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Contents Preface The Wave Function Time-Independent Schrödinger Equation 14 Formalism 62 Quantum Mechanics in Three Dimensions 87 Identical Particles 132 Time-Independent Perturbation Theory 154 The Variational Principle 196 The WKB Approximation 219 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 www.pdfgrip.com 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 www.pdfgrip.com CHAPTER THE WAVE FUNCTION Chapter The Wave Function Problem 1.1 (a) j = 212 = 441 j2 = = N j N (j) = (142 ) + (152 ) + 3(162 ) + 2(222 ) + 2(242 ) + 5(252 ) 14 6434 (196 + 225 + 768 + 968 + 1152 + 3125) = = 459.571 14 14 j 14 15 16 22 24 25 (b) σ2 = = σ= N (∆j)2 N (j) = ∆j = j − j 14 − 21 = −7 15 − 21 = −6 16 − 21 = −5 22 − 21 = 24 − 21 = 25 − 21 = (−7)2 + (−6)2 + (−5)2 · + (1)2 · + (3)2 · + (4)2 · 14 260 (49 + 36 + 75 + + 18 + 80) = = 18.571 14 14 √ 18.571 = 4.309 (c) j2 − j = 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 www.pdfgrip.com CHAPTER THE WAVE FUNCTION Problem 1.2 (a) h 1 x2 √ dx = √ hx h x2 = σ = x2 − x = h2 − h 5/2 x = h h2 = 2h h ⇒ σ = √ = 0.2981h 45 (b) x+ P =1− x− √ 1 √ dx = − √ (2 x) hx h x+ x− √ √ =1− √ x+ − x− h x+ ≡ x + σ = 0.3333h + 0.2981h = 0.6315h; √ P =1− 0.6315 + √ x− ≡ x − σ = 0.3333h − 0.2981h = 0.0352h 0.0352 = 0.393 Problem 1.3 (a) ∞ Ae−λ(x−a) dx 1= Let u ≡ x − a, du = dx, u : −∞ → ∞ −∞ ∞ π λ e−λu du = A 1=A −∞ ⇒ A= λ π (b) ∞ x =A −∞ ∞ =A ∞ ue−λu du + a e−λu du = A + a −∞ π λ = a x2 e−λ(x−a) dx −∞ ∞ u2 e−λu du + 2a =A −∞ =A (u + a)e−λu du −∞ ∞ −∞ x2 = A ∞ xe−λ(x−a) dx = A 2λ σ = x2 − x ∞ ue−λu du + a2 −∞ π + + a2 λ = a2 + π λ = a2 + 1 − a2 = ; 2λ 2λ ∞ e−λu du −∞ 2λ σ=√ 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 www.pdfgrip.com CHAPTER THE WAVE FUNCTION (c) ρ(x) A x a Problem 1.4 (a) |A|2 a2 1= = |A|2 |A|2 a x2 dx + b (b − a) a a b−a b + = |A|2 ⇒ A = 3 (b) x3 a2 (b − x)2 dx = |A|2 a + (b − a)2 − (b − x)3 b a b Ψ A a b x (c) At x = a (d) a |Ψ|2 dx = P = |A|2 a2 a x2 dx = |A|2 a a = b P = if b = a, P = 1/2 if b = 2a (e) x|Ψ|2 dx = |A|2 x = = b a2 x4 a + a2 a x3 dx + (b − a)2 b2 (b − a)2 b x(b − x)2 dx a x2 x3 x4 − 2b + b a = a2 (b − a)2 + 2b4 − 8b4 /3 + b4 − 2a2 b2 + 8a3 b/3 − a4 4b(b − a)2 = 4b(b − a)2 b4 − a2 b2 + a3 b 3 = 2a + b (b3 − 3a2 b + 2a3 ) = 4(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 www.pdfgrip.com CHAPTER THE WAVE FUNCTION Problem 1.5 (a) ∞ |Ψ| dx = 2|A| 1= −2λx e e−2λx −2λ dx = 2|A| ∞ = |A|2 ; λ A= √ λ (b) x = x|Ψ|2 dx = |A|2 ∞ x2 = 2|A|2 ∞ xe−2λ|x| dx = [Odd integrand.] −∞ x2 e−2λx dx = 2λ = (2λ) 2λ2 (c) σ = x2 − x = ; 2λ2 √ σ=√ 2λ |Ψ(±σ)|2 = |A|2 e−2λσ = λe−2λ/ λ 2λ √ = λe− = 0.2431λ |Ψ| 24λ −σ x +σ Probability outside: ∞ |Ψ|2 dx = 2|A|2 σ ∞ e−2λx dx = 2λ σ e−2λx −2λ ∞ √ = e−2λσ = e− = 0.2431 