page 11, Problem 1.14(a), line 1: Ψ(x, t) 2 Ψ(x, t) 2 ; line 2: h 2 2mk B page 13, Problem 1.18(b), line 2, first inequality: 3mk B

Transcription

1 Corrections to the Instructor s Soution Manua Introduction to Quantum Mechanics, nd ed by David Griffiths Cumuative errata for the print version corrected in the current eectronic version I especiay thank Kenny Scott and Aain Thys for catching many of these errors August, 4 page, Probem 4(a), ine : Ψ(x, t) Ψ(x, t) ; ine : tj(x, t) xj(x, t) page 3, Probem 8(b), ine, first inequaity: page 3, Probem 8(b): change ines 5 and 6 to read: h mk B h 3mk B For atomic hydrogen (m m p 7 7 kg) with d m: T < (66 34 ) 3(7 7 )(4 3 )( ) 6 4 K page 4, Probem (b), ines,, and 4: d (6 times); part (c), ines and : d (5 times) page 5, Probem, ine 6: requuire require page 5, Probem 3, ine 4: e iκa e κa page 6, Probem 4, second ine in the cacuation of x : y 3 /4 shoud be y /4 page 6, Probem 5(a): in the first ine, change Ψ Ψ to Ψ Ψ page 8, Probem 7(b), ast ine (in box): e Ent/ h e ient/ h page, Probem (a), ine five, first integra: e ξ / e ξ page, Probem (a), second ine of p : e ξ / e ξ page 3, Probem 3(b), ine : E and E E and E page 3, Probem 3: move (With ψ in pace ω) from the midde of part (c) to the end of part (b) page 5, Probem 7(d): in the first box, change H to H ; in the second box change H to H ; in the ast ine change H to H 3 ; aso, in the first ine, ( z + ξ) ( z + ξ) page 7, Probem (d), ast term: e ikx e ikx page 8, end of ast ine: d (twice) page 9, ines, 3, and 4: d (8 times) page 3, Probem 7(b), ine : (x < a) (x > a) page 3, Probem 7(b), first ine after Now ook for odd soutions : (x < a) (x > a) page 33, Probem 8, after () Continuity at a : Ae ika Ae ika

2 page 33, second ine after Sove these for F : [4 γ + γ] [4 γ + γ ] page 4, Probem 36: at the end of the paragraph starting If B, change A / A to A a A / a ; at the end of the paragraph starting If A, change a / B to B a B / a page 4, Probem 37: add the foowing, at the end: 9 π h 9π h Using Eq 39, H p E + p 3 E 3 ma + ma 9π h ma page 4, Probem 38(a), end of ine : remove period page 43, Probem 4(b), end of ine : z a; ine : interms in terms page 46, Probem 43(d), beginning of ast ine of p : h h page 55, ine : S A + S B S A + S G page 56, Probem 53(d): in the first ine, switch the indices (three times); in a the rest, switch the sign of a page 63, Probem 3(d), ine 4: differenting differentiating ; Probem 33, ine 6: particar particuar page 65, Probem 38(b), remove the fina sentence ( But notice one another ) page 65, Probem 3: remove the ast sentence page 65, Probem 3, first ine: e iω/ e iωt/ page 68, Probem 37(d), first chain of equations: page 68, Probem 38, end of ine : E n E dv dx V x page 68, Probem 38, ine after Simiary, : remove E at the very end page 69, Probem 38, mid-page, second ine after Meanwhie, : in the second expression a a, and in the ast expression π π (in the denominators) page 7, ine 4: Prob (a) Prob (b) page 7, Probem 39, ine 7 (the one starting with Φ(p, t) ): e a e a in the term after the second equas sign, page 7, Probem 39, first ine of σ H : h h 4 page 75, Probem 39, 4 ines from the end: zero finite (π h/λ) page 76, Probem 33, ines and 3: dv dx V x page 79, Probem 335(e), ine : form from (four times)

