Lindhard theory

Not to be confused with with the classical theory of channelling, also by Jens Lindhard.

In condensed matter physics, Lindhard theory is a method of calculating the effects of electric field screening by electrons in a solid. It is based on quantum mechanics (first-order perturbation theory) and the random phase approximation. It is named after Danish physicist Jens Lindhard, who first developed the theory in 1954.

Thomas–Fermi screening and the plasma oscillations can be derived as a special case of the more general Lindhard formula. In particular, Thomas–Fermi screening is the limit of the Lindhard formula when the wavevector (the reciprocal of the length-scale of interest) is much smaller than the Fermi wavevector, i.e. the long-distance limit. The Lorentz–Drude expression for the plasma oscillations are recovered in the dynamic case (long wavelengths, finite frequency).

This article uses cgs-Gaussian units.

Formula

The Lindhard formula for the longitudinal dielectric function is given by

$\epsilon (\mathbf {q} ,\omega )=1-V_{\mathbf {q} }\sum _{\mathbf {k} }{\frac {f_{\mathbf {k} -\mathbf {q} }-f_{\mathbf {k} }}{\hbar (\omega i\delta ) E_{\mathbf {k} -\mathbf {q} }-E_{\mathbf {k} }}}.$

Here, $\delta$ is a positive infinitesimal constant, $V_{\mathbf {q} }$ is $V_{\text{eff}}(\mathbf {q} )-V_{\text{ind}}(\mathbf {q} )$ and $f_{\mathbf {k} }$ is the carrier distribution function which is the Fermi–Dirac distribution function for electrons in thermodynamic equilibrium. However this Lindhard formula is valid also for nonequilibrium distribution functions. It can be obtained by first-order perturbation theory and the random phase approximation (RPA).

Limiting cases

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 ($\mathbf {q} \to 0$), Lindhard function reduces to

$\epsilon (\mathbf {q} =0,\omega )\approx 1-{\frac {\omega _{\rm {pl}}^{2}}{\omega ^{2}}},$

where $\omega _{\rm {pl}}^{2}={\frac {4\pi e^{2}N}{L^{3}m}}$ is the three-dimensional plasma frequency (in SI units, replace the factor $4\pi$ by $1/\epsilon _{0}$.) For two-dimensional systems,

$\omega _{\rm {pl}}^{2}(\mathbf {q} )={\frac {2\pi e^{2}nq}{\epsilon m}}$.

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

$E_{\mathbf {k} -\mathbf {q} }-E_{\mathbf {k} }={\frac {\hbar ^{2}}{2m}}(k^{2}-2\mathbf {k} \cdot \mathbf {q} q^{2})-{\frac {\hbar ^{2}k^{2}}{2m}}\simeq -{\frac {\hbar ^{2}\mathbf {k} \cdot \mathbf {q} }{m}}$,

and for the numerator of the Lindhard formula, we get

$f_{\mathbf {k} -\mathbf {q} }-f_{\mathbf {k} }=f_{\mathbf {k} }-\mathbf {q} \cdot \nabla _{\mathbf {k} }f_{\mathbf {k} } \cdots -f_{\mathbf {k} }\simeq -\mathbf {q} \cdot \nabla _{\mathbf {k} }f_{\mathbf {k} }$.

Inserting these into the Lindhard formula and taking the $\delta \to 0$ limit, we obtain

