griffiths quantum mechanics solutions, Exercises of Quantum Mechanics

Download griffiths quantum mechanics solutions and more Exercises Quantum Mechanics in PDF only on Docsity! Corrections to the Instructor’s Solution Manual Introduction to Quantum Mechanics, 2nd ed. by David Griffiths Cumulative errata for the print version—corrected in the current electronic version. I especially thank Kenny Scott and Alain Thys for catching many of these errors. August 1, 2014 • page 11, Problem 1.14(a), line 1: |Ψ(x, t)2 → |Ψ(x, t)|2; line 2: − ∂∂tJ(x, t) → − ∂ ∂xJ(x, t). • page 13, Problem 1.18(b), line 2, first inequality: h 2 2mkB → h 2 3mkB . • page 13, Problem 1.18(b): change lines 5 and 6 to read: For atomic hydrogen (m = mp = 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. • page 14, Problem 2.1(b), lines 1, 2, and 4: ∂ → d (6 times); part (c), lines 1 and 2: ∂ → d (5 times). • page 15, Problem 2.2, line 6: “requuire” → “require”. • page 15, Problem 2.3, line 4: eiκa → e−κa. • page 16, Problem 2.4, second line in the calculation of 〈x2〉: y3/4 should be y2/4. • page 16, Problem 2.5(a): in the first line, change Ψ2Ψ to Ψ∗Ψ. • page 18, Problem 2.7(b), last line (in box): e−Ent/h̄ → e−iEnt/h̄. • page 20, Problem 2.11(a), line five, first integral: e−ξ2/2 → e−ξ2 . • page 21, Problem 2.11(a), second line of 〈p2〉: e−ξ2/2 → e−ξ2 . • page 23, Problem 2.13(b), line 2: “E1 and E2” → “E0 and E1”. • page 23, Problem 2.13: move “(With ψ2 in place . . . 2ω.)” from the middle of part (c) to the end of part (b). • page 25, Problem 2.17(d): in the first box, change H0 to H1; in the second box change H1 to H2; in the last line change H2 to H3; also, in the first line, (−2z + ξ) → (−2z + 2ξ). • page 27, Problem 2.20(d), last term: eikx → e−ikx. • page 28, end of last line: d→ ∂ (twice). • page 29, lines 2, 3, and 4: d→ ∂ (8 times). • page 31, Problem 2.27(b), line 2: “(x < a)” → “(x > a)”. • page 32, Problem 2.27(b), first line after “Now look for odd solutions”: “(x < a)” → “(x > a)”. • page 33, Problem 2.28, after “(1) Continuity at −a”: Aeika → Ae−ika. 1 • page 33, second line after “Solve these for F . . . ”: [4− γ2 + γ] → [4− γ2 + γ2]. • page 40, Problem 2.36: at the end of the paragraph starting “If B = 0”, change “|A|2/2 ⇒ A = √ 2 ” to “|A|2a ⇒ A = 1/ √ a ”; at the end of the paragraph starting “If A = 0”, change “|a|2/2 ⇒ B = √ 2 ” to “|B|2a⇒ B = 1/ √ a ”. • page 41, Problem 2.37: add the following, at the end: Using Eq. 2.39, 〈H〉 = p1E1 + p3E3 = ( 9 10 ) π2h̄2 2ma2 + ( 1 10 ) 9π2h̄2 2ma2 = 9π2h̄2 10ma2 . • page 42, Problem 2.38(a), end of line 2: remove period. • page 43, Problem 2.40(b), end of line 2: z → a; line 11: “interms” → “in terms”. • page 46, Problem 2.43(d), beginning of last line of 〈p2〉: h̄→ h̄2. • page 55, line 