Introduction to quantum mechanics 2nd ed. - Solutions

303 Pages·2005·2.75 MB·english

by  Griffiths D.J.

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
ThesearemyownsolutionstotheproblemsinIntroduction to Quantum Mechanics, 2nd ed. Ihavemadeevery
efforttoinsurethattheyareclearandcorrect,buterrorsareboundtooccur,andforthisIapologizeinadvance.
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[e-mail]). 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
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CHAPTER 1. THE WAVE FUNCTION 3
Chapter 1
The Wave Function
Problem 1.1
(a)
(cid:1)j(cid:2)2 =212 = 441.
(cid:2) (cid:3) (cid:4)
1 1
(cid:1)j2(cid:2) = 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−(cid:1)j(cid:2)
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
(cid:2) (cid:3) (cid: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)
(cid:1)j2(cid:2)−(cid:1)j(cid:2)2 =459.571−441=18.571. [Agrees with (b).]
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4 CHAPTER 1. THE WAVE FUNCTION
Problem 1.2
(a)
(cid:5) (cid:6) (cid:7)(cid:8)
(cid:1)x2(cid:2)= hx2 √1 dx= √1 2x5/2 (cid:8)(cid:8)(cid:8)h = h2.
2 hx 2 h 5 5
0 0
(cid:6) (cid:7)
h2 h 2 4 2h
σ2 =(cid:1)x2(cid:2)−(cid:1)x(cid:2)2 = − = h2 ⇒ σ = √ =0.2981h.
5 3 45 3 5
(b)
(cid:5) (cid:8)
P =1− x+ √1 dx=1− √1 (2√x)(cid:8)(cid:8)(cid:8)x+ =1− √1 (cid:9)√x+−√x−(cid:10).
2 hx 2 h h
x− x−
x+ ≡(cid:1)x(cid:2)+σ =0.3333h+0.2981h=0.6315h; x− ≡(cid:1)x(cid:2)−σ =0.3333h−0.2981h=0.0352h.
√ √
P =1− 0.6315+ 0.0352= 0.393.
Problem 1.3
(a)
(cid:5)
∞
1= Ae−λ(x−a)2dx. Let u≡x−a, du=dx, u:−∞→∞.
−∞
(cid:5) (cid:11) (cid:11)
∞
π λ
1=A e−λu2du=A ⇒ A= .
−∞ λ π
(b)
(cid:5) (cid:5)
∞ ∞
(cid:1)x(cid:2)=A xe−λ(x−a)2dx=A (u+a)e−λu2du
(cid:12)(cid:5)−∞ (cid:5) −∞ (cid:13) (cid:6) (cid:11) (cid:7)
∞ ∞
π
=A ue−λu2du+a e−λu2du =A 0+a = a.
−∞ −∞ λ
(cid:5)
∞
(cid:1)x2(cid:2)=A x2e−λ(x−a)2dx
(cid:14)−(cid:5)∞ (cid:5) (cid:5) (cid:15)
∞ ∞ ∞
=A u2e−λu2du+2a ue−λu2du+a2 e−λu2du
−∞ −∞ −∞
(cid:12) (cid:11) (cid:11) (cid:13)
1 π π 1
=A +0+a2 = a2+ .
2λ λ λ 2λ
1 1 1
σ2 =(cid:1)x2(cid:2)−(cid:1)x(cid:2)2 =a2+ −a2 = ; σ = √ .
2λ 2λ 2λ
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CHAPTER 1. THE WAVE FUNCTION 5
(c)
ρ(x)
A
a x
Problem 1.4
(a)
(cid:5) (cid:5) (cid:16) (cid:6) (cid:7)(cid:8) (cid:6) (cid:7)(cid:8) (cid:17)
1= |A|2 ax2dx+ |A|2 b(b−x)2dx=|A|2 1 x3 (cid:8)(cid:8)(cid:8)a+ 1 −(b−x)3 (cid:8)(cid:8)(cid:8)b
a2 0 (b−a)2 a a2 3 0 (b−a)2 3 a
(cid:12) (cid:13) (cid:11)
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)
(cid:5) (cid:5) (cid:14)
a |A|2 a a a P =1 if b=a, (cid:1)
P = |Ψ|2dx= x2dx=|A|2 = .