${\begin{alignedat}{2}\epsilon (\mathbf {q} =0,\omega _{0})&\simeq 1 V_{\mathbf {q} }\sum _{\mathbf {k} ,i}{\frac {q_{i}{\frac {\partial f_{\mathbf {k} }}{\partial k_{i}}}}{\hbar \omega _{0}-{\frac {\hbar ^{2}\mathbf {k} \cdot \mathbf {q} }{m}}}}\\&\simeq 1 {\frac {V_{\mathbf {q} }}{\hbar \omega _{0}}}\sum _{\mathbf {k} ,i}{q_{i}{\frac {\partial f_{\mathbf {k} }}{\partial k_{i}}}}(1 {\frac {\hbar \mathbf {k} \cdot \mathbf {q} }{m\omega _{0}}})\\&\simeq 1 {\frac {V_{\mathbf {q} }}{\hbar \omega _{0}}}\sum _{\mathbf {k} ,i}{q_{i}{\frac {\partial f_{\mathbf {k} }}{\partial k_{i}}}}{\frac {\hbar \mathbf {k} \cdot \mathbf {q} }{m\omega _{0}}}\\&=1-V_{\mathbf {q} }{\frac {q^{2}}{m\omega _{0}^{2}}}\sum _{\mathbf {k} }{f_{\mathbf {k} }}\\&=1-V_{\mathbf {q} }{\frac {q^{2}N}{m\omega _{0}^{2}}}\\&=1-{\frac {4\pi e^{2}}{\epsilon q^{2}L^{3}}}{\frac {q^{2}N}{m\omega _{0}^{2}}}\\&=1-{\frac {\omega _{\rm {pl}}^{2}}{\omega _{0}^{2}}}.\end{alignedat}}$,

where we used $E_{\mathbf {k} }=\hbar \omega _{\mathbf {k} }$ and $V_{\mathbf {q} }={\frac {4\pi e^{2}}{\epsilon q^{2}L^{3}}}$.

Derivation in 2D

First, consider the long wavelength limit ($q\to 0$).

For the denominator of the Lindhard formula,

$E_{\mathbf {k} -\mathbf {q} }-E_{\mathbf {k} }={\frac {\hbar ^{2}}{2m}}(k^{2}-2\mathbf {k} \cdot \mathbf {q} q^{2})-{\frac {\hbar ^{2}k^{2}}{2m}}\simeq -{\frac {\hbar ^{2}\mathbf {k} \cdot \mathbf {q} }{m}}$,

and for the numerator,

$f_{\mathbf {k} -\mathbf {q} }-f_{\mathbf {k} }=f_{\mathbf {k} }-\mathbf {q} \cdot \nabla _{\mathbf {k} }f_{\mathbf {k} } \cdots -f_{\mathbf {k} }\simeq -\mathbf {q} \cdot \nabla _{\mathbf {k} }f_{\mathbf {k} }$.

Inserting these into the Lindhard formula and taking the limit of $\delta \to 0$, we obtain

${\begin{alignedat}{2}\epsilon (0,\omega )&\simeq 1 V_{\mathbf {q} }\sum _{\mathbf {k} ,i}{\frac {q_{i}{\frac {\partial f_{\mathbf {k} }}{\partial k_{i}}}}{\hbar \omega _{0}-{\frac {\hbar ^{2}\mathbf {k} \cdot \mathbf {q} }{m}}}}\\&\simeq 1 {\frac {V_{\mathbf {q} }}{\hbar \omega _{0}}}\sum _{\mathbf {k} ,i}{q_{i}{\frac {\partial f_{\mathbf {k} }}{\partial k_{i}}}}(1 {\frac {\hbar \mathbf {k} \cdot \mathbf {q} }{m\omega _{0}}})\\&\simeq 1 {\frac {V_{\mathbf {q} }}{\hbar \omega _{0}}}\sum _{\mathbf {k} ,i}{q_{i}{\frac {\partial f_{\mathbf {k} }}{\partial k_{i}}}}{\frac {\hbar \mathbf {k} \cdot \mathbf {q} }{m\omega _{0}}}\\&=1 {\frac {V_{\mathbf {q} }}{\hbar \omega _{0}}}2\int d^{2}k({\frac {L}{2\pi }})^{2}\sum _{i,j}{q_{i}{\frac {\partial f_{\mathbf {k} }}{\partial k_{i}}}}{\frac {\hbar k_{j}q_{j}}{m\omega _{0}}}\\&=1 {\frac {V_{\mathbf {q} }L^{2}}{m\omega _{0}^{2}}}2\int {\frac {d^{2}k}{(2\pi )^{2}}}\sum _{i,j}{q_{i}q_{j}k_{j}{\frac {\partial f_{\mathbf {k} }}{\partial k_{i}}}}\\&=1 {\frac {V_{\mathbf {q} }L^{2}}{m\omega _{0}^{2}}}\sum _{i,j}{q_{i}q_{j}2\int {\frac {d^{2}k}{(2\pi )^{2}}}k_{j}{\frac {\partial f_{\mathbf {k} }}{\partial k_{i}}}}\\&=1-{\frac {V_{\mathbf {q} }L^{2}}{m\omega _{0}^{2}}}\sum _{i,j}{q_{i}q_{j}n\delta _{ij}}\\&=1-{\frac {2\pi e^{2}}{\epsilon qL^{2}}}{\frac {L^{2}}{m\omega _{0}^{2}}}q^{2}n\\&=1-{\frac {\omega _{\rm {pl}}^{2}(\mathbf {q} )}{\omega _{0}^{2}}},\end{alignedat}}$

