Application to the free electron model
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Derivation to second order in temperature
edit
We seek an expansion that is second order in temperature, i.e., to
τ
2
{\displaystyle \tau ^{2}}
, where
β
−
1
=
τ
=
k
B
T
{\displaystyle \beta ^{-1}=\tau =k_{B}T}
is the product of temperature and Boltzmann's constant . Begin with a change variables to
τ
x
=
ε
−
μ
{\displaystyle \tau x=\varepsilon -\mu }
:
I
=
∫
−
∞
∞
H
(
ε
)
e
β
(
ε
−
μ
)
+
1
d
ε
=
τ
∫
−
∞
∞
H
(
μ
+
τ
x
)
e
x
+
1
d
x
,
{\displaystyle I=\int _{-\infty }^{\infty }{\frac {H(\varepsilon )}{e^{\beta (\varepsilon -\mu )}+1}}\,\mathrm {d} \varepsilon =\tau \int _{-\infty }^{\infty }{\frac {H(\mu +\tau x)}{e^{x}+1}}\,\mathrm {d} x\,,}
Divide the range of integration,
I
=
I
1
+
I
2
{\displaystyle I=I_{1}+I_{2}}
, and rewrite
I
1
{\displaystyle I_{1}}
using the change of variables
x
→
−
x
{\displaystyle x\rightarrow -x}
:
I
=
τ
∫
−
∞
0
H
(
μ
+
τ
x
)
e
x
+
1
d
x
⏟
I
1
+
τ
∫
0
∞
H
(
μ
+
τ
x
)
e
x
+
1
d
x
⏟
I
2
.
{\displaystyle I=\underbrace {\tau \int _{-\infty }^{0}{\frac {H(\mu +\tau x)}{e^{x}+1}}\,\mathrm {d} x} _{I_{1}}+\underbrace {\tau \int _{0}^{\infty }{\frac {H(\mu +\tau x)}{e^{x}+1}}\,\mathrm {d} x} _{I_{2}}\,.}
I
1
=
τ
∫
−
∞
0
H
(
μ
+
τ
x
)
e
x
+
1
d
x
=
τ
∫
0
∞
H
(
μ
−
τ
x
)
e
−
x
+
1
d
x
{\displaystyle I_{1}=\tau \int _{-\infty }^{0}{\frac {H(\mu +\tau x)}{e^{x}+1}}\,\mathrm {d} x=\tau \int _{0}^{\infty }{\frac {H(\mu -\tau x)}{e^{-x}+1}}\,\mathrm {d} x\,}
Next, employ an algebraic 'trick' on the denominator of
I
1
{\displaystyle I_{1}}
,
1
e
−
x
+
1
=
1
−
1
e
x
+
1
,
{\displaystyle {\frac {1}{e^{-x}+1}}=1-{\frac {1}{e^{x}+1}}\,,}
to obtain:
I
1
=
τ
∫
0
∞
H
(
μ
−
τ
x
)
d
x
−
τ
∫
0
∞
H
(
μ
−
τ
x
)
e
x
+
1
d
x
{\displaystyle I_{1}=\tau \int _{0}^{\infty }H(\mu -\tau x)\,\mathrm {d} x-\tau \int _{0}^{\infty }{\frac {H(\mu -\tau x)}{e^{x}+1}}\,\mathrm {d} x\,}
Return to the original variables with
−
τ
d
x
=
d
ε
{\displaystyle -\tau \mathrm {d} x=\mathrm {d} \varepsilon }
in the first term of
I
1
{\displaystyle I_{1}}
. Combine
I
=
I
1
+
I
2
{\displaystyle I=I_{1}+I_{2}}
to obtain:
I
=
∫
−
∞
μ
H
(
ε
)
d
ε
+
τ
∫
0
∞
H
(
μ
+
τ
x
)
−
H
(
μ
−
τ
x
)
e
x
+
1
d
x
{\displaystyle I=\int _{-\infty }^{\mu }H(\varepsilon )\,\mathrm {d} \varepsilon +\tau \int _{0}^{\infty }{\frac {H(\mu +\tau x)-H(\mu -\tau x)}{e^{x}+1}}\,\mathrm {d} x\,}
The numerator in the second term can be expressed as an approximation to the first derivative, provided
τ
{\displaystyle \tau }
is sufficiently small and
H
(
ε
)
{\displaystyle H(\varepsilon )}
is sufficiently smooth:
Δ
H
=
H
(
μ
+
τ
x
)
−
H
(
μ
−
τ
x
)
≈
2
τ
x
H
′
(
μ
)
+
⋯
,
{\displaystyle \Delta H=H(\mu +\tau x)-H(\mu -\tau x)\approx 2\tau xH'(\mu )+\cdots \,,}
to obtain,
I
=
∫
−
∞
μ
H
(
ε
)
d
ε
+
2
τ
2
H
′
(
μ
)
∫
0
∞
x
d
x
e
x
+
1
{\displaystyle I=\int _{-\infty }^{\mu }H(\varepsilon )\,\mathrm {d} \varepsilon +2\tau ^{2}H'(\mu )\int _{0}^{\infty }{\frac {x\mathrm {d} x}{e^{x}+1}}\,}
The definite integral is known[3] to be:
∫
0
∞
x
d
x
e
x
+
1
=
π
2
12
{\displaystyle \int _{0}^{\infty }{\frac {x\mathrm {d} x}{e^{x}+1}}={\frac {\pi ^{2}}{12}}}
.
