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236
CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
Chapter 9
Time-Dependent Perturbation Theory
Problem 9.1
ψnlm = Rnl Ylm .
ψ100 = √
ψ210 = √
1
πa3
From Tables 4.3 and 4.7:
e−r/a ;
1
32πa3
ψ200 = √
r −r/2a
cos θ;
e
a
1
8πa3
1−
r
e−r/2a ;
2a
ψ21±1 = ∓ √
1
64πa3
r r/2a
sin θ e±iφ .
e
a
But r cos θ = z and r sin θe±iφ = r sin θ(cos φ ± i sin φ) = r sin θ cos φ ± ir sin θ sin φ = x ± iy. So |ψ|2 is an
even function of z in all cases, and hence z|ψ|2 dx dy dz = 0, so Hii = 0. Moreover, ψ100 is even in z, and
so are ψ200 , ψ211 , and ψ21−1 , so Hij = 0 for all except
H100,210 = −eE √
eE
=− √
4 2πa4
or
∞
1
πa3
√
1
32πa3
1
a
π
r4 e−3r/2a dr
0
eE
e−r/a e−r/2a z 2 d3 r = − √
4 2πa4
2π
cos2 θ sin θ dθ
0
0
e−3r/2a r2 cos2 θ r2 sin θ dr dθ dφ
eE
dφ = − √
4!
4 2πa4
2a
3
5
2
2π = −
3
28
√
35 2
eEa,
−0.7449 eEa.
Problem 9.2
i
c˙a = − Hab e−iω0 t cb ;
�
i
c˙b = − Hba eiω0 t ca .
�
Differentiating with respect to t :
i
i
i
i
c¨b = − Hba iω0 eiω0 t ca + eiω0 t c˙a = iω0 − Hba eiω0 t ca − Hba eiωo t − Hab e−iω0 t cb , or
�
�
�
�
c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.
237
CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
c¨b = iω0 c˙b −
1
|H |2 cb .
�2 ab
Let α2 ≡
1
|H |2 .
�2 ab
Then c¨b − iω0 c˙b + α2 cb = 0.
This is a linear differential equation with constant coefficients, so it can be solved by a function of the form
cb = eλt :
λ2 − iω0 λ + α2 = 0 =⇒ λ =
1
iω0 ±
2
−ω02 − 4α2 =
i
(ω0 ± ω) , where ω ≡
2
ω02 + 4α2 .
The general solution is therefore
cb (t) = Aei(ω0 +ω)/2 + Bei(ω0 −ω)/2 = eiω0 t/2 Aeiωt/2 + Be−iωt/2 , or
cb (t) = eiω0 t/2 [C cos (ωt/2) + D sin (ωt/2)] .
But cb (0) = 0,
so C = 0,
and hence
cb (t) = Deiω0 t/2 sin (ωt/2) . Then
c˙b = D
ca =
ω0
i
iω0 iω0 t/2
ω
ω
e
sin (ωt/2) + eiω0 t/2 cos (ωt/2) = Deiω0 t/2 cos (ωt/2) + i sin (ωt/2) = − Hba eiω0 t ca .
2
2
2
ω
�
i� ω −iω0 t/2
ω0
D cos (ωt/2) + i sin (ωt/2) .
e
Hba 2
ω
ω0
ca (t) = e−iω0 t/2 cos (ωt/2) + i sin (ωt/2) ,
ω
2Hba iω0 t/2
cb (t) =
e
sin (ωt/2) ,
i�ω
But ca (0) = 1,
where ω ≡
ω02 + 4
so
i� ω
D = 1.
Hba 2
Conclusion:
|Hab |2
�2 .
ω02
4|Hab |2
2
sin
(ωt/2)
+
sin2 (ωt/2)
ω2
�2 ω 2
1
|H |2
= cos2 (ωt/2) + 2 ω02 + 4 ab
sin2 (ωt/2) = cos2 (ωt/2) + sin2 (ωt/2) = 1.
ω
�2
|ca |2 + |cb |2 = cos2 (ωt/2) +
Problem 9.3
This is a tricky problem, and I thank Prof. Onuttom Narayan for showing me the correct solution. The safest
approach is to represent the delta function as a sequence of rectangles:
δ (t) =
Then Eq. 9.13 ⇒

t<− :








t > :
(1/2 ), − < t < ,
0,
otherwise.
ca (t) = 1, cb (t) = 0,
ca (t) = a, cb (t) = b,



 c˙a = − 2iα� e−iω0 t cb ,




− <t< :


