 + V Ψ:

V Ψ = i(−ia)Ψ +  2

2m

( 2am



) ( 1 2amx

2



) Ψ

= [ a− a

( 1 2amx

2



)] Ψ = 2a2mx, 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〉 = ∫ ∞ −∞

x|Ψ|2dx = 0. [Odd integrand.]

〈x2= 2|A|2 ∫

0

x2e−2amx 2/dx = 2|A|2 1

22(2am/)

π

2am =



4am .

〈p〉 = md〈x〉 dt

= 0.

〈p2= ∫

Ψ( 

i

∂x

)2 Ψdx = 2

∫ Ψ

∂x2

dx

= 2 ∫

Ψ[ 2am



( 1 2amx

2



) Ψ

] dx = 2am

{∫ |Ψ|2dx− 2am



x2|Ψ|2dx

} = 2am

( 1 2am

 〈x2

) = 2am

( 1 2am





4am

) = 2am

( 1 2

) = am.

(d)

σ2x = 〈x2〉 − 〈x〉2 = 

4am =⇒ σx =

√ 

4am ; σ2p = 〈p2〉 − 〈p〉2 = am =⇒ σp =

√ am.

σxσp = √



4am

√ 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) 〈j2= 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.

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.

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) = 1 π

π 0

θdθ = 1 π

( θ2

2

)∣∣∣∣π 0

= π

2 [of course].

〈θ2= 1 π

π 0

θ2 = 1 π

( θ3

3

)∣∣∣∣π 0

= π2

3 .

σ2 = 〈θ2〉 − 〈θ〉2 = π 2

3 − π

2

4 =

π2

12 ; σ =

π

2

3 .

(c)

sin θ〉 = 1 π

π 0

sin θ dθ = 1 π

(cos θ)0 = 1 π

(1 (1)) = 2 π

.

cos θ〉 = 1 π

π 0

cos θ dθ = 1 π

(sin θ)0 = 0.

cos2 θ〉 = 1 π

π 0

cos2 θ dθ = 1 π

π 0

(1/2)= 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 is ρ(θ)= 1πdθ, 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 √ r2 − x2

.

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.

10 CHAPTER 1. THE WAVE FUNCTION

ρ(x)

xr 2r-r-2r

ρ(x) = { 1

π √ r2−x2 , if − r < x < r,

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

Total: ∫ r −r

1 π √ r2−x2 dx =

2 π

r 0

1√ r2−x2 dx =

2 π sin

1 x r

∣∣r 0

= 2π sin 1(1) = 2π · π2 = 1.

(b)

〈x〉 = 1 π

r −r

x 1

r2 − x2 dx = 0 [odd integrand, even interval].

〈x2= 2 π

r 0

x2√ r2 − x2

dx = 2 π

[ −x

2

r2 − x2 + r

2

2 sin1

(x r

)]∣∣∣∣r 0

= 2 π

r2

2 sin1(1) =

r2

2 .

σ2 = 〈x2〉 − 〈x〉2 = r2/2 =⇒ σ = r/ √

2.

To get 〈x〉 and 〈x2from Problem 1.11(c), use x = r cos θ, so 〈x〉 = r〈cos θ〉 = 0, 〈x2= r2cos2 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

P (y) = ∫ −y −l

ρ(x)dx + ∫ l l−y

ρ(x)dx = 1 π

{∫ −y −l

1√ l2 − x2

dx + ∫ l l−y

1√ l2 − x2

dx

}

= 1 π

{ sin1

(x l

)∣∣∣−y −l

+ sin1 (x l

)∣∣∣l l−y

} =

1 π

[ sin1(y/l) + 2 sin1(1) sin1(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 − sin1

(y l

) sin1

( l − y l

)] dy =

1 πl

l 0

[ π − 2 sin1(y/l)

] dy

= 1 πl

[ πl − 2

( y sin1(y/l) + l

√ 1 (y/l)2

)∣∣∣l 0

] = 1 2

πl [l sin1(1) − l] = 1 1 + 2

π =

2 π

.

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 11

Problem 1.14

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

b a

∂ ∂t |Ψ|2dx. But (Eq. 1.25):

∂|Ψ|2 ∂t

=

∂x

[ i

2m

( 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 seconds1.