a2 3 b P =1/2 if b=2a. (cid:1)
0 0
(e)
(cid:5) (cid:14) (cid:5) (cid:5) (cid:15)
1 a 1 b
(cid:1)x(cid:2)= x|Ψ|2dx=|A|2 x3dx+ x(b−x)2dx
a2 (b−a)2
(cid:16) (cid:6) (cid:7)(cid:8) 0 (cid:6) a (cid:7)(cid:8) (cid:17)
= 3 1 x4 (cid:8)(cid:8)(cid:8)a+ 1 b2x2 −2bx3 + x4 (cid:8)(cid:8)(cid:8)b
b a2 4 (b−a)2 2 3 4
0 a
(cid:3) (cid:4)
3
= a2(b−a)2+2b4−8b4/3+b4−2a2b2+8a3b/3−a4
4b(b−a)2
(cid:6) (cid:7)
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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6 CHAPTER 1. THE WAVE FUNCTION
Problem 1.5
(a)
(cid:5) (cid:5) (cid:6) (cid:7)(cid:8)
1= |Ψ|2dx=2|A|2 ∞e−2λxdx=2|A|2 e−2λx (cid:8)(cid:8)(cid:8)∞ = |A|2; A=√λ.
−2λ λ
0 0
(b)
(cid:5) (cid:5)
∞
(cid:1)x(cid:2)= x|Ψ|2dx=|A|2 xe−2λ|x|dx= 0. [Odd integrand.]
−∞
(cid:5) (cid:12) (cid:13)
∞
2 1
(cid:1)x2(cid:2)=2|A|2 x2e−2λxdx=2λ = .
(2λ)3 2λ2
0
(c)
1 1 √ √
σ2 =(cid:1)x2(cid:2)−(cid:1)x(cid:2)2 = ; σ = √ . |Ψ(±σ)|2 =|A|2e−2λσ =λe−2λ/ 2λ =λe− 2 =0.2431λ.
2λ2 2λ
|Ψ|2
λ
.24λ
x
−σ +σ
Probability outside:
(cid:5) (cid:5) (cid:6) (cid:7)(cid:8)
2 ∞|Ψ|2dx=2|A|2 ∞e−2λxdx=2λ e−2λx (cid:8)(cid:8)(cid:8)∞ =e−2λσ = e−√2 =0.2431.
−2λ
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:
(cid:5) (cid:5)
bx∂ |Ψ|2dx= b ∂ (x|Ψ|2)dx(cid:12)= (x|Ψ|2)(cid:8)(cid:8)b.
∂t ∂t a
a a
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CHAPTER 1. THE WAVE FUNCTION 7
Problem 1.7
(cid:18) (cid:9) (cid:10)
From Eq. 1.33, d(cid:6)p(cid:7) =−i(cid:1) ∂ Ψ∗∂Ψ dx. But, noting that ∂2Ψ = ∂2Ψ and using Eqs. 1.23-1.24:
dt ∂t ∂x ∂x∂t ∂t∂x
(cid:6) (cid:7) (cid:6) (cid:7) (cid:12) (cid:13) (cid:12) (cid:13)
∂ ∂Ψ ∂Ψ∗∂Ψ ∂ ∂Ψ i(cid:1) ∂2Ψ∗ i ∂Ψ ∂ i(cid:1) ∂2Ψ i
Ψ∗ = +Ψ∗ = − + VΨ∗ +Ψ∗ − VΨ
∂t ∂x ∂t ∂x ∂x ∂t 2m ∂x2 (cid:1) ∂x ∂x 2m ∂x2 (cid:1)
(cid:12) (cid:13) (cid:12) (cid:13)
i(cid:1) ∂3Ψ ∂2Ψ∗∂Ψ i ∂Ψ ∂
= Ψ∗ − + VΨ∗ −Ψ∗ (VΨ)
2m ∂x3 ∂x2 ∂x (cid:1) ∂x ∂x
The first term integrates to zero, using integration by parts twice, and the second term can be simplified to
VΨ∗∂Ψ −Ψ∗V ∂Ψ −Ψ∗∂VΨ=−|Ψ|2∂V. So
∂x ∂x ∂x ∂x
(cid:6) (cid:7)(cid:5)
d(cid:1)p(cid:2) i ∂V ∂V
=−i(cid:1) −|Ψ|2 dx=(cid:1)− (cid:2). QED
dt (cid:1) ∂x ∂x
Problem 1.8
Suppose Ψ satisfies the Schro¨dinger equation without V : i(cid:1)∂Ψ =−(cid:1)2 ∂2Ψ+VΨ. We want to find the solution
0 ∂t 2m ∂x2
Ψ with V : i(cid:1)∂Ψ0 =−(cid:1)2 ∂2Ψ0 +(V +V )Ψ .