∗


iω0 t
ca .
c˙b = − iα
2 �e
c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.
240
CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
Similarly,
i
d˙b = e �
t
0
Hbb (t ) dt
i i
= − e�
�
i i
= − e�
�
t
0
t
0
i
i
H
cb + e �
� bb
Hbb (t ) dt
ca Hba eiω0 t .
t
0
Hbb (t ) dt
c˙b .
i
But ca = e− �
But c˙b = −
t
0
Haa (t ) dt
i
cb Hbb + ca Hba eiω0 t .
�
da .
[Hbb (t )−Haa (t )]dt H eiω0 t d = − i e−iφ H eiω0 t d .
a
a
ba
ba
�
QED
(c)
Initial conditions:
Zeroth order:
ca (0) = 1 =⇒ da (0) = 1;
da (t) = 1,
cb (0) = 0 =⇒ db (0) = 0.
db (t) = 0.
i
d˙a = 0 =⇒ da (t) = 1 =⇒ ca (t) = e− �
First order:
i
i
d˙b = − e−iφ Hba eiω0 t =⇒ db = −
�
�
i i
cb (t) = − e− �
�
t
0
t
Hbb (t )dt
0
t
t
0
Haa (t ) dt
.
e−iφ(t ) Hba (t )eiω0 t dt =⇒
0
e−iφ(t ) Hba (t )eiω0 t dt .
These don’t look much like the results in (a), but remember, we’re only working to first order in H ,
t
so ca (t) ≈ 1 − �i 0 Haa (t ) dt (to this order), while for cb , the factor Hba in the integral means it is
already first order and hence both the exponential factor in front and e−iφ should be replaced by 1. Then
t
cb (t) ≈ − �i 0 Hba (t )eiω0 t dt , and we recover the results in (a).
Problem 9.5
Zeroth order:
First order:
c(0)
a (t) = a,

i
ib


 c˙a = − Hab e−iω0 t b =⇒ c(1)
a (t) = a −
�
�
i
ia

(1)
iω0 t

 c˙b = − Hba e
a =⇒ cb (t) = b −
�
�
t
i
ia
c˙a = − Hab e−iω0 t b −
�
�
Second order:
c(2)
a (t) = a −
(0)
cb (t) = b.
ib
�
t
0
Hab (t )e−iω0 t dt −
0
t
a
�2
0
t
0
t
0
Hab (t )e−iω0 t dt .
Hba (t )eiω0 t dt .
Hba (t )eiω0 t dt
=⇒
t
Hab (t )e−iω0 t
0
Hba (t )eiω0 t dt
dt .
To get cb , just switch a ↔ b (which entails also changing the sign of ω0 ):
(2)
cb (t) = b −
ia
�
t
0
Hba (t )eiω0 t dt −
b
�2
t
0
Hba (t )eiω0 t
t
0
Hab (t )e−iω0 t dt
dt .
c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.
241
CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
Problem 9.6
i
(2)
(1)
For H independent of t, Eq. 9.17 =⇒ cb (t) = cb (t) = − Hba
�
(2)
cb (t)
eiω0 t
i
= − Hba
�
iω0
c(2)
a (t) = 1 −
1
|H |2
�2 ab
t
0
= −
t
t
e−iω0 t
eiω0 t dt
0
0
i
= 1+
|H |2
ω0 �2 ab
Hba iω0 t
−1 .
e
�ω0
e−iω0 t
t +
iω0
t
eiω0 t dt =⇒
0
Meanwhile Eq. 9.18 =⇒
dt = 1 −
t
= 1+
0
1
1
|H |2
�2 ab iω0
t
0
1 − e−iω0 t
dt
1
i
|H |2 t +
e−iω0 t − 1 .
ω0 �2 ab
iω0
For comparison with the exact answers (Problem 9.2), note first that cb (t) is already first order (because of
the Hba in front), whereas ω differs from ω0 only in second order, so it suffices to replace ω → ω0 in the exact
formula to get the second-order result:
cb (t) ≈
2Hba iω0 t/2
2Hba iω0 t/2 1
H
e
sin (ω0 t/2) =
e
eiω0 t/2 − e−iω0 t/2 = − ba eiω0 t − 1 ,
i�ω0
i�ω0
2i
�ω0
in agreement with the result above. Checking ca is more difficult. Note that
ω = ω0
1+
4|Hab |2
|Hab |2
1
+
2
≈
ω
0
2
ω 0 �2
ω02 �2
= ω0 + 2
Taylor expansion:



 cos(x + ) = cos x − sin x =⇒ cos (ωt/2) = cos


 sin(x + ) = sin x + cos x =⇒ sin (ωt/2) = sin
|Hab |2
;
ω 0 �2
ω0
|H |2
≈ 1 − 2 2ab 2 .
ω
ω0 �
ω0 t |Hab |2 t
+
2
ω 0 �2
ω0 t |Hab |2 t
+
2
ω 0 �2
|Hab |2 t
sin (ω0 t/2)
ω 0 �2
2
|H | t
≈ sin (ω0 t/2) + ab 2 cos (ω0 t/2)
ω0 �
≈ cos (ω0 t/2) −
ω0 t
ω0 t
ω0 t
ω0 t
|H |2 t
|H |2
|H |2 t
sin
− ab 2 sin
+ i 1 − 2 2ab 2
+ ab 2 cos
2
ω0 �
2
ω0 �
2
ω0 �
2
2
ω0 t
ω0 t
ω0 t
ω0 t
|H |
ω0 t
2i
= e−iω0 t/2 cos
sin
+ i sin
− ab2 t sin
− i cos
+
2
2
ω0 �
2
2
ω0
2
2
|H |
2i 1
= e−iω0 t/2 eiω0 t/2 − ab2 −iteiω0 t/2 +
eiω0 t/2 − e−iω0 t/2
ω0 �
ω 2i
|H |2
1
1
i
= 1 − ab2 −it +
|H |2 t +
1 − e−iω0 t = 1 +
e−iω0 t − 1 , as above.
ω0 �
ω0
ω0 �2 ab
iω0
ca (t) ≈ e−iω0 t/2 cos
Problem 9.7
(a)
c˙a = −
i
Vab eiωt e−iω0 t cb ;
2�
c˙b = −
i
Vba e−iωt eiω0 t ca .
2�
c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.
242
CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
Differentiate the latter, and substitute in the former:
Vba
i(ω0 − ω)ei(ω0 −ω)t ca + ei(ω0 −ω)t c˙a
2�
Vba i(ω0 −ω)t
Vab −i(ω0 −ω)t
Vba i(ω0 −ω)t
|Vab |2
= i(ω0 − ω) −i
e
e
e
−i
ca − i
cb = i(ω0 − ω)c˙b −
cb .
2�
2�
2�
(2�)2
c¨b = −i
d2 cb
dcb
|Vab |2
+ i(ω − ω0 )
cb = 0. Solution is of the form cb = eλt :
+
2
dt
dt
4�2
λ=
1
−i(ω − ω0 ) ±
2
−(ω − ω0 )2 −
i −
General solution: cb (t) = Ae
|Vab |2
�2
(ω−ω0 )
+ωr
2
=i −
t
|Vab |2
= 0.
4�2
(ω − ω0 )
± ωr , with ωr defined in Eq. 9.30.
2
i −
+ Be
λ2 + i(ω − ω0 )λ +
(ω−ω0 )
+ωr
2
t
= e−i(ω−ω0 )t/2 Aeiωr t + Be−iωr t ,
or, more conveniently: cb (t) = e−i(ω−ω0 )t/2 [C cos(ωr t) + D sin(ωr t)] .
But cb (0) = 0, so C = 0 :
ω0 − ω
ei(ω0 −ω)t/2 sin(ωr t) + ωr ei(ω0 −ω)t/2 cos(ωr t) ;
2
2� i(ω−ω0 )t
2� i(ω−ω0 )t/2
ω0 − ω
ca (t) = i
e
c˙b = i
e
D i
sin(ωr t) + ωr cos(ωr t) . But ca (0) = 1 :
Vba
Vba
2
2�
−iVba
1=i
Dωr , or D =
.
Vba
2�ωr
cb (t) = Dei(ω0 −ω)t/2 sin(ωr t).
cb (t) = −
c˙b = D i
i
Vba ei(ω0 −ω)t/2 sin(ωr t),
2�ωr
ca (t) = ei(ω−ω0 )t/2 cos(ωr t) + i
ω0 − ω
2ωr
sin(ωr t) .
(b)
Pa→b (t) = |cb (t)|2 =
and the denominator,
|Vab |
2�ωr
2
sin2 (ωr t).
The largest this gets (when sin2 = 1) is
|Vab |2 /�2
,
4ωr2
4ωr2 = (ω − ω0 )2 + |Vab |2 /�2 , exceeds the numerator, so P ≤ 1 (and 1 only if ω = ω0 ).
2
2
ω0 − ω
|Vab |
sin2 (ωr t) +
sin2 (ωr t)
2ωr
2�ωr
(ω − ω0 )2 + (|Vab |/�)2
= cos2 (ωr t) +
sin2 (ωr t) = cos2 (ωr t) + sin2 (ωr t) = 1.
4ωr2
|ca |2 + |cb |2 = cos2 (ωr t) +
(c) If
|Vab |2
Eq. 9.28.
�2 (ω − ω0 )2 ,
then ωr ≈
1
|ω − ω0 |,
2
and Pa→b ≈
0
|Vab |2 sin2 ω−ω
2 t
,
2
�
(ω − ω0 )2
confirming
(d) ωr t = π =⇒ t = π/ωr .
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.