1: S21A+ S22B → S21A+ S22G. • page 56, Problem 2.53(d): in the first line, switch the indices 1 ↔ 2 (three times); in all the rest, switch the sign of a. • page 63, Problem 3.2(d), line 4: “differenting” → “differentiating”; Problem 3.3, line 6: “particlar” → “partic- ular”. • page 65, Problem 3.8(b), remove the final sentence (“But notice . . . one another.”). • page 65, Problem 3.10: remove the last sentence. • page 65, Problem 3.11, first line: e−iω/2 → e−iωt/2. • page 68, Problem 3.17(d), first chain of equations: dV dx → ∂V ∂x . • page 68, Problem 3.18, end of line 1: En → E2. • page 68, Problem 3.18, line after “Similarly, . . . ”: remove −E22 at the very end. • page 69, Problem 3.18, mid-page, second line after “Meanwhile, . . . ”: in the second expression a2 → a, and in the last expression π → π2 (in the denominators). • page 70, line 4: “Prob. 2.22(a)” → “Prob. 2.22(b)”. • page 70, Problem 3.19, line 7 (the one starting with |Φ(p, t)|2): in the term after the second equals sign, e 1 2a ... → e− 12a .... • page 70, Problem 3.19, first line of σ2H : h̄ 2 → h̄4. • page 75, Problem 3.29, 4 lines from the end: “zero” → “finite (2πh̄/λ)”. • page 76, Problem 3.31, lines 2 and 3: dV dx → ∂V ∂x (four times). • page 79, Problem 3.35(e), line 2: “form” → “from”. 2 Begin a new paragraph with “Now, Problem 4.22”. • page 123, Problem 4.54, last box: (−1)l+m → (−1)l; next line: “overall sign, which of course” → “factor of (−1)l, which”. • page 124, Problem 4.55(e), line 3: s→ j. • page 125, Problem 4.55(h): in the integral, sin2 θ → sin θ. • page 125, Problem 4.56(e): at the end of the second line the minus sign should be plus. • page 126, Problem 4.57(b), end of first line: a should be squared. • page 128, Problem 4.59(b), line 8: ∂Ax∂y → ∂Ax ∂z . • page 129, Problem 4.60(b), line 4 (the boxed equation): ∇ · (Aψ) → (∇ ·A)ψ. • page 129, Problem 4.60(b), line 12 (starting with “Now let . . . ”): before the colon, insert “(you can also do it in Cartesian coordinates)”. • page 130, Problem 4.60(b), beginning of penultimate line: open parentheses after “E =”. • page 133, Problem 5.1(b), end of penultimate line: + 1m1+m2 → = 1 m1+m2 . • page 134, Problem 5.2(b), beginning of penultimate line: insert equals sign before 536R. • page 138, bottom line: −ar1a → − ar1 4 . • page 139, Problem 5.12(b), second line of last box: 2F3/2 →2 F7/2. • page 142, Problem 5.20, line beginning “In the Figure”: “postive” → “positive”. • page 144, Problem 5.23(c), first line: end the boxed answer with a comma, and after the box insert “so the three configurations are all equally likely.” • page 146, Problem 5.25, N = 4, second line of “Total”: 124 → d 24 . • page 151, Problem 5.35(c), line 2: h̄ should be squared, in the last expression. • page 155, Problem 6.3(b), end of first line: δ(x2 − x2) → δ(x1 − x2). • page 159, Problem 6.7(b), line 4: 6.26 → 6.27. • page 159, Problem 6.7(c), line 1: E1− → E1. • page 161, first line: sin ( 3π 4 ) → sin2 ( 3π 4 ) . • page 161, Problem 6.9(c), line 4: (0 1 0) 00 1  → (0 1 0) −1 0 00 0 1 0 1 0  00 1 . • page 162, Problem 6.10, first line: “orthonornal” → “orthonormal”. • page 165, Problem 6.14, last line: 3n2 → 2n2. 