(b) Here Ψ(x, t) = f(x)e−iat, where f(x) ≡ Ae−amx2/, so ΨΨ∗∂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 = − i2m ∂

∂x2 + i  V ∗Ψ, and Eq. 1.25 picks up an extra term:

∂t |Ψ|2 = · · · + i

 |Ψ|2(V ∗ − V ) = · · · + i

 |Ψ|2(V0 + iΓ − V0 + iΓ) = · · · −

2Γ  |Ψ|2,

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

2Γ  P . QED

(b)

dP

P =

 dt =lnP =

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

.

Problem 1.16

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

d

dt

∞ −∞

Ψ1Ψ2 dx = ∫ ∞ −∞

∂t 1Ψ2) dx =

∞ −∞

( Ψ1 ∂t

Ψ2 + Ψ1 Ψ2 ∂t

) dx

= ∫ ∞ −∞

[(−i 2m

1 ∂x2

+ i

 V Ψ1

) Ψ2 + Ψ1

( i

2m ∂2Ψ2 ∂x2

− i  V Ψ2

)] dx

= − i 2m

∞ −∞

( 1 ∂x2

Ψ2 Ψ1 2Ψ2 ∂x2

) dx

= − i 2m

[ Ψ1 ∂x

Ψ2

∣∣∣∣∞ −∞

∞ −∞

Ψ1 ∂x

Ψ2 ∂x

dx− Ψ1 Ψ2 ∂x

∣∣∣∣∞ −∞

+ ∫ ∞ −∞

Ψ1 ∂x

Ψ2 ∂x

dx

] = 0. QED

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.

12 CHAPTER 1. THE WAVE FUNCTION

Problem 1.17

(a)

1 = |A|2 ∫ a −a

( a2 − x2

)2 dx = 2|A|2

a 0

( a4 2a2x2 + x4

) dx = 2|A|2

[ a4x− 2a2x

3

3 +

x5

5

]∣∣∣∣a 0

= 2|A|2a5 (

1 2 3

+ 1 5

) =

16 15

a5|A|2, so A = √

15 16a5

.

(b)

〈x〉 = ∫ a −a

x|Ψ|2 dx = 0. (Odd integrand.)

(c)

〈p〉 =  i A2

a −a

( a2 − x2

) d dx

( a2 − x2

) ︸ ︷︷ ︸

2x

dx = 0. (Odd integrand.)

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

(d)

〈x2= A2 ∫ a −a

x2 ( a2 − x2

)2 dx = 2A2

a 0

( a4x2 2a2x4 + x6

) dx

= 2 15

16a5

[ a4

x3

3 2a2x

5

5 +

x7

7

]∣∣∣∣a 0

= 15 8a5

( a7

)(1 3 2

5 +

1 7

)

= ✚ 15a2

8

( 35 42 + 15

✁3 · ✁5 · 7

) =

a2

8 · 8 7

= a2

7 .

(e)

〈p2= −A22 ∫ a −a

( a2 − x2

) d2 dx2

( a2 − x2

) ︸ ︷︷ ︸

2

dx = 2A222 ∫ a

0

( a2 − x2

) dx

= 4 · 15 16a5

 2

( a2x− x

3

3

)∣∣∣∣a 0

= 152

4a5

( a3 − a

3

3

) =

152

4a2 · 2 3

= 5 2 

2

a2 .

(f)

σx = √ 〈x2〉 − 〈x〉2 =

√ 1 7 a2 =

a√ 7 .

(g)

σp = √ 〈p2〉 − 〈p〉2 =

√ 5 2 2

a2 =

√ 5 2 

a .

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 13

(h)

σxσp = a√ 7 ·

5 2 

a =

√ 5 14  =

√ 10 7 

2 > 

2 . 

Problem 1.18

h√ 3mkBT

> d ⇒ T < h 2

3mkBd2 .

(a) Electrons (m = 9.1 × 1031 kg):

T < (6.6 × 1034)2

3(9.1 × 1031)(1.4 × 1023)(3 × 1010)2 = 1.3 × 10 5 K.

