Errata for Quantum Mechanics by Ernest Abers Intermediate Level Errata as of July 1, 2007 (3rd printing)

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1 Chapter I Errata for Quantum Mechanics by Ernest Abers Intermediate Level Errata as of July 1, 007 (3rd printing) Page, just above Equation (1.5): Change third to second : With these definitions Newton s second law [Thanks to J. Schilling, 1/11/005] Page 4, Equation (1.16), in the second equation, first term on the right, r i should have only one dot. L ṙ i = mṙ i + q c A i (1.16) [Thanks to J. May, 10/11/004] Page 5, Equation (1.4): Delete the factor m that appears before the first summation sign: H = k p k q k L = m k ( pk ) L = T T + V = T + V (1.4) m [Thanks to D. Matlock, 10/06/004] Page 1, Equation (1.58): Insert i before the first ɛ. [10/07/004] ( J i ) jk = J(ˆn i ) jk = iɛ ikj = iɛ ijk (1.58) Page 1: Delete the first sentence immediately below equation (1.64): δl = d δω(q, q, t) (1.64) dt This rule must hold as a consequence of the functional form of L, for any q k (t) whether or not they satisfy the equations of motion. Let [Thanks to M. Eides, 8/6/005] Page 5, Equation (.6b): β should be α: [ ] α ψ 1 + ψ = α ψ 1 + α ψ (.6b) [Thanks to D. Staszak, 11/09/004] Page 8, Equation (.35): Change the subscript i to n: ψ n ψ n ψ = ψ (.35) [Thanks to D. Gangadharan, 11/01/004] n

2 Quantum Mechanics Page 4, Equation (.17): the sign is wrong: [Thanks to J. Ma, 11/14/005] i d dt A = [A, H(t)] (.17) Page 50. equation (.177): In the last matrix element, the operator x should be just x: ψ x +a x ψ x +a = (x + a) ψ x ψ x = ψ x x + ai ψ x (.177) [10/0/004] Page 54, Equation (.16): In the last term, in the denominator, replace k by k : ωt kx = ω o t k o x + (k k o )[v g (k o )t x] + 1 (k k o) d ω(k o ) dk t + (.16) [10/0/004] Page 54, equation (.18): In the last expression, delete π in the denominator: [Thanks to T. Tao, 10/0/004] Chapter III = e i(ωo kovg)t Ψ(x v g t) Page 81, Equation (3.119): Insert an equals sign before F(r, θ): [11/01/004] ψ(r, θ, φ) = F(r, θ)e imφ (3.119) Page 81, Equations (3.10) and (3.13): In the denominator just before the closing parenthesis, change φ to φ : ( 1 L ψ Elm (r) = sin θ θ sinθ θ + 1 ) sin ψ θ φ Elm (r, θ, φ) (3.10) ( 1 L Yl m (θ, φ) = sin θ θ sin θ θ + 1 ) sin θ φ Yl m (θ, φ) (3.13) [Thanks to D. Matlock, 11/01/004] Page 94, Problem 3.8, part (a), first line: Delete Hermitean. Let A be any well-defined operator, and [Thanks to A. Kao, 11/08/006] Page 101, Problem 3., part (e): Change see Problem.1 to see Section.6. [11/07/004]

3 ERRATA 3 Chapter IV Page 114, Equation (4.60): On the right, change the summation index from m to m : R(ˆn 1, θ 1 )R(ˆn, θ ) r, m = m R(ˆn 1, θ 1 ) r, m D m m(ˆn 1, θ 1 ) = m r, m [D(ˆn, θ )D(ˆn 1, θ 1 )] m m (4.60) [Thanks to J. May, 4/11/005] Page 19, above Equation (4.146): Change left-hand side to right-hand side : On the right-hand side, use [Thanks to J. May, 4/11/005] Page 133, Problem 4.6, second equation: Delete the summation sign on the far left. [Thanks to A. Forrester, 1/08/005] J i J i ψ m1m m 3 = j i (j i + 1) ψ m1m m 3 Page 14, footnote: Y l m(ˆn) should read Y m 1 (ˆn): This is because the spherical harmonics Y1 m (ˆn) are the spherical components of the unit vector ˆn. [Thanks to Y. Guo, 1/3/006] Page 157, Equation (5.109). Replace the last m by m: T α, j, m = ( 1) j m e iδ T α, j, m = ( 1) j α, j, m (5.109) [Thanks to Y. Guo, 3/16/006] Page 163, Equation (5.131): Insert j just after the equals sign: N i [ N j D ( 1 ) (ˆn, θ) ] = [ D ( 1 ) (ˆn, θ) ] ji ij j j N j (5.131) [Thanks to Y. Guo, 3/16/006] Chapter VI Page 179, just above Equation (6.53): a should be λ a : The eigenvalue λ a cannot be negative: [1/09/005] Page 180, in the first sentence of the second paragraph: Replace each way of distributing the particles is equally likely with when the n i correspond to the most probable distribution : The collection of particles is in thermal equilibrium when the n i correspond to the most probable distribution, subject [1/09/005]