5 • page 168, Problem 6.17, penultimate line: − 3 + → +3−. • page 168, Problem 6.18, under “For n = 3”, line j = 1/2: after the second equals sign, −9 → −92 . • page 169, box at bottom, line 2: “ν3 − ν3” → “ν3 − ν2”. • page 176, 3 lines up from the bottom: remove γ in front of the final parentheses. • page 178, figure, right end of 4th line up: 2/5 → 2/3. • page 178, bottom line, last term: [3(Sp · r̂)(Se · r̂)] → [3(Sp · r̂)(Se · r̂)− Sp · Se]. • page 180, Problem 6.29, 3 lines from end: ( 10−15 5×10−11 ) → ( 10−15 5×10−11 )2 . • page 181, Problem 6.31(c), last line: 12h̄ω0 → h̄ω0 2 . • page 182, at end of Problem 6.31, insert the following: [There is an interesting fraud in this well-known problem. If you expand H ′ to order 1/R5, the extra term has a nonzero expectation value in the ground state of H0, so there is a non-zero first-order perturbation, and the dominant contribution goes like 1/R5, not 1/R6 (as desired). The model gets the power “right” in three dimensions (where the expectation value is zero), but not in one. See A. C. Ipsen and K. Splittorff, ArXiv: 1401.8144.] • page 188, last line: R31 → R32. • page 190, Problem 6.38, line 6: “(Eq. 6.98)” → “(Eq. 6.93)”. • page 192, line 3 of “Off-diagonal elements”: 〈|y2〉 → 〈|y2|〉. • page 194, Problem 6.40(a), penultimate line of (i): “second-order” → “first order”; 2 lines below: 6.11 → 6.14. • page 197, line 3: 〈T 〉 = − h̄ 2 2m 2b3 π → 〈T 〉 = − h̄2 2m 2b3 π 4. • page 197, Problem 7.3, line 2: ( a 3 )3 → (a2 )3. • page 198, Problem 7.4(b), 3 lines from the end: in the expression for 〈H〉, ω → ω2. • page 200, line 1: in front of both integrals, a3 → a2. • page 200, line 4, inside first integral: e2R/a → e2r/a. • page 203, last line, first term inside square brackets: 4n− 1 → 4n+ 1 (twice). • page 210, Problem 7.18, line 1: remove spurious extra equals sign. • page 211, mid-page, line beginning “The term in . . . ”: (Z22 + Z21 ) → (Z22 + Z21 )E1. • page 211, last line: second (Z1 − 1) → (Z2 − 1). • page 212, line 2 of 〈H〉: 4E1A2 → −4E1A2. • page 212, mid-page (three lines following 〈Vee〉): e→ e2 (3 times); in the first of these three lines, ψ2(r1)+ψ1(r2) → ψ2(r1)ψ1(r2) (i.e. remove the plus sign). 