Sodium nuclei (m = 23mp = 23(1.7 × 1027) = 3.9 × 1026 kg):

T < (6.6 × 1034)2

3(3.9 × 1026)(1.4 × 1023)(3 × 1010)2 = 3.0 K.

(b) PV = NkBT ; volume occupied by one molecule (N = 1, V = d3) ⇒ d = (kBT/P )1/3.

T < h2

2mkB

( P

kBT

)2/3 ⇒ T 5/3 < h

2

3m P 2/3

k 5/3 B

⇒ T < 1 kB

( h2

3m

)3/5 P 2/5.

For helium (m = 4mp = 6.8 × 1027 kg) at 1 atm = 1.0 × 105 N/m2:

T < 1

(1.4 × 1023)

( (6.6 × 1034)2 3(6.8 × 1027)

)3/5 (1.0 × 105)2/5 = 2.8 K.

For hydrogen (m = 2mp = 3.4 × 1027 kg) with d = 0.01 m:

T < (6.6 × 1034)2

3(3.4 × 1027)(1.4 × 1023)(102)2 = 3.1 × 10 14 K.

At 3 K it is definitely in the classical regime.

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.

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(E0+iΓ)t/ = ψ(x)eΓt/e−iE0t/ =⇒ |Ψ|2 = |ψ|2et/.

∞ −∞

|Ψ(x, t)|2dx = et/ ∫ ∞ −∞

|ψ|2dx.

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

(b) If ψ satisfies Eq. 2.5, 22m ∂2ψ dx2 + V ψ = , then (taking the complex conjugate and noting that V and

E are real): 22m ∂2ψ∗

dx2 + V ψ ∗ = Eψ∗, so ψ∗ also satisfies Eq. 2.5. Now, if ψ1 and ψ2 satisfy Eq. 2.5, so

too does any linear combination of them (ψ3 ≡ c1ψ1 + c2ψ2):

 2

2m ∂2ψ3 dx2

+ V ψ3 = 

2

2m

( c1

2ψ1 dx2

+ c2 2ψ2 ∂x2

) + V (c1ψ1 + c2ψ2)

= c1

[ 

2

2m d2ψ1 dx2

+ V ψ1

] + c2

[ 

2

2m d2ψ2 dx2

+ V ψ2

] = c1(1) + c2(2) = E(c1ψ1 + c2ψ2) = 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/∂x2,

 2

2m ∂2ψ(−x)

dx2 + V (−x)ψ(−x) = (−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.

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 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 2ψ

dx2 = 2m 2

[V (x)−E]ψ, if E < Vmin, 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 2ψ

dx2 = 2mE2 ψ; Eq. 2.23 says ψ(0) = ψ(a) = 0. If E = 0, d2ψ/dx2 = 0, so ψ(x) = A + Bx; ψ(0) = A = 0 ⇒ ψ = Bx; ψ(a) = Ba = 0 ⇒ B = 0, so ψ = 0. If E < 0, d2ψ/dx2 = κ2ψ, with κ ≡

√ −2mE/

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 e2κa = 1, so 2κa = 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|2dx = 2 a

a 0

x sin2 (

a x ) dx. Let y ≡ nπ

a x, so dx =

a

nπ dy; y : 0 → nπ.

= 2 a

( a nπ

)2 ∫ 0

y sin2 y dy = 2a

n2π2

[ y2

4 − y sin 2y

4 cos 2y

8

]∣∣∣∣0

= 2a

n2π2

[ n2π2

4 cos 2

8 +

1 8

] =

a

2 . (Independent of n.)

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.

16 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION

〈x2= 2 a

a 0

x2 sin2 (

a x ) dx =

2 a

( a nπ

)3 ∫ 0

y2 sin2 y dy

= 2a2

()3

[ y3

6

( y3

4 1

8

) sin 2y − y cos 2y

4

]0

= 2a2

()3

[ ()3

6 − nπ cos(2)

4

] = a2

[ 1 3 1

2()2

] .

〈p〉 = md〈x〉 dt

= 0. (Note : Eq. 1.33 is much faster than Eq. 1.35.)

〈p2= ∫

ψ∗n

( 

i

d

dx

)2 ψn dx = 2

ψ∗n

( d2ψn dx2

) dx

= (2) ( 2mEn

2

) ∫ ψ∗nψn dx = 2mEn =

( 

a

)2 .

