,Solutions to Exercises in Modern Condensed Matter Physics
S. M. Girvin and Kun Yang
c 2019
[Compiled: August 12, 2019]
,Note to Instructors
For a few of the more difficult problems, we include notes to the instructor suggesting simplifications,
specializations and hints that the instructor may wish to give the students when assigning those
problems.
Note also that there are some useful exercises within the appendices of the textbook.
1
, Chapter 2
Ex. 2.1
(i)
The radiated electric field is
eikRD h i
~ in ) e−iωt e−i~q·~r .
~a ≈ re n̂ × (n̂ × E (1)
RD
~ in ) is perpendicular to both n̂ and to E
Replace e−i~q·~r by he−i~q·~r i = f (~q ). The vector (n̂ × E ~ in and
~ 2 2 2
has length |Ein sin θ|. Hence |n̂ × (n̂ × Ein )| = Ein sin θ. Thus
re2 2
|a |2 = Ein
2
q )|2 .
2 sin θ|f (~ (2)
RD
The total radiated power passing through a sphere of radius RD is
Z 2
2 a
P = cRD dΩ ×2 (3)
8π
Z
c 2 2
= r E dΩ sin2 θ|f (~q )|2 . (4)
4π e in
Let us normalize the incident electric field to that associated with a single photon in the normal-
ization volume L3
2
Ein ~ω ~ck
= 3 = 3 (5)
4π L L
which yields
r2
Z
P = ~c k e32
dΩ sin2 θ|f (~q )|2 . (6)
L
(ii)
Now compare this to the quantum result using the photon scattering matrix element in Eq. (2.28)
M = re f (~q ) ∧2k ˆ~kλ · ˆ~k0 λ . (7)
2
S. M. Girvin and Kun Yang
c 2019
[Compiled: August 12, 2019]
,Note to Instructors
For a few of the more difficult problems, we include notes to the instructor suggesting simplifications,
specializations and hints that the instructor may wish to give the students when assigning those
problems.
Note also that there are some useful exercises within the appendices of the textbook.
1
, Chapter 2
Ex. 2.1
(i)
The radiated electric field is
eikRD h i
~ in ) e−iωt e−i~q·~r .
~a ≈ re n̂ × (n̂ × E (1)
RD
~ in ) is perpendicular to both n̂ and to E
Replace e−i~q·~r by he−i~q·~r i = f (~q ). The vector (n̂ × E ~ in and
~ 2 2 2
has length |Ein sin θ|. Hence |n̂ × (n̂ × Ein )| = Ein sin θ. Thus
re2 2
|a |2 = Ein
2
q )|2 .
2 sin θ|f (~ (2)
RD
The total radiated power passing through a sphere of radius RD is
Z 2
2 a
P = cRD dΩ ×2 (3)
8π
Z
c 2 2
= r E dΩ sin2 θ|f (~q )|2 . (4)
4π e in
Let us normalize the incident electric field to that associated with a single photon in the normal-
ization volume L3
2
Ein ~ω ~ck
= 3 = 3 (5)
4π L L
which yields
r2
Z
P = ~c k e32
dΩ sin2 θ|f (~q )|2 . (6)
L
(ii)
Now compare this to the quantum result using the photon scattering matrix element in Eq. (2.28)
M = re f (~q ) ∧2k ˆ~kλ · ˆ~k0 λ . (7)
2