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| + x |Ψ|2 = x |Ψ|2 , ∂t ∂t ∂t ∂t but this does not allow us to perform the integration: b x a ∂ |Ψ|2 dx = ∂t b a ∂ b (x|Ψ|2 )dx = (x|Ψ|2 ) a ∂t 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 www.pdfgrip.com CHAPTER THE WAVE FUNCTION Problem 1.7 d p dt From Eq 1.33, ∂ ∂t Ψ∗ = −i ∂ ∂t ∂2Ψ ∂x∂t Ψ∗ ∂Ψ ∂x dx But, noting that = ∂2Ψ ∂t∂x and using Eqs 1.23-1.24: ∂ ∂Ψ ∂Ψ∗ ∂Ψ i ∂ Ψ∗ i ∂2Ψ i i ∗ ∂Ψ ∗ ∂ + Ψ∗ = − V Ψ + Ψ + − VΨ ∂t ∂x ∂x ∂t 2m ∂x2 ∂x ∂x 2m ∂x2 i ∂ Ψ ∂ Ψ∗ ∂Ψ ∂Ψ ∂ i = Ψ∗ − + V Ψ∗ − Ψ∗ (V Ψ) 2m ∂x ∂x ∂x ∂x ∂x ∂Ψ ∂x = The first term integrates to zero, using integration by parts twice, and the second term can be simplified to ∗ ∂Ψ ∗ ∂V ∂V V Ψ∗ ∂Ψ ∂x − Ψ V ∂x − Ψ ∂x Ψ = −|Ψ| ∂x So dp = −i dt i −|Ψ|2 ∂V ∂V dx = − ∂x ∂x QED Problem 1.8 Suppose satises the Schră odinger equation without V0 : i ∂Ψ0 ∂ Ψ0 Ψ0 with V0 : i ∂t = − 2m ∂x2 + (V + V0 )Ψ0 2 = − 2m ∂∂xΨ2 + V Ψ We want to find the solution ∂Ψ ∂t Claim: Ψ0 = Ψe−iV0 t/ Proof: i ∂Ψ0 ∂t ∂Ψ −iV0 t/ ∂t e =i 2 + i Ψ − iV0 e−iV0 t/ = − 2m ∂∂xΨ2 + V Ψ e−iV0 t/ + V0 Ψe−iV0 t/ = − 2m ∂∂xΨ20 + (V + V0 )Ψ0 QED This has no effect on the expectation value of a dynamical variable, since the extra phase factor, being independent of x, cancels out in Eq 1.36 Problem 1.9 (a) ∞ = 2|A|2 e−2amx / dx = 2|A|2 π = |A|2 (2am/ ) π ; 2am A= 2am π 1/4 (b) ∂Ψ = −iaΨ; ∂t ∂Ψ 2amx =− Ψ; ∂x ∂2Ψ 2am = x2 Plug these into the Schră odinger equation, i V Ψ = i (−ia)Ψ + = − 2am 2m 2amx2 a− a 1− 1− ∂Ψ ∂t Ψ+x ∂Ψ ∂x =− 2am 1− 2amx2 Ψ = − 2m ∂∂xΨ2 + V Ψ: 2amx2 Ψ = 2a2 mx2 Ψ, Ψ so V (x) = 2ma2 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 www.pdfgrip.com CHAPTER THE WAVE FUNCTION (c) ∞ x|Ψ|2 dx = x = [Odd integrand.] −∞ ∞ x2 = 2|A|2 x2 e−2amx / dx = 2|A|2 p =m 22 (2am/ ) π = 2am 4am dx = dt p2 = =− ∂ i ∂x Ψ∗ Ψdx = − Ψ∗ − 1− = 2am 2am 2am 1− x2 Ψ∗ 2amx2 = 2am ∂2Ψ dx ∂x2 |Ψ|2 dx − Ψ dx = 2am 1− 2am = 2am 4am 2am x2 |Ψ|2 dx = am (d) σx2 = x2 − x √ σx σp = 4am = 4am =⇒ σx = 4am ; σp2 = p2 − p = am =⇒ σp = √ am am = This is (just barely) consistent with the uncertainty principle Problem 1.10 From Math Tables: π = 3.141592653589793238462643 · · · (a) P (0) = P (5) = 3/25 P (1) = 2/25 P (6) = 3/25 In general, P (j) = = 25 [0 (c) j = = 25 [0 P (3) = 5/25 P (8) = 2/25 25 [0 Median: 13 are ≤ 4, 12 are ≥ 5, so median is · + · + · + · + · + · + · + · + · + · 3] + + + 15 + 12 + 15 + 18 + + 16 + 27] = 25 [0 P (4) = 3/25 P (9) = 3/25 N (j) N (b) Most probable: Average: j = P (2) = 3/25 P (7) = 1/25 118 25 = 4.72 + 12 · + 22 · + 32 · + 42 · + 52 · + 62 · + 72 · + 82 · + 92 · 3] + + 12 + 45 + 48 + 75 + 108 + 49 + 128 + 243] = σ2 = j − j 710 25 = 28.4 − 4.722 = 28.4 − 22.2784 = 6.1216; = 28.4 √ 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 www.pdfgrip.com