σ2x = 〈x2〉 − 〈x〉2 = a2 (

1 3 1

2()2 1

4

) =

a2

4

( 1 3 2

()2

) ; σx =

a

2

√ 1 3 2

()2 .

σ2p = 〈p2〉 − 〈p〉2 = ( 

a

)2 ; σp =



a . ∴ σxσp =



2

√ ()2

3 2.

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

π2

3 2 = (1.136)/2 > /2. 

Problem 2.5

(a)

|Ψ|2 = Ψ2Ψ = |A|2(ψ∗1 + ψ∗2)(ψ1 + ψ2) = |A|2[ψ∗1ψ1 + ψ∗1ψ2 + ψ∗2ψ1 + ψ∗2ψ2].

1 = ∫

|Ψ|2dx = |A|2 ∫

[1|2 + ψ∗1ψ2 + ψ∗2ψ1 + 2|2]dx = 2|A|2 ⇒ A = 1/ √

2.

(b)

Ψ(x, t) = 12

[ ψ1e

−iE1t/ + ψ2e−iE2t/ ]

(but En 

= n2ω)

= 12

√ 2 a

[ sin

(π a x ) e−iωt + sin

( 2π a x

) e−i4ωt

] =

1√ a e−iωt

[ sin

(π a x )

+ sin (

2π a x

) e−3iωt

] .

|Ψ(x, t)|2 = 1 a

[ sin2

(π a x )

+ sin (π a x )

sin (

2π a x

) ( e−3iωt + e3iωt

) + sin2

( 2π a x

)] =

1 a

[ sin2

(π a x )

+ sin2 (

2π a x

) + 2 sin

(π a x )

sin (

2π a x

) cos(3ωt)

] .

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 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION 17

(c)

〈x〉 = ∫

x|Ψ(x, t)|2dx

= 1 a

a 0

x

[ sin2

(π a x )

+ sin2 (

2π a x

) + 2 sin

(π a x )

sin (

2π a x

) cos(3ωt)

] dx

a 0

x sin2 (π a x ) dx =

[ x2

4 − x sin

( 2π a x

) 4π/a

cos (

2π a x

) 8(π/a)2

]∣∣∣∣∣ a

0

= a2

4 =

a 0

x sin2 (

2π a x

) dx.

a 0

x sin (π a x )

sin (

2π a x

) dx =

1 2

a 0

x

[ cos

(π a x ) cos

( 3π a x

)] dx

= 1 2

[ a2

π2 cos

(π a x )

+ ax

π sin

(π a x ) − a

2

9π2 cos

( 3π a x

) − ax

3π sin

( 3π a x

)]a 0

= 1 2

[ a2

π2 ( cos(π) cos(0)

) − a

2

9π2 ( cos(3π) cos(0)

)] = −a

2

π2

( 1 1

9

) = 8a

2

9π2 .

〈x〉 = 1 a

[ a2

4 +

a2

4 16a

2

9π2 cos(3ωt)

] =

a

2

[ 1 32

9π2 cos(3ωt)

] .

Amplitude: 32 9π2

(a 2

) = 0.3603(a/2); angular frequency: 3ω =

3π2 2ma2

.

(d)

〈p〉 = md〈x〉 dt

= m (a

2

) ( 32

9π2

) (3ω) sin(3ωt) = 8

3a sin(3ωt).

(e) You could get either E1 = π22/2ma2 or E2 = 2π22/ma2, with equal probability P1 = P2 = 1/2.

So 〈H〉 = 1 2 (E1 + E2) =

5π22

4ma2 ; it’s the average of E1 and E2.

Problem 2.6

From Problem 2.5, we see that

Ψ(x, t) = 1√ a e−iωt

[ sin

( π ax

) + sin

( 2π a x

) e−3iωteiφ

] ;

|Ψ(x, t)|2 = 1a [ sin2

( π ax

) + sin2

( 2π a x

) + 2 sin

( π ax

) sin

( 2π a x

) cos(3ωt− φ)

] ;

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.

18 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION

and hence 〈x〉 = a2 [ 1 329π2 cos(3ωt− 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[ψ1(x) + 2(x)], then cos(3ωt− φ) = sin(3ωt); 〈x〉 starts at

a

2 .

