Errata for Quantum Mechanics
by Ernest Abers
August 1, 2004
Preface
• Page xv: Add a sentence at the end of the next-to-last paragraph:
Rudaz (University of Minnesota), and several anonymous reviewers. My special
thanks to Kenneth Lane and his students at Boston University.
I thank the Department of Physics and Astronomy at UCLA for granting me
Chapter I
• Page 14, Equation (1.78): The last expression needs an in front:
X
X
X
G=
pm,i δrm,i = ijk pm,i nj rm,k = (rm × pm ) · n̂ = L · n̂
m,i
(1.78)
m
ijk
Chapter II
• Page 22, Equation (2.2): In the leftmost expression, change ~ to h:
h
2π~
=λ= √
≈ 1.2 Å
p
2mE
• Page 25, Equation (2.6a): Change the last α to β:
h
i
α + β |ψi = α|ψi + β|ψi
(2.2)
(2.6a)
Chapter III
• Page 75, Equation (3.80): Put an ~ in front of the left-hand side:
X
~[ri , Lj ] = i~
ijk rk
(3.80)
k
• Page 84, Equation (3.137): The last term should be Ze2 /r:
H=
p2
Ze2
−
2m
r
(3.137)
• Page 90, Equation (3.172). Change m to M .
n−l−1 =M
(3.172)
2
Quantum Mechanics
Chapter IV
• Page 122, just above Equation (4.105), insert “solutions” between “the” and “to”:
The eigenstates are linear combinations of |1, −1/2i and |0, 1/2i. The eigenvalues are
the solutions to the characteristic equation:
• Page 126, third paragraph, third line: change 1/2f to 1/2:
Unless l = 0, there are two states with m = l − 1/2. One linear combination
will be in the j = l + 1/2 ladder. The remaining one must be the top of a new
ladder, this time with j = l − 1/2. The number of states in these two ladders adds
• Page 131, Problem 4.2, part (b): Change the superscript j to 1:
X
†
Uam D(1) (J¯i )mn Unb
= J¯i ab
mn
• Page 132, Problem 4.4, part (c). The last equation on the page should be changed to
m±
ψ(r) = R±
(θ, φ)
El (r)Yl
Chapter V
• Page 140, line below Equation (5.6): Change g(a) to g():
For a one-parameter subgroup of continuous symmetries g() = exp(−iG),
• Page 161, Equation (5.123): Divide the exponent by ~:
U (tf , ti) = e−iH(tf −ti )/~
(5.123)
Chapter VII
• Page 203, Equation (7.8): In the first line change Eon to Eno (twice):
(H − Eno + Eno − Ho ) |ψn i = H 0|ψni
(Ho − Eno ) |ψn i = (∆n − H 0) |ψni
(7.8)
• Page 205, second bulleted paragraph, seventh line: Change “so” to “but”:
Two states in one of these ladders will have the same unperturbed energy, but we
• Page 214, Equations (7.69) to (7.70): Change the subscript “3” to “D”:
HD = −
1
∇2V (r)
8m2
ED =
α4 m
δl0
2n3
(7.69)
(7.70)
ERRATA
• Page 226. Equation (7.133): Delete the subscript “i” (four times):
2
Z
p
σe2
ψ100(r)
−
ψ100(r) d3r = E1o (σ) = σ2 E1o
2m
r
3
(7.133)
Chapter VIII
• Page 270, Equation (8.40): change the left-hand side to f (1) (Θ):
Z
1
(1)
f (Θ) = −
eiq·rU (r) d3r
4π
Z 1
Z
Z
1 ∞ 2
1 ∞
iqr cos θ
=
r dr
e
U (r)d cos θ = −
r sin(qr)U (r)dr
2 0
q 0
−1
• Page 270, Equation (8.39), replace r by r0 inside the integral.
