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: − (^) ∂t∂ J(x, t) → − (^) ∂x∂ J(x, t).
- page 13, Problem 1.18(b), line 2, first inequality: h^2 2 mkB
h^2 3 mkB
- 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 〈x^2 〉: y^3 /4 should be y^2 /4.
- page 16, Problem 2.5(a): in the first line, change Ψ^2 Ψ to Ψ∗Ψ.
- page 18, Problem 2.7(b), last line (in box): e−Ent/¯h^ → e−iEnt/¯h.
- page 20, Problem 2.11(a), line five, first integral: e−ξ^2 /^2 → e−ξ^2.
- page 21, Problem 2.11(a), second line of 〈p^2 〉: e−ξ^2 /^2 → e−ξ^2.
- page 23, Problem 2.13(b), line 2: “E 1 and E 2 ” → “E 0 and E 1 ”.
- page 23, Problem 2.13: move “(With ψ 2 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 H 0 to H 1 ; in the second box change H 1 to H 2 ; in the last line change H 2 to H 3 ; also, in the first line, (− 2 z + ξ) → (− 2 z + 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.
- page 33, second line after “Solve these for F... ”: [4 − γ^2 + γ] → [4 − γ^2 + γ^2 ].
- page 40, Problem 2.36: at the end of the paragraph starting “If B = 0”, change “|A|^2 / 2 ⇒ A =
2 ” to “|A|^2 a ⇒ A = 1/
a ”; at the end of the paragraph starting “If A = 0”, change “|a|^2 / 2 ⇒ B =
2 ” to “|B|^2 a ⇒ B = 1/
a ”.
- page 41, Problem 2.37: add the following, at the end:
Using Eq. 2.39, 〈H〉 = p 1 E 1 + p 3 E 3 =
π^2 ¯h^2 2 ma^2
9 π^2 ¯h^2 2 ma^2 = 9 π
(^2) ¯h^2 10 ma^2
- 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 〈p^2 〉: ¯h → ¯h^2.
- page 55, line 1: S 21 A + S 22 B → S 21 A + S 22 G.
- 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 → E 2.
- page 68, Problem 3.18, line after “Similarly,... ”: remove −E 22 at the very end.
- page 69, Problem 3.18, mid-page, second line after “Meanwhile,... ”: in the second expression a^2 → a, and in the last expression π → π^2 (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 21 a^ ...^ → e−^21 a^ ....
- page 70, Problem 3.19, first line of σ^2 H : ¯h^2 → ¯h^4.
- page 75, Problem 3.29, 4 lines from the end: “zero” → “finite (2π¯h/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”.
- page 116, Problem 4.44(a): remove cos θ in the middle term; Problem 4.44(c): change 4(5) to 3(4) and 11 to 3.
- page 118, Problem 4.48(b): 4. 113 → 4 .118.
- page 119, Problem 4.51(a), line 1: “Eqs. 4.136 and 4.144” → “Eq. 4.136 and line after Eq. 4.146”; line 7: in the second square brackets, S 2 → s 2.
- page 120, Problem 4.51(a), line beginning “where a ≡... ”: “Multiply by (a − b)” → “Multiply by (a − m)”.
- page 121, Problem 4.51(a), penultimate line: insert “= 1” before the period.
- page 123, Problem 4.54, line 1: begin with “For positive upper index:”; Y (^) lm ±^1 → Y (^) lm +1.
- page 123-124, Problem 4.54, bottom of page 123 to top of page 124, replace “For m < 0” to just before “Now, Problem 4.22” with the following: For negative upper index: write Y (^) l− m= B−l me−imφP (^) l− mand (Problem 4.18)
L−Y (^) l− m= A−l mY (^) l− m−^1 = ¯h
(l − m)(l + m + 1) Y (^) l− m−^1. Noting (Eq. 4.27) that P (^) l− m= P (^) lm , we have (Eq. 4.130)
−¯he−iφ
∂θ − i cot θ ∂ ∂φ
B−l me−imφP (^) lm = ¯h
(l − m)(l + m + 1)B l− m−^1 e−i(m+1)φP (^) lm +1,
or (^) ( d dθ − m cot θ
P (^) lm B−l m= −
(l − m)(l + m + 1) B l− m−^1 P (^) lm +1.
As before, (^) ( d dθ − m cot θ
P (^) lm = −P (^) lm +1, so B− l m−^1 =
√^1
(l − m)(l + m + 1)
Bl− m.