3 page 83, Probem 338(c), first ine of B: e iωt e iωt page 83, Probem 338(c), ast ine: (, ) (, 3) page 87, Probem 4(a), penutimate ine: f y f x ; space after since page 88, Probem 4(a), ine 3, second term: d X dy d Y dy page 9, Probem 44, attach the foowing comment at the end: [In truth, this is not as decisive as it appears, since π [(n[tan(θ/)]) sin θ dθ π /6, which is perfecty finite; Θ itsef bows up at and π, but sin θ tames it] page 9, beginning of ine : P 3 P 3 (x) page 95, Probem 4, ine 3: Eq 45 Eq 474 page 98, Probem 45(b), second ine, first term in big parentheses: a a in denominator page, Probem 49(d): insert period after p, and repace the rest with the foowing: L z aso commutes with r, as we can show using a test function f(r): [L z, r]f h [ x i y y ] x, r f h ( x (rf) y (rf) ) i y x rx f f + ry y x But r x x + y x + z x (x r, so r y y r ), x and hence [L z, r] (and the same goes for the other two components) H p /m + V (r) QED page, Probem 43, first ine: i cot θ θ i cot θ φ page 4, Probem 47(b), first ine: (i + i) (i + i) page 4, Probem 47(c), second ine, after the second equas sign: h 4 h 4 page 7, Probem 43(b), ine, third term: sin α eiγbt/ sin α e iγbt/ page 8, Probem 434(a), ine : (Eq 443) (ine above Eq 446) ( i h x r y y r ) f x So L commutes with page 9, Probem 436 (a): after the second comma in the box, it shoud read or (probabiity 6/5) page, Probem 439, third ine from bottom: in the second term, ( + ) ( + ) page, Probem 44(a), ine 3: ( iδ ij ) ( i hδ ij ) page, Probem 44(b), ast ine: change T V to T + V 5 3a page 5, Probem 443(c), second ine: in the ast two terms, and 3 3 3a s 3a

4 page 6, Probem 444(a): remove cos θ in the midde term; Probem 444(c): change 4(5) to 3(4) and to 3 page 8, Probem 448(b): page 9, Probem 45(a), ine : Eqs 436 and 444 Eq 436 and ine after Eq 446 ; ine 7: in the second square brackets, S s page, Probem 45(a), ine beginning where a : Mutipy by (a b) Mutipy by (a m) page, Probem 45(a), penutimate ine: insert before the period page 3, Probem 454, ine : begin with For positive upper index: ; Y m± Y m+ page 3-4, Probem 454, bottom of page 3 to top of page 4, repace For m < to just before Now, Probem 4 with the foowing: For negative upper index: write Y m L Y m Noting (Eq 47) that P m or As before, so he iφ ( θ i cot θ φ B m e imφ P m and (Probem 48) A m Y m h ( m)( + m + ) Y m P m, we have (Eq 43) ) B m e imφ P m h ( m)( + m + )B m e i(m+)φ P m+, d dθ m cot θ P m B m ( m)( + m + ) B m P m+ Thus (using m, m, m, ): d dθ m cot θ P m P m+, B m B m ( m)( + m + ) Evidenty and in genera, where B B ; ( + ) (as in Eq 43) B ( )( + ) B B m ɛ B m () m B m, ( m )! ɛ ( + m )! C(), { () m, for m,, for m, ( + )( + )( ) B, 4

5 Begin a new paragraph with Now, Probem 4 page 3, Probem 454, ast box: () +m () ɛ; next ine: overa sign, which of course factor of (), which page 4, Probem 455(e), ine 3: s j page 5, Probem 455(h): in the integra, sin θ sin θ page 5, Probem 456(e): at the end of the second ine the minus sign shoud be pus page 6, Probem 457(b), end of first ine: a shoud be squared page 8, Probem 459(b), ine 8: A x y Ax z page 9, Probem 46(b), ine 4 (the boxed equation): (Aψ) ( A)ψ page 9, Probem 46(b), ine (starting with Now et ): before the coon, insert (you can aso do it in Cartesian coordinates) page 3, Probem 46(b), beginning of penutimate ine: open parentheses after E page 33, Probem 5(b), end of penutimate ine: + m +m m +m page 34, Probem 5(b), beginning of penutimate ine: insert equas sign before 5 36 R page 38, bottom ine: ar a ar 4 page 39, Probem 5(b), second ine of ast box: F 3/ F 7/ page 4, Probem 5, ine beginning In the Figure : postive positive page 44, Probem 53(c), first ine: end the boxed answer with a comma, and after the box insert so the three configurations are a equay ikey page 46, Probem 55, N 4, second ine of Tota : 4 d 4 page 5, Probem 535(c), ine : h shoud be squared, in the ast expression page 55, Probem 63(b), end of first ine: δ(x x ) δ(x x ) page 59, Probem 67(b), ine 4: page 59, Probem 67(c), ine : E E page 6, first ine: sin 3π 4 sin 3π 4 page 6, Probem 69(c), ine 4: ( ) ( ) page 6, Probem 6, first ine: orthonorna orthonorma page 65, Probem 64, ast ine: 3n n 5

