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# Lindhard theory

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### Lindhard theory

Lindhard theory[1][2] is a method of calculating the effects of electric field screening by electrons in a solid. It is based on quantum mechanics and the random phase approximation.

Thomas-Fermi screening 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.[2]

## Formula

Lindhard formula for the longitudinal dielectric function is given by

 $\epsilon\left(q,\omega\right) = 1 - V_q \sum_k\left\{\frac\left\{f_\left\{k-q\right\}-f_k\right\}\left\{\hbar\left(\omega+i\delta\right)+E_\left\{k-q\right\}-E_k\right\}\right\}.$

Here, $V_q$ is $V_\left\{eff\right\}\left(q\right) - V_\left\{ind\right\}\left(q\right)$ and $f_k$ is the carrier distribution function which is the Fermi-Dirac distribution function(see also Fermi–Dirac statistics) for electrons in thermodynamic equilibrium. However this Lindhard formula is valid also for nonequilibrium distribution functions.

## Analysis of the Lindhard formula

For understanding the Lindhard formula, let's consider some limiting cases in 3 dimensions and 2 dimensions. 1 dimension case is also considered in other way.

### Three Dimensions

#### Long Wave-length Limit

First, consider the long wavelength limit ($q\to0$).

For denominator of Lindhard formula,

$E_\left\{k-q\right\} - E_k = \frac\left\{\hbar^2\right\}\left\{2m\right\}\left(k^2-2\vec\left\{k\right\}\cdot\vec\left\{q\right\}+q^2\right) - \frac\left\{\hbar^2 k^2\right\}\left\{2m\right\} \simeq -\frac\left\{\hbar^2 \vec\left\{k\right\}\cdot\vec\left\{q\right\}\right\}\left\{m\right\}$,

and for numerator of Lindhard formula,

$f_\left\{k-q\right\} - f_k = f_k - \vec\left\{q\right\}\cdot\nabla_k f_k + \cdots - f_k \simeq - \vec\left\{q\right\}\cdot\nabla_k f_k$.

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



\begin{alignat}{2} \epsilon(0,\omega) & \simeq 1 + V_q \sum_{k,i}{ \frac{q_i \frac{\partial f_k}{\partial k_i}}{\hbar \omega_0 - \frac{\hbar^2 \vec{k}\cdot\vec{q}}{m}} }\\ & \simeq 1 + \frac{V_q}{\hbar \omega_0} \sum_{k,i}{q_i \frac{\partial f_k}{\partial k_i}}(1+\frac{\hbar \vec{k}\cdot\vec{q}}{m \omega_0})\\ & \simeq 1 + \frac{V_q}{\hbar \omega_0} \sum_{k,i}{q_i \frac{\partial f_k}{\partial k_i}}\frac{\hbar \vec{k}\cdot\vec{q}}{m \omega_0}\\ & = 1 - V_q \frac{q^2}{m \omega_0^2} \sum_k{f_k}\\ & = 1 - V_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_{pl}^2}{\omega_0^2} \end{alignat} , where we used $E_k = \hbar \omega_k$, $V_q = \frac\left\{4 \pi e^2\right\}\left\{\epsilon q^2 L^3\right\}$ and $\omega_\left\{pl\right\}^2 = \frac\left\{4 \pi e^2 N\right\}\left\{\epsilon L^3 m\right\}$.

(In SI units, replace the factor $4\pi$ by $1/\epsilon_\left\{0\right\}$.)

This result is same as the classical dielectric function.

#### Static Limit

Second, consider the static limit ($\omega + i\delta \to 0$). The Lindhard formula becomes

$\epsilon\left(q,0\right) = 1 - V_q \sum_k\left\{\frac\left\{f_\left\{k-q\right\}-f_k\right\}\left\{E_\left\{k-q\right\}-E_k\right\}\right\}$.

Inserting above equalities for denominator and numerator to this, we obtain

$\epsilon\left(q,0\right) = 1 - V_q \sum_\left\{k,i\right\}\left\{\frac\left\{-q_i \frac\left\{\partial f\right\}\left\{\partial k_i\right\} \right\}\left\{ -\frac\left\{\hbar^2 \vec\left\{k\right\}\cdot\vec\left\{q\right\}\right\}\left\{m\