where we used $E_{\mathbf {k} }=\hbar \epsilon _{\mathbf {k} }$, $V_{\mathbf {q} }={\frac {2\pi e^{2}}{\epsilon qL^{2}}}$ and $\omega _{\rm {pl}}^{2}(\mathbf {q} )={\frac {2\pi e^{2}nq}{\epsilon m}}$.

Static limit

Consider the static limit ($\omega i\delta \to 0$).

The Lindhard formula becomes

$\epsilon (\mathbf {q} ,\omega =0)=1-V_{\mathbf {q} }\sum _{\mathbf {k} }{\frac {f_{\mathbf {k} -\mathbf {q} }-f_{\mathbf {k} }}{E_{\mathbf {k} -\mathbf {q} }-E_{\mathbf {k} }}}$.

Inserting the above equalities for the denominator and numerator, we obtain

$\epsilon (\mathbf {q} ,0)=1-V_{\mathbf {q} }\sum _{\mathbf {k} ,i}{\frac {-q_{i}{\frac {\partial f}{\partial k_{i}}}}{-{\frac {\hbar ^{2}\mathbf {k} \cdot \mathbf {q} }{m}}}}=1-V_{\mathbf {q} }\sum _{\mathbf {k} ,i}{\frac {q_{i}{\frac {\partial f}{\partial k_{i}}}}{\frac {\hbar ^{2}\mathbf {k} \cdot \mathbf {q} }{m}}}$.

Assuming a thermal equilibrium Fermi–Dirac carrier distribution, we get

$\sum _{i}{q_{i}{\frac {\partial f_{\mathbf {k} }}{\partial k_{i}}}}=-\sum _{i}{q_{i}{\frac {\partial f_{\mathbf {k} }}{\partial \mu }}{\frac {\partial E_{\mathbf {k} }}{\partial k_{i}}}}=-\sum _{i}{q_{i}k_{i}{\frac {\hbar ^{2}}{m}}{\frac {\partial f_{\mathbf {k} }}{\partial \mu }}}$

here, we used $E_{\mathbf {k} }={\frac {\hbar ^{2}k^{2}}{2m}}$ and ${\frac {\partial E_{\mathbf {k} }}{\partial k_{i}}}={\frac {\hbar ^{2}k_{i}}{m}}$.

