To understand the Lindhard formula, consider some limiting cases in 2 and 3 dimensions. The 1-dimensional case is also considered in other ways.
Long wavelength limit
In the long wavelength limit (
q
→
0
{\displaystyle \mathbf {q} \to 0}
), Lindhard function reduces to
ϵ
(
q
=
0
,
ω
)
≈
1
−
ω
p
l
2
ω
2
,
{\displaystyle \epsilon (\mathbf {q} =0,\omega )\approx 1-{\frac {\omega _{\rm {pl}}^{2}}{\omega ^{2}}},}
where
ω
p
l
2
=
4
π
e
2
N
L
3
m
{\displaystyle \omega _{\rm {pl}}^{2}={\frac {4\pi e^{2}N}{L^{3}m}}}
is the three-dimensional plasma frequency (in SI units, replace the factor
4
π
{\displaystyle 4\pi }
by
1
/
ϵ
0
{\displaystyle 1/\epsilon _{0}}
.) For two-dimensional systems,
ω
p
l
2
(
q
)
=
2
π
e
2
n
q
ϵ
m
{\displaystyle \omega _{\rm {pl}}^{2}(\mathbf {q} )={\frac {2\pi e^{2}nq}{\epsilon m}}}
.
This result recovers the plasma oscillations from the classical dielectric function from Drude model and from quantum mechanical free electron model .
Derivation in 3D
For the denominator of the Lindhard formula, we get
E
k
−
q
−
E
k
=
ℏ
2
2
m
(
k
2
−
2
k
⋅
q
+
q
2
)
−
ℏ
2
k
2
2
m
≃
−
ℏ
2
k
⋅
q
m
{\displaystyle E_{\mathbf {k} -\mathbf {q} }-E_{\mathbf {k} }={\frac {\hbar ^{2}}{2m}}(k^{2}-2\mathbf {k} \cdot \mathbf {q} +q^{2})-{\frac {\hbar ^{2}k^{2}}{2m}}\simeq -{\frac {\hbar ^{2}\mathbf {k} \cdot \mathbf {q} }{m}}}
,
and for the numerator of the Lindhard formula, we get
f
k
−
q
−
f
k
=
f
k
−
q
⋅
∇
k
f
k
+
⋯
−
f
k
≃
−
q
⋅
∇
k
f
k
{\displaystyle f_{\mathbf {k} -\mathbf {q} }-f_{\mathbf {k} }=f_{\mathbf {k} }-\mathbf {q} \cdot \nabla _{\mathbf {k} }f_{\mathbf {k} }+\cdots -f_{\mathbf {k} }\simeq -\mathbf {q} \cdot \nabla _{\mathbf {k} }f_{\mathbf {k} }}
.
Inserting these into the Lindhard formula and taking the
δ
→
0
{\displaystyle \delta \to 0}
limit, we obtain
ϵ
(
q
=
0
,
ω
0
)
≃
1
+
V
q
∑
k
,
i
q
i
∂
f
k
∂
k
i
ℏ
ω
0
−
ℏ
2
k
⋅
q
m
≃
1
+
V
q
ℏ
ω
0
∑
k
,
i
q
i
∂
f
k
∂
k
i
(
1
+
ℏ
k
⋅
q
m
ω
0
)
≃
1
+
V
q
ℏ
ω
0
∑
k
,
i
q
i
∂
f
k
∂
k
i
ℏ
k
⋅
q
m
ω
0
=
1
−
V
q
q
2
m
ω
0
2
∑
k
f
k
=
1
−
V
q
q
2
N
m
ω
0
2
=
1
−
4
π
e
2
ϵ
q
2
L
3
q
2
N
m
ω
0
2
=
1
−
ω
p
l
2
ω
0
2
.