CHAPTER THE WAVE FUNCTION Problem 1.11 (a) Constant for ≤ θ ≤ π, otherwise zero In view of Eq 1.16, the constant is 1/π ρ(θ) = 1/π, if ≤ θ ≤ π, 0, otherwise ρ(θ) 1/π −π/2 π θ 3π/2 (b) θ = θ2 = θρ(θ) dθ = π π θ2 dθ = σ2 = θ2 − θ = π π π θ2 = π2 θdθ = π π θ3 π2 π2 π2 − = ; 12 π = π [of course] π σ= √ (c) sin θ = cos θ = π π cos2 θ = π π sin θ dθ = 1 π (− cos θ)|0 = (1 − (−1)) = π π π cos θ dθ = π (sin θ)|0 = π π π cos2 θ dθ = π π (1/2)dθ = [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 = probability that it’s in the range dx is ρ(x)dx = dx = π r sin θ πr dx 1− (x/r)2 = π dθ, so the 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 www.pdfgrip.com 10 CHAPTER THE WAVE FUNCTION ρ(x) -r -2r r √ dx −r π r −x2 Total: 0, if − r < x < r, otherwise π r √ dx r −x2 √ , π r −x2 ∴ ρ(x) = = r = π x 2r [Note: We want the magnitude of dx here.] sin−1 x r r sin−1 (1) = π = π · π = (b) x = r π −r π x2 = x√ r √ σ = x2 − x r2 dx = − x2 [odd integrand, even interval] x2 x dx = − π r2 − x2 r2 − x2 + r x r2 sin−1 r = r2 r2 sin−1 (1) = π 2 √ = r2 /2 =⇒ σ = r/ 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 < (i.e x < −y) So for a given value of y, the probability of crossing (using Problem 1.12) is −y P (y) = l ρ(x)dx + −l = π sin−1 l−y x l −y −l + sin−1 x l −y π ρ(x)dx = −l l = l−y √ dx + l − x2 l l−y √ l2 dx − x2 − sin−1 (y/l) + sin−1 (1) − sin−1 (1 − y/l) π 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− P = = πl l π − sin−1 y − sin−1 l πl − y sin−1 (y/l) + l πl l−y l − (y/l)2 dy = l πl =1− l π − sin−1 (y/l) dy 2 [l sin−1 (1) − l] = − + = πl π 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 www.pdfgrip.com APPENDIX LINEAR ALGEBRA (d) 289 cos θ ˜ a Ta = − sin θ T 010 001 ˜ b Tb = 0 1 1 0 = 0 T 100 010 sin θ cos θ − sin θ cos θ 0 sin θ cos θ 0 = 0 1 0 0 0 0 ˜ c Tc = 0 0 0 T 0 −1 det Ta = cos2 θ + sin2 θ = det Tb = 0 0 0 = 0 0 −1 0 det Tc = -1 Problem A.15 y y' θ x, x' θ z z' ˆi = ˆi; ˆ ˆj = cos θ ˆj + sin θ k; 0 Tx (θ) = 0 cos θ − sin θ sin θ cos θ kˆ = cos θ kˆ − sin θ ˆj y, y' x' θ x z θ z' ˆ ˆi = cos θ ˆi − sin θ k; ˆj = ˆj; kˆ = cos θ kˆ + sin θ ˆi ˆi = ˆj; ˆj = −ˆi; ˆ kˆ = k −1 S = 1 0 0 cos θ sin θ Ty (θ) = − sin θ cos θ S−1 = −1 0 0 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 www.pdfgrip.com 290 APPENDIX LINEAR ALGEBRA STx S−1 STy S−1 −1 0 0 0 cos θ sin θ −1 0 0 − cos θ − sin θ −1 cos θ sin θ 0 0 −1 0 − sin θ cos θ 0 −1 0 cos θ sin θ 0 0 −1 0 = 0 cos θ sin θ = Tx (−θ) − sin θ cos θ − sin θ cos θ = 1 = 1 0 = 1 = 1 0 − sin θ −1 0 cos θ 0 cos θ sin θ − sin θ = = Ty (θ) cos θ − sin θ cos θ Is this what we would expect? Yes, for rotation about the x axis now means rotation about the y axis, and rotation about the y axis has become rotation about the −x axis—which is to say, rotation