Therefore,

${\begin{alignedat}{2}\epsilon (\mathbf {q} ,0)&=1 V_{\mathbf {q} }\sum _{\mathbf {k} ,i}{\frac {q_{i}k_{i}{\frac {\hbar ^{2}}{m}}{\frac {\partial f_{\mathbf {k} }}{\partial \mu }}}{\frac {\hbar ^{2}\mathbf {k} \cdot \mathbf {q} }{m}}}=1 V_{\mathbf {q} }\sum _{\mathbf {k} }{\frac {\partial f_{\mathbf {k} }}{\partial \mu }}=1 {\frac {4\pi e^{2}}{\epsilon q^{2}}}{\frac {\partial }{\partial \mu }}{\frac {1}{L^{3}}}\sum _{\mathbf {k} }{f_{\mathbf {k} }}\\&=1 {\frac {4\pi e^{2}}{\epsilon q^{2}}}{\frac {\partial }{\partial \mu }}{\frac {N}{L^{3}}}=1 {\frac {4\pi e^{2}}{\epsilon q^{2}}}{\frac {\partial n}{\partial \mu }}\equiv 1 {\frac {\kappa ^{2}}{q^{2}}}.\end{alignedat}}$

Here, $\kappa$ is the 3D screening wave number (3D inverse screening length) defined as

$\kappa ={\sqrt {{\frac {4\pi e^{2}}{\epsilon }}{\frac {\partial n}{\partial \mu }}}}$.

Then, the 3D statically screened Coulomb potential is given by

$V_{\rm {s}}(\mathbf {q} ,\omega =0)\equiv {\frac {V_{\mathbf {q} }}{\epsilon (\mathbf {q} ,0)}}={\frac {\frac {4\pi e^{2}}{\epsilon q^{2}L^{3}}}{\frac {q^{2} \kappa ^{2}}{q^{2}}}}={\frac {4\pi e^{2}}{\epsilon L^{3}}}{\frac {1}{q^{2} \kappa ^{2}}}$.

And the inverse Fourier transformation of this result gives

$V_{\rm {s}}(r)=\sum _{\mathbf {q} }{{\frac {4\pi e^{2}}{L^{3}(q^{2} \kappa ^{2})}}e^{i\mathbf {q} \cdot \mathbf {r} }}={\frac {e^{2}}{r}}e^{-\kappa r}$

known as the Yukawa potential. Note that in this Fourier transformation, which is basically a sum over all $\mathbf {q}$, we used the expression for small $|\mathbf {q} |$ for every value of $\mathbf {q}$ 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

$E_{\rm {F}}={\frac {\hbar ^{2}}{2m}}(3\pi ^{2}n)^{\frac {2}{3}}$,

So the density is

$n={\frac {1}{3\pi ^{2}}}\left({\frac {2m}{\hbar ^{2}}}E_{\rm {F}}\right)^{\frac {3}{2}}$.

At T=0, $E_{\rm {F}}\equiv \mu$, so ${\frac {\partial n}{\partial \mu }}={\frac {3}{2}}{\frac {n}{E_{\rm {F}}}}$.

Inserting this into the above 3D screening wave number equation, we obtain

$\kappa ={\sqrt {{\frac {4\pi e^{2}}{\epsilon }}{\frac {\partial n}{\partial \mu }}}}={\sqrt {\frac {6\pi e^{2}n}{\epsilon E_{\rm {F}}}}}$.

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 $\kappa ={\sqrt {\frac {4\pi e^{2}n\beta }{\epsilon }}}$, known as the 3D Debye–Hückel screening wave number.

In two dimensions, the screening wave number is

$\kappa ={\frac {2\pi e^{2}}{\epsilon }}{\frac {\partial n}{\partial \mu }}={\frac {2\pi e^{2}}{\epsilon }}{\frac {m}{\hbar ^{2}\pi }}(1-e^{-\hbar ^{2}\beta \pi n/m})={\frac {2me^{2}}{\hbar ^{2}\epsilon }}f_{k=0}.$

Note that this result is independent of n.

Derivation in 2D

Consider the static limit ($\omega i\delta \to 0$). The Lindhard formula becomes

$\epsilon (\mathbf {q} ,0)=1-V_{\mathbf {q} }\sum _{\mathbf {k} }{\frac {f_{\mathbf {k} -\mathbf {q} }-f_{\mathbf {k} }}{E_{\mathbf {k} -\mathbf {q} }-E_{\mathbf {k} }}}$.