Hence,
I
=
∫
−
∞
∞
H
(
ε
)
e
β
(
ε
−
μ
)
+
1
d
ε
≈
∫
−
∞
μ
H
(
ε
)
d
ε
+
π
2
6
β
2
H
′
(
μ
)
{\displaystyle I=\int _{-\infty }^{\infty }{\frac {H(\varepsilon )}{e^{\beta (\varepsilon -\mu )}+1}}\,\mathrm {d} \varepsilon \approx \int _{-\infty }^{\mu }H(\varepsilon )\,\mathrm {d} \varepsilon +{\frac {\pi ^{2}}{6\beta ^{2}}}H'(\mu )\,}
Higher order terms and a generating function
edit
We can obtain higher order terms in the Sommerfeld expansion by use of a
generating function for moments of the Fermi distribution. This is given by
∫
−
∞
∞
d
ϵ
2
π
e
τ
ϵ
/
2
π
{
1
1
+
e
β
(
ϵ
−
μ
)
−
θ
(
−
ϵ
)
}
=
1
τ
{
(
τ
T
2
)
sin
(
τ
T
2
)
e
τ
μ
/
2
π
−
1
}
,
0
<
τ
T
/
2
π
<
1.
{\displaystyle \int _{-\infty }^{\infty }{\frac {d\epsilon }{2\pi }}e^{\tau \epsilon /2\pi }\left\{{\frac {1}{1+e^{\beta (\epsilon -\mu )}}}-\theta (-\epsilon )\right\}={\frac {1}{\tau }}\left\{{\frac {({\frac {\tau T}{2}})}{\sin({\frac {\tau T}{2}})}}e^{\tau \mu /2\pi }-1\right\},\quad 0<\tau T/2\pi <1.}
Here
k
B
T
=
β
−
1
{\displaystyle k_{\rm {B}}T=\beta ^{-1}}
and Heaviside step function
−
θ
(
−
ϵ
)
{\displaystyle -\theta (-\epsilon )}
subtracts the divergent zero-temperature contribution.
Expanding in powers of
τ
{\displaystyle \tau }
gives, for example [4]
∫
−
∞
∞
d
ϵ
2
π
{
1
1
+
e
β
(
ϵ
−
μ
)
−
θ
(
−
ϵ
)
}
=
(
μ
2
π
)
,
{\displaystyle \int _{-\infty }^{\infty }{\frac {d\epsilon }{2\pi }}\left\{{\frac {1}{1+e^{\beta (\epsilon -\mu )}}}-\theta (-\epsilon )\right\}=\left({\frac {\mu }{2\pi }}\right),}
∫
−
∞
∞
d
ϵ
2
π
(
ϵ
2
π
)
{
1
1
+
e
β
(
ϵ
−
μ
)
−
θ
(
−
ϵ
)
}
=
1
2
!
(
μ
2
π
)
2
+
T
2
4
!
,
{\displaystyle \int _{-\infty }^{\infty }{\frac {d\epsilon }{2\pi }}\left({\frac {\epsilon }{2\pi }}\right)\left\{{\frac {1}{1+e^{\beta (\epsilon -\mu )}}}-\theta (-\epsilon )\right\}={\frac {1}{2!}}\left({\frac {\mu }{2\pi }}\right)^{2}+{\frac {T^{2}}{4!}},}
∫
−
∞
∞
d
ϵ
2
π
1
2
!
(
ϵ
2
π
)
2
{
1
1
+
e
β
(
ϵ
−
μ
)
−
θ
(
−
ϵ
)
}
=
1
3
!
(
μ
2
π
)
3
+
(
μ
2
π
)
T
2
4
!
