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
edit
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}}}
.
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}}}}
.
Experiments on one dimensional systems
edit
This time, consider some generalized case for lowering the dimension.
The lower the dimension is, the weaker the screening effect.
In lower dimension, some of the field lines pass through the barrier material wherein the screening has no effect.
For the 1-dimensional case, we can guess that the screening affects only the field lines which are very close to the wire axis.
In real experiment, we should also take the 3D bulk screening effect into account even though we deal with 1D case like the single filament. The Thomas–Fermi screening has been applied to an electron gas confined to a filament and a coaxial cylinder.[ 5] For a K2 Pt(CN)4 Cl0.32 ·2.6H2 0 filament, it was found that the potential within the region between the filament and cylinder varies as
e
−
k
e
f
f
r
/
r
{\displaystyle e^{-k_{\rm {eff}}r}/r}
and its effective screening length is about 10 times that of metallic platinum .[ 5]
Haug, Hartmut; W. Koch, Stephan (2004). Quantum Theory of the Optical and Electronic Properties of Semiconductors (4th ed.) . World Scientific Publishing Co. Pte. Ltd. ISBN 978-981-238-609-0 .