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Introduction to Quantum Mechanics, 3rd Edition Solution Manual

Introduction to Quantum Mechanics, 3rd Edition Solution Manual is your guide to textbook mastery, offering detailed solutions to every chapter's exercises.

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    Instructors’ Solution Manual Introduction to Quantum Mechanics, 3rd ed. David Griffiths, Darrell Schroeter Reed College August 3, 2018 SOLUTIONS GUIDE

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    2 Contents 1 The Wave Function 4 2 The Time-Independent Schr¨ odinger 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

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    4 CHAPTER 1. THE WAVE FUNCTION Chapter 1 The Wave Function Problem 1.1 (a) 〈 j 〉 2 = 21 2 = 441. 〈 j 2 〉 = 1 N ∑ j 2 N ( j ) = 1 14 [ (14 2 ) + (15 2 ) + 3(16 2 ) + 2(22 2 ) + 2(24 2 ) + 5(25 2 ) ] = 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 σ 2 = 1 N ∑ (∆ j ) 2 N ( 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) 〈 j 2 〉 − 〈 j 〉 2 = 459 . 571 − 441 = 18 . 571 . [Agrees with (b).]

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    CHAPTER 1. THE WAVE FUNCTION 5 Problem 1.2 (a) 〈 x 2 〉 = ∫ h 0 x 2 1 2 √ hx dx = 1 2 √ h ( 2 5 x 5 / 2 )∣ ∣ ∣ ∣ h 0 = h 2 5 . σ 2 = 〈 x 2 〉 − 〈 x 〉 2 = h 2 5 − ( h 3 ) 2 = 4 45 h 2 ⇒ σ = 2 h 3 √ 5 = 0 . 2981 h. (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 . 3333 h + 0 . 2981 h = 0 . 6315 h ; x − ≡ 〈 x 〉 − σ = 0 . 3333 h − 0 . 2981 h = 0 . 0352 h. 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 . 〈 x 2 〉 = A ∫ ∞ −∞ x 2 e − λ ( x − a ) 2 dx = A {∫ ∞ −∞ u 2 e − λu 2 du + 2 a ∫ ∞ −∞ ue − λu 2 du + a 2 ∫ ∞ −∞ e − λu 2 du } = A [ 1 2 λ √ π λ + 0 + a 2 √ π λ ] = a 2 + 1 2 λ . σ 2 = 〈 x 2 〉 − 〈 x 〉 2 = a 2 + 1 2 λ − a 2 = 1 2 λ ; σ = 1 √ 2 λ .

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    6 CHAPTER 1. THE WAVE FUNCTION (c) A x a l ( x) Problem 1.4 (a) 1 = | A | 2 a 2 ∫ a 0 x 2 dx + | A | 2 ( b − a ) 2 ∫ b a ( b − x ) 2 dx = | A | 2 { 1 a 2 ( x 3 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) x a A b ^ (c) At x = a . (d) P = ∫ a 0 | Ψ | 2 dx = | A | 2 a 2 ∫ a 0 x 2 dx = | A | 2 a 3 = a b . { P = 1 if b = a, X P = 1 / 2 if b = 2 a. X (e) 〈 x 〉 = ∫ x | Ψ | 2 dx = | A | 2 { 1 a 2 ∫ a 0 x 3 dx + 1 ( b − a ) 2 ∫ b a x ( b − x ) 2 dx } = 3 b { 1 a 2 ( x 4 4 )∣ ∣ ∣ ∣ a 0 + 1 ( b − a ) 2 ( b 2 x 2 2 − 2 b x 3 3 + x 4 4 )∣ ∣ ∣ ∣ b a } = 3 4 b ( b − a ) 2 [ a 2 ( b − a ) 2 + 2 b 4 − 8 b 4 / 3 + b 4 − 2 a 2 b 2 + 8 a 3 b/ 3 − a 4 ] = 3 4 b ( b − a ) 2 ( b 4 3 − a 2 b 2 + 2 3 a 3 b ) = 1 4( b − a ) 2 ( b 3 − 3 a 2 b + 2 a 3 ) = 2 a + b 4 .