4 4 Quantum Mechanics Page 183, Equation (6.81): Change B on the left to B 1 (t) and B 1 on the right to B 1. Just above Equation (6.8) add (with ω 1 = gµ B B 1 ) : B 1 (t) = B 1 (ˆn x cos ωt + ˆn y sin ωt) (6.81) the form (6.71), with H o = 0 and (with ω 1 = gµ B B 1 ) [1/0/005]...The Hamiltonian matrix has Page 183, in the paragraph below Equation (6.84): Replace H with H twice. The last line of that paragraph should read P o = ˆn z also. In equation (6.88) delete both primes on the right-hand side, and add a prime after the closing bracket, above the subscript i : In the rotating frame H has constant components: H = ω oˆn z + ω 1ˆn x... P o = ˆn z also. [1/0/005] d dt P i = [ (H ω) P ] i (6.88) Page 184, at the very top: Add where the prime attached to the bracket means that the vector is to be resolved along the i-th rotating axis. : where the prime attached to the bracket means that the vector is to be resolved along the i-th rotating axis. Since H ω is a constant vector... [1/0/005] Page 186, fourth line: Replace with : with only one chance in that it interacts at all. [Thanks to A. Teymourian, 4/10/007] Page 198, Problem 6.6: In the second equation, in the first term on the right-hand side, replace R o with R o: L H = MR o [Thanks to A. Zhitnitski, 10/0/004] + ω o ˆn(φ) σ Page 198, Problem 6.6: In the fourth equation, there should be a second equals sign before tanθ: B = ω = tanθ B 1 ω 1 [10/0/004]

5 ERRATA 5 Chapter VII Page 04, two lines above Equation (7.17): This sentence should read: The solution to Equation (7.8) is not unique, since you can always add a solution of the homogeneous equation [Thanks to J. degrassie, 1/6/005] Page 16. Figure 7.3: replace Balmer α and Balmer β with Lyman β and Balmer α respectively. n = 3 Lyman β 1.1 ev Balmer α 1.89 ev ev ev 3D 5 3D 3,3P 3 3P 1,3S 1 n = ev P 3 P 1,S 1 Lyman α 10. ev n = 1 1S 1 [Thanks to N. Kugland, 1/30/006] Page 17, line below Equation (7.84): ψ should be φ: L z +s z is not diagonal in the φ nl jm basis, but its diagonal elements can be computed [Thanks to X. Han, 10/7/004] Page 31, Equation (7.161): On the left, add d n r (and also on the first line of Equation (7.165), and on the right, the second and third terms should begin with +i : ψ o (r,r) Pi ψ o(r,r)φ(r)d n r = Pi φ(r) + ia i(r) i φ(r) + i i A i (R)φ(R) [Thanks to Y. Guo, 3/08/006] (7.161)

6 6 Quantum Mechanics Page 3, Equation (7.169), on the right-hand side: Replace A(R) R with A(R) dr [/09/005] λ(r) = R R o A(R) dr (7.169) Page 33, Equation (7.173): Replace M with M in the denominator in the first term on the right: [/08/006] H eff = P M + Ee o(r) + U(R) (7.173) Page 37, Equation (7.199): There is a ψ(x) missing at the end of the second term: [Thanks to T. Butler, /14/005] Page 4, Equation (7.36). x should be t: [ t Bi( t) = J [1/17/006] ψ (x) + m[e V (x)]ψ(x) = 0 (7.199) ( 3 t 3 ) ( )] J t 3 Page 44, Equation (7.50), the lower limit of the integral should be b: ψ(x) A [ x ] exp κ(x )dx κ(x) [Thanks to Y. Guo, /7/006] b (7.36) (7.50) Page 51, Problem 7.8, Part (c), in the displayed equation there should a superscrtipt after the partial derivative symbol in the numerator. And in the next to last line, so should be to : H = c (s i ) V (r) r i i r=0 Hint: Be careful here. This is not a transformation of the quantum-mechanical states; the idea is to show that... [Thanks to C. Cooper, A. Goodhue, and N. Kugland, /14/006] Page 58, Problem 7.18: In the third and the last displayed equation, delete the minus sign: The magnetic perturbation is H mag = µ B... Make a catalog of matrix elements of H mag : ψ nl Hmag ψ nl m lm s m lm s = µb B(m l + m s ) [Thanks to Y. Wang, 3/0/006]