{\displaystyle {\begin{alignedat}{2}\epsilon (\mathbf {q} =0,\omega _{0})&\simeq 1+V_{\mathbf {q} }\sum _{\mathbf {k} ,i}{\frac {q_{i}{\frac {\partial f_{\mathbf {k} }}{\partial k_{i}}}}{\hbar \omega _{0}-{\frac {\hbar ^{2}\mathbf {k} \cdot \mathbf {q} }{m}}}}\\&\simeq 1+{\frac {V_{\mathbf {q} }}{\hbar \omega _{0}}}\sum _{\mathbf {k} ,i}{q_{i}{\frac {\partial f_{\mathbf {k} }}{\partial k_{i}}}}(1+{\frac {\hbar \mathbf {k} \cdot \mathbf {q} }{m\omega _{0}}})\\&\simeq 1+{\frac {V_{\mathbf {q} }}{\hbar \omega _{0}}}\sum _{\mathbf {k} ,i}{q_{i}{\frac {\partial f_{\mathbf {k} }}{\partial k_{i}}}}{\frac {\hbar \mathbf {k} \cdot \mathbf {q} }{m\omega _{0}}}\\&=1-V_{\mathbf {q} }{\frac {q^{2}}{m\omega _{0}^{2}}}\sum _{\mathbf {k} }{f_{\mathbf {k} }}\\&=1-V_{\mathbf {q} }{\frac {q^{2}N}{m\omega _{0}^{2}}}\\&=1-{\frac {4\pi e^{2}}{\epsilon q^{2}L^{3}}}{\frac {q^{2}N}{m\omega _{0}^{2}}}\\&=1-{\frac {\omega _{\rm {pl}}^{2}}{\omega _{0}^{2}}}.\end{alignedat}}}
,
where we used
E
k
=
ℏ
ω
k
{\displaystyle E_{\mathbf {k} }=\hbar \omega _{\mathbf {k} }}
and
V
q
=
4
π
e
2
ϵ
q
2
L
3
{\displaystyle V_{\mathbf {q} }={\frac {4\pi e^{2}}{\epsilon q^{2}L^{3}}}}
.
Derivation in 2D
First, consider the long wavelength limit (
q
→
0
{\displaystyle q\to 0}
).
For the denominator of the Lindhard formula,
E
k
−
q
−
E
k
=
ℏ
2
2
m
(
k
2
−
2
k
⋅
q
+
q
2
)
−
ℏ
2
k
2
2
m
≃
−
ℏ
2
k
⋅
q
m
{\displaystyle E_{\mathbf {k} -\mathbf {q} }-E_{\mathbf {k} }={\frac {\hbar ^{2}}{2m}}(k^{2}-2\mathbf {k} \cdot \mathbf {q} +q^{2})-{\frac {\hbar ^{2}k^{2}}{2m}}\simeq -{\frac {\hbar ^{2}\mathbf {k} \cdot \mathbf {q} }{m}}}
,
and for the numerator,
f
k
−
q
−
f
k
=
f
k
−
q
⋅
∇
k
f
k
+
⋯
−
f
k
≃
−
q
⋅
∇
k
f
k
{\displaystyle f_{\mathbf {k} -\mathbf {q} }-f_{\mathbf {k} }=f_{\mathbf {k} }-\mathbf {q} \cdot \nabla _{\mathbf {k} }f_{\mathbf {k} }+\cdots -f_{\mathbf {k} }\simeq -\mathbf {q} \cdot \nabla _{\mathbf {k} }f_{\mathbf {k} }}
.