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    CHAPTER 1. THE WAVE FUNCTION 7 Problem 1.5 (a) 1 = ∫ | Ψ | 2 dx = 2 | A | 2 ∫ ∞ 0 e − 2 λx dx = 2 | A | 2 ( e − 2 λx − 2 λ )∣ ∣ ∣ ∣ ∞ 0 = | A | 2 λ ; A = √ λ . (b) 〈 x 〉 = ∫ x | Ψ | 2 dx = | A | 2 ∫ ∞ −∞ xe − 2 λ | x | dx = 0. [Odd integrand . ] 〈 x 2 〉 = 2 | A | 2 ∫ ∞ 0 x 2 e − 2 λx dx = 2 λ [ 2 (2 λ ) 3 ] = 1 2 λ 2 . (c) σ 2 = 〈 x 2 〉 − 〈 x 〉 2 = 1 2 λ 2 ; σ = 1 √ 2 λ . | Ψ( ± σ ) | 2 = | A | 2 e − 2 λσ = λe − 2 λ/ √ 2 λ = λe − √ 2 = 0 . 2431 λ. | ^ | 2 h m <m + x .24 h Probability outside: 2 ∫ ∞ σ | Ψ | 2 dx = 2 | A | 2 ∫ ∞ σ e − 2 λx dx = 2 λ ( e − 2 λx − 2 λ )∣ ∣ ∣ ∣ ∞ σ = e − 2 λσ = 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 | 2 dx = ∫ b a ∂ ∂t ( x | Ψ | 2 ) dx 6 = ( x | Ψ | 2 ) ∣ ∣ b a .

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    8 CHAPTER 1. THE WAVE FUNCTION Problem 1.7 From Eq. 1.33, d 〈 p 〉 dt = − i ℏ ∫ ∂ ∂t ( Ψ ∗ ∂ Ψ ∂x ) dx . But, noting that ∂ 2 Ψ ∂x∂t = ∂ 2 Ψ ∂t∂x and using Eqs. 1.23-1.24: ∂ ∂t ( Ψ ∗ ∂ Ψ ∂x ) = ∂ Ψ ∗ ∂t ∂ Ψ ∂x + Ψ ∗ ∂ ∂x ( ∂ Ψ ∂t ) = [ − i ℏ 2 m ∂ 2 Ψ ∗ ∂x 2 + i ℏ V Ψ ∗ ] ∂ Ψ ∂x + Ψ ∗ ∂ ∂x [ i ℏ 2 m ∂ 2 Ψ ∂x 2 − i ℏ V Ψ ] = i ℏ 2 m [ Ψ ∗ ∂ 3 Ψ ∂x 3 − ∂ 2 Ψ ∗ ∂x 2 ∂ Ψ ∂x ] + i ℏ [ 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 ℏ ( i ℏ ) ∫ −| Ψ | 2 ∂V ∂x dx = 〈− ∂V ∂x 〉 . QED Problem 1.8 Suppose Ψ satisfies the Schr¨ odinger equation without V 0 : i ℏ ∂ Ψ ∂t = − ℏ 2 2 m ∂ 2 Ψ ∂x 2 + V Ψ. We want to find the solution Ψ 0 with V 0 : i ℏ ∂ Ψ 0 ∂t = − ℏ 2 2 m ∂ 2 Ψ 0 ∂x 2 + ( V + V 0 )Ψ 0 . Claim : Ψ 0 = Ψ e − iV 0 t/ ℏ . Proof : i ℏ ∂ Ψ 0 ∂t = i ℏ ∂ Ψ ∂t e − iV 0 t/ ℏ + i ℏ Ψ ( − iV 0 ℏ ) e − iV 0 t/ ℏ = [ − ℏ 2 2 m ∂ 2 Ψ ∂x 2 + V Ψ ] e − iV 0 t/ ℏ + V 0 Ψ e − iV 0 t/ ℏ = − ℏ 2 2 m ∂ 2 Ψ 0 ∂x 2 + ( V + V 0 )Ψ 0 . 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 − 2 amx 2 / ℏ dx = 2 | A | 2 1 2 √ π (2 am/ ℏ ) = | A | 2 √ π ℏ 2 am ; A = ( 2 am π ℏ ) 1 / 4 . (b) ∂ Ψ ∂t = − ia Ψ; ∂ Ψ ∂x = − 2 amx ℏ Ψ; ∂ 2 Ψ ∂x 2 = − 2 am ℏ ( Ψ + x ∂ Ψ ∂x ) = − 2 am ℏ ( 1 − 2 amx 2 ℏ ) Ψ . Plug these into the Schr¨ odinger equation, i ℏ ∂ Ψ ∂t = − ℏ 2 2 m ∂ 2 Ψ ∂x 2 + V Ψ: V Ψ = i ℏ ( − ia )Ψ + ℏ 2 2 m ( − 2 am ℏ ) ( 1 − 2 amx 2 ℏ ) Ψ = [ ℏ a − ℏ a ( 1 − 2 amx 2 ℏ )] Ψ = 2 a 2 mx 2 Ψ , so V ( x ) = 2 ma 2 x 2 .