6 page 68, Probem 67, penutimate ine: page 68, Probem 68, under For n 3, ine j /: after the second equas sign, 9 9 page 69, box at bottom, ine : ν 3 ν 3 ν 3 ν page 76, 3 ines up from the bottom: remove γ in front of the fina parentheses page 78, figure, right end of 4th ine up: /5 /3 page 78, bottom ine, ast term: [3(S p ˆr)(S e ˆr)] [3(S p ˆr)(S e ˆr) S p S e ] page 8, Probem 69, 3 ines from end: page 8, Probem 63(c), ast ine: hω hω page 8, at end of Probem 63, insert the foowing: [There is an interesting fraud in this we-known probem If you expand H to order /R 5, the extra term has a nonzero expectation vaue in the ground state of H, so there is a non-zero first-order perturbation, and the dominant contribution goes ike /R 5, not /R 6 (as desired) The mode gets the power right in three dimensions (where the expectation vaue is zero), but not in one See A C Ipsen and K Spittorff, ArXiv: 4844] page 88, ast ine: R 3 R 3 page 9, Probem 638, ine 6: (Eq 698) (Eq 693) page 9, ine 3 of Off-diagona eements : y y page 94, Probem 64(a), penutimate ine of (i): second-order first order ; ines beow: 6 64 page 97, ine 3: T h b 3 m π T h b 3 m π 4 page 97, Probem 73, ine : ( a 3 ) 3 ( a ) 3 page 98, Probem 74(b), 3 ines from the end: in the expression for H, ω ω page, ine : in front of both integras, a 3 a page, ine 4, inside first integra: e R/a e r/a page 3, ast ine, first term inside square brackets: 4n 4n + (twice) page, Probem 78, ine : remove spurious extra equas sign page, mid-page, ine beginning The term in : (Z + Z ) (Z + Z )E page, ast ine: second (Z ) (Z ) page, ine of H : 4E A 4E A page, mid-page (three ines foowing V ee ): e e (3 times); in the first of these three ines, ψ (r )+ψ (r ) ψ (r )ψ (r ) (ie remove the pus sign) 6

7 page 3, mid-page (ine starting with V ee ): e e page 6, ine 3: cance the second in the numerator page 7, ine 6: between ] and } insert [ y a θ(a x) + αθ(x a) ( y a)] page 4, Probem 89, bottom part of second ine: p(x) p(x) page 5, ine : whereα where α page 5, ines into Overap region, in the denominator of ψ WKB : α 3/ α 3/4 page 5, first ine of Overap region : p(x ) p(x ) page 6-7, Probem 8 The statement of the probem has now been corrected (switching the signs in the exponents of the two terms in the first ine of Eq 85) Accordingy, the soution shoud be changed as foows: ψ WKB (x) [Ae ī R h x p(x ) dx + Be ī R h x p(x ) dx ] (x < ) p(x) [Ce h R x p(x ) dx + De h R x p(x ) dx ] (x > ) p(x) [ In overap region, Eq 843 becomes ψ WKB Ae i h / 3 ( αx)3/ + Be i ( αx)3/] 3, α 3/4 ( x) /4 hα ia + b hα ia + b A e iπ/4 ; B e iπ/4 Putting in the expressions above for a and b : π π C C A + id e iπ/4 ; B id e iπ/4 A C i e + id e iπ/4 4 e γ e iπ/4 F + ie γ e iπ/4 F e iπ/4 γ 4 + eγ F T F A e γ (e γ + e γ 4 ) [ + (e γ /4)] page 8, penutimate ine, Limits, top ine: z z ; bottom ine: z z page 9, Probem 83, ine 4: e e e x e page 3, Probem 84, 3 ines from end: 4πɛ e 4πɛ ; same at beginning of the next ine page 35, Probem 86(d), ines and 3: 9 6 7