in the opposite direction about the +x axis Problem A.16 From Eq A.64 we have Af Bf = SAe S−1 SBe S−1 = S(Ae Be )S−1 = SCe S−1 = Cf Suppose S† = S−1 and He = He† (S unitary, He hermitian) Then Hf † = (SHe S−1 )† = (S−1 )† He† S† = SHe S−1 = Hf , so Hf is hermitian In an orthonormal basis, α|β = a† b (Eq A.50) So if {|fi } is orthonormal, α|β = af † bf But bf = Sbe (Eq A.63), and also af † = ae† S† So α|β = ae† S† Sbe This is equal to ae† be (and hence {|ei } is also orthonormal), for all vectors |α and |β ⇔ S† S = I, i.e S is unitary Problem A.17 n Tr(T1 T2 ) = n i=1 Is n n (T1 T2 )ii = i=1 j=1 Tr(T1 T2 T3 ) = Tr(T2 T1 T3 )? No T1 = n n (T1 )ij (T2 )ji = (T2 )ji (T1 )ij = j=1 i=1 (T2 T1 )jj = Tr(T2 T1 ) j=1 Counterexample: 01 , 00 0 , T2 = T3 = 0 T1 T2 T3 = 01 00 00 10 10 00 = 0 0 = 0 =⇒ Tr(T1 T2 T3 ) = T2 T1 T3 = 00 10 01 00 10 00 = 0 0 0 = 0 0 =⇒ Tr(T2 T1 T3 ) = 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 www.pdfgrip.com APPENDIX LINEAR ALGEBRA 291 Problem A.18 Eigenvalues: (cos θ − λ) − sin θ = (cos θ − λ)2 + sin2 θ = cos2 θ − 2λ cos θ + λ2 + sin2 θ = 0, or λ2 − 2λ cos θ + = sin θ (cos θ − λ) λ= cos θ ± √ cos2 θ − = cos θ ± − sin2 θ = cos θ ± i sin θ = e±iθ So there are two eigenvalues, both of them complex Only if sin θ = does this matrix possess real eigenvalues, i.e., only if θ = or π Eigenvectors: cos θ − sin θ sin θ cos θ a(1) = √ α β ; −i = e±iθ α β a(2) = √ =⇒ cos θ α − sin θ β = (cos θ ± i sin θ)α ⇒ β = ∓iα Normalizing: i Diagonalization: (1) (S−1 )11 = a1 = √ ; S−1 = √ 1 ; −i i = STS−1 = (2) (S−1 )12 = a1 = √ ; inverting: S = √ i −i cos θ − sin θ sin θ cos θ i −i eiθ e−iθ −ieiθ ie−iθ 1 −i i = −i (1) (S−1 )21 = a2 = √ ; i (2) (S−1 )22 = a2 = √ i −i = (cos θ + i sin θ) (cos θ − i sin θ) (sin θ − i cos θ) (sin θ + i cos θ) i −i 2eiθ 0 2e−iθ = eiθ e−iθ Problem A.19 (1 − λ) = (1 − λ)2 = =⇒ λ = (1 − λ) 11 01 α β = α β =⇒ α + β = α =⇒ β = 0; (only one eigenvalue) a= (only one eigenvector—up to an arbitrary constant factor) Since the eigenvectors not span the space, this 10 matrix cannot be diagonalized [If it could be diagonalized, the diagonal form would have to be , since 01 the only eigenvalue is But in that case I = SMS−1 Multiplying from the left by S−1 and on the right by S : S−1 IS = S−1 SMS−1 S = M But S−1 IS = S−1 S = I So M = I, which is false.] 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 www.pdfgrip.com 292 APPENDIX LINEAR ALGEBRA Problem A.20 Expand the determinant (Eq A.72) by minors, using the first column: (T22 − λ) n det(T − λ1) = (T11 − λ) Tj1 cofactor(Tj1 ) + j=2 (Tnn − λ) But the cofactor of Tj1 (for j > 1) is missing two of the original diagonal elements: (T11 − λ) (from the first column), and (Tjj − λ) (from the j-th row) So its highest power of λ will be (n − 2) Thus