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    CHAPTER 1. THE WAVE FUNCTION 9 (c) 〈 x 〉 = ∫ ∞ −∞ x | Ψ | 2 dx = 0. [Odd integrand.] 〈 x 2 〉 = 2 | A | 2 ∫ ∞ 0 x 2 e − 2 amx 2 / ℏ dx = 2 | A | 2 1 2 2 (2 am/ ℏ ) √ π ℏ 2 am = ℏ 4 am . 〈 p 〉 = m d 〈 x 〉 dt = 0 . 〈 p 2 〉 = ∫ Ψ ∗ ( ℏ i ∂ ∂x ) 2 Ψ dx = − ℏ 2 ∫ Ψ ∗ ∂ 2 Ψ ∂x 2 dx = − ℏ 2 ∫ Ψ ∗ [ − 2 am ℏ ( 1 − 2 amx 2 ℏ ) Ψ ] dx = 2 am ℏ {∫ | Ψ | 2 dx − 2 am ℏ ∫ x 2 | Ψ | 2 dx } = 2 am ℏ ( 1 − 2 am ℏ 〈 x 2 〉 ) = 2 am ℏ ( 1 − 2 am ℏ ℏ 4 am ) = 2 am ℏ ( 1 2 ) = am ℏ . (d) σ 2 x = 〈 x 2 〉 − 〈 x 〉 2 = ℏ 4 am = ⇒ σ x = √ ℏ 4 am ; σ 2 p = 〈 p 2 〉 − 〈 p 〉 2 = am ℏ = ⇒ σ p = √ am ℏ . σ x σ p = √ ℏ 4 am √ am ℏ = ℏ 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 〉 = 1 25 [0 · 0 + 1 · 2 + 2 · 3 + 3 · 5 + 4 · 3 + 5 · 3 + 6 · 3 + 7 · 1 + 8 · 2 + 9 · 3] = 1 25 [0 + 2 + 6 + 15 + 12 + 15 + 18 + 7 + 16 + 27] = 118 25 = 4.72. (c) 〈 j 2 〉 = 1 25 [0 + 1 2 · 2 + 2 2 · 3 + 3 2 · 5 + 4 2 · 3 + 5 2 · 3 + 6 2 · 3 + 7 2 · 1 + 8 2 · 2 + 9 2 · 3] = 1 25 [0 + 2 + 12 + 45 + 48 + 75 + 108 + 49 + 128 + 243] = 710 25 = 28.4. σ 2 = 〈 j 2 〉 − 〈 j 〉 2 = 28 . 4 − 4 . 72 2 = 28 . 4 − 22 . 2784 = 6 . 1216; σ = √ 6 . 1216 = 2.474.