Inserting these into the Lindhard formula and taking the limit of
δ
→
0
{\displaystyle \delta \to 0}
, we obtain
ϵ
(
0
,
ω
)
≃
1
+
V
q
∑
k
,
i
q
i
∂
f
k
∂
k
i
ℏ
ω
0
−
ℏ
2
k
⋅
q
m
≃
1
+
V
q
ℏ
ω
0
∑
k
,
i
q
i
∂
f
k
∂
k
i
(
1
+
ℏ
k
⋅
q
m
ω
0
)
≃
1
+
V
q
ℏ
ω
0
∑
k
,
i
q
i
∂
f
k
∂
k
i
ℏ
k
⋅
q
m
ω
0
=
1
+
V
q
ℏ
ω
0
2
∫
d
2
k
(
L
2
π
)
2
∑
i
,
j
q
i
∂
f
k
∂
k
i
ℏ
k
j
q
j
m
ω
0
=
1
+
V
q
L
2
m
ω
0
2
2
∫
d
2
k
(
2
π
)
2
∑
i
,
j
q
i
q
j
k
j
∂
f
k
∂
k
i
=
1
+
V
q
L
2
m
ω
0
2
∑
i
,
j
q
i
q
j
2
∫
d
2
k
(
2
π
)
2
k
j
∂
f
k
∂
k
i
=
1
−
V
q
L
2
m
ω
0
2
∑
i
,
j
q
i
q
j
n
δ
i
j
=
1
−
2
π
e
2
ϵ
q
L
2
L
2
m
ω
0
2
q
2
n
=
1
−
ω
p
l
2
(
q
)
ω
0
2
,
{\displaystyle {\begin{alignedat}{2}\epsilon (0,\omega )&\simeq 1+V_{\mathbf {q} }\sum _{\mathbf {k} ,i}{\frac {q_{i}{\frac {\partial f_{\mathbf {k} }}{\partial k_{i}}}}{\hbar \omega _{0}-{\frac {\hbar ^{2}\mathbf {k} \cdot \mathbf {q} }{m}}}}\\&\simeq 1+{\frac {V_{\mathbf {q} }}{\hbar \omega _{0}}}\sum _{\mathbf {k} ,i}{q_{i}{\frac {\partial f_{\mathbf {k} }}{\partial k_{i}}}}(1+{\frac {\hbar \mathbf {k} \cdot \mathbf {q} }{m\omega _{0}}})\\&\simeq 1+{\frac {V_{\mathbf {q} }}{\hbar \omega _{0}}}\sum _{\mathbf {k} ,i}{q_{i}{\frac {\partial f_{\mathbf {k} }}{\partial k_{i}}}}{\frac {\hbar \mathbf {k} \cdot \mathbf {q} }{m\omega _{0}}}\\&=1+{\frac {V_{\mathbf {q} }}{\hbar \omega _{0}}}2\int d^{2}k({\frac {L}{2\pi }})^{2}\sum _{i,j}{q_{i}{\frac {\partial f_{\mathbf {k} }}{\partial k_{i}}}}{\frac {\hbar k_{j}q_{j}}{m\omega _{0}}}\\&=1+{\frac {V_{\mathbf {q} }L^{2}}{m\omega _{0}^{2}}}2\int {\frac {d^{2}k}{(2\pi )^{2}}}\sum _{i,j}{q_{i}q_{j}k_{j}{\frac {\partial f_{\mathbf {k} }}{\partial k_{i}}}}\\&=1+{\frac {V_{\mathbf {q} }L^{2}}{m\omega _{0}^{2}}}\sum _{i,j}{q_{i}q_{j}2\int {\frac {d^{2}k}{(2\pi )^{2}}}k_{j}{\frac {\partial f_{\mathbf {k} }}{\partial k_{i}}}}\\&=1-{\frac {V_{\mathbf {q} }L^{2}}{m\omega _{0}^{2}}}\sum _{i,j}{q_{i}q_{j}n\delta _{ij}}\\&=1-{\frac {2\pi e^{2}}{\epsilon qL^{2}}}{\frac {L^{2}}{m\omega _{0}^{2}}}q^{2}n\\&=1-{\frac {\omega _{\rm {pl}}^{2}(\mathbf {q} )}{\omega _{0}^{2}}},\end{alignedat}}}
where we used
E
k
=
ℏ
ϵ
k
{\displaystyle E_{\mathbf {k} }=\hbar \epsilon _{\mathbf {k} }}
,
V
q
=
2
π
e
2
ϵ
q
L
2
{\displaystyle V_{\mathbf {q} }={\frac {2\pi e^{2}}{\epsilon qL^{2}}}}
and
ω
p
l
2
(
q
)
=
2
π
e
2
n
q
ϵ
m
{\displaystyle \omega _{\rm {pl}}^{2}(\mathbf {q} )={\frac {2\pi e^{2}nq}{\epsilon m}}}
.