terms in λn and λn−1 come exclusively from the first term above Indeed, the same argument applied now to the cofactor of (T11 − λ) – and repeated as we expand that determinant – shows that only the product of the diagonal elements contributes to λn and λn−1 : (T11 − λ)(T22 − λ) · · · (Tnn − λ) = (−λ)n + (−λ)n−1 (T11 + T22 + · · · + Tnn ) + · · · Evidently then, Cn = (−1)n , and Cn−1 = (−1)n−1 Tr(T) To get C0 – the term with no factors of λ – we simply set λ = Thus C0 = det(T) For a × matrix: (T11 − λ) T12 T13 T21 (T22 − λ) T23 T31 T32 (T33 − λ) = (T11 − λ)(T22 − λ)(T33 − λ) + T12 T23 T31 + T13 T21 T32 − T31 T13 (T22 − λ) − T32 T23 (T11 − λ) − T12 T21 (T33 − λ) = −λ3 + λ2 (T11 + T22 + T33 ) − λ(T11 T22 + T11 T33 + T22 T33 ) + λ(T13 T31 + T23 T32 + T12 T21 ) + T11 T22 T33 + T12 T23 T31 + T13 T21 T32 − T31 T13 T22 − T32 T23 T11 − T12 T21 T33 = −λ3 + λ2 Tr(T) + λC1 + det(T), with C1 = (T13 T31 + T23 T32 + T12 T21 ) − (T11 T22 + T11 T33 + T22 T33 ) Problem A.21 The characteristic equation is an n-th order polynomial, which can be factored in terms of its n (complex) roots: (λ1 − λ)(λ2 − λ) · · · (λn − λ) = (−λ)n + (−λ)n−1 (λ1 + λ2 + · · · + λn ) + · · · + (λ1 λ2 · · · λn ) = Comparing Eq A.84, it follows that Tr(T) = λ1 + λ2 + · · · λn and det(T) = λ1 λ2 · · · λn QED Problem A.22 (a) [Tf1 , Tf2 ] = Tf1 Tf2 −Tf2 Tf1 = STe1 S−1 STe2 S−1 −STe2 S−1 STe1 S−1 = STe1 Te2 S−1 −STe2 Te1 S−1 = S[Te1 , Te2 ]S−1 = 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 www.pdfgrip.com APPENDIX LINEAR ALGEBRA 293 (b) Suppose SAS−1 = D and SBS−1 = E, where D and E are diagonal: d1 · · · e1 · · · d2 · · · e2 · · · D= , E= . 0 · · · dn 0 ··· 0 en Then [A, B] = AB − BA = (S−1 DS)(S−1 ES) − (S−1 ES)(S−1 DS) = S−1 DES − S−1 EDS = S−1 [D, E]S But diagonal matrices always commute: d1 e1 · · · d2 e2 · · · DE = = ED, 0 · · · dn en so [A, B] = QED Problem A.23 (a) M† = 1 ; −i MM† = (1 − i) , (1 + i) M† M = (1 + i) ; (1 − i) [M, M† ] = −2i 2i = No (b) Find the eigenvalues: (1 − λ) = (1 − λ)(i − λ) − = i − λ(1 + i) + λ2 − = 0; (i − λ) √ (1 + i) ± (1 + i)2 − 4(i − 1) (1 + i) ± − 2i λ= = 2 Since there are two distinct eigenvalues, there must be two linearly independent eigenvectors, and that’s enough to span the space So this matrix is diagonalizable, even though it is not normal Problem A.24 Let |γ = |α + c|β , for some complex number c Then γ|Tˆγ = α|Tˆα + c α|Tˆβ + c∗ β|Tˆα + |c|2 β|Tˆβ , and Tˆγ|γ = Tˆα|α + c∗ Tˆβ|α + c Tˆα|β + |c|2 Tˆβ|β Suppose Tˆγ|γ = γ|Tˆγ for all vectors For instance, Tˆα|α = α|Tˆα and Tˆβ|β = β|Tˆβ ), so c α|Tˆβ + c∗ β|Tˆα = c Tˆα|β + c∗ Tˆβ|α , and this holds for any complex number c In particular, for c = 1: α|Tˆβ + β|Tˆα = Tˆα|β + Tˆβ|α , while for c = i: α|Tˆβ − β|Tˆα = Tˆα|β − Tˆβ|α (I canceled the i’s) Adding: α|Tˆβ = Tˆα|β 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 www.pdfgrip.com 294 APPENDIX LINEAR ALGEBRA Problem A.25 (a) ˜∗ = T† = T (b) 1−i 1+i = T √ 1± 1+8 1±3 (1 − λ) (1 − i) = −(1 − λ)λ − − = 0; λ − λ − = 0; λ = = (1 + i) (0 − λ) 2 λ1 = 2, λ2 = −1 (c) (1 − i) (1 + i) α β =2 α β =⇒ α + (1 − i)β = 2α =⇒ α = (1 − i)β |α|2 + |β|2 = =⇒ 2|β|2 + |β|2 = =⇒ β = √ (1 − i) (1 + i) α β =− α β (i − 1) 2 a(2) = √ 1−i 1 α = − (1 − i)β =⇒ α + (1 − i)β = −α; 2|β|2 + |β|2 = =⇒ |β|2 = 1; β = a(1)† a(2) = √ (1 + i) a(1) = √ i−1 = √ (i − − − i + 2) = (d) (1) Eq A.81 =⇒ (S−1 )11 = a1 = √ (1 − i); (1) (S−1 )21 = a2 = √ ; =√ −1 S −1 STS √ (1 − i) (i −√ 1)/ ; = (1 + i)√ √1 −(1 + i)/ 2 (1 + i)√ √1 −(1 + i)/ 2 = (2) (S−1 )12 = a1 = √ (i − 1); (2) (S−1 )22 = a2 = √ S = (S−1 )† = √ (1 + i)√ √1 (−i − 1)/ 2 √ (1 − i) (i −√ 1)/ 2 √ 2(1 − i) (1 − √ i)/ = = − −3 (1 − i) (1 + i) −1 (e) Tr(T) = 1; det(T) = − (1 + i)(1 − i) = −2 Tr(STS−1 ) = − = det(STS−1 ) = −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 www.pdfgrip.com APPENDIX LINEAR ALGEBRA 295 Problem A.26 (a) det(T) = − − − − − = Tr(T) = + + = (b) (2 − λ) i −i (2 − λ) i = (2 − λ)3 − − − (2 − λ) − (2 − λ) − (2 − λ) = − 12λ + 6λ2 − λ3 − + 3λ = −i (2 − λ) −λ3 + 6λ2 − 9λ = −λ(λ2 − 6λ + 9) = −λ(λ − 3)2 = λ1 = 0, λ2 = λ3 = λ1 + λ2 + λ3 = = Tr(T) λ1 λ2 λ3 = = det(T) 0 Diagonal form: 0 0 0 (c) i α −i i β = =⇒ −i γ 2α + iβ + γ = −iα + 2β + iγ = =⇒ α + 2iβ − γ = Add the two equations: 3α + 3iβ = =⇒ β = iα; 2α − α + γ = =⇒ γ = −α (1) a α = iα Normalizing: |α|2 + |α|2 + |α|2 = =⇒ α = √ −α (1) a 1 i =√ −1 i α α =⇒ −α + iβ + γ = 0, 2α + iβ + γ = 3α −i i β = β =⇒ −iα + 2β + iγ = 3β =⇒ α − iβ − γ = 0, −i γ γ α − iβ + 2γ = 3γ =⇒ α − iβ − γ = The three equations are redundant – there is only one condition here: α − iβ − γ = We could pick γ = 0, β = −iα, or β = 0, γ = α Then α α (2) (3) a0 = −iα ; a0 = α (2) But these are not orthogonal, so we use the Gram-Schmidt procedure (Problem A.4); first normalize a0 : a(2) 1 −i =√ † (3) a(2) a0 1/2 1 † (3) α α α (3) = √ i 0 = √ So a0 − (a(2) a0 ) a(2) = α 0 − −i = α i/2 2 1 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 www.pdfgrip.com 296 APPENDIX LINEAR ALGEBRA |α|2 Normalize: 1 + +1 4 = |α|2 = =⇒ α= a(3) 1 i =√ Check orthogonality: † a(1) a(2) (1) † (3) a a 1 = √ −i −1 −i = √ (1 − + 0) = 6 1 = √ −i −1 i = √ (1 + − 2) = 3 2 (d) S−1 is the matrix whose columns are the eigenvectors of T (Eq A.81): S−1 √ √ √3 √ =√ i − 3i −√2 STS−1 √ √ √ −√ i − √ S = (S−1 )† = √ 3i −i i ; √ √ √ √ √ − 2i − 2 i √2 √3 √ √ = 2i − 3i 3i −i i √ −i −i − √ √3 −3 i 3i 0 0 0 0 i = 0 18 = 0 0 0 18 0 Problem A.27 ˆ α|U ˆβ = U ˆ †U ˆ α|β = α|β (a) U ˆ |α = λ|α =⇒ U ˆ α|U ˆ α = |λ|2 α|α But from (a) this is also α|α So |λ| = (b) U ˆ |α = λ|α , U ˆ |β = µ|β =⇒ |β = µU ˆ −1 |β , so U ˆ † |β = |β = µ∗ |β (from (b)) (c) U µ † ˆ ˆ β|U α = λ β|α = U β|α = µ β|α , or (λ − µ) β|α = So if λ = µ, then β|α = QED Problem A.28 (a) (i) 0 M2 = 0 0 ; 0 0 M3 = 0 0 , 0 so 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 www.pdfgrip.com APPENDIX LINEAR ALGEBRA 297 1 100 013 0 eM = 0 0 + 0 4 + 0 0 = 0 4 0 001 000 0 (ii) M2 = −θ2 0 −θ2 = −θ2 I; M3 = −θ3 M; M4 = θ4 I; etc θ3 θ4 − θ2 I − + I + ··· −1 3! −1 4! θ θ θ θ 1− + − ··· I + θ − + − ··· −1 4! 3! 5! eM = I + θ = 10 + sin θ 01 −1 = cos θ cos θ sin θ − sin θ cos θ = (b) d1 SMS−1 = D = for some S dn 1 SeM S−1 = S I + M + M2 + M3 + · · · S−1 Insert SS−1 = I : 3! 1 SeM S−1 = I + SMS−1 + SMS−1 SMS−1 + SMS−1 SMS−1 SMS−1 + · · · 3! D = I + D + D + D + · · · = e Evidently 3! det(eD ) = det(SeM S−1 ) = det(S) det(eM ) det(S−1 ) = det(eM ) But d21 D2 = 0 d2n , d31 D3 = , d3n Dk = dk1 , so dkn 0 d e d21 d31 0 0 d1 eD = I + + + + ··· = 3! dn dn dn dn e det(eD ) = ed1 ed2 · · · edn = e(d1 +d2 +···dn ) = eTr D = eTr M (Eq A.68), so det(eM ) = eTr M QED (c) Matrices that commute obey the same algebraic rules as ordinary numbers, so the standard proofs of ex+y = ex ey will the job Here are two: 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 www.pdfgrip.com 298 APPENDIX LINEAR ALGEBRA m n (i) Combinatorial Method: Use the binomial theorem (valid if multiplication is commutative): ∞ eM+N = ∞ ∞ n n 1 n Mm Nn−m = (M + N)n = Mm Nn−m m n! n! m!(n − m)! n=0 n=0 m=0 n=0 m=0 Instead of summing vertically first, for fixed n (m : → n), sum horizontally first, for fixed m (n : m → ∞, or k ≡ n − m : → ∞)—see diagram (each dot represents a term in the double sum) ∞ eM+N = m M m! m=0 ∞ k=0 k N = eM eN QED k! (ii) Analytic Method: Let S(λ) ≡ eλM eλN ; dS = MeλM eλN + eλM NeλN = (M + N)eλM eλN = (M + N)S dλ (The second equality, in which we pull N through eλM , would not hold if M and N did not commute.) Solving the differential equation: S(λ) = Ae(M+N)λ , for some constant A But S(0) = I, so A = 1, and hence eλM eλN = eλ(M+N) , and (setting λ = 1) we conclude that eM eN = e(M+N) [This method generalizes most easily when M and N not commute—leading to the famous Baker-CampbellHausdorf lemma.] As a counterexample when [M, N] = 0, let M = eM = I + M = But (M + N) = 11 , eN = I + N = 01 01 , N= 00 ; −1 eM eN = , so (from a(ii)): eM+N = −1 0 Then M2 = N2 = 0, so −1 11 01 −1 = −1 cos(1) sin(1) − sin(1) cos(1) The two are clearly not equal (d) ∞ eiH = ∞ ∞ n n 1 i H =⇒ (eiH )† = (−i)n (H† )n = (−i)n Hn = e−iH (for H hermitian) n! n! n! n=0 n=0 n=0 (eiH )† (eiH ) = e−iH eiH = ei(H−H) = I, using (c) So eiH is unitary 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 www.pdfgrip.com 299 2nd Edition – 1st Edition Problem Correlation Grid N = New M = 1/e problem number (modified for 2/e) X = 2/e problem number (unchanged from 1/e) Chapter 2/e 1/e Chapter 2/e 1/e 2N 10 11 12 13 14 15 16N 17N 18N 7N 8N 9N 10 11 12 13 14N 15 16 17 18 19N 20 21N 22 23 24 25N 26 27 28 29 30 31 32 33 34 35 36 37 38 39N 40N 41N 42 43 44 11 12 13 14 9M 10 6M 13M 14 37 17M 15 16 18 19M 20 22 23 24 25 26 27 28 29 30 31 32 33 41M 4M 36 3.48 38 40 39 www.pdfgrip.com Chapter (cont.) 2/e 1/e 45 46 47 48N 49 50 51 52 53 54N 55N 56N 42 43 44 45 47 48M 34M, 35M 49 300 2nd Edition – 1st Edition Problem Correlation Grid N = New M = 1/e problem number (modified for 2/e) (M) = 1/e problem number (distant model for 2/e) X = 2/e problem number (unchanged from 1/e) Chapter 2/e 1/e Chapter 2/e 1/e 1N 2N 3N 4N 5N 6N 7N 8N 9N 10N 11 12 13 14 15N 16 17 18 19 20 21 22N 23N 24 25 26N 27N 28 29N 30N 31 32 33 34 35N 36N 37N 38N 39 40N 10 11 12 13 14N 15N 16 17 18 19 20 21N 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41N 42 (33M) (21M) (12M) 38 51 41M 39 42 43 44 45 46 57M 57M 25M 52M 53 56 50 49M 55 7M 9M 10 11 12 13 17 16 19 20 21 22 23 25 26 27 28 29 30 31M 32 33 34 35 36 37 38 39 40 41 42 www.pdfgrip.com Chapter (cont.) 2/e 1/e 43 44N 45 46 47N 48N 49N 50 51 52 53N 54 55 56 57 58N 59 60 61 43 14 15 44 45M 46 47 48 49 50 51 52M 53 301 2nd Edition – 1st Edition Problem Correlation Grid N = New M = 1/e problem number (modified for 2/e) X = 2/e problem number (unchanged from 1/e) Chapter 2/e 1/e Chapter 2/e 1/e Chapter 2/e 1/e 3N 10 11 12 13 14 15N 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32N 33 34 35 36 37 10N 11 12 13 14 15N 16 17 18 19 20 21 22 23 24 25 26 27 28 29N 30N 31N 32 33 34 35 36 37 38 39 40N 10 11N 12N 13 14 15 16 17 18 19 20N 10 11M 11M 12 13 14 15M 16M 17M 18 19M 20 21M 22 23 24 25 26 27M 28 29 30 31 32 33 1M 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 www.pdfgrip.com 2M 3M 10 11 12 13 14 15 16 17 302 2nd Edition – 1st Edition Problem Correlation Grid N = New M = 1/e problem number (modified for 2/e) X = 2/e problem number (unchanged from 1/e) Chapter 2/e 1/e Chapter 2/e 1/e Chapter 10 2/e 1/e 10 11 12 13 14 15 16N 17N 9N 10 11 12 13 14 15 16 17 18 19 20 21 22N 8N 10 10 11 12 13 14 15 3M 10 11 12 13 14 15 16 17 21 19M 20 www.pdfgrip.com 3M 10 11M 303 2nd Edition – 1st Edition Problem Correlation Grid N = New M = 1/e problem number (modified for 2/e) X = 2/e problem number (unchanged from 1/e) Chapter 11 2/e 1/e Chapter 12 2/e 1/e Appendix 2/e 1/e 5N 6N 7N 10 11 12 13 14 15 16 17 18 19N 20N 1N 10 11 12 13N 14N 15 16 17 18 19 20 21 22 23N 24 25 26 27 28 10 11 12 13 14 15 www.pdfgrip.com 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.9 3.10 3.11 3.12 3.16 3.13 3.14 3.15 3.17 3.18 3.19 3.20 3.40M 3.21M 3.22 3.23 3.24 3.47
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