Thus (using m = 0, m = 1, m = 2,... ):
B l− 1 =
√^1
l(l + 1)
B^0 l ; B− l 2 =
√^1
(l − 1)(l + 2)
B l− 1 =
√^1
(l + 2)(l + 1)l(l − 1)
B l^0 ,....
Evidently Bl− m= (−1)mBml , and in general,
Bml =
(l − |m|)! (l + |m|)! C(l),
where ≡
(−1)m, for m ≥ 0 , 1 , for m ≤ 0 , (as in Eq. 4.32).
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, sin^2 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: ∂A ∂yx → ∂A ∂zx.
- 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: + (^) m 1 +^1 m 2 → = (^) m 1 +^1 m 2.
- page 134, Problem 5.2(b), beginning of penultimate line: insert equals sign before 365 R.
- page 138, bottom line: − ar a^1 → − ar 41.
- page 139, Problem 5.12(b), second line of last box: 2 F 3 / 2 →^2 F 7 / 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”: 241 → 24 d.
- page 151, Problem 5.35(c), line 2: ¯h should be squared, in the last expression.
- page 155, Problem 6.3(b), end of first line: δ(x 2 − x 2 ) → δ(x 1 − x 2 ).
- page 159, Problem 6.7(b), line 4: 6.26 → 6.27.
- page 159, Problem 6.7(c), line 1: E^1 − → E^1.
- page 161, first line: sin
( (^3) π 4
→ sin^2
( (^3) π 4
- page 161, Problem 6.9(c), line 4: (0 1 0)
- page 162, Problem 6.10, first line: “orthonornal” → “orthonormal”.
- page 165, Problem 6.14, last line: 3n^2 → 2 n^2.
- page 213, mid-page (line starting with 〈Vee〉): e → e^2.
- page 216, line 3: cancel the second 2 in the numerator.
- page 217, line 6: between
]
and
insert
[ (^) y a θ(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: α^3 /^2 → α^3 /^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−^ ¯hi^ R^ x^0 p(x′)^ dx′
- Be ¯hi^ R^ x^0 p(x′)^ dx′^ ] (x < 0)
√^1 |p(x)|
[
Ce (^1) ¯h^ R^0 x |p(x′)| dx′
- De−^ h^1 ¯^ R^0 x |p(x′)|^ dx′^ ] (x > 0)
... In overlap region 1, Eq. 8.43 becomes ψWKB ≈ 1 ¯h^1 /^2 α^3 /^4 (−x)^1 /^4
[
Ae−i^23 (−αx) 3 / 2
3 / 2 ]
A =
¯hα π
ia + b 2
e−iπ/^4 ; B =
¯hα π
−ia + b 2
eiπ/^4. Putting in the expressions above for a and b :
A =
C
e−iπ/^4 ; B =
C
− iD
eiπ/^4.
...
A =
C
2 +^ iD
e−iπ/^4 =
i 4 e
−γ (^) e−iπ/ (^4) F + ieγ (^) e−iπ/ (^4) F
e−iπ/^4 =
e−γ 4 +^ e
γ
F.
T =
∣∣^ F
A
2
(eγ^ + e− 4 γ )^2
e−^2 γ [1 + (e−^2 γ^ /4)]^2
- page 228, penultimate line, “Limits”, top line: z → z 1 ; bottom line: z → z 2.
- page 229, Problem 8.13, line 4: e−e^ → e−x.
- page 230, Problem 8.14, 3 lines from end: 4 eπ^20 → − (^4) πe^20 ; same at beginning of the next line.
- page 235, Problem 8.16(d), lines 1 and 3: 2 × 10 −^19 → 2 × 10 −^16.
- page 235, Problem 8.17, insert the following at the end: [Actually, to tunnel all the way through the classically forbidden region, the center of mass must not only rise from 0 to x 0 , but also drop back to 0. This doubles γ, and makes the final exponent 1 × 1031 .]
- page 236, Problem 9.1, line 3, last expression: er/^2 a^ → e−r/^2 a.
- page 236, Problem 9.1, last three lines: remove the minus signs in front of all 6 expressions.
- page 237, line 6: Aei(ω^0 +ω)/^2 + Bei(ω^0 −ω)/^2 → Aei(ω^0 +ω)t/^2 + Bei(ω^0 −ω)t/^2.