8 page 35, Probem 87, insert the foowing at the end: [Actuay, to tunne a the way through the cassicay forbidden region, the center of mass must not ony rise from to x, but aso drop back to This doubes γ, and makes the fina exponent 3 ] page 36, Probem 9, ine 3, ast expression: e r/a e r/a page 36, Probem 9, ast three ines: remove the minus signs in front of a 6 expressions page 37, ine 6: Ae i(ω+ω)/ + Be i(ω ω)/ Ae i(ω+ω)t/ + Be i(ω ω)t/ page 37, end of Probem 9, add the foowing: [In ight of the Comment you might question the initia conditions If the perturbation incudes a factor θ(t), are we sure this doesn t ater c a () and c b ()? That is, are we sure c a (t) and c b (t) are continuous at a step function potentia? The answer is yes, for if we integrate Eq 93 from ɛ to ɛ, c a (ɛ) c a ( ɛ) ī h H ab ɛ e iωt c b (t) dt But c b (t), so the integra goes to zero as ɛ, and hence c a ( ɛ) c a (ɛ) The same goes for c b, of course] page 38, 5 ines up from end: c b (t) iα ɛ hω c b(t) α ɛ hω page 4, Probem 94, 3 ines from end: H ba H ba page 4, Probem 96, penutimate ine: ω ω page 4, ine 6 ( Genera soution ), second exponent: +ω r ω r page 43, Probem 98, ine 6: e hω/kbt e hω/kbt [ ) 5 5 a a page 44, ine 5: 3 3 page 44, ine 8: page 45, ine 5: insert i h right after { 4V 4V page 5, Probem 98, ine 4: 3π 3π B a B page 5, Probem 9(b), ine, ast term: rf e iωt b B rf e iωt a B b page 53, Probem 9: + B a + B rf e iωt b B rf e iωt a B b 8

9 From Eq 969, n m z nm uness m m, so the ony nonzero z term is n m z nm R n (Y m ) r cos θ R n Y m r dr sin θ dθ dφ ( I + ) ( m )! ( + ) ( m )! π 4π ( + m )! 4π ( + m )! π P m P m cos θ sin θ dθ () This is independent of the sign of m, so we might as we assume m The integra (changing variabes to x cos θ) is (using Eq 999) Int θ [ x P+(x)P m m ] (x) dx ( + m) P m ( + ) +P m dx + ( m + ) P+P m + m dx Now, it foows from Eq 433 that the associated Legendre functions satisfy the orthogonaity reation so Int θ ( m + ) ( + ) P m (x)p m (x) dx ( + m )! ( + ) ( m )! δ, () ( + + m)! ( + 3) ( + m)! ( + )( + 3) Putting this into Eq(): n ( + )m z nm I ( + m)! ( m)! ( + 3) ( + ) ( + + m)! ( + m)! I ( + + m)! ( m)! ( + + m)! ( + )( + 3) ( m)! ( + ) m ( + )( + 3) (3) From Eq 97, n m x nm uness m m ± ; et s start with m + : n m+ (m + ) x nm R n (Y ) r sin θ cos φ R n Y m r dr sin θ dθ dφ ( I + ) ( m )! ( + ) ( m)! π π 4π ( P m+ P m + m + )! 4π ( + m)! sin θ dθ cos φe i(m+)φ e imφ dφ (4) Int φ π (e iφ + e iφ )e iφ dφ Changing variabes (x cos θ), and using Eqs 9 and (): Int θ π ( + e iφ ) dφ π (5) [ x P m+ + (x)p m (x) dx P m+ + ( + ) P m+ + dx P m+ ( + )( + 3) ( + m + )! ( m)! + P m+ ] dx 9

10 Thus (4) becomes n ( + )(m + ) x nm I ( m)! ( m)! ( + 3) ( + ) ( + m + )! ( + m)! I Now we do the same for m m : n (m ) x nm ( I + ) ( m + )! 4π ( + m )! Int φ ( + m + )! ( + )( + 3) ( m)! ( + m + )( + m + ) (6) ( + )( + 3) m R n (Y π ( + ) ( m)! 4π ( + m)! ) r sin θ cos φ R n Y m r dr sin θ dθ dφ π (e iφ + e iφ )e iφ dφ Changing variabes (x cos θ), and using Eqs 9 and (): Int θ and (7) becomes P m π P m sin θ dθ [ x P m + (x)p m (x) dx P m ( + 3) ( + m)! ( + )( + 3) ( m)!, n ( + )(m ) x nm I ( + 3) 4 I ( m + )! ( + m)! Meanwhie, Eq 97 says n m y nm n m y nm, so π cos φe i(m)φ e imφ dφ (7) (e iφ + ) dφ π (8) +P m dx P m P m ] dx ( m)! ( + m)! ( + ) ( + m)! ( + )( + 3) ( m)! ( m + )( m + ) (9) ( + )( + 3) n ( + )(m + ) r nm + n ( + )m r nm + n ( + )(m ) r nm [ ] [ ] [ I ( + m + )( + m + ) ( + ) + I m + I ( m + )( m + ) ( + )( + 3) ( + )( + 3) ( + )( + 3) { I ( + m + )( + m + ) + [( + ) m } ] + ( m + )( m + ) ( + )( + 3) I ( ) ( + ) I ( + )( + 3) ( + ) () ]