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    10 CHAPTER 1. THE WAVE FUNCTION Problem 1.11 (a) 1 2 mv 2 + V = E → v ( x ) = √ 2 m ( E − V ( x )) . (b) T = ∫ b a 1 √ 2 m ( E − 1 2 kx 2 ) dx = √ m k ∫ b a 1 √ (2 E/k ) − x 2 dx. Turning points: v = 0 ⇒ E = V = 1 2 kb 2 ⇒ b = √ 2 E/k ; a = − b. T = 2 √ m k ∫ b 0 1 √ b 2 − x 2 dx = 2 √ m k sin − 1 ( x b ) ∣ ∣ ∣ b 0 = 2 √ m k sin − 1 (1) = 2 √ m k ( π 2 ) = π √ m k . ρ ( x ) = 1 π √ m k √ 2 m ( E − 1 2 kx 2 1 ) = 1 π √ b 2 − x 2 . ∫ b a ρ ( x ) dx = 2 π ∫ b 0 1 √ b 2 − x 2 dx = 2 π ( π 2 ) = 1 . X ρ( x) x b -b (c) 〈 x 〉 = 0 . 〈 x 2 〉 = 1 π ∫ b − b x 2 √ b 2 − x 2 dx = 2 π ∫ b 0 x 2 √ b 2 − x 2 dx = 2 π [ − x 2 √ b 2 − x 2 + b 2 2 sin − 1 ( x b ) ] ∣ ∣ ∣ ∣ b 0 = b 2 π sin − 1 (1) = b 2 π π 2 = b 2 2 = E k . σ x = √ 〈 x 2 〉 − 〈 x 〉 2 = √ 〈 x 2 〉 = b √ 2 = √ E k . Problem 1.12 (a) ρ ( p ) dp = dt T = | dt/dp | dp 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 p 2 2 m + 1 2 kx 2 = E ⇒ x = ± √ 2 k ( E − p 2 2 m ) ,

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    CHAPTER 1. THE WAVE FUNCTION 11 so ρ ( p ) = 1 π √ m k k √ 2 k ( E − p 2 2 m ) = 1 π √ 2 mE − p 2 = 1 π √ c 2 − p 2 , where c ≡ √ 2 mE. 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 ). (b) From Problem 1.11(c), 〈 p 〉 = 0 , 〈 p 2 〉 = c 2 2 , σ p = c √ 2 = √ mE . (c) σ x σ p = √ E k √ mE = √ m k E = E ω . If E ≥ 1 2 ℏ ω , then σ x σ p ≥ 1 2 ℏ , which is precisely the Heisenberg uncertainty principle! Problem 1.13 x @ t_ D : = Cos @ t D snapshots = Table @ x @ p RandomReal @ j DD , 8 j, 10 000 <D 8 - 0.977661, 0.371147, - 0.0389023, - 0.0257086, - 0.968682, - 0.681951, - 0.874704, Ü 9987 á , 0.584236, - 0.974667, 0.10798, 0.996311, - 0.767106, 0.926913 < Histogram @ snapshots, 100, "PDF", PlotRange Æ 8 0, 2 <D - 0.5 0.0 0.5 1.0 0.5 1.0 1.5 2.0 r @ x_ D : = 1 p 1 - x 2 Plot @ r @ x D , 8 x, - 1, 1 < , PlotRange -> 8 0, 2 <D x @ t_ D : = Cos @ t D snapshots = Table @ x @ p RandomReal @ j DD , 8 j, 10 000 <D 8 - 0.977661, 0.371147, - 0.0389023, - 0.0257086, - 0.968682, - 0.681951, - 0.874704, Ü 9987 á , 0.584236, - 0.974667, 0.10798, 0.996311, - 0.767106, 0.926913 < Histogram @ snapshots, 100, "PDF", PlotRange Æ 8 0, 2 <D - 0.5 0.0 0.5 1.0 0.5 1.0 1.5 2.0 r @ x_ D : = 1 p 1 - x 2 Plot @ r @ x D , 8 x, - 1, 1 < , PlotRange -> 8 0, 2 <D 1.5 2.0