- page 237, end of Problem 9.2, add the following:
[In light of the Comment you might question the initial conditions. If the perturbation includes a factor θ(t), are we sure this doesn’t alter ca(0) and cb(0)? That is, are we sure ca(t) and cb(t) are continuous at a step function potential? The answer is “yes”, for if we integrate Eq. 9.13 from − to , ca() − ca(−) = − i ¯h H ab′
0
e−iω^0 tcb(t) dt.
But |cb(t)| ≤ 1, so the integral goes to zero as → 0, and hence ca(−) = ca(). The same goes for cb, of course.]
- page 238, 5 lines up from end: cb(t) = − iα ∗ 2 ¯hω →^ cb(t) =^ −^ α∗ 2 ¯hω.
- page 240, Problem 9.4, 3 lines from end: Hba → H ba′.
- page 241, Problem 9.6, penultimate line: ω → ω 0.
- page 242, line 6 (“General solution”), second exponent: +ωr → −ωr.
- page 243, Problem 9.8, line 6: e¯hω/kB^ t^ → e¯hω/kB^ T^.
- page 244, line 5:
[
2 a 3
2 a 3
- page 244, line 8: 1 0 0〉 → |1 0 0〉.
- page 245, line 5: insert “2i¯h” right after “{”.
- page 250, Problem 9.18, line 4:
4 V 0
3 π
4 V 0
3 π
- page 250, Problem 9.20(b), line 2, last term:
B 0 a Brf eiωtb Brf e−iωta −B 0 b
B 0 a + Brf eiωtb Brf e−iωta − B 0 b
l′^ = l + 1
Thus (4) becomes
〈n′(l + 1)(m + 1)|x|nlm〉 =
I
(2l + 3) (l − m)! (l + m + 2)!
(2l + 1) (l − m)! (l + m)!
(2l + 1)(2l + 3)
(l + m + 2)! (l − m)!
=
I
(l + m + 2)(l + m + 1) (2l + 1)(2l + 3)
Now we do the same for m′^ = m − 1:
〈n′l′(m − 1)|x|nlm〉 =
Rn′l′^ (Y (^) lm′ −^1 )∗r sin θ cos φ RnlY (^) lm r^2 dr sin θ dθ dφ
= I
(2l′^ + 1) 4 π
(l′^ − m + 1)! (l′^ + m − 1)!
(2l + 1) 4 π
(l − m)! (l + m)!
∫ (^) π
0
P (^) lm′ −^1 P (^) lm sin^2 θ dθ
∫ (^2) π
0
cos φe−i(m−1)φeimφ^ dφ. (7)
Intφ =
∫ (^2) π
0
(eiφ^ + e−iφ)eiφ^ dφ =
∫ (^2) π
0
(e^2 iφ^ + 1) dφ = π. (8)
Changing variables (x ≡ cos θ), and using Eqs. 9.100 and (2):
Intθ =
− 1
1 − x^2 P (^) lm+1− 1 (x)P (^) lm (x) dx =
(2l + 3)
[∫ 1
− 1
P (^) lm+2P (^) lm dx −
− 1
P (^) lm P (^) lm dx
]
(2l + 1)(2l + 3)
(l + m)! (l − m)! ,
and (7) becomes
〈n′(l + 1)(m − 1)|x|nlm〉 = −
I
(2l + 3) (l − m + 2)! (l + m)!
(2l + 1) (l − m)! (l + m)!
(2l + 1)(2l + 3)
(l + m)! (l − m)!
= −
I
(l − m + 2)(l − m + 1) (2l + 1)(2l + 3)
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
(l + m + 2)(l + m + 1) (2l + 1)(2l + 3)
] 2
[
I
(l + 1)^2 − m^2 (2l + 1)(2l + 3)
] 2
[
I
(l − m + 2)(l − m + 1) (2l + 1)(2l + 3)
] 2
I^2
(l + m + 2)(l + m + 1) + 2[(l + 1)^2 − m^2 ] + (l − m + 2)(l − m + 1) (2l + 1)(2l + 3)
= I^2
(2l^2 + 5l + 3) (2l + 1)(2l + 3)
= I^2
(l + 1) (2l + 1)
Therefore, |℘℘℘|^2 (summed over the three allowed transitions) is e^2 I^2 (l + 1)/(2l + 1), and the spontaneous emission rate (Eq. 9.56) is
Al→l+1 = e
(^2) ω (^3) I 2 3 π 0 ¯hc^3
(l + 1) (2l + 1)
l′^ = l − 1
Return to Eq. (1). This time the integral is
Intθ =
− 1
x P (^) lm− 1 (x)P (^) lm (x) dx =
(2l + 1)
[
(l + m)
− 1
P (^) lm− 1 P (^) lm− 1 dx + (l − m + 1)
− 1
P (^) lm− 1 P (^) lm+1 dx
]
(l + m) (2l + 1)
(2l − 1)
(l − 1 + m)! (l − 1 − m)! =^
(2l − 1)(2l + 1)
(l + m)! (l − m − 1)!.