7 ERRATA 7 Page 59, Problem 7.18, Part (e), second paragraph, second line: Delete the stray f after m = 1/ : Take the basis to be the two m = 1/ states, with j = 3/ or 1/. [Thanks to E. Osoba, 3/1/005] Chapter VIII Page 70, Equation (8.38a): π should be (π) 3/ /4π. And in Equation (8.40), on the bottom line, insert a minus sign after the first equals sign: [ ψ (n) (r) = 1 ] e ik r (π)3/ d 3 r e r ik r (π) 3 4π 4π r r U(r )ψ (n 1) (r ) (8.38a)... = 1 0 [Thanks to Y. Guo, 3/06/006] 1 r dr e iqr cos θ U(r)d cosθ = 1 1 q 0 r sin(qr)u(r)dr (8.40) Page 74, in the first line of Equation (8.60), and again in Eqation (8.6), replace ψ sc with dψ sc. And on the first line of Equation (8.60) insert f(θ) to the left of the vertical bar. e ikr ɛr dψ sc = πndzφ o ik ɛ f(θ) 1 πndzφ o z o ik ɛ z o ikr ɛr df(θ) e dr (8.60) dr dψ sc πindz φ o e ikzo f(0) (8.6) k [Thanks to Y. Guo, 3/06/006] Page 80, last sentence: Replace in that number by in that limit. For small k there are only a few, and as k 0 only l = 0 survives. In that limit the scattering is characterized by a single number. [Thanks to M. Gutperle, 5/05/004] Page 307, Equation (9.105), in the second expression change φ + π to φ + π: [Thanks to Y. Guo, 5/03/006] T exchange (ω a, θ, φ) = T direct (ω a, π θ, φ + π) (9.105) Page 309, Equation (9.11): Delete the integral sign: [Thanks to A. Young, 4/6/005] Γ ba = π H ba δ (ω a ω b ) (9.11)

8 8 Quantum Mechanics Page 314, just below Equation (9.14): Change closer to to farther from : which is farther from unity than the same probability [Thanks to M. Gutperle, 5/16/004] Page 33, Problem 9.10, Part (c): replace Λ polarization with proton spin : F(ˆn) is some function of the proton direction, independent of the proton spin. [5/16/005] Chapter X Page 36, on the second line below Equation (10.16), change positive to negative. In Equation (10.17), in the numerator of the exponent, change q to q n. And in Equation (10.18), second line, change q n to q twice: A better way to evaluate them is continuing ɛ to a negative imaginary value ( ) ( ) dp n exp iɛp n q n iɛ p n mπ iɛm q = exp n m iɛ = [Thanks to Y. Guo, 5//006] [ tb ( ) ] m q Dq exp i t a V (q) dt (10.17) (10.18) Page 38, Equation (10.6): The final exponent should be inside the bracket instead of outside, and followed by dt. In Equation (10.31), in the first line, inside the large parenthesis, in the second term, delete the stray t in the numerator and in the denominator: K(q b, t b ; q a, t a ) = exp (is[q o ]) Dδq exp [ i tb L( q o ) t a q ] δ q(t) dt (10.6) S[q] = S[q o ] + 1 tb ( ) L( q o ) dt [δ q(t)] + L(q o ) [δq(t)] 0 q q = S[q o ] + m tb (10.31) dt ([δ q(t)] ω [δq(t)] ) 0 [Thanks to N. Robles, 1/30/005] Page 39, Equation (10.38): In the second term interchange q and q, and in the last term in the numerator, change ω to ω. S[q o ] = m tb 0 = mωq 4 [Thanks to Y. Guo, 5//006] ( q(t) ω q(t) ) dt = mω Q (sin(ωt b + φ) sin φ) tb 0 cos (ωt + φ)dt (10.38)