Static limit
Consider the static limit (
ω
+
i
δ
→
0
{\displaystyle \omega +i\delta \to 0}
).
The Lindhard formula becomes
ϵ
(
q
,
ω
=
0
)
=
1
−
V
q
∑
k
f
k
−
q
−
f
k
E
k
−
q
−
E
k
{\displaystyle \epsilon (\mathbf {q} ,\omega =0)=1-V_{\mathbf {q} }\sum _{\mathbf {k} }{\frac {f_{\mathbf {k} -\mathbf {q} }-f_{\mathbf {k} }}{E_{\mathbf {k} -\mathbf {q} }-E_{\mathbf {k} }}}}
.
Inserting the above equalities for the denominator and numerator, we obtain
ϵ
(
q
,
0
)
=
1
−
V
q
∑
k
,
i
−
q
i
∂
f
∂
k
i
−
ℏ
2
k
⋅
q
m
=
1
−
V
q
∑
k
,
i
q
i
∂
f
∂
k
i
ℏ
2
k
⋅
q
m
{\displaystyle \epsilon (\mathbf {q} ,0)=1-V_{\mathbf {q} }\sum _{\mathbf {k} ,i}{\frac {-q_{i}{\frac {\partial f}{\partial k_{i}}}}{-{\frac {\hbar ^{2}\mathbf {k} \cdot \mathbf {q} }{m}}}}=1-V_{\mathbf {q} }\sum _{\mathbf {k} ,i}{\frac {q_{i}{\frac {\partial f}{\partial k_{i}}}}{\frac {\hbar ^{2}\mathbf {k} \cdot \mathbf {q} }{m}}}}
.
Assuming a thermal equilibrium Fermi–Dirac carrier distribution, we get
∑
i
q
i
∂
f
k
∂
k
i
=
−
∑
i
q
i
∂
f
k
∂
μ
∂
E
k
∂
k
i
=
−
∑
i
q
i
k
i
ℏ
2
m
∂
f
k
∂
μ
{\displaystyle \sum _{i}{q_{i}{\frac {\partial f_{\mathbf {k} }}{\partial k_{i}}}}=-\sum _{i}{q_{i}{\frac {\partial f_{\mathbf {k} }}{\partial \mu }}{\frac {\partial E_{\mathbf {k} }}{\partial k_{i}}}}=-\sum _{i}{q_{i}k_{i}{\frac {\hbar ^{2}}{m}}{\frac {\partial f_{\mathbf {k} }}{\partial \mu }}}}
here, we used
E
k
=
ℏ
2
k
2
2
m
{\displaystyle E_{\mathbf {k} }={\frac {\hbar ^{2}k^{2}}{2m}}}
and
∂
E
k
∂
k
i
=
ℏ
2
k
i
m
{\displaystyle {\frac {\partial E_{\mathbf {k} }}{\partial k_{i}}}={\frac {\hbar ^{2}k_{i}}{m}}}
.
Therefore,
ϵ
(
q
,
0
)
=
1
+
V
q
∑
k
,
i
q
i
k
i
ℏ
2
m
∂
f
k
∂
μ
ℏ
2
k
⋅
q
m
=
1
+
V
q
∑
k
∂
f
k
∂
μ
=
1
+
4
π
e
2
ϵ
q
2
∂
∂
μ
1
L
3
∑
k
f
k
=
1
+
4
π
e
2
ϵ
q
2
∂
∂
μ
N
L
3
=
1
+
4
π
e
2
ϵ
q
2
∂
n
∂
μ
≡
1
+
κ
2
q
2
.