Z
0
0
1
(1)
f (θ, φ) = −
ei(k−k )·r U (r0) d3r0
4π
(8.40)
(8.39)
• Page 273, Equation (8.56): In the second line, second term inside the large parentheses,
move the asterisk from χ to φ. In the fourth line, first expression, second term inside
the parentheses, change plus to minus:
Z kσtot
∗ ∂χ
∗ ∂φ
=
−Im
φ
(
(
)
+
χ
)
r2dΩ
(2π)3
∂r
∂r
Z ∂χ
∂φ∗
= −Im
φ∗( ) − χ(
) r2dΩ
∂r
∂r
Z ∂ψ
∂φ∗
= −Im
φ∗ (
) − ψ(
) r2dΩ
∂r
∂r
Z
Z
= −Im (φ∗ ∇ψ − ψ∇φ∗ ) · n̂dS = −Im
φ∗ ∇2 ψ − ψ∇2φ∗ d3r
Z
1
= −Im φ∗ (r)U (r)ψ(r)d3 r =
Im f(0)
(8.56)
2π2
• Page 277, Equation (8.80), in the first line, in the numerator of first factor, change
2l + l to 2l + 1:
Z
2l l!
2l + 1 X n ρn 2n(n!)2 1
Al ρl
i
Pn(z)Pl (z)dz
=
(2l + 1)!
2
n! (2n)! −1
n
(8.80)
l l
2
l ρ 2 (l!)
=i
l! (2l)!
• Page 280, Equation (8.97): Change the last r to r0. Move the asterisk to after the
second Ylm factor.
4
Quantum Mechanics
• Page 280, Equation (8.98) Change the arguments of the first Ylm to (θ0 , φ0).
• Page 280, Equation (8.99) second line, change n̂ to θ0 , φ0, and change n̂z to 0, φ:
4π X l m
φk0 (r0 ) =
i Yl (θ, φ)Ylm (θ0 , φ0)∗ jl (kr0)
(8.97)
3
(2π) 2 l,m
and
ψk (r0 ) =
4π X l m 0 0 ∗ m
i Yl (θ , φ ) Yl (0, φ)Rl (r0 )
3
(2π) 2 l,m
(8.98)
Therefore the scattering amplitude is


Z X
f(θ, φ) = −4π  (−i)l Ylm (θ, φ)∗ Ylm (θ0 , φ0)jl (kr0 )
l,m
0
× U (r )
X
i
l0
Ylo0 (θ0 , φ0)Rl0 (r0)Ylo0 (0, φ)
!
r02dr0dΩ0
(8.99)
l0
Z
=−
∞
0
"
X
0
0
0
#
(2l + 1)jl (kr )U (r )Rl (r )Pl (cos θ) r02dr0
l
• Page 280, Equation (8.100) On the left, k should be the denominator of all the rest,
not just δl :
Z ∞
eiδl sin δl
jl (kr)U (r)Rl (r)r2 dr
(8.100)
=−
k
0
• page 280, last sentence, replace “in that number” by “in that limit”.
• Page 284, Problem 8.1, in the line just following the unnumbered displayed equation,
on the right-hand-side of the in-line equation, remove the middle vertical line:
where ρ(r) = ψ100(r)2 , the probability distribution of the bound electron.
• Page 287, Reference [3], Replace the comma after K by a period:
[3] K. Gottfried,
Appendix D.
Quantum Mechanics, Volume I, Benjamin, 1966.
See also
Chapter IX
• Page 288, in the subsubsection heading just below the 9.1.2 subsection heading, change
Transistions to Transitions.
• Page 289, Equation (9.10), second line, 4m2 in denominator should be just m2 .
i(ω −ω)t
2
ba
− 1 2 e
Pb (t) ≈ |Vba | ωba − ω (9.10)
2
sin(|ω
|
−
ω)t/2
α 2 2
ba
= 2 Ao ψb p · ε̂e−ik·r ψa m
|ωba| − ω
ERRATA
5
• Page 290, Equation (9.16), replace 4m2 by m2 and 8π2 by 4π2.
P (t) −→
t→∞
α 4π2
ψb p · ε̂e−ik·r ψn 2 t
ρ(ω
)
ba
2
m2 ωba
(9.16)
• Page 291, Equation (9.17), replace 4m2 by m2 and 8π2 by 4π2.