If φ = π, so Ψ(x, 0) = A[ψ1(x) − ψ2(x)], then cos(3ωt− φ) = cos(3ωt); 〈x〉 starts at a

2

( 1 +

32 9π2

) .

Problem 2.7

Ψ(x,0)

xaa/2

Aa/2

(a)

1 = A2 ∫ a/2

0

x2dx + A2 ∫ a a/2

(a− x)2dx = A2 [ x3

3

∣∣∣∣a/2 0

(a− x) 3

3

∣∣∣∣a a/2

]

= A2

3

( a3

8 +

a3

8

) =

A2a3

12 ⇒ A = 2

3√ a3

.

(b)

cn =

√ 2 a

2

3 a √ a

[ ∫ a/2 0

x sin (

a x

) dx +

a a/2

(a− x) sin (

a x

) dx

] =

2

6 a2

{[( a

)2 sin

(

a x

) − xa

cos

(

a x

)]∣∣∣∣a/2 0

+ a [ − a

cos

(

a x

)]∣∣∣∣a a/2

[(

a

)2 sin

(

a x

)

( ax

) cos

(

a x

)]∣∣∣∣a a/2

}

= 2

6 a2

[( a

)2 sin

(

2

) ✘✘✘

✘✘✘ ✘a2

2cos

(

2

) ✟✟

✟✟ ✟a2

cos+

✟✟ ✟✟

✟✟a2

cos

(

2

) +

( a

)2 sin

(

2

) + ✟✟

✟✟ ✟a2

cosnπ −✘✘✘

✘✘✘ ✘a2

2cos

(

2

)] =

2

6

a2 2 

a2

()2 sin

(

2

) =

4

6 ()2

sin (

2

) =

{ 0, n even, (1)(n−1)/2 4

6

()2 , n odd.

So Ψ(x, t) = 4

6 π2

√ 2 a

n=1,3,5,...

(1)(n−1)/2 1 n2

sin (

a x

) e−Ent/, where En =

n2π22

2ma2 .

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 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION 19

(c)

P1 = |c1|2 = 16 · 6 π4

= 0.9855.

(d)

〈H〉 = ∑

|cn|2En = 96 π4

π22

2ma2

( 1 1

+ 1 32

+ 1 52

+ 1 72

+ · · ·︸ ︷︷ ︸ π2/8

) =

482

π2ma2 π2

8 =

62

ma2 .

Problem 2.8

(a)

Ψ(x, 0) =

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

1 = A2 ∫ a/2

0

dx = A2(a/2) ⇒ A = √

2 a .

(b) From Eq. 2.37,

c1 = 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 .

P1 = |c1|2 = (2)2 = 0.4053.

Problem 2.9

ĤΨ(x, 0) =  2

2m ∂2

∂x2 [Ax(a− x)] = −A 

2

2m ∂

∂x (a− 2x) = A

2

m .

∫ Ψ(x, 0)∗ĤΨ(x, 0) dx = A2

 2

m

a 0

x(a− x) dx = A2  2

m

( a x2

2 − x

3

3

) ∣∣∣∣a 0

= A2 

2

m

( a3

2 − a

3

3

) =

30 a5 

2

m

a3

6 =

52

ma2

(same as Example 2.3).

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.

20 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION

Problem 2.10

(a) Using Eqs. 2.47 and 2.59,

a+ψ0 = 1

2

(  d

dx + mωx

) (mω π

)1/4 e−

2 x

2

= 1

2

(mω π

)1/4 [ 

( −mω

2

) 2x + mωx

] e−

2 x

2 =

12

(mω π

)1/4 2mωxe−

2 x

2 .

(a+)2ψ0 = 1

2

(mω π

)1/4 2

(  d

dx + mωx

) xe−

2 x

2

= 1 

(mω π

)1/4 [ 

( 1 − xmω

2 2x

) + mωx2

] e−

2 x

2 =

(mω π

)1/4 (2

x2 1 ) e−

2 x

2 .

Therefore, from Eq. 2.67,

ψ2 = 12 (a+)2ψ0 =

12

(mω π

)1/4 (2

x2 1 ) e−

2 x

2 .