11 Therefore, (summed over the three aowed transitions) is e I ( + )/( + ), and the spontaneous emission rate (Eq 956) is A + e ω 3 I 3πɛ hc 3 ( + ) ( + ) () Return to Eq () This time the integra is Int θ [ x P(x)P m m ] (x) dx ( + m) P m ( + ) P m dx + ( m + ) PP m + m dx ( + m) ( + m)! ( + ) ( ) ( m)! ( + m)! ( )( + ) ( m )! Therefore n ( )m z nm I ( m)! ( m)! ( + m)! ( ) ( + ) ( + m)! ( + m)! ( )( + ) ( m )! I m ( )( + ) () From n m x nm with m m +, Eqs (4) and (5) are unchanged; this time Int θ and (4) becomes [ x P m+ (x)p m (x) dx P m+ ( + ) P m+ + dx P m+ ( + m)! ( )( + ) ( m )!, n ( )(m + ) x nm I ( ) I ( m )! ( + m)! P m+ ] dx ( m)! ( + m)! ( + ) ( + m)! ( )( + ) ( m )! ( m)( m ) (3) ( )( + ) Now we do the same for m m Eqs (7) and (8) are unchanged, the θ integra is Int θ x P m (x)p m (x) dx ( + m)! ( )( + ) ( m)!, [ ( ) P m P m dx P m P m ] dx

12 and (7) becomes n ( )(m ) x nm I ( m)! ( m)! ( + m)! ( ) ( + ) ( + m )! ( + m)! ( )( + ) ( m)! I ( + m)( + m ) (4) ( )( + ) Thus n ( )(m + ) r nm + n ( )m r nm + n ( )(m ) r nm [ ] [ ] [ ] I ( m)( m ) + I m I ( + m)( + m ) + ( )( + ) ( )( + ) ( )( + ) { I ( m)( m ) + ( m } ) + ( + m)( + m ) ( )( + ) I ( ) ( )( + ) I ( + ), (5) and the emission rate is A e ω 3 I 3πɛ hc 3 ( + ) (6) (Of course, I is different for the two cases ± ) page 54, 4th ine from bottom: imb hw imv hw page 55, Probem (b), ine 3, upper imit of second integra: π a; end of that ine: c n c n page 55, penutimate ine, sin ( nπ a sin nπ a x) page 56, in the first box: T e a/v T e a/v page 56, Probem (c), penutimate ine: mva ha page 57, ine, ower row: first cos sin page 57, ine 8: ω (ω ω) λ iω(ω ω) λ page 59, Probem 4, fina box: h h 3 mva h mv(a) h(a) mva h page 65, Probem 9(b), ine 4: remove h after (n + ) page 65, 4 ines from end: dψ n dz dψn dz page 66, Probem 9(e), ine 4: remove h after (n + ) page 67, Probem (b), part (), ine 3:, and (Eq 39) ; Eqs 39 and 93 yied ; ine 4, insert t after hωt: mω h [f(t )] dt

13 page 68, Probem (a), first ine: square ṙ page 7, Probem (c), in the expression for σ: 8 6 page 7, ine beginning () ψ continuous : sinka sin ka page 7, Probem 5(a), ine : B ikx Be ikx ; ine : B ika Be ika page 7, Probem 5(a), ine 4: k ψ (k ) ψ page 73, Probem 6, ine 3, in the denominator: j (x) + in (x) j (ka) + in (ka) page 75, ine 3: dr dr page 75, Probem, ine 4: cos(κa) cos(κa) page 76, Probem, fina box: k h page 78, ine 3: insert pus sign at beginning page 79, ine 6: before G insert page 8, Probem 8, end of ine 4: α α page 8, Probem 8, 4 ines from end: R [ ) m ( V h k k sin(ka) ] [ ( R m ) ( V h k k sin(ka) )] page 88, Probem A3, end of ine : det U det U page 89-9, Probem A5: starting with the ast ine on page 89, we shoud have î ĵ ; ĵ î ; ˆk ˆk, so (Eq A6) S S ST x S cos θ sin θ sin θ cos θ cos θ sin θ cos θ sin θ T y ( θ) sin θ cos θ sin θ cos θ ST y S cos θ sin θ sin θ cos θ cos θ sin θ cos θ sin θ T x (θ) sin θ cos θ sin θ cos θ Is this what we woud 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 page 97, Probem A8(a), first ine of part (ii): M 3 θ 3 M M 3 θ M 3