1
The life is not so simple. One can imagine also
collective
excitations which may have no gap. As has
been shown, it is not the case because of the finite electron charge.
360
CHAPTER 18.
MICROSCOPIC THEORY
18.4
Temperature Dependence of the Energy Gap
Now we are prepared to discuss the temperature dependence of the gap. From Eq. (18.10)
we get
1 =
λ
2
Z
(
dk
)
1

2
n
0
(
ε
k
)
ε
k
=
λg
(
F
)
2
Z
~
ω
D
0
dξ
tanh(
p
ξ
2
+ ∆
2
/
2
k
B
T
)
p
ξ
2
+ ∆
2
.
(18.13)
At
T
→
0
tanh(
p
ξ
2
+ ∆
2
/
2
k
B
T
)
→
1 and
1 =
λg
(
F
)
2
ln
2
~
ω
D
∆(0)
→
∆(0) = 2
~
ω
D
exp (

2
/λg
(
F
))
(18.14)
(note the extra factor 2). At
T
→
T
c
∆
→
0, and
1 =
λg
(
F
)
2
Z
~
ω
D
0
dξ
tanh(
ξ/
2
k
B
T
c
)
ξ
=
λg
(
F
)
2
ln
2
~
ω
D
γ
πk
B
T
c
,
γ
= 1
.
78
.
Thus
k
B
T
c
=
2
~
ω
D
γ
π
exp (

2
/λg
(
F
))
(18.15)
and
∆(0) =
π
γ
k
B
T
c
= 1
.
76
k
B
T
c
.
(18.16)
To discuss the temperature dependence one can simply calculate the integral in (18.13).
It is interesting to show the analytical formula, so we analyze briefly this integral. At low
temperatures we can rewrite the formula as
ln
∆(0)
∆
=
Z
∞
0
1

tanh(
p
ξ
2
+ ∆
2
/
2
k
B
T
)
p
ξ
2
+ ∆
2
dξ
= 2
f
∆
k
B
T
where
f
(
x
) =
Z
∞
1
dy
(1 +
e
yx
)
p
y
2

1
,
y
=
p
ξ
2
+ ∆
2
∆
.
Here we have used the expression (18.14) and expanded the integration region to infinity
(the important region is
∼
∆). We are interested in large values of
x,
f
(
x
) =
Z
∞
1
dy
∞
X
n
=1
(

1)
n
+1
e

nyx
p
y
2

1
.
Then we use the integral representation for the McDonald function
K
ν
(
z
) =
Γ(1
/
2)
Γ(
ν
+ 1
/
2)
z
2
ν
Z
∞
1
e

yz
(
y
2

1)
v

1
/
2
dy
18.4.
TEMPERATURE DEPENDENCE ...
361
and get
f
(
x
) =
∞
X
n
=1
(

1)
n
+1
K
0
(
nx
)
∼
r
π
2
x
e

x
.
Thus, at
T
T
c
∆(
T
) = ∆(0)

p
2
π
∆(0)
k
B
T e

∆(0)
/k
B
T

{z
}
.
∼
number of quasiparticles
(18.17)
At
T
→
T
c
it is convenient to expand over ∆
→
0
.
To do this we also divide the expression
(18.13) by (
λg
(
F
)
/
2) and subtract its limit at ∆ = 0
.
We get
ln
T
c
T
=
Z
∞
0
dξ
"
tanh
ξ
ξ

tanh(
p
ξ
2
+ ∆
2
/
2
k
B
T
)
p
ξ
2
+ ∆
2
#
.
Then we use the formula
tanh
πx
2
=
4
x
π
∞
X
k
=0
1
(2
k
+ 1)
2
+
x
2
.
Substituting this formula into the previous one, expanding over ∆ and integrating over
ξ
we get
ln
T
c
T
= 2
∞
X
n
=1
(

1)
n
+1
(2
n

1)!!
(2
n
)!!
∆
πk
B
T
2
n
∞
X
k
=0
1
(2
k
+ 1)
2
n
+1
.
The first item leads to
∆
≈
3
.
06
r
T
c
T
c

T
.
The graph of the dependence ∆(
T
) is given in Fig. 18.3.
Figure 18.3: Temperature dependence of the energy gap.
362
CHAPTER 18.
MICROSCOPIC THEORY
18.5
Thermodynamics of a Superconductor
According to statistical physics, the thermodynamic potential Ω depending on the variables
T,
V
,
and chemical potential
is
Ω =

k
B
T
ln
Z,
with the partition function
Z
=
X
s,N
exp

E
sN

N
k
B
T
= Tr
exp

H
N
k
B
T
.
Splitting the Hamiltonian as
H
0
+
H
int
and differentiating with respect to the interaction
constant
λ
we get
∂
Ω
∂λ
=
1
λ
Tr
h
H
int
exp

H
N
k
B
T
i
Tr
h
exp

H
N
k
B
T
i
=
1
λ
hH
int
i
.
Substituting the expression for the interaction Hamiltonian
2
H
int
=

λ
X
k
1
+
k
2
=
k
0
1
+
k
0
2
k
1
6
=
k
0
1
a
†
k
0
1
↑
a
†
k
0
2
↓
a
k
2
↓
a
k
1
↑
and expressing the electron operators through the quasiparticle ones as
a
k
↑
=
u
k
α
k
↑
+
v
k
α
†

k
↓
,
a
k
↓
=
u
k
α
k
↓

v
k
α
†

k
↑
we get (
Problem 18.2
)
∂
Ω
∂λ
=

∆
2
λ
2
.
(18.18)
You've reached the end of your free preview.
Want to read all 477 pages?
 Spring '10
 Unknow
 Physics, Cubic crystal system, periodic structures, Reciprocal lattice, Lattice Vibrations