(b) ψ ψ ψ1 20

(c) Since ψ0 and ψ2 are even, whereas ψ1 is odd, ∫ ψ∗0ψ1dx and

ψ∗2ψ1dx vanish automatically. The only one

we need to check is ∫ ψ∗2ψ0 dx:∫

ψ∗2ψ0 dx = 12

π

∞ −∞

( 2

x2 1 ) e−



x2dx

=

2π

( ∫ ∞ −∞

e− mω 

x2dx− 2

∞ −∞

x2e− mω 

x2dx

) =

2π

(√ π

mω − 2





2

π

) = 0. 

Problem 2.11

(a) Note that ψ0 is even, and ψ1 is odd. In either case |ψ|2 is even, so 〈x〉 = ∫ x|ψ|2dx = 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, ψ0 = αe−ξ

2/2, ψ1 =

2αξe−ξ 2/2. So

n = 0:

〈x2= α2 ∫ ∞ −∞

x2e−ξ 2/2dx = α2

( 

)3/2 ∫ ∞ −∞

ξ2e−ξ 2 =

1√ π

( 

)√ π

2 =



2mω .

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 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION 21

〈p2= ∫

ψ0

( 

i

d

dx

)2 ψ0 dx = 2α2



∞ −∞

e−ξ 2/2

( d2

2 e−ξ

2/2

)

= −mω√ π

∞ −∞

( ξ2 1

) e−ξ

2/2= −mω√ π

(√ π

2 − √ π

) =

mω

2 .

n = 1:

〈x2= 2α2 ∫ ∞ −∞

x2ξ2e−ξ 2 dx = 2α2

( 

)3/2 ∫ ∞ −∞

ξ4e−ξ 2 =

2√ πmω

3 √ π

4 =

3 2

.

〈p2= 22α2 √



∞ −∞

ξe−ξ 2/2

[ d2

2 ( ξe−ξ

2/2 )]

= 2√ π

∞ −∞

( ξ4 3ξ2

) e−ξ

2 = 2

π

( 3 4 √ π − 3

√ π

2

) =

3mω 2

.

(b) n = 0:

σx = √ 〈x2〉 − 〈x〉2 =

√ 

2; σp =

〈p2〉 − 〈p〉2 =

mω

2 ;

σxσp =

√ 

2



2 = 

2 . (Right at the uncertainty limit.)

n = 1:

σx =

√ 3

2; σp =

√ 3mω

2 ; σxσp = 3



2 > 

2 . 

(c)

〈T 〉 = 1 2m

〈p2=

 1 4ω (n = 0)

3 4ω (n = 1)

 ; 〈V 〉 = 122〈x2= 

1 4ω (n = 0)

3 4ω (n = 1)

 .

〈T 〉 + 〈V 〉 = 〈H〉 =

 1 2ω (n = 0) = E0

3 2ω (n = 1) = E1

 , as expected.

Problem 2.12

From Eq. 2.69,

x =

√ 

2(a+ + a−), p = i

√ 

2 (a+ − a−),

so

〈x〉 = √



2

ψ∗n(a+ + a−)ψn dx.

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.

22 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION

But (Eq. 2.66) a+ψn =

√ n + 1ψn+1, a−ψn =

√ nψn−1.

So

〈x〉 = √



2

[√ n + 1

ψ∗nψn+1 dx +

√ n

ψ∗nψn−1 dx

] = 0 (by orthogonality).

〈p〉 = md〈x〉 dt

= 0. x2 = 

2(a+ + a−)2 =



2( a2+ + a+a− + a−a+ + a

2

) .

〈x2=  2

ψ∗n

( a2+ + a+a− + a−a+ + a

2

) ψn. But

 a2+ψn = a+

(√ n + 1ψn+1

) =

√ n + 1

√ n + 2ψn+2 =

√ (n + 1)(n + 2)ψn+2.

a+a−ψn = a+ (

nψn−1 )

= √ n √ nψn = nψn.

a−a+ψn = a− (

n + 1ψn+1 )

= √ n + 1)

√ n + 1ψn = (n + 1)ψn.

a2−ψn = a− (

nψn−1 )

= √ n √ n− 1ψn−2 =

√ (n− 1)nψn−2.