9 ERRATA 9 Page 330, Equation (10.41): On the second line, the minus sign should be inside the bracket, before iɛ, not before the integral sign. And on the third line, delete k=1 and dy k : [ I N dy n exp iɛ m n=1 n=1 [ = dy n exp iɛ m n=1 [Thanks to Y. Guo, 5//006] n=1 Page 331, Equation (10.54): Replace 1/ by m/: [Thanks to Y. Guo, 5//006] ( L E = L q, i dq ) = m dw (ẏ n ω yn ) ] ( yn ÿ n + ω yn ) ] (10.41) ( ) dq + V (q) (10.54) dw Page 335, Equation (10.77): Insert i before the integrals on the last two lines: i i tb dt 1 dq 1 K o (q b, t b ; q 1, t 1 )V (q 1 )K o (q 1, t 1 ; q a, 0) 0 dt 1 dq 1 K o (q b, t b ; q 1, t 1 )V (q 1 )K o (q 1, t 1 ; q a, 0) (10.77) [Thanks to Y. Guo, 5//006] Page 337, Equation (10.90), second line: Change dr to ṙ : K(r b, t b ;r a, t a ) = [ tb ( Dr exp i Lo (r,ṙ) ea(r, t) ṙ + eφ(r, t) ) ] dt t a [ = Dr exp is o [r] exp ie [Thanks to N. Robles, /16/005] tb t a ( ) ] (10.90) A(r, t) ṙ φ(r, t) dt Page 338, Equation (10.91a): Change A (r) to A (r, t) In Equation (10.9), in the second line change dr to ṙ here also, and in the third line add dt after φ(r, t) : A(r, t) A (r, t) = A(r, t) + Λ(r, t) (10.91a)...

10 10 Quantum Mechanics In the new gauge the propagator is K (r b, t b ;r a, t a ) = Dr exp is o [r] [ rb ( exp ie A(r, t) dr dr φ(r, t) + Λ(r, t) r a dt dt + ) ] Λ(r, t) dt t [ rb = Dr exp is o [r] exp ie (A(r, t) dr φ(r, t)dt + ddt )] Λ(r, t)dt [Thanks to N. Robles, /15/005] r a = K(r b, t b ;r a, t a )exp [ ie ( Λ(r a, t a ) Λ(r b, t b ) )] (10.9) Page 347, Equations (10.143) and (10.144): Delete the minus sign before the last term in both equations: = 1 B ˆn z (10.143)... A(B) = ˆB B (10.144) [Thanks to Y. Guo, 5/3/006] Chapter XI Page 36, Equation (11.36), delete the δ preceding the last two integration symbols. Also, add dt at the end of both these integrals. t t 0 = δ L(t)dt = [ẋα (k, t)δẋ α (k, t) ω x α (k, t)δx α (k, t) ] d 3 k dt t 1 t 1 α t [ = d ] dtẋα(k, t) ω x α (k, t) δx α (k, t)d 3 k dt (11.36) t 1 [Thanks to J. Ma, 4/17/006] α Page 38, Equation (11.144): Insert the factor δ (E f + ω E i ) at the end: α Γ = d 3 k ψ πm f ˆε α (k) p e ik r ψ i δ (Ef + ω E i ) (11.144) ω α [4/6/005] Page 384, in Equation (11.164), inside the integral delete the factor e iωˆn r : dσ(ω) = 4π αk f 1 1 dω f mω πa 3 (π) 3 ˆε k f e iq r e r/a d 3 r (11.164) [Thanks to Y. Guo, 5/04/006]