Γb =
2
α 4π2
p · ε̂e−ik·r ψn ρ(ω
)
ψ
ba
b
2
m2 ωba
(9.17)
• Page 291, in the line just above Equation (9.21) change θ to cos2 θ. In Equation (9.21)
change cos θ to cos2 θ:
where θ is the angle between the constant vector r and the polarization. If the radiation
is unpolarized and isotropic, just replace cos2 θ by its average value:
Z
1
1
cos2 θ →
cos2 θ dΩ =
(9.21)
4π
3
• Page 292, change the text between Equations (9.25) and (9.26) to read: “Then aba(ω)
is related to Aba (ω) by”
• Page 294: There is no Equation (3.33). Latex just skipped an equation number! I
don’t really understand how this could have happened, but it is under control.
• Page 294, in the text just below Equation (9.30), change “has an inverse” to “exists”:
Even though the operator H − ω has no inverse for positive real ω, G(ω) exists for
real ω and any finite real .
• Page 295, in the sentence above equation (9.42), change the first (9.40) to (9.39):
From equations (9.39) and (9.40) it also follows that
• Page 301, just before the “Optical Theorem” heading, add “The relativistically correct
forms of Equations (9.71) and (9.72) are obtained replacing m2 by ωa2 .”
• Page 304, Equation (9.90): Change the left hand side to hφb| T |φa i:
hφb | T |φa i = δ3 (K0 − K) φb T̄ φa
(9.90)
• Page 304, Equation (9.93): The integrals should end in d3k10 d3k20 and d3K 0 d3k0:
Z
2
Γ̄ = 2π |Tba | δ (ωa − ωb )d3k10 d3k20
Z
2
= 2π δ3 (K0 − K)T̄ba δ (ωa − ωb )d3K 0 d3k0
(9.93)
Z
2
V
≈ 2π T̄ba δ (ωa − ωb)δ3 (K0 − K)
d3K 0 d3k0
(2π)3
6
Quantum Mechanics
• Pages 304-305, In equations (9.90) through (9.95) change T 0 to T everywhere.
• Page 306, Equation (9.103). In the first line exchange r1 and r2 in the exponent. In
the second line change the sign of the exponent:
Z
0
0
1 1
0
Hba,2 = ±
V (|r1 − r2 |)ei(k2−k1 )·r2 ei(k1 −k2 )·r1 d3r1 d3r2
2 (2π)6
Z
(9.103)
1 1
i(k+k0 )·r
0
3
V
(r)e
=±
δ
(K
−
K
)
d
r
3
2 (2π)3
• Page 307: In equation (9.104) delete the delta function δ3 (K − K0 ) in the first line.
Change T to T̄ everywhere in (9.104) and (9.105). And replace p by k in these and
the next two equations:
Z
0
1
T¯direct (ωa , θ, φ) =
V (r)ei(k−k )·r d3r
3
(2π)
(9.104)
e2
e2
α
=
=
=
2π2q2
4π2 k2(1 − cos θ)
8π2 k2 sin2 (θ/2)
T̄exchange (ωa , θ, φ) = T̄direct (ωa , π − θ, φ + 2π)
=
e2
16π2k2 cos2 (θ/2)
=
α
(9.105)
8π2 k2 cos2 (θ/2)
2
2
dσ
1
1
α2 α2µ2 1
1
4 2
=
= (2π) µ
±
±
2
2
dΩ
64π4 k4 sin (θ/2) cos2 (θ/2)
4k4 sin (θ/2) cos2 (θ/2) 1
α 2 µ2
1
1
1
=
+
±
2
4k4 cos4 (θ/2) sin4(θ/2)
sin2 (θ/2) cos2(θ/2)
(9.106)
dσ
1
α 2 µ2
1
1
1
=
−
(9.107)
+
dΩ unpolarized
4k4 sin4 (θ/2) cos4(θ/2) sin2 (θ/2) cos2 (θ/2)
• Page 312, Equation (9.132): On the left, change ω0 to ωa0 . On the right, change the
denominator to ωb − ωa . In Equation (9.133) replace ω by ωb. In the text below
Equation (9.134) delete “with ωa0 = Re ω0 above”:
0
ωa0 = ωa + Haa
+ Re
X
b6=a
−π
X
0 2
|Hba
|
1
+···
ωb − ωa
0 2
|Hba
| δ (ωb − ωa) = −Γ/2
(9.132)
(9.133)
b6=a
G(ω)aa =
1
ω − ωa0 + iΓ/2
Evidently T (ω) has a singularity in the lower half plane,
(9.134)
ERRATA
7
• Page 314 just below Equation (9.142) change “closer to” to “farther from”:
which is farther from unity than the same probability
0
0
• Page 315, Equation (9.148) change absHba
to |H|ba :
0
0
d
0 2 2(ωa − ωb ) sin(ωa − ωb )t
|
|aba(t)|2 → |Hba
0
2
dt
(ωa − ωb) + Γ2/4
(9.148)
Chapter X
• Page 340, Figure 10.1. The arcs should form a circle, whose center is at the base of
the arrow. [Some glitch is a graphics program printed some of the arcs in the wrong
place.]