So

〈x2=  2

[ 0 + n

|ψn|2dx + (n + 1)

|ψn|2 dx + 0

] =



2(2n + 1) =

( n +

1 2

) 

mω .

p2 = 2

(a+ − a−)2 = 

2 ( a2+ − a+a− − a−a+ + a2

)

〈p2= 2

[0 − n− (n + 1) + 0] = 2

(2n + 1) = ( n +

1 2

) mω.

〈T 〉 = 〈p2/2m〉 = 1 2

( n +

1 2

) ω .

σx = √ 〈x2〉 − 〈x〉2 =

n +

1 2

√ 

; σp =

〈p2〉 − 〈p〉2 =

n +

1 2

√ mω; σxσp =

( n +

1 2

)  

2 . 

Problem 2.13

(a)

1 = ∫

|Ψ(x, 0)|2dx = |A|2 ∫ (

90|2 + 12ψ∗0ψ1 + 12ψ∗1ψ0 + 161|2 ) dx

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

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 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION 23

(b)

Ψ(x, t) = 1 5

[ 3ψ0(x)e−iE0t/ + 4ψ1(x)e−iE1t/

] =

1 5

[ 3ψ0(x)e−iωt/2 + 4ψ1(x)e−3iωt/2

] .

(Here ψ0 and ψ1 are given by Eqs. 2.59 and 2.62; E1 and E2 by Eq. 2.61.)

|Ψ(x, t)|2 = 1 25

[ 9ψ20 + 12ψ0ψ1e

iωt/2e−3iωt/2 + 12ψ0ψ1e−iωt/2e3iωt/2 + 16ψ21 ]

= 1 25

[ 9ψ20 + 16ψ

2 1 + 24ψ0ψ1 cos(ωt)

] .

(c)

〈x〉 = 1 25

[ 9

20 dx + 16

21 dx + 24 cos(ωt)

0ψ1 dx

] .

But ∫ 20 dx =

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

0ψ1 dx = √

π

√ 2

xe−

2 x

2 xe−

2 x

2 dx =

√ 2 π

(

) ∫ ∞ −∞

x2e− mω 

x2dx

=

√ 2 π

(

) 2 √ π2

( 1 2

√ 

)3 =

√ 

2mω .

So

〈x〉 = 24 25

√ 

2cos(ωt); 〈p〉 = m d

dt 〈x〉 = 24

25



2 sin(ωt).

(With ψ2 in place of ψ1 the frequency would be (E2 − E0)/ = [(5/2)ω − (1/2)ω]/ = 2ω.) Ehrenfest’s theorem says d〈p〉/dt = −〈∂V/∂x〉. Here

d〈p〉 dt

= 24 25



2 ω cos(ωt), V =

1 2 2x2 ⇒ ∂V

∂x = 2x,

so

∂V ∂x

〉 = −mω2〈x〉 = −mω2 24

25

√ 

2cos(ωt) = 24

25

√ 

2 ω cos(ωt),

so Ehrenfest’s theorem holds.

(d) You could get E0 = 12ω, with probability |c0|2 = 9/25, or E1 = 32ω, with probability |c1|2 = 16/25.

Problem 2.14

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

= 2(n + 12 )ω = ω, 3ω, 5ω, . . . . So the probability of getting 12ω is zero. The probability of getting ω (the new ground state energy) is P0 = |c0|2, where c0 =∫

Ψ(x, 0)ψ′0 dx, with

Ψ(x, 0) = ψ0(x) = (mω π

)1/4 e−

2 x

2 , ψ0(x)=

( m2ω π

)1/4 e−

m2ω 2 x

2 .

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.

24 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION

So

c0 = 21/4 √

π

∞ −∞

e− 32 x

2 dx = 21/4

π 2 √ π

( 1 2

√ 2

3

) = 21/4

√ 2 3 .

Therefore

P0 = 2 3

2 = 0.9428.

Problem 2.15

ψ0 = (mω π

)1/4 e−ξ

2/2, so P = 2 √

π

∞ x0

e−ξ 2 dx = 2

π

√ 

∞ ξ0

e−ξ 2 dξ.

