As a result,
e
j
1
=
r
M
1
M
1
+
M
2
i
j
,
e
j
2
=

r
M
2
M
1
+
M
2
i
j
.
Finally we substitute these expressions to Eqs. (6.45) for
u
±
and get (
Problem 6.9
)
φ
=

i
r
4
πω
2
l
*
X
q
1
q
(
b
q
e
i
qr

b
*
q
e

i
qr
)
(6.46)
Using the second quantization procedure (page 41) we arrive at the famous
Fr¨
olich Hamil
tonian
H
e

ph
=
1
√
V
X
q
,j
M
j
q
a
†
p

~
q
a
p
(
b
q
j
+
b
†
q
j
)
(6.47)
with
M
2
j
= 2
πe
2
~
ω
l,j
/
*
.
Sometimes it is expressed as
M
2
=
4
πα
~
(
~
ω
l
)
3
/
2
√
2
m
,
α
=
e
2
~
*
r
m
2
~
ω
l
.
The dimensionless constant
α
is called the
polaron constant.
Now we can analyze the conservation laws
ε
k
±
q
=
ε
k
±
~
ω
l
which can be rewritten as
~
2
q
2
2
m
±
~
2
kq
cos
ϑ
2
m
∓
~
ω
l
= 0
.
The roots of this equation are
q
1
=

k
cos
ϑ
±
q
k
2
cos
2
ϑ
+
k
2
0
,
q
2
=

k
cos
ϑ
±
q
k
2
cos
2
ϑ

k
2
0
where
~
2
k
2
0
/
2
m
=
~
ω
l
.
1. If
k
k
0
we get for both absorption and emission the old conditions
q
min
= 0
, q
max
=
2
k.
At high temperatures,
k
B
T
~
ω
l
the scattering is elastic, and we return to the same
expression as for acoustic phonons. Using Eq. (6.40) we get
τ
tr
=
√
2
2
~
2
*
e
2
√
mk
B
T
ε
3
/
2
∝
T

1
ε
3
/
2
.
2. The case of low temperatures,
~
ω
l
k
B
T ,
6.7.
ELECTRONPHONON INTERACTION IN SEMICONDUCTORS
129
is more tricky. In this case only absorption processes can take place, and
q
min
=
q
k
2
+
k
2
0

k
(
ϑ
= 0)
,
q
max
=
q
k
2
+
k
2
0
+
k
(
ϑ
=
π
)
.
It is clear, that the scattering is strongly inelastic and in general one cannot use the
relaxation time approximation. Nevertheless, some simplification does exist. Indeed, note
that the ratio of the emission and absorption probabilities is
(
N
q
+ 1)
/N
q
≈
exp (
~
ω
l
/k
B
T
)
1
.
So if the electron absorbs an optical phonon it should immediately emit another one. As
a result, the change of energy appears small while the change of the quasimomentum is
large.
One can get the result taking into account only the absorption processes.
The
corresponding
δ
function is
δ
~
2
q
2
2
m
+
~
2
kq
cos
ϑ
2
m

~
2
k
2
0
2
m
=
m
~
2
kq
δ
q
2

k
2
0
2
kq
+ cos
ϑ
while the relaxation time can be obtained as
1
τ
tr
=
1
8
π
2
m
~
2
k
3
Z
q
max
q
min
w
(
q
)
N
q
q
2

k
2
0
q
q
2
dq
≈
(
e
2
mω
l
/
~
2
*
k
3
) exp(

~
ω
l
/k
B
T
)
2
k
q
k
2
+
k
2
0

k
0
ln
p
k
2
+
k
2
0
+
k
p
k
2
+
k
2
0

k

{z
}
≈
(4
/
3)
k
3
/k
0
.
Expanding this expression in powers of
k/k
0
we get/k
0
τ
tr
=
2
√
2
2
~
2
*
e
2
√
m
~
ω
l
exp
~
ω
l
k
B
T
.
This scattering is very weak. Note that it is not the case of the socalled
hot
electrons with
high energies, which can emit optical phonons.
The polaron.
4
We take the opportunity to demonstrate the role of interaction when it cannot be considered
as weak. Let us consider the interaction of an electron with optical phonons. According
to quantum mechanics, the change of the ground state energy due to interaction is
E
n
 E
(0)
n
=
h
n
H
int

n
i
+
X
m
6
=
n
h
m
H
int

n
i
2
E
(0)
n
 E
(0)
0
.
4
Optional section
130
CHAPTER 6.
CLASSICAL DC TRANSPORT ...
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 Physics, Cubic crystal system, periodic structures, Reciprocal lattice, Lattice Vibrations