,
{\displaystyle \int _{-\infty }^{\infty }{\frac {d\epsilon }{2\pi }}{\frac {1}{2!}}\left({\frac {\epsilon }{2\pi }}\right)^{2}\left\{{\frac {1}{1+e^{\beta (\epsilon -\mu )}}}-\theta (-\epsilon )\right\}={\frac {1}{3!}}\left({\frac {\mu }{2\pi }}\right)^{3}+\left({\frac {\mu }{2\pi }}\right){\frac {T^{2}}{4!}},}
∫
−
∞
∞
d
ϵ
2
π
1
3
!
(
ϵ
2
π
)
3
{
1
1
+
e
β
(
ϵ
−
μ
)
−
θ
(
−
ϵ
)
}
=
1
4
!
(
μ
2
π
)
4
+
1
2
!
(
μ
2
π
)
2
T
2
4
!
+
7
8
T
4
6
!
,
{\displaystyle \int _{-\infty }^{\infty }{\frac {d\epsilon }{2\pi }}{\frac {1}{3!}}\left({\frac {\epsilon }{2\pi }}\right)^{3}\left\{{\frac {1}{1+e^{\beta (\epsilon -\mu )}}}-\theta (-\epsilon )\right\}={\frac {1}{4!}}\left({\frac {\mu }{2\pi }}\right)^{4}+{\frac {1}{2!}}\left({\frac {\mu }{2\pi }}\right)^{2}{\frac {T^{2}}{4!}}+{\frac {7}{8}}{\frac {T^{4}}{6!}},}
∫
−
∞
∞
d
ϵ
2
π
1
4
!
(
ϵ
2
π
)
4
{
1
1
+
e
β
(
ϵ
−
μ
)
−
θ
(
−
ϵ
)
}
=
1
5
!
(
μ
2
π
)
5
+
1
3
!
(
μ
2
π
)
3
T
2
4
!
+
(
μ
2
π
)
7
8
T
4
6
!
,
{\displaystyle \int _{-\infty }^{\infty }{\frac {d\epsilon }{2\pi }}{\frac {1}{4!}}\left({\frac {\epsilon }{2\pi }}\right)^{4}\left\{{\frac {1}{1+e^{\beta (\epsilon -\mu )}}}-\theta (-\epsilon )\right\}={\frac {1}{5!}}\left({\frac {\mu }{2\pi }}\right)^{5}+{\frac {1}{3!}}\left({\frac {\mu }{2\pi }}\right)^{3}{\frac {T^{2}}{4!}}+\left({\frac {\mu }{2\pi }}\right){\frac {7}{8}}{\frac {T^{4}}{6!}},}
∫
−
∞
∞
d
ϵ
2
π
1
5
!
(
ϵ
2
π
)
5
{
1
1
+
e
β
(
ϵ
−
μ
)
−
θ
(
−
ϵ
)
}
=
1
6
!
(
μ
2
π
)
6
+
1
4
!
(
μ
2
π
)
4
T
2
4
!
+
1
2
!
(
μ
2
π
)
2
7
8
T
4
6
!
+
31
24
T
6
8
!
.
{\displaystyle \int _{-\infty }^{\infty }{\frac {d\epsilon }{2\pi }}{\frac {1}{5!}}\left({\frac {\epsilon }{2\pi }}\right)^{5}\left\{{\frac {1}{1+e^{\beta (\epsilon -\mu )}}}-\theta (-\epsilon )\right\}={\frac {1}{6!}}\left({\frac {\mu }{2\pi }}\right)^{6}+{\frac {1}{4!}}\left({\frac {\mu }{2\pi }}\right)^{4}{\frac {T^{2}}{4!}}+{\frac {1}{2!}}\left({\frac {\mu }{2\pi }}\right)^{2}{\frac {7}{8}}{\frac {T^{4}}{6!}}+{\frac {31}{24}}{\frac {T^{6}}{8!}}.}
A similar generating function for the odd moments of the Bose function is
∫
0
∞
d
ϵ
2
π
sinh
(
ϵ
τ
/
π
)
1
e
β
ϵ
−
1
=
1
4
τ
{
1
−
τ
T
tan
τ
T
}
,
0
<
τ
T
<
π
.
{\displaystyle \int _{0}^{\infty }{\frac {d\epsilon }{2\pi }}\sinh(\epsilon \tau /\pi ){\frac {1}{e^{\beta \epsilon }-1}}={\frac {1}{4\tau }}\left\{1-{\frac {\tau T}{\tan \tau T}}\right\},\quad 0<\tau T<\pi .}
References
edit