243
CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
Problem 9.8
Spontaneous emission rate (Eq. 9.56): A =
R=
π
|℘|2 ρ(ω),
3 0 �2
with
ρ(ω) =
ω 3 |℘|2
.
3π 0 �c3
Thermally stimulated emission rate (Eq. 9.47):
ω3
�
π 2 c3 (e�ω/kB T − 1)
(Eq. 9.52).
So the ratio is
A
ω 3 |℘|2 3 0 �2 π 2 c3 e�ω/kB T − 1
=
·
·
= e�ω/kB T − 1.
R
3π 0 �c3 π|℘|2
�ω 3
The ratio is a monotonically increasing function of ω, and is 1 when
e�ω/kb t = 2,
or
�ω
= ln 2,
kB T
ν=
ω=
kB T
ln 2,
�
or
ν=
ω
kB T
=
ln 2. For T = 300 K,
2π
h
(1.38 × 10−23 J/K)(300 K)
ln 2 = 4.35 × 1012 Hz.
(6.63 × 10−34 J · s)
For higher frequencies, (including light, at 1014 Hz), spontaneous emission dominates.
Problem 9.9
(a) Simply remove the factor e�ω/kB T − 1 in the denominator of Eq. 5.113:
ρ0 (ω) =
�ω 3
.
π 2 c3
(b) Plug this into Eq. 9.47:
Rb→a =
�ω 3
ω 3 |℘|2
π
|℘|2 2 3 =
,
2
3 0�
π c
3π 0 �c3
reproducing Eq. 9.56.
Problem 9.10
N (t) = e−t/τ N (0) (Eqs. 9.58 and 9.59).
so t/τ = ln 2, or t1/2 = τ ln 2.
After one half-life, N (t) = 12 N (0), so
1
2
= e−t/τ ,
or
2 = et/τ ,
Problem 9.11
28
√ a. As
35 2
for x and y, we noted that |1 0 0 , |2 0 0 , and |2 1 0 are even (in x, y), whereas |2 1 ± 1 is odd. So the only
In Problem 9.1 we calculated the matrix elements of z; all of them are zero except 1 0 0|z|2 1 0 =
c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.
244
CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
non-zero matrix elements are 1 0 0|x|2 1 ± 1 and 1 0 0|y|2 1 ± 1 . Using the wave functions in Problem 9.1:
1 0 0|x|2 1 ± 1 = √
=∓
=
1 0 0|y|2 1 ± 1 =
=
1
∞
1
8πa4
A=−
=
π
r4 e−3r/2a dr
2a
3
5
∓1
2a
4!
4
8πa
3
5
2a
3
5
∓1
4!
8πa4
4
3
4
3
4
3
0
(π) = ∓
2π
0
(cos φ ± i sin φ) sin φ dφ
(±iπ) = −i
√
27 2 ˆ
1 0 0|r|2 1 0 =
a k;
35
E2 − E 1
1
=
�
�
and |℘|2 = (qa)2
E1
− E1
4
13.6
0.511 × 106
2
(cos φ ± i sin φ) cos φ dφ
27
a.
35
=−
27
a.
35
1 0 0|r|2 1 ± 1 =
27
a ∓ˆi − i ˆj , and hence
35
215
(for |2 1 0 → 1 0 0 and |2 1 ± 1 → |1 0 0 ).
310
3E1
,
4�
33 E13 (ea)2 215
1
29 E13 e2 a2
210
=
−
=
26 �3
310 3π 0 �c3
38 π 0 �4 c3
38
210
38
2π
sin3 θ dθ
0
℘2 = 0 (for |2 0 0 → |1 0 0 ),
ω=
e−r/a re−r/2a sin θ e±iφ (r sin θ cos φ)r2 sin θ dr dθ dφ
0
∓1
4!
8πa4
1 0 0|r|2 0 0 = 0;
Meanwhile,
1
a
∓1
√
8 πa3
πa3
so for the three l = 1 states:
E1
mc2
(3.00 × 108 m/s)
= 6.27 × 108 /s;
(0.529 × 10−10 m)
for the three l = 1 states (all have the same lifetime); τ = ∞
2
c
a
τ=
1
= 1.60 × 10−9 s
A
for the l = 0 state.
Problem 9.12
[L2 , z] = [L2x , z] + [L2y , z] + [L2z , z] = Lx [Lx , z] + [Lx , z]Lx + Ly [Ly , z] + [Ly , z]Ly + Lz [Lz , z] + [Lz , z]Lz