Therefore
〈n′(l − 1)m|z|nlm〉 =
I
(2l − 1) (l − 1 − m)! (l − 1 + m)!
(2l + 1) (l − m)! (l + m)!
(2l − 1)(2l + 1)
(l + m)! (l − m − 1)!
= I
l^2 − m^2 (2l − 1)(2l + 1)
From 〈n′l′m′|x|nlm〉 with m′^ = m + 1, Eqs. (4) and (5) are unchanged; this time
Intθ =
− 1
1 − x^2 P (^) lm−+1 1 (x)P (^) lm (x) dx =
(2l + 1)
[∫ 1
− 1
P (^) lm−+1 1 P (^) lm+1+1 dx −
− 1
P (^) lm−+1 1 P (^) lm−+1 1 dx
]
(2l − 1)(2l + 1)
(l + m)! (l − m − 2)!
and (4) becomes
〈n′(l − 1)(m + 1)|x|nlm〉 = −
I
(2l − 1) (l − m − 2)! (l + m)!
(2l + 1) (l − m)! (l + m)!
(2l − 1)(2l + 1)
(l + m)! (l − m − 2)!
= −
I
(l − m)(l − m − 1) (2l − 1)(2l + 1)
Now we do the same for m′^ = m − 1. Eqs. (7) and (8) are unchanged, the θ integral is
Intθ =
− 1
1 − x^2 P (^) lm−− 1 1 (x)P (^) lm (x) dx =
(2l − 1)
[∫ 1
− 1
P (^) lm P (^) lm dx −
− 1
P (^) lm− 2 P (^) lm dx
]
(2l − 1)(2l + 1)
(l + m)! (l − m)!
- page 268, Problem 11.1(a), first line: square ˙r.
- page 270, Problem 11.1(c), in the expression for σ: 8 → 16.
- page 271, line beginning “(1) ψ continuous”: sinka → sin ka”.
- page 272, Problem 11.5(a), line 2: B−ikx^ → Be−ikx; line 10: Bika^ → Beika.
- page 272, Problem 11.5(a), line 4: −k′ψ → −(k′)^2 ψ.
- page 273, Problem 11.6, line 3, in the denominator: jl(x) + inl(x) → jl(ka) + inl(ka).
- page 275, line 3: dr → dr 0.
- page 275, Problem 11.10, line 4: cos(κa) = → cos(κa) ≈.
- page 276, Problem 11.12, final box: k → ¯h.
- page 278, line 3: insert plus sign at beginning.
- page 279, line 6: before G insert
- page 280, Problem 11.18, end of line 4: α → α^2.
- page 280, Problem 11.18, 4 lines from end: R =
[ (^) m ¯h^2 k
) 2 ( V 0
k sin(2ka)
] 2
→ R =
[(
m ¯h^2 k
) 2 ( V 0
k sin(2ka)
)]^2
- page 288, Problem A.13, end of line 2: det U = 1 → | det U| = 1.
- page 289-290, Problem A.15: starting with the last line on page 289, we should have
ˆi = −ˆj′; ˆj = ˆi′; kˆ = ˆk′, so (Eq. A.61) S =
. S−^1 =
STxS−^1 =
0 cos θ − sin θ 0 sin θ cos θ
cos θ 0 − sin θ sin θ 0 cos θ
cos θ 0 − sin θ 0 1 0 sin θ 0 cos θ
(^) = Ty (−θ).
STy S−^1 =
cos θ 0 sin θ 0 1 0 − sin θ 0 cos θ
0 − cos θ sin θ 1 0 0 0 sin θ cos θ
0 cos θ − sin θ 0 sin θ cos θ
(^) = Tx(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.
- page 297, Problem A.28(a), first line of part (ii): M^3 = −θ^3 M → M^3 = −θ^2 M.
