Sakurai_solucoes

Modern\nQuantum Mechanics\nSolutions Manual\nRevised Edition\nJ.J. Sakurai\nLate, University of California, Los Angeles\nBy San Fu Tuan\nUniversity of Hawaii, Manoa\n\n 2\nADDISON-WESLEY PUBLISHING COMPANY\nReading, Massachusetts • Menlo Park, California • New York\nDon Mills, Ontario • Wokingham, England • Amsterdam • Bonn\nSydney • Singapore • Tokyo • Madrid • San Juan • Milan • Paris Copyright © 1994 by Addison-Wesley Publishing Company, Inc.\nAll rights reserved. No part of this publication may be reproduced,\nstored in a retrieval system, or transmitted, in any form or by any\nmeans, electronic, mechanical, photocopying, recording, or otherwise,\nwithout the prior written permission of the publisher. Printed in the\nUnited States of America.\n\nISBN 0-201-53730-3\n2 3 4 5 6 7 8 9 10 CRW 979695594 Contents\n1 Fundamental Concepts 1\n2 Quantum Dynamics 18\n3 Theory of Angular Momentum 45\n4 Symmetry in Quantum Mechanics 63\n5 Approximation Methods 70\n6 Identical Particles 110\n7 Scattering Theory 115 Chapter 1\n\n1. [AB,CD] = ABCD - CDAB = ABCD + ACBD - ACDB + ACBD + CABD - CDAB = A(C,B)D - AC(D,A)B - C(D,A)B.\nCDAB = A(C,B)D - AC(D,B) + (C,A)DB - C(D,A)B.\n\n2. (a) x = x_0 + t_a x_o^2, tr(X) = 2a because tr(σo) = 0. Next evaluate tr(σo X) = tr(σo a_k x_o x_o^2) = I_{ij} a_1 a_2 26 x_k = 2a_k (where we have used tr(σk σ_j) = tr(σk σ_j + σ_j σ_k)) = 26 I_j. Hence a_o = 1/2 tr(X), a_k = 1/2 tr(σo X).\n\n(b) a_o = 1/2 (x_11 + x_22), while a_k can be explicitly evaluated from a_k = 1/2 tr(σo X) with X = [x_{ij}] and i,j = 1,2. The result is a_1 = 1/2 (x_{12} + x_{21}), a_2 = 1/2 (-x_{11} + x_{22}), and a_3 = 1/2 (x_{11} - x_{22}).\n\n3. oz = gx_0 + gy_0 + oz_z = ( az - ia_y ) ( ax + ia_y ) - az - det ( oz.a ) = - |d|^2.\nWithout loss of generality, choose z along positive z-direction, then exp(iσ_α/2) = 1 cos(φ/2) + iσ_z sin(φ/2), and if B is defined to be B ≡ cos(φ/2) + i sin(φ/2), then exp(iσ^B) = ( a^z B ( ax - ia_y )^2 ( az - x ) )\n\texp(-iσ_z φ/2) = - ( a_z^2 + a_x^2 + a_y^2 - |d|^2 )^2,\nthat is determinant is exp(-iσ_z φ/2) = - ( a_z^2 + a_x^2 + a_y^2 - |d|^2 )^2.\n\nhence ( a^B )^B = cos^2(φ/2) + sin^2(φ/2) = 1, det [ exp(iσ^2/2) ] . Modern Quantum Mechanics - Solutions\n\ninvariant under specified operation. Next we note\noz. d = ( a_z i - a_x i )\n =\n( ax + i a_y ) - az )\n =\n( a_x + i a_y ) ( cos(θ) + i sin(θ))\n\nhence a_1' = a_2', a_x = a_x cos(θ) + a_y sin(θ), a_y' = a_x cos(θ) - a_y sin(θ). This is a counter-clockwise rotation about z-axis through angle θ in x-y plane.\n\n4. (a) Note tr(XY) = Σ_a <a|XY|a> = Σ_a <|X|a>|<a|Y|a> (by closure property) = Σ_a <a|Y|a>|<a|X|a> (by rearrangement) = Σ_a <a|XY|a> = tr(XY) = tr(YX).\n\n(b) <(XY)*a| a'> = <a'| (XY)†|a> = <a'|X|Y|a> = <a|Y|X> = rY*.\nTherefore (XY)† = XY†.\n\n(c) Take exp(if(A))|α> = (1 + if(A) + (if(A))^2/2! + ... )|α> = (1 + if(α) + (if(α))^2/2! = (1 + if(α) + (......)|α> = exp(if(α))|α>, where we assume that |α> = q|α>. Modern Quantum Mechanics - Solutions\n\n(a) |α><β| = Σ_a |a><a| <α|a>|<β|a>| = Σ_a <a|a>|<a|a>| = <α|a| <β|a> = |α><β|**.