{\displaystyle {\begin{alignedat}{2}\epsilon (\mathbf {q} ,0)&=1+V_{\mathbf {q} }\sum _{\mathbf {k} ,i}{\frac {q_{i}k_{i}{\frac {\hbar ^{2}}{m}}{\frac {\partial f_{\mathbf {k} }}{\partial \mu }}}{\frac {\hbar ^{2}\mathbf {k} \cdot \mathbf {q} }{m}}}=1+V_{\mathbf {q} }\sum _{\mathbf {k} }{\frac {\partial f_{\mathbf {k} }}{\partial \mu }}=1+{\frac {4\pi e^{2}}{\epsilon q^{2}}}{\frac {\partial }{\partial \mu }}{\frac {1}{L^{3}}}\sum _{\mathbf {k} }{f_{\mathbf {k} }}\\&=1+{\frac {4\pi e^{2}}{\epsilon q^{2}}}{\frac {\partial }{\partial \mu }}{\frac {N}{L^{3}}}=1+{\frac {4\pi e^{2}}{\epsilon q^{2}}}{\frac {\partial n}{\partial \mu }}\equiv 1+{\frac {\kappa ^{2}}{q^{2}}}.\end{alignedat}}}
Here,
κ
{\displaystyle \kappa }
is the 3D screening wave number (3D inverse screening length) defined as
κ
=
4
π
e
2
ϵ
∂
n
∂
μ
{\displaystyle \kappa ={\sqrt {{\frac {4\pi e^{2}}{\epsilon }}{\frac {\partial n}{\partial \mu }}}}}
.
Then, the 3D statically screened Coulomb potential is given by
V
s
(
q
,
ω
=
0
)
≡
V
q
ϵ
(
q
,
0
)
=
4
π
e
2
ϵ
q
2
L
3
q
2
+
κ
2
q
2
=
4
π
e
2
ϵ
L
3
1
q
2
+
κ
2
{\displaystyle V_{\rm {s}}(\mathbf {q} ,\omega =0)\equiv {\frac {V_{\mathbf {q} }}{\epsilon (\mathbf {q} ,0)}}={\frac {\frac {4\pi e^{2}}{\epsilon q^{2}L^{3}}}{\frac {q^{2}+\kappa ^{2}}{q^{2}}}}={\frac {4\pi e^{2}}{\epsilon L^{3}}}{\frac {1}{q^{2}+\kappa ^{2}}}}
.
And the inverse Fourier transformation of this result gives
V
s
(
r
)
=
∑
q
4
π
e
2
L
3
(
q
2
+
κ
2
)
e
i
q
⋅
r
=
e
2
r
e
−
κ
r
{\displaystyle V_{\rm {s}}(r)=\sum _{\mathbf {q} }{{\frac {4\pi e^{2}}{L^{3}(q^{2}+\kappa ^{2})}}e^{i\mathbf {q} \cdot \mathbf {r} }}={\frac {e^{2}}{r}}e^{-\kappa r}}
known as the Yukawa potential . Note that in this Fourier transformation, which is basically a sum over all
q
{\displaystyle \mathbf {q} }
, we used the expression for small
|
q
|
{\displaystyle |\mathbf {q} |}
for every value of
q
{\displaystyle \mathbf {q} }
which is not correct.
Statically screened potential(upper curved surface) and Coulomb potential(lower curved surface) in three dimensions
For a degenerated Fermi gas (T =0), the Fermi energy is given by
E
F
=
ℏ
2
2
m
(
3
π
2
n
)
2
3
{\displaystyle E_{\rm {F}}={\frac {\hbar ^{2}}{2m}}(3\pi ^{2}n)^{\frac {2}{3}}}
,
So the density is
n
=
1
3
π
2
(
2
m
ℏ
2
E
F
)
3
2
{\displaystyle n={\frac {1}{3\pi ^{2}}}\left({\frac {2m}{\hbar ^{2}}}E_{\rm {F}}\right)^{\frac {3}{2}}}
.