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    12 CHAPTER 1. THE WAVE FUNCTION - 0.5 0.0 0.5 1.0 0.5 1.0 1.5 r @ x_ D : = 1 p 1 - x 2 Plot @ r @ x D , 8 x, - 1, 1 < , PlotRange -> 8 0, 2 <D - 1.0 - 0.5 0.5 1.0 0.5 1.0 1.5 2.0 Show @ Histogram @ snapshots, 100, "PDF", PlotRange Æ 8 0, 2 <D , Plot @ r @ x D , 8 x, - 1, 1 < , PlotRange -> 8 0, 2 <DD - 0.5 0.0 0.5 1.0 0.5 1.0 1.5 2.0 2 1p13.nb Problem 1.14 (a) P ab ( t ) = ∫ b a | Ψ( x, t ) | 2 dx, so dP ab dt = ∫ b a ∂ ∂t | Ψ | 2 dx. But (Eq. 1.25): ∂ | Ψ | 2 ∂t = ∂ ∂x [ i ℏ 2 m ( Ψ ∗ ∂ Ψ ∂x − ∂ Ψ ∗ ∂x Ψ )] = − ∂ ∂x J ( x, t ) . ∴ dP ab dt = − ∫ b a ∂ ∂x J ( x, t ) dx = − [ J ( x, t )] | b a = J ( a, t ) − J ( b, t ) . QED Probability is dimensionless, so J has the dimensions 1/time, and units seconds − 1 . (b) Here Ψ( x, t ) = f ( x ) e − iat , where f ( x ) ≡ Ae − amx 2 / ℏ , so Ψ ∂ Ψ ∗ ∂x = f e − iat df dx e iat = f df dx , and Ψ ∗ ∂ Ψ ∂x = f df dx too, so J ( x, t ) = 0.

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    CHAPTER 1. THE WAVE FUNCTION 13 Problem 1.15 Use Eqs. [1.23] and [1.24], and integration by parts: d dt ∫ ∞ −∞ Ψ ∗ 1 Ψ 2 dx = ∫ ∞ −∞ ∂ ∂t (Ψ ∗ 1 Ψ 2 ) dx = ∫ ∞ −∞ ( ∂ Ψ ∗ 1 ∂t Ψ 2 + Ψ ∗ 1 ∂ Ψ 2 ∂t ) dx = ∫ ∞ −∞ [( − i ℏ 2 m ∂ 2 Ψ ∗ 1 ∂x 2 + i ℏ V Ψ ∗ 1 ) Ψ 2 + Ψ ∗ 1 ( i ℏ 2 m ∂ 2 Ψ 2 ∂x 2 − i ℏ V Ψ 2 )] dx = − i ℏ 2 m ∫ ∞ −∞ ( ∂ 2 Ψ ∗ 1 ∂x 2 Ψ 2 − Ψ ∗ 1 ∂ 2 Ψ 2 ∂x 2 ) dx = − i ℏ 2 m [ ∂ Ψ ∗ 1 ∂x Ψ 2 ∣ ∣ ∣ ∣ ∞ −∞ − ∫ ∞ −∞ ∂ Ψ ∗ 1 ∂x ∂ Ψ 2 ∂x dx − Ψ ∗ 1 ∂ Ψ 2 ∂x ∣ ∣ ∣ ∣ ∞ −∞ + ∫ ∞ −∞ ∂ Ψ ∗ 1 ∂x ∂ Ψ 2 ∂x dx ] = 0 . QED Problem 1.16 (a) 1 = | A | 2 ∫ a − a ( a 2 − x 2 ) 2 dx = 2 | A | 2 ∫ a 0 ( a 4 − 2 a 2 x 2 + x 4 ) dx = 2 | A | 2 [ a 4 x − 2 a 2 x 3 3 + x 5 5 ]∣ ∣ ∣ ∣ a 0 = 2 | A | 2 a 5 ( 1 − 2 3 + 1 5 ) = 16 15 a 5 | A | 2 , so A = √ 15 16 a 5 . (b) 〈 x 〉 = ∫ a − a x | Ψ | 2 dx = 0. (Odd integrand.) (c) 〈 p 〉 = ℏ i A 2 ∫ a − a ( a 2 − x 2 ) d dx ( a 2 − x 2 ) ︸ ︷︷ ︸ − 2 x dx = 0. (Odd integrand.) Since we only know 〈 x 〉 at t = 0 we cannot calculate d 〈 x 〉 /dt directly. (d) 〈 x 2 〉 = A 2 ∫ a − a x 2 ( a 2 − x 2 ) 2 dx = 2 A 2 ∫ a 0 ( a 4 x 2 − 2 a 2 x 4 + x 6 ) dx = 2 15 16 a 5 [ a 4 x 3 3 − 2 a 2 x 5 5 + x 7 7 ]∣ ∣ ∣ ∣ a 0 = 15 8 a 5 ( a 7 ) ( 1 3 − 2 5 + 1 7 ) =   15 a 2 8 ( 35 − 42 + 15  3 ·  5 · 7 ) = a 2 8 · 8 7 = a 2 7 .