\nHence <α|<β|** = <a(1)|<b(3)|, where <(a'|<c|B>*. Hence |α><β| = <a(1)|<b(3)|B>*, hence <|α| < β > * = <a(1)|<c(3)| < a'| >*.\n\n(b) Given |ψ> = a_1 |a_1> + |ψ> = a_2 |a_2>. The normalized state vector |i> + |j> is of form |ψ> = (1/√2)(|1> + |2>). Hence |A|ψ> = (1/√2)(a_1 |a_1> + a_2 |a_2>) where a_1, a_2 are real numbers if A is Hermitian; but for a_1 ≠ a_2 |ψ> is a state vector distinct from |ψ>. However under the condition that |i> and |j> are degenerate (i.e. a_1 = a_2 = a), then |A|ψ> = a(1/√2)(|1> + |2>)|a> + |ψ> or |i> + |j> is also an eigenket of A. Modern Quantum Mechanics - Solutions\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n(c) Let A = S z , than I . (S -a') = (S z - W/2)(S z + W/2). Hence evidently\na = a†(S z - a')/2 = 0. This verifies (a) above. For case (b)\nwe have ⊕ z = (S z +W/2)/N, ⊖ = -(S z -W/2)/N and S z = M/2(|+⟩<|+⟩ - |−⟩<|−⟩)\nx<|−⟩ = |+⟩<|+⟩ + |−⟩<|−⟩. Hence ⊕ z |⟩ to |⟩ and \n⊖ |⟩ = <|−⟩|−⟩ are the projection operators of |⟩ to |⟩ states.\n8. The orthonormality property is <+|+> = <−|−> = 1, <+|−> = <−|+> = 0.\nHence using the explicit representations of S 1 in terms of linear\ncombinations of bra-ket products, we obtain by elementary calculation\n{S 1, S 1} = iℏI M S k and {S 1, S 2} = ℏ2/2δij.\n9. Let n = n x i + n y j + n k, then x n = sinθcosϕ, n y = sinθsinϕ, n z = cosθ and S n = sinθS x + sinθS n + cosβ S z . Also due to\ncompleteness property of the ket space |S 3.n;±⟩ = a|+⟩ + b|−⟩ where |a|2 + |b|2 = 1 (normalization). Therefore the relation S 3.S 4.xtaking advantage of explicit representations S x = 1/2(|+⟩<|+|<−|+⟩), S y = iℏ/2(-|+⟩<|−| + |−⟩<|+⟩), S z = 1/2(|+⟩<|+| − |−⟩<|−|)\nleads to :\n\n(sinθcosϕ - isinθsinϕ) + cosθ a = a\n(sinθcosϕ + isinθsinϕ) - cosθ b = b\n(1a)\n(1b)\nTogether with the normalization condition |a|2 + |b|2 = 1, we find\na = cos(θ/2) e^{iθ 0} and b = sin(θ/2)e^{iϕ b}. From equation (1a) we have\na = = sinθ e^{-iϕ 0}, hence ie^{i(θ 0 -θ a )} = iα. Choose θ a = 0, then θ b = α, and\n|S 3.n;±⟩ = cos(θ/2)|+⟩ + sin(θ/2)e^{-iγ−>. Modern Quantum Mechanics - Solutions\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n10. H = a|1⟩<|1⟩ - |2⟩<|2⟩ + |1⟩<|2⟩ + |2⟩<|1⟩. Let |1⟩ = (1 0 ), |2⟩ = (0 1 ),\n|1⟩ = (1,0) and |2⟩ = (0,1) , H can be explicitly written using\nouter product of matrices as\nH = a ( 1 1 )\n ( 1 -1 )\nThe eigenvalues and corresponding eigenkets are obtained from\n(H - λI)X = 0 where X = ( x1\n x2 ) are eigenvectors and λ are corresponding\neigenvalues determined from secular equation det (H - λI) = 0. This leads\nto λ = ±√2a and x2 = (±√2 - 1)x1, hence X = x1(√2 - 1)\n1 of X we have x1 = 1/√2(2/√2). Thus eigenvectors and eigenvalues are\n|ψ1⟩ = |1⟩ + (√2 - 1)|2⟩,\nλ = ±√2\n|ψ2⟩ = |1⟩ - (√2 - 1)|2⟩\n√2(2 + √2),λ = −√2.\n\n11. Rewrite H as H = H_{11}( |1⟩<|1⟩ + |2⟩<|2⟩) + H_{12}(|1⟩<|2⟩ + |2⟩<|1⟩), where the three operator terms on\nr.h.s. behave like I, S z , and S x respectively. Note that (1/2)(H_{11} + H_{22}) is\nsimply the \"center of gravity\" of the two levels. Because the identity\noperator I remains the same under any change of basis, we ignore the\n1/2(H_{11} + H_{22}) term for the moment. Compare now with the spin 1/2 problem\n[problem 9 above]. |S 3.n.