At T =0,
E
F
≡
μ
{\displaystyle E_{\rm {F}}\equiv \mu }
, so
∂
n
∂
μ
=
3
2
n
E
F
{\displaystyle {\frac {\partial n}{\partial \mu }}={\frac {3}{2}}{\frac {n}{E_{\rm {F}}}}}
.
Inserting this into the above 3D screening wave number equation, we obtain
κ
=
4
π
e
2
ϵ
∂
n
∂
μ
=
6
π
e
2
n
ϵ
E
F
{\displaystyle \kappa ={\sqrt {{\frac {4\pi e^{2}}{\epsilon }}{\frac {\partial n}{\partial \mu }}}}={\sqrt {\frac {6\pi e^{2}n}{\epsilon E_{\rm {F}}}}}}
.
This result recovers the 3D wave number from Thomas–Fermi screening .
For reference, Debye–Hückel screening describes the non-degenerate limit case. The result is
κ
=
4
π
e
2
n
β
ϵ
{\displaystyle \kappa ={\sqrt {\frac {4\pi e^{2}n\beta }{\epsilon }}}}
, known as the 3D Debye–Hückel screening wave number.
In two dimensions, the screening wave number is
κ
=
2
π
e
2
ϵ
∂
n
∂
μ
=
2
π
e
2
ϵ
m
ℏ
2
π
(
1
−
e
−
ℏ
2
β
π
n
/
m
)
=
2
m
e
2
ℏ
2
ϵ
f
k
=
0
.
{\displaystyle \kappa ={\frac {2\pi e^{2}}{\epsilon }}{\frac {\partial n}{\partial \mu }}={\frac {2\pi e^{2}}{\epsilon }}{\frac {m}{\hbar ^{2}\pi }}(1-e^{-\hbar ^{2}\beta \pi n/m})={\frac {2me^{2}}{\hbar ^{2}\epsilon }}f_{k=0}.}
Note that this result is independent of n .
Derivation in 2D
Consider the static limit (
ω
+
i
δ
→
0
{\displaystyle \omega +i\delta \to 0}
).
The Lindhard formula becomes
ϵ
(
q
,
0
)
=
1
−
V
q
∑
k
f
k
−
q
−
f
k
E
k
−
q
−
E
k
{\displaystyle \epsilon (\mathbf {q} ,0)=1-V_{\mathbf {q} }\sum _{\mathbf {k} }{\frac {f_{\mathbf {k} -\mathbf {q} }-f_{\mathbf {k} }}{E_{\mathbf {k} -\mathbf {q} }-E_{\mathbf {k} }}}}
.
Inserting the above equalities for the denominator and numerator, we obtain
ϵ
(
q
,
0
)
=
1
−
V
q
∑
k
,
i
−
q
i
∂
f
∂
k
i
−
ℏ
2
k
⋅
q
m
=
1
−
V
q
∑
k
,
i
q
i
∂
f
∂
k
i
ℏ
2
k
⋅
q
m
{\displaystyle \epsilon (\mathbf {q} ,0)=1-V_{\mathbf {q} }\sum _{\mathbf {k} ,i}{\frac {-q_{i}{\frac {\partial f}{\partial k_{i}}}}{-{\frac {\hbar ^{2}\mathbf {k} \cdot \mathbf {q} }{m}}}}=1-V_{\mathbf {q} }\sum _{\mathbf {k} ,i}{\frac {q_{i}{\frac {\partial f}{\partial k_{i}}}}{\frac {\hbar ^{2}\mathbf {k} \cdot \mathbf {q} }{m}}}}
.