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    14 CHAPTER 1. THE WAVE FUNCTION (e) 〈 p 2 〉 = − A 2 ℏ 2 ∫ a − a ( a 2 − x 2 ) d 2 dx 2 ( a 2 − x 2 ) ︸ ︷︷ ︸ − 2 dx = 2 A 2 ℏ 2 2 ∫ a 0 ( a 2 − x 2 ) dx = 4 · 15 16 a 5 ℏ 2 ( a 2 x − x 3 3 )∣ ∣ ∣ ∣ a 0 = 15 ℏ 2 4 a 5 ( a 3 − a 3 3 ) = 15 ℏ 2 4 a 2 · 2 3 = 5 2 ℏ 2 a 2 . (f ) σ x = √ 〈 x 2 〉 − 〈 x 〉 2 = √ 1 7 a 2 = a √ 7 . (g) σ p = √ 〈 p 2 〉 − 〈 p 〉 2 = √ 5 2 ℏ 2 a 2 = √ 5 2 ℏ a . (h) σ x σ p = a √ 7 · √ 5 2 ℏ a = √ 5 14 ℏ = √ 10 7 ℏ 2 > ℏ 2 . X Problem 1.17 (a) Eq. 1.24 now reads ∂ Ψ ∗ ∂t = − i ℏ 2 m ∂ 2 Ψ ∗ ∂x 2 + i ℏ V ∗ Ψ ∗ , and Eq. 1.25 picks up an extra term: ∂ ∂t | Ψ | 2 = · · · + i ℏ | Ψ | 2 ( V ∗ − V ) = · · · + i ℏ | Ψ | 2 ( V 0 + i Γ − V 0 + i Γ) = · · · − 2Γ ℏ | Ψ | 2 , and Eq. 1.27 becomes dP dt = − 2Γ ℏ ∫ ∞ −∞ | Ψ | 2 dx = − 2Γ ℏ P . QED (b) dP P = − 2Γ ℏ dt = ⇒ ln P = − 2Γ ℏ t + constant = ⇒ P ( t ) = P (0) e − 2Γ t/ ℏ , so τ = ℏ 2Γ . Problem 1.18 h √ 3 mk B T > d ⇒ T < h 2 3 mk B d 2 . (a) Electrons ( m = 9 . 1 × 10 − 31 kg): T < (6 . 6 × 10 − 34 ) 2 3(9 . 1 × 10 − 31 )(1 . 4 × 10 − 23 )(3 × 10 − 10 ) 2 = 1 . 3 × 10 5 K. Silicon nuclei ( m = 28 m p = 28(1 . 7 × 10 − 27 ) = 4 . 8 × 10 − 26 kg): T < (6 . 6 × 10 − 34 ) 2 3(4 . 8 × 10 − 26 )(1 . 4 × 10 − 23 )(3 × 10 − 10 ) 2 = 2 . 4 K.