0 0 ∂t 2m ∂x2 0 0
Claim: Ψ0 =Ψe−iV0t/(cid:1). (cid:9) (cid:10) (cid:19) (cid:20)
Proof: i(cid:1)∂∂Ψt0 =i(cid:1)∂∂Ψte−iV0t/(cid:1)+i(cid:1)Ψ −iV(cid:1)0 e−iV0t/(cid:1) = −2(cid:1)m2 ∂∂2xΨ2 +VΨ e−iV0t/(cid:1)+V0Ψe−iV0t/(cid:1)
=−(cid:1)2 ∂2Ψ0 +(V +V )Ψ . QED
2m ∂x2 0 0
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)
(cid:5) (cid:11) (cid:11) (cid:6) (cid:7)
∞ 1 π π(cid:1) 2am 1/4
1=2|A|2 e−2amx2/(cid:1)dx=2|A|2 =|A|2 ; A= .
2 (2am/(cid:1)) 2am π(cid:1)
0
(b)
(cid:6) (cid:7) (cid:6) (cid:7)
∂Ψ ∂Ψ 2amx ∂2Ψ 2am ∂Ψ 2am 2amx2
=−iaΨ; =− Ψ; =− Ψ+x =− 1− Ψ.
∂t ∂x (cid:1) ∂x2 (cid:1) ∂x (cid:1) (cid:1)
Plug these into the Schro¨dinger equation, i(cid:1)∂Ψ =−(cid:1)2 ∂2Ψ +VΨ:
∂t 2m ∂x2
(cid:6) (cid:7)(cid:6) (cid:7)
(cid:1)2 2am 2amx2
VΨ=i(cid:1)(−ia)Ψ+ − 1− Ψ
2m (cid:1) (cid:1)
(cid:12) (cid:6) (cid:7)(cid:13)
2amx2
= (cid:1)a−(cid:1)a 1− Ψ=2a2mx2Ψ, so V(x)=2ma2x2.
(cid:1)
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8 CHAPTER 1. THE WAVE FUNCTION
(c)
(cid:5)
∞
(cid:1)x(cid:2)= x|Ψ|2dx= 0. [Odd integrand.]
−∞
(cid:5) (cid:11)
∞ 1 π(cid:1) (cid:1)
(cid:1)x2(cid:2)=2|A|2 x2e−2amx2/(cid:1)dx=2|A|2 = .
22(2am/(cid:1)) 2am 4am
0
d(cid:1)x(cid:2)
(cid:1)p(cid:2)=m = 0.
dt
(cid:5) (cid:6) (cid:7) (cid:5)
(cid:1) ∂ 2 ∂2Ψ
(cid:1)p2(cid:2)= Ψ∗ Ψdx=−(cid:1)2 Ψ∗ dx
i ∂x ∂x2
(cid:5) (cid:12) (cid:6) (cid:7) (cid:13) (cid:14)(cid:5) (cid:5) (cid:15)
2am 2amx2 2am
=−(cid:1)2 Ψ∗ − 1− Ψ dx=2am(cid:1) |Ψ|2dx− x2|Ψ|2dx
(cid:1) (cid:1) (cid:1)
(cid:6) (cid:7) (cid:6) (cid:7) (cid:6) (cid:7)
2am 2am (cid:1) 1
=2am(cid:1) 1− (cid:1)x2(cid:2) =2am(cid:1) 1− =2am(cid:1) = am(cid:1).
(cid:1) (cid:1) 4am 2
(d)
(cid:11)
(cid:1) (cid:1) √
σ2 =(cid:1)x2(cid:2)−(cid:1)x(cid:2)2 = =⇒ σ = ; σ2 =(cid:1)p2(cid:2)−(cid:1)p(cid:2)2 =am(cid:1)=⇒ σ = am(cid:1).
x 4am x 4am p p
(cid:21)
√
σ σ = (cid:1) am(cid:1)= (cid:1). This is (just barely) consistent with the uncertainty principle.
x p 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: (cid:1)j(cid:2)= 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) (cid:1)j2(cid:2)= 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 =(cid:1)j2(cid:2)−(cid:1)j(cid:2)2 =28.4−4.722 =28.4−22.2784=6.1216; σ = 6.1216= 2.474.
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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.
(cid:14)
1/π, if 0≤θ ≤π,
ρ(θ)=
0, otherwise.