Assuming a thermal equilibrium Fermi–Dirac carrier distribution, we get
∑
i
q
i
∂
f
k
∂
k
i
=
−
∑
i
q
i
∂
f
k
∂
μ
∂
E
k
∂
k
i
=
−
∑
i
q
i
k
i
ℏ
2
m
∂
f
k
∂
μ
{\displaystyle \sum _{i}{q_{i}{\frac {\partial f_{\mathbf {k} }}{\partial k_{i}}}}=-\sum _{i}{q_{i}{\frac {\partial f_{\mathbf {k} }}{\partial \mu }}{\frac {\partial E_{\mathbf {k} }}{\partial k_{i}}}}=-\sum _{i}{q_{i}k_{i}{\frac {\hbar ^{2}}{m}}{\frac {\partial f_{\mathbf {k} }}{\partial \mu }}}}
.
Therefore,
ϵ
(
q
,
0
)
=
1
+
V
q
∑
k
,
i
q
i
k
i
ℏ
2
m
∂
f
k
∂
μ
ℏ
2
k
⋅
q
m
=
1
+
V
q
∑
k
∂
f
k
∂
μ
=
1
+
2
π
e
2
ϵ
q
L
2
∂
∂
μ
∑
k
f
k
=
1
+
2
π
e
2
ϵ
q
∂
∂
μ
N
L
2
=
1
+
2
π
e
2
ϵ
q
∂
n
∂
μ
≡
1
+
κ
q
.
{\displaystyle {\begin{alignedat}{2}\epsilon (\mathbf {q} ,0)&=1+V_{\mathbf {q} }\sum _{\mathbf {k} ,i}{\frac {q_{i}k_{i}{\frac {\hbar ^{2}}{m}}{\frac {\partial f_{\mathbf {k} }}{\partial \mu }}}{\frac {\hbar ^{2}\mathbf {k} \cdot \mathbf {q} }{m}}}=1+V_{\mathbf {q} }\sum _{\mathbf {k} }{\frac {\partial f_{\mathbf {k} }}{\partial \mu }}=1+{\frac {2\pi e^{2}}{\epsilon qL^{2}}}{\frac {\partial }{\partial \mu }}\sum _{\mathbf {k} }{f_{\mathbf {k} }}\\&=1+{\frac {2\pi e^{2}}{\epsilon q}}{\frac {\partial }{\partial \mu }}{\frac {N}{L^{2}}}=1+{\frac {2\pi e^{2}}{\epsilon q}}{\frac {\partial n}{\partial \mu }}\equiv 1+{\frac {\kappa }{q}}.\end{alignedat}}}
κ
{\displaystyle \kappa }
is 2D screening wave number(2D inverse screening length) defined as
κ
=
2
π
e
2
ϵ
∂
n
∂
μ
{\displaystyle \kappa ={\frac {2\pi e^{2}}{\epsilon }}{\frac {\partial n}{\partial \mu }}}
.
Then, the 2D statically screened Coulomb potential is given by
V
s
(
q
,
ω
=
0
)
≡
V
q
ϵ
(
q
,
0
)
=
2
π
e
2
ϵ
q
L
2
q
q
+
κ
=
2
π
e
2
ϵ
L
2
1
q
+
κ
{\displaystyle V_{\rm {s}}(\mathbf {q} ,\omega =0)\equiv {\frac {V_{\mathbf {q} }}{\epsilon (\mathbf {q} ,0)}}={\frac {2\pi e^{2}}{\epsilon qL^{2}}}{\frac {q}{q+\kappa }}={\frac {2\pi e^{2}}{\epsilon L^{2}}}{\frac {1}{q+\kappa }}}
.
It is known that the chemical potential of the 2-dimensional Fermi gas is given by
μ
(
n
,
T
)
=
1
β
ln
(
e
ℏ
2
β
π
n
/
m
−
1
)
{\displaystyle \mu (n,T)={\frac {1}{\beta }}\ln {(e^{\hbar ^{2}\beta \pi n/m}-1)}}
,
and
∂
μ
∂
n
=
ℏ
2
π
m
1
1
−
e
−
ℏ
2
β
π
n
/
m
{\displaystyle {\frac {\partial \mu }{\partial n}}={\frac {\hbar ^{2}\pi }{m}}{\frac {1}{1-e^{-\hbar ^{2}\beta \pi n/m}}}}
.