• page 355, Reference [14], change Tycho to Tycko, here and in the index.
[14] R. Tycko, Phys. Rev. Letters 58 (1987) 2281.
Chapter XI
• Page 365, Equations (11.57) and (11.57): In both equations change 4pi to 4π.
Z
∂A
1
(4π)1/2
E(r, t) = −
i √ ω a(k, t) − a† (−k, t) eik·r d3k
(11.57)
=
∂t
(2π)3/2
2ω
and
B(r, t) = ∇ × A = −
(4π)1/2
i
(2π)3/2
Z
1
√ ωn̂ × a(−k, t) + a† (k, t) e−ik·r d3k
2ω
(11.58)
• Page 377, Equation (11.119). Replace 25 by 26, and therefore replace 9.49 × 10−31 by
1.87 × 10−30 and replace 1.44 × 10−15 by 2.84 × 10−15. The the next line needs to read
“The lifetime is 3.53 × 1014 seconds, over 100,000 centuries....”. Also, delete Equation
(11.117e) – there is nothing there:
The J = 0 state is
1 1 1 1
, − − − ,
2
2
2 2
√
2
Γ=
M =0
26 13 3
α gp (m/M )3 m = 1.87 × 10−30eV = 2.84 × 10−15 sec−1
34
(11.117d)
(11.109)
The lifetime is 3.53 × 1014 seconds, over 100,000 centuries. It should be clear
• Page 403 Problem 11.5: Delete the sentence “What is the width of these components
of the Lyman-β line?”
Compute the rate for the spontaneous decay of a 3P state to the 1S state. (Here
you can ignore spin—these are electric transitions.)
8
Quantum Mechanics
Chapter XII
• Page 409, Equation (12.14c): Change x to z:
z0 = z
(12.14c)
• Page 410, Equations (12.20): The exponents should be boldface:
¯
R̄ = e−iθn̂·J
(12.20a)
B̄ = e−iωn̂·K̄
(12.20b)
• Page 416, Equation (12.67): Change the sign of the last term inside the large parentheses from minus to plus:
1 d2
l(l + 1) − α2
ωα ω2 − m2
−
+
+
u(r) = 0
(12.67)
2m dr2
2mr2
mr
2m
Chapter XIII
• Page 439, Equations (13.9) through (13.11): Change pi to π four times:
Z
1
Ψ(r) =
eip·ra(p)d3p
(2π)3/2
(13.9)
Then
Ψ† (r)|i =
1
(2π)3/2
Z
e−ip·ra† (p)|i d3p =
1
(2π)3/2
Z
e−ip·r|pi d3 p
This is a one-particle state whose wave function is
Z
0 † 1
0
ψ(r ) = r Ψ (r) =
e−ip·r hr0| pi d3p = δ3 (r − r0)
(2π)3/2
(13.10)
(13.11)