Inserting the above equalities for the denominator and numerator, we obtain

$\epsilon (\mathbf {q} ,0)=1-V_{\mathbf {q} }\sum _{\mathbf {k} ,i}{\frac {-q_{i}{\frac {\partial f}{\partial k_{i}}}}{-{\frac {\hbar ^{2}\mathbf {k} \cdot \mathbf {q} }{m}}}}=1-V_{\mathbf {q} }\sum _{\mathbf {k} ,i}{\frac {q_{i}{\frac {\partial f}{\partial k_{i}}}}{\frac {\hbar ^{2}\mathbf {k} \cdot \mathbf {q} }{m}}}$.

Assuming a thermal equilibrium Fermi–Dirac carrier distribution, we get

$\sum _{i}{q_{i}{\frac {\partial f_{\mathbf {k} }}{\partial k_{i}}}}=-\sum _{i}{q_{i}{\frac {\partial f_{\mathbf {k} }}{\partial \mu }}{\frac {\partial E_{\mathbf {k} }}{\partial k_{i}}}}=-\sum _{i}{q_{i}k_{i}{\frac {\hbar ^{2}}{m}}{\frac {\partial f_{\mathbf {k} }}{\partial \mu }}}$.

Therefore,

${\begin{alignedat}{2}\epsilon (\mathbf {q} ,0)&=1 V_{\mathbf {q} }\sum _{\mathbf {k} ,i}{\frac {q_{i}k_{i}{\frac {\hbar ^{2}}{m}}{\frac {\partial f_{\mathbf {k} }}{\partial \mu }}}{\frac {\hbar ^{2}\mathbf {k} \cdot \mathbf {q} }{m}}}=1 V_{\mathbf {q} }\sum _{\mathbf {k} }{\frac {\partial f_{\mathbf {k} }}{\partial \mu }}=1 {\frac {2\pi e^{2}}{\epsilon qL^{2}}}{\frac {\partial }{\partial \mu }}\sum _{\mathbf {k} }{f_{\mathbf {k} }}\\&=1 {\frac {2\pi e^{2}}{\epsilon q}}{\frac {\partial }{\partial \mu }}{\frac {N}{L^{2}}}=1 {\frac {2\pi e^{2}}{\epsilon q}}{\frac {\partial n}{\partial \mu }}\equiv 1 {\frac {\kappa }{q}}.\end{alignedat}}$

$\kappa$ is 2D screening wave number(2D inverse screening length) defined as

$\kappa ={\frac {2\pi e^{2}}{\epsilon }}{\frac {\partial n}{\partial \mu }}$.

Then, the 2D statically screened Coulomb potential is given by

$V_{\rm {s}}(\mathbf {q} ,\omega =0)\equiv {\frac {V_{\mathbf {q} }}{\epsilon (\mathbf {q} ,0)}}={\frac {2\pi e^{2}}{\epsilon qL^{2}}}{\frac {q}{q \kappa }}={\frac {2\pi e^{2}}{\epsilon L^{2}}}{\frac {1}{q \kappa }}$.

It is known that the chemical potential of the 2-dimensional Fermi gas is given by

$\mu (n,T)={\frac {1}{\beta }}\ln {(e^{\hbar ^{2}\beta \pi n/m}-1)}$,

and ${\frac {\partial \mu }{\partial n}}={\frac {\hbar ^{2}\pi }{m}}{\frac {1}{1-e^{-\hbar ^{2}\beta \pi n/m}}}$.

Experiments on one dimensional systems

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. For a K₂Pt(CN)₄Cl_0.32·2.6H₂0 filament, it was found that the potential within the region between the filament and cylinder varies as $e^{-k_{\rm {eff}}r}/r$ and its effective screening length is about 10 times that of metallic platinum.

See also

• Kohn anomaly
• Pomeranchuk instability
• Friedel oscillations

References

General

• 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.