ρ(θ)
1/π
θ
−π/2 0 π 3π/2
(b)
(cid:5) (cid:5) (cid:6) (cid:7)(cid:8)
(cid:1)θ(cid:2)= θρ(θ)dθ = 1 πθdθ = 1 θ2 (cid:8)(cid:8)(cid:8)π = π [of course].
π π 2 2
0 0
(cid:5) (cid:6) (cid:7)(cid:8)
(cid:1)θ2(cid:2)= 1 πθ2dθ = 1 θ3 (cid:8)(cid:8)(cid:8)π = π2.
π π 3 3
0 0
π2 π2 π2 π
σ2 =(cid:1)θ2(cid:2)−(cid:1)θ(cid:2)2 = − = ; σ = √ .
3 4 12 2 3
(c)
(cid:5)
1 π 1 1 2
(cid:1)sinθ(cid:2)= sinθdθ = (−cosθ)|π = (1−(−1))= .
π π 0 π π
0
(cid:5)
1 π 1
(cid:1)cosθ(cid:2)= cosθdθ = (sinθ)|π = 0.
π π 0
0
(cid:5) (cid:5)
1 π 1 π 1
(cid:1)cos2θ(cid:2)= cos2θdθ = (1/2)dθ = .
π π 2
0 0
[Because sin2θ+cos2θ =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 = rcosθ ⇒ dx = −rsinθdθ. The probability that the needle lies in range dθ is ρ(θ)dθ = 1dθ, so the
π
probability that it’s in the range dx is
1 dx 1 dx dx
ρ(x)dx= = (cid:22) = √ .
πrsinθ πr 1−(x/r)2 π r2−x2
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10 CHAPTER 1. THE WAVE FUNCTION
ρ(x)
-2r -r r 2r x
(cid:14)
√ 1 , if −r <x<r,
∴ ρ(x)= π r2−x2 [Note: We want the magnitude of dx here.]
0, otherwise.
(cid:18) (cid:18) (cid:8)
Total: r √ 1 dx= 2 r √ 1 dx= 2 sin−1 x(cid:8)r = 2 sin−1(1)= 2 · π =1.(cid:1)
−r π r2−x2 π 0 r2−x2 π r 0 π π 2
(b)
(cid:5)
1 r 1
(cid:1)x(cid:2)= x√ dx= 0 [odd integrand, even interval].
π −r r2−x2
(cid:1)x2(cid:2)= 2 (cid:5) r √ x2 dx= 2 (cid:12)−x(cid:22)r2−x2+ r2 sin−1(cid:23)x(cid:24)(cid:13)(cid:8)(cid:8)(cid:8)(cid:8)r = 2r2 sin−1(1)= r2.
π r2−x2 π 2 2 r π 2 2
0 0
√
σ2 =(cid:1)x2(cid:2)−(cid:1)x(cid:2)2 =r2/2=⇒ σ =r/ 2.
To get (cid:1)x(cid:2) and (cid:1)x2(cid:2) from Problem 1.11(c), use x=rcosθ, so (cid:1)x(cid:2)=r(cid:1)cosθ(cid:2)=0,(cid:1)x2(cid:2)=r2(cid:1)cos2θ(cid:2)=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
(cid:16) (cid:17)
(cid:5) (cid:5) (cid:5) (cid:5)
−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 (cid:14) (cid:23)x(cid:24)(cid:8)(cid:8)−y (cid:23)x(cid:24)(cid:8)(cid:8)l (cid:15) 1 (cid:3) (cid:4)
= sin−1 (cid:8) + sin−1 (cid:8) = −sin−1(y/l)+2sin−1(1)−sin−1(1−y/l)
π l −l l l−y π
sin−1(y/l) sin−1(1−y/l)
=1− − .
π π
Now, all values of y are equally likely, so ρ(y)=1/l, and hence the probability of crossing is
1 (cid:5) l(cid:12) (cid:23)y(cid:24) (cid:6)l−y(cid:7)(cid:13) 1 (cid:5) l(cid:3) (cid:4)
P = π−sin−1 −sin−1 dy = π−2sin−1(y/l) dy
πl l l πl
0 0
(cid:12) (cid:23) (cid:22) (cid:24)(cid:8) (cid:13)
1 (cid:8)l 2 2 2
= πl−2 ysin−1(y/l)+l 1−(y/l)2 (cid:8) =1− [lsin−1(1)−l]=1−1+ = .
πl 0 πl π π
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