6 • page 213, mid-page (line starting with 〈Vee〉): e→ e2. • page 216, line 3: cancel the second 2 in the numerator. • page 217, line 6: between ] and } insert [ y aθ(a− x) + αθ(x− a) ( 1− ya )] . • page 224, Problem 8.9, bottom part of second line: |p(x)| → p(x). • page 225, line 2: “whereα” → “where α”. • page 225, 2 lines into “Overlap region 1”, in the denominator of ψWKB: α3/2 → α3/4. • page 225, first line of “Overlap region 2”: |p(x′)| → p(x′). • page 226-227, Problem 8.10. The statement of the problem has now been corrected (switching the signs in the exponents of the two terms in the first line of Eq. 8.52). Accordingly, the solution should be changed as follows: . . . ψWKB(x) =  1√ p(x) [ Ae− i h̄ R 0 x p(x′) dx′ +Be i h̄ R 0 x p(x′) dx′ ] (x < 0) 1√ |p(x)| [ Ce 1 h̄ R x 0 |p(x ′)| dx′ +De− 1 h̄ R x 0 |p(x ′)| dx′ ] (x > 0) . . . In overlap region 1, Eq. 8.43 becomes ψWKB ≈ 1 h̄1/2α3/4(−x)1/4 [ Ae−i 2 3 (−αx) 3/2 +Bei 2 3 (−αx) 3/2 ] , . . . A = √ h̄α π ( ia+ b 2 ) e−iπ/4; B = √ h̄α π ( −ia+ b 2 ) eiπ/4. Putting in the expressions above for a and b : A = ( C 2 + iD ) e−iπ/4; B = ( C 2 − iD ) eiπ/4. . . . A = ( C 2 + iD ) e−iπ/4 = ( i 4 e−γe−iπ/4F + ieγe−iπ/4F ) e−iπ/4 = ( e−γ 4 + eγ ) F. T = ∣∣∣∣FA ∣∣∣∣2 = 1(eγ + e−γ4 )2 = e −2γ [1 + (e−2γ/4)]2 . . . . • page 228, penultimate line, “Limits”, top line: z → z1; bottom line: z → z2. • page 229, Problem 8.13, line 4: e−e → e−x. • page 230, Problem 8.14, 3 lines from end: e 2 4π0 → − e 2 4π0 ; same at beginning of the next line. • page 235, Problem 8.16(d), lines 1 and 3: 2× 10−19 → 2× 10−16. 7 Thus (4) becomes 〈n′(l + 1)(m+ 1)|x|nlm〉 = I 2 √ (2l + 3) (l −m)! (l +m+ 2)! √ (2l + 1) (l −m)! (l +m)! 1 (2l + 1)(2l + 3) (l +m+ 2)! (l −m)! = I 2 √ (l +m+ 2)(l +m+ 1) (2l + 1)(2l + 3) . (6) Now we do the same for m′ = m− 1: 〈n′l′(m− 1)|x|nlm〉 = ∫ Rn′l′(Y m−1l′ ) ∗r sin θ cosφRnlY ml r 2 dr sin θ dθ dφ = I √ (2l′ + 1) 4π (l′ −m+ 1)! (l′ +m− 1)! √ (2l + 1) 4π (l −m)! (l +m)! ∫ π 0 Pm−1l′ P m l sin 2 θ dθ ∫ 2π 0 cosφe−i(m−1)φeimφ dφ. (7) Intφ = 1 2 ∫ 2π 0 (eiφ + e−iφ)eiφ dφ = 1 2 ∫ 2π 0 (e2iφ + 1) dφ = π. (8) Changing variables (x ≡ cos θ), and using Eqs. 9.100 and (2): Intθ = ∫ 1 −1 √ 1− x2 Pm−1l+1 (x)P m l (x) dx = 1 (2l + 3) [∫ 1 −1 Pml+2P m l dx− ∫ 1 −1 Pml P m l dx ] = − 2 (2l + 1)(2l + 3) (l +m)! (l −m)! , and (7) becomes 〈n′(l + 1)(m− 1)|x|nlm〉 = −I 4 √ (2l + 3) (l −m+ 2)! (l +m)! √ (2l + 1) (l −m)! (l +m)! 2 (2l + 1)(2l + 3) (l +m)! (l −m)! = −I 2 √ (l −m+ 2)(l −m+ 1) (2l + 1)(2l + 3) . (9) Meanwhile, Eq. 9.70 says |〈n′l′m′|y|nlm〉|2 = |〈n′l′m′|y|nlm〉|2, so |〈n′(l + 1)(m+ 1)|r|nlm〉|2 + |〈n′(l + 1)m|r|nlm〉|2 + |〈n′(l + 1)(m− 1)|r|nlm〉|2 = 2 [ I 2 √ (l +m+ 2)(l +m+ 1) (2l + 1)(2l + 3) ]2 + [ I √ (l + 1)2 −m2 (2l + 1)(2l + 3) ]2 + 2 [ −I 2 √ (l −m+ 2)(l −m+ 1) (2l + 1)(2l + 3) ]2 = I2 2 { (l +m+ 2)(l +m+ 1) + 2[(l + 1)2 −m2] + (l −m+ 2)(l −m+ 1) (2l + 1)(2l + 3) } = I2 (2l2 + 5l + 3) (2l + 1)(2l + 3) = I2 (l + 1) (2l + 1) . (10) 10 Therefore, |℘|2 (summed over the three allowed transitions) is e2I2(l+1)/(2l+1), and the spontaneous emission rate (Eq. 9.56) is Al→l+1 = e2ω3I2 3π0h̄c3 (l + 1) (2l + 1) . (11) l′ = l − 1 Return to Eq. (1). This time the integral is Intθ = ∫ 1 −1 xPml−1(x)P m l (x) dx = 1 (2l + 1) [ (l +m) ∫ 1 −1 Pml−1P m l−1 dx+ (l −m+ 1) ∫ 1 −1 Pml−1P m l+1 dx ] = (l +m) (2l + 1) 2 (2l − 1) (l − 1 +m)! (l − 1−m)! = 2 (2l − 1)(2l + 1) (l +m)! (l −m− 1)! . Therefore 〈n′(l − 1)m|z|nlm〉 = I 2 √ (2l − 1)(l − 1−m)! (l − 1 +m)! √ (2l + 1) (l −m)! (l +m)! 2 (2l − 1)(2l + 1) (l +m)! (l −m− 1)! = I √ l2 −m2 (2l − 1)(2l + 1) . (12) From 〈n′l′m′|x|nlm〉 with m′ = m+ 1, Eqs. (4) and (5) are unchanged; this time Intθ = ∫ 1 −1 √ 1− x2 Pm+1l−1 (x)P m l (x) dx = 1 (2l + 1) [∫ 1 −1 Pm+1l−1 P m+1 l+1 dx− ∫ 1 −1 Pm+1l−1 P m+1 l−1 dx ] = − 2 (2l − 1)(2l + 1) (l +m)! (l −m− 2)! , and (4) becomes 〈n′(l − 1)(m+ 1)|x|nlm〉 = −I 2 √ (2l − 1)(l −m− 2)! (l +m)! √ (2l + 1) (l −m)! (l +m)! 1 (2l − 1)(2l + 1) (l +m)! (l −m− 2)! = −I 2 √ (l −m)(l −m− 1) (2l − 1)(2l + 1) . (13) Now we do the same for m′ = m− 1. Eqs. (7) and (8) are unchanged, the θ integral is Intθ = ∫ 1 −1 √ 1− x2 Pm−1l−1 (x)P m l (x) dx = 1 (2l − 1) [∫ 1 −1 Pml P m l dx− ∫ 1 −1 Pml−2P m l dx ] = 2 (2l − 1)(2l + 1) (l +m)! (l −m)! , 11 and (7) becomes 〈n′(l − 1)(m− 1)|x|nlm〉 = I 2 √ (2l − 1) (l −m)! (l +m− 2)! √ (2l + 1) (l −m)! (l +m)! 1 (2l − 1)(2l + 1) (l +m)! (l −m)! = I 2 √ (l +m)(l +m− 1) (2l − 1)(2l + 1) . (14) Thus |〈n′(l − 1)(m+ 1)|r|nlm〉|2 + |〈n′(l − 1)m|r|nlm〉|2 + |〈n′(l − 1)(m− 1)|r|nlm〉|2 = 2 [ −I 2 √ (l −m)(l −m− 1) (2l − 1)(2l + 1) ]2 + [ I √ l2 −m2 (2l − 1)(2l + 1) ]2 + 2 [ I 2 √ (l +m)(l +m− 1) (2l − 1)(2l + 1) . ]2 = I2 2 { (l −m)(l −m− 1) + 2(l2 −m2) + (l +m)(l +m− 1) (2l − 1)(2l + 1) } = I2 (2l2 − l) (2l − 1)(2l + 1) = I2 l (2l + 1) , (15) and the emission rate is Al→l−1 = e2ω3I2 3π0h̄c3 l (2l + 1) . (16) (Of course, I is different for the two cases l→ l ± 1.) • page 254, 4th line from bottom: imbh̄w → imv h̄w . • page 255, Problem 10.1(b), line 3, upper limit of second integral: π → a; end of that line: c′n → cn′ . • page 255, penultimate line, sin ( nπ a ) → sin ( nπ a x ) . • page 256, in the first box: “Tea/v” → “Te = a/v”. • page 256, Problem 10.1(c), penultimate line: mva 2 2h̄a = mva 2h̄ → mv(2a)2 2h̄(2a) = mva h̄ . • page 257, line 2, lower row: first cos → sin. • page 257, line 8: ω1(ω1−ω)λ → iω1(ω1−ω) λ . • page 259, Problem 10.4, final box: h̄2 → h̄3. • page 265, Problem 10.9(b), line 4: remove h̄ after (n+ 12 ). • page 265, 4 lines from end: dΨndz → dψn dz . • page 266, Problem 10.9(e), line 4: remove h̄ after (n+ 12 ). • page 267, Problem 10.10(b), part (1), line 3: “, and (Eq. 10.39)” → “; Eqs. 10.39 and 10.93 yield”; line 4, insert after h̄ωt: −mω 2 2h̄ ∫ t 0 [f(t′)]2 dt′. 12