 [Lx , z] = [ypz − zpy , z] = [ypz , z] − [zpy , z] = y[pz , z] = −i�y,
[Ly , z] = [zpx − xpz , z] = [zpx , z] − [xpz , z] = −x[pz , z] = i�x,
But

[Lz , z] = [xpy − ypx , z] = [xpy , z] − [ypx , z] = 0.
So:
But
[L2 , z] = Lx (−i�y) + (−i�y)Lx + Ly (i�x) + (i�x)Ly = i�(−Lx y − yLx + Ly x + xLy ).
Lx y = Lx y − yLx + yLx = [Lx , y] + yLx = i�z + yLx ,
Ly x = Ly x − xLy + xLy = [Ly , x] + xLy = −i�z + xLy .
c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.
245
CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
So: [L2 , z] = i�(2xLy − i�z − 2yLx − i�z) =⇒ [L2 , z] = 2i�(xLy − yLx − i�z).
L2 , [L2 , z] = 2i� [L2 , xLy ] − [L2 , yLx ] − i�[L2 , z]
= 2i� [L2 , x]Ly + x[L2 , Ly ] − [L2 , y]Lx − y[L2 , Lx ] − i�(L2 z − zL2 ) .
But
[L2 , Ly ] = [L2 , Lx ] = 0
(Eq. 4.102),
so
, or
L2 , [L2 , z] = 2i� (yLz − zLy − i�x) Ly − 2i� (zLx − xLz − i�y) Lx − i� L2 z − zL2
L2 , [L2 , z] = −2�2 2yLz Ly
−2zL2y − 2zL2x
−2i�xLy + 2xLz Lx + 2i�yLx − L2 z + zL2
−2z(L2x +L2y +L2z )+2zL2z
= −2�
2
2yLz Ly − 2i�xLy + 2xLz Lx + 2i�yLx + 2zL2z − 2zL2 − L2 z + zL2
= −2�2 zL2 + L2 z − 4�2 (yLz − i�x) Ly + (xLz + i�y) Lx + zLz Lz
Lz y
Lz x
= 2�2 zL2 + L2 z − 4�2 (Lz yLy + Lz xLx + Lz zLz ) = 2�2 (zL2 + L2 z).
QED
Lz (r·L)=0
Problem 9.13
1
|n 0 0 = Rn0 (r)Y00 (θ, φ) = √ Rn0 (r),
4π
so
n 0 0|r|n 0 0 =
1
4π
ˆ dx dy dz.
Rn 0 (r)Rn0 (r)(x ˆi + y ˆj + z k)
But the integrand is odd in x, y, or z, so the integral is zero.
Problem 9.14
(a)


 |2 1 0 
|2 1 1
|3 0 0 →
→ |1 0 0 .


|2 1−1
(|3 0 0 → |2 0 0 and |3 0 0 → |1 0 0 violate ∆l = ±1 rule.)
(b)
From Eq. 9.72:
ˆ
2 1 0|r|3 0 0 = 2 1 0|z|3 0 0 k.
From Eq. 9.69:
2 1 ± 1|r|3 0 0 = 2 1 ± 1|x|3 0 0 ˆi + 2 1 ± 1|y|3 0 0 ˆj.
From Eq. 9.70:
± 2 1 ± 1|x|3 0 0 = i 2 1 ± 1|y|3 0 0 .
Thus
| 2 1 0|r|3 0 0 |2 = | 2 1 0|z|3 0 0 |2
and | 2 1 ± 1|r|3 0 0 |2 = 2| 2 1 ± 1|x|3 0 0 |2 ,
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