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    CHAPTER 1. THE WAVE FUNCTION 15 (b) P V = N k B T ; volume occupied by one molecule ( N = 1 , V = d 3 ) ⇒ d = ( k B T /P ) 1 / 3 . T < h 2 3 mk B ( P k B T ) 2 / 3 ⇒ T 5 / 3 < h 2 3 m P 2 / 3 k 5 / 3 B ⇒ T < 1 k B ( h 2 3 m ) 3 / 5 P 2 / 5 . For helium ( m = 4 m p = 6 . 8 × 10 − 27 kg) at 1 atm = 1 . 0 × 10 5 N/m 2 : T < 1 (1 . 4 × 10 − 23 ) ( (6 . 6 × 10 − 34 ) 2 3(6 . 8 × 10 − 27 ) ) 3 / 5 (1 . 0 × 10 5 ) 2 / 5 = 2.8 K. For atomic hydrogen ( m = m p = 1 . 7 × 10 − 27 kg) with d = 0 . 01 m: T < (6 . 6 × 10 − 34 ) 2 3(1 . 7 × 10 − 27 )(1 . 4 × 10 − 23 )(10 − 2 ) 2 = 6 . 2 × 10 − 14 K. At 3 K it is definitely in the classical regime.

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    16 CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ ODINGER EQUATION Chapter 2 The Time-Independent Schr¨ odinger Equation Problem 2.1 (a) Ψ( x, t ) = ψ ( x ) e − i ( E 0 + i Γ) t/ ℏ = ψ ( x ) e Γ t/ ℏ e − iE 0 t/ ℏ = ⇒ | Ψ | 2 = | ψ | 2 e 2Γ t/ ℏ . ∫ ∞ −∞ | Ψ( x, t ) | 2 dx = e 2Γ t/ ℏ ∫ ∞ −∞ | ψ | 2 dx. The second term is independent of t , so if the product is to be 1 for all time, the first term ( e 2Γ t/ ℏ ) must also be constant, and hence Γ = 0. QED (b) If ψ satisfies Eq. 2.5, − ℏ 2 2 m d 2 ψ dx 2 + V ψ = Eψ , then (taking the complex conjugate and noting that V and E are real): − ℏ 2 2 m d 2 ψ ∗ dx 2 + V ψ ∗ = Eψ ∗ , so ψ ∗ also satisfies Eq. 2.5. Now, if ψ 1 and ψ 2 satisfy Eq. 2.5, so too does any linear combination of them ( ψ 3 ≡ c 1 ψ 1 + c 2 ψ 2 ): − ℏ 2 2 m d 2 ψ 3 dx 2 + V ψ 3 = − ℏ 2 2 m ( c 1 d 2 ψ 1 dx 2 + c 2 d 2 ψ 2 dx 2 ) + V ( c 1 ψ 1 + c 2 ψ 2 ) = c 1 [ − ℏ 2 2 m d 2 ψ 1 dx 2 + V ψ 1 ] + c 2 [ − ℏ 2 2 m d 2 ψ 2 dx 2 + V ψ 2 ] = c 1 ( Eψ 1 ) + c 2 ( Eψ 2 ) = E ( c 1 ψ 1 + c 2 ψ 2 ) = Eψ 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 = 1 2 [( 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 d 2 /d ( − x ) 2 = d 2 /dx 2 , − ℏ 2 2 m d 2 ψ ( − x ) dx 2 + V ( − x ) ψ ( − x ) = Eψ ( − 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. 2.5. But ψ ( x ) = 1 2 ( ψ + ( x ) + ψ − ( x )), so any solution can be expressed as a linear combination of even and odd solutions. QED

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