11 ERRATA 11 Page 389, Equations (11.184) and (11.185), the subscripts of ˆε should be α, not γ. And in Equation (11.186), third line: Change ˆε α to ˆε α :1 e 4m n γ 1γ γ 3γ 4 4π (π) 3 1 ω δ γ 3γδ γ1γ 4 δ γγ 3 δ γ1γ [ ψn p ˆε α 4 (k 4 )e ik4 r 1 ψo ψo p ˆεα (k )e ] ik r ψn ω + E o ω 1 ω E n = e 4π 1 4m (π) 3 ω n [ p ψ n ˆε α (k )e ik r 1 ψ o ψo p ˆε α (k)e ] ik r ψ n ω + E o ω ω E n [4/19/006] n ψo ˆε α pe ik r 1 ψ n ψ n ˆε E on ω α pe ik r = e m ψ o (11.184) (11.185) 4π 1 (π) 3 4ω ω ˆε α ˆε α (11.186) Page 399, Equation (11.30), last line: Change π 3 to π = 1 [ 6a π πx 4 B o B π 3 4 a ]x=1/ωc 3 + = 3aω4 c π 1 π 70 a 3 + (11.30) [4/7/005] Page 404, Problem 11.8, first line of Part (a): delete the factor e ik r : (a) First convince yourself that the matrix element of p A(r) indeed vanishes to all orders in the expansion of e ik r. [Thanks to D. Staszak, 4/9/005] Chapter XII Page 410, Equations (.1): The equations should be more general: where all the diagonal elements vanish, and (J i ) 0 j = (J i ) j 0 = 0 (1.1a) (J i ) j k = iɛ ijk (1.1b) (K i ) j k = 0 (1.1c) (K i ) j 0 = (K i ) 0 j = iδ ij (1.1d) 1 There are similar corrections to the subscript of ˆε on the next three pages also.

12 1 Quantum Mechanics [Thanks to J. May, A. Forrester, and others, 5/0/006] Page 416, Equation (1.64): change qa o to +qa o : [ ] (p + qa) + qa o φ(r) = E φ(r) + (E qa o ) φ(r) (1.64) m m [Thanks to Y. Guo, 5/4/006] Page 416, Equation (1.67): Change the sign of the last term inside the large parentheses from minus to plus: ( 1 d l(l + 1) α m dr mr + ωα mr + ω m ) u(r) = 0 (1.67) m [Thanks to M. Gutperle, 6/09/004] Page 44, Equation (1.1) replace D( K i ) with D( K i ): Σ 0i = iα i = D( K i ) and Σ ij = k ɛ ijk Σ k (1.1) [Thanks to Y. Guo, 5/4/006] Page 46, Equations(1.14). On the second line, change p /8m to p /4m and p 4 /8m to p 4 /8m 3 : = 1 ] [ ] φ 1 (r) [1 p p p φ m 4m 1 (r)d 3 r = φ 1 (r) m p4 φ 8m 3 1 (r)d 3 r (1.14) [Thanks to F. O shea, 5/0/006] Page 47, Equation (1.148): In the denominator, replace r with r 3 : = α m φ 1 (r) 1 r 3s L φ 1(r)d 3 r (1.148) [Thanks to Y. Guo, 5/4/006] Chapter XIII Page 441, Equation (13.8): On the right-hand-side, delete the minus sign before the first term: H = 1 Ψ (r) Ψ(r)d 3 r m + 1 v(r 1,r )Ψ (r 1 )Ψ (r )Ψ(r 1 )Ψ(r )d 3 r 1 d 3 r (13.8) [Thanks to D. Ramunno-Johnson, 5/18/005]

13 ERRATA 13 Page 444, Equation (13.49), insert after b j : [Thanks to D. Staszak, 6/14/005] i, j = b i b j (i j) (13.49) Page 444, two lines above Equation (13.53), change kt to 1/kT : the relative population of any two states is exp[β(e (n1) E (n))], where β = 1/kT. [Thanks to A. Collette, 5/18/005] Page 464, in Equation (13.183), the second H should be H o, and in Equation (13.187) on the left, Hba should be H ba. H = H(x)d 3 x = H o + H = H o (x)d 3 x + H (x)d 3 x (13.183)... [6/01/006] H ba = δ 3 (p 1 + p p 3 p 4 ) H ba (13.187) Appendix Page 484, Equation (A.109): There should be a factor 1/ l l! in front of the last two terms: d [ ] [ ] dρ ρh (1) l(l + 1) 1+i l (ρ) = ρ 1 ρh (1) l (ρ) + i(l + 1) ρl l ze iρz (1 z ) l dz l! 1 + ρl+1 l l! 1+i 1 e iρz (1 z ) l+1 dz (A.111) [Thanks to E. Brown, /13/007] Page 494, Equation (1.166): A factor r is missing from the denominators in the last two terms: f(r) = 1 f r r r r + 1 r sin θ θ sinθ f θ + 1 f r sin θ φ (A.166) [Thanks to M. Mecklenberg, 11/16/005]