<|+⟩<|+|> + |−⟩<|+|−>. Modern Quantum Mechanics - Solutions\n\n\n\n\n\n\n\n\n\n\n\n\n6 0 = K/2 = 0 (α=0), K/2 = k(H_{11} + H_{22}). So one of the energy eigenkets is cos(θ/2)|1⟩ + sin(θ/2)|2⟩ where β, analogous to tan^{-1}(χ_n/n_z), is given by β = tan^{-1}_{H_{11} - H_{22}}\n\nThe other energy eigenket can be written down by the orthogonality requirement\n(or by letting β = β + π) as -sin(β/2)|1⟩ + cos(β/2)|2⟩. The energy eigenvalues can\nbe obtained by diagonalizing \n\\[𝑕(𝐻_{11} - H_{22}) & H_{12} \\\\ H_{12} & \\to (H_{11} - H_{22})\\] But they can also be obtained by comparing with the spin ½ problem;\nwhere \\[𝑔_{2x}^2 & ( 𝑔_{12}^2 \\] = 𝑔_{2x.n}^2 = K^2/4 (eigenvalue W/2),\nsome analogy in our case is \\[𝑓(𝑔_{11}+H_{22})^2.\\] We must still\naadd to this the center of gravity energy. The final answer is\n\\[𝑔(𝐻_{11} +H_{22}) = 𝑔(𝐻_{11} - H_{22}) + H_{22} \\] a very reasonable result.\n12. Here |S 3.n;±⟩ = 1/2|n 1⟩; and |1;±⟩ = cos(γ/2)|+⟩ + sin(γ/2)|−⟩\nIt is easily seen that the eigenket of S x belonging to eigenvalue H/2 is 1/2(1 1)\nThus (a) probability of getting +H/2 when S x is measured is A. = 1/2(p(cos(γ/2))^2 = |1/2(1,1)|/cos(γ/2)^2.\n\n= <S 3.x | S x; |2>x = K^{2}/4 - (K^{2}/4)s^{2}\cdot cos^{2} γ. Answers are entirely reasonable for γ = 0,π (parallel and anti-parallel to OZ), and for γ = π/2 (along OX). 13. Choosing the S_z diagonal basis, the first measurement corresponds to the operator M(+)= |+><+|. The second measurement is expressed by the operator M(+;n) = |+;n><+| where |+;n> = cos(β/2)|+> + sin(β/2)|-> with α = 0. Therefore M(+;n) = (cos(β/2) + sin(β/2))(cos(β/2)|+> + sin(β/2)|->)\n= cos^{2}(β/2)|+><+| + cos(β/2)sin(β/2)(|+><-| + |-><+|) + sin^{2}(β/2)|-><-|.\nThe final measurement corresponds to the operator M(-) = |-><-|, and the total measurement M_{T} = M(-)M(+;n)M(+)+ = |-><-|cos^{2}(β/2)|+><+| + cos^{2}(β/2)|+><+| + sin^{2}(β/2)|-><-| + sin^{2}(β/2)|-><-| = cos^{2}(β/2) |+><+| + cos^{2}(β/2)|+><+| + sin^{2}(β/2)|-><-| + sin^{2}(β/2)|-><-|.\nThe intensity of the final S_z = -H/2 beam, when the S_z = H/2 beam surviving the first measurement is normalized to unity, is thus cos^{2}(β/2)sin^{2}(β/2) = (sin^{2}(β/2)/4. To maximize S_z = -H/2 final beam, set β = π/2, i.e. along OX, and intensity is k.\n14. (a) The eigenvalues and eigenvectors of 3x3 matrix representation\nA = (1/√2) ( 0 1 0 )\n ( 1 0 1 )\n ( 0 1 0 )\ncan be obtained by solving det[A - λI] = 0 and normalized eigenvectors x = (x₁\nx₂\nx₃)\nwhere [A-λI]x = 0 and x₁²+x₂²+x₃² = 1. The eigenvalues are: +1, 0, -1 and the corresponding eigenvectors are respectively\n(1/√2)\n(1/√2) (0)\n(1)\n ( -1)\n(1)\nThe eigenvalues and eigenvectors of J_x = ħA for a spin 1 particle.\n15. Yes! Proof uses completeness and orthonormality of {|a',b'>}, hence [A,B] = a'∈(b')^T*b'*<a',b'|(|AB-BA|a',b'> < a',b'|;\nbut (AB-BA)|a',b'> = (ab'-ba')|a',b'> = 0, hence [A,B] = 0. An alternative