Classically allowed region extends out to: 122x20 = E0 =

1 2ω, or x0 =

√ 

mω , so ξ0 = 1.

P = 2√ π

1

e−ξ 2 = 2(1 − F (

2)) (in notation of CRC Table) = 0.157.

Problem 2.16

n = 5: j = 1 ⇒ a3 = 2(51)(1+1)(1+2)a1 = 43a1; j = 3 ⇒ a5 = 2(53)

(3+1)(3+2)a3 = 15a3 = 415a1; j = 5 ⇒ a7 = 0. So H5(ξ) = a1ξ − 43a1ξ3 + 415a1ξ5 =

a1 15 (15ξ − 20ξ3 + 4ξ5). By convention the coefficient of ξ5 is 25, so a1 = 15 · 8,

and H5(ξ) = 120ξ − 160ξ3 + 32ξ5 (which agrees with Table 2.1).

n = 6: j = 0 ⇒ a2 = 2(60)(0+1)(0+2)a0 = 6a0; j = 2 ⇒ a4 = 2(62)

(2+1)(2+2)a2 = 23a2 = 4a0; j = 4 ⇒ a6 = 2(64)

(4+1)(4+2)a4 = 215a4 = 815a0; j = 6 ⇒ a8 = 0. So H6(ξ) = a0 6a0ξ2 + 4a0ξ4 815ξ6a0. The coefficient of ξ6

is 26, so 26 = 815a0 ⇒ a0 = 15 · 8 = 120. H6(ξ) = 120 + 720ξ 2 480ξ4 + 64ξ6.

Problem 2.17

(a)

d

(e−ξ

2 ) = 2ξe−ξ2 ;

( d

)2 e−ξ

2 =

d

(2ξe−ξ2) = (2 + 4ξ2)e−ξ2 ;

( d

)3 e−ξ

2 =

d

[ (2 + 4ξ2)e−ξ2

] =

[ 8ξ + (2 + 4ξ2)(2ξ)

] e−ξ

2 = (12ξ − 8ξ3)e−ξ2 ;

( d

)4 e−ξ

2 =

d

[ (12ξ − 8ξ3)e−ξ2

] =

[ 12 24ξ2 + (12ξ − 8ξ3)(2ξ)

] e−ξ

2 = (12 48ξ2 + 16ξ4)e−ξ2 .

H3(ξ) = −eξ 2 (

d

)3 e−ξ

2 = 12ξ + 8ξ3; H4(ξ) =

2 (

d

)4 e−ξ

2 = 12 48ξ2 + 16ξ4.

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 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION 25

(b)

H5 = 2ξH4 8H3 = 2ξ(12 48ξ2 + 16ξ4) 8(12ξ + 8ξ3) = 120ξ − 160ξ3 + 32ξ5.

H6 = 2ξH5 10H4 = 2ξ(120ξ − 160ξ3 + 32ξ5) 10(12 48ξ2 + 16ξ4) = 120 + 720ξ2 480ξ4 + 64ξ6.

(c)

dH5

= 120 480ξ2 + 160ξ4 = 10(12 48ξ2 + 16ξ4) = (2)(5)H4. 

dH6

= 1440ξ − 1920ξ3 + 384ξ5 = 12(120ξ − 160ξ3 + 32ξ5) = (2)(6)H5. 

(d)

d

dz (e−z

2+2) = (2z + ξ)e−z2+2; setting z = 0, H0(ξ) = 2ξ.

( d

dz

)2 (e−z

2+2) = d

dz

[ (2z + 2ξ)e−z2+2

] =

[ 2 + (2z + 2ξ)2

] e−z

2+2; setting z = 0, H1(ξ) = 2 + 4ξ2.

( d

dz

)3 (e−z

2+2) = d

dz

{[ 2 + (2z + 2ξ)2

] e−z

2+2

} =

{ 2(2z + 2ξ)(2) +

[ 2 + (2z + 2ξ)2

] (2z + 2ξ)

} e−z

2+2;

setting z = 0, H2(ξ) = 8ξ + (2 + 4ξ2)(2ξ) = 12ξ + 8ξ3.

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

2i

) =

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).

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.

comments (1)

This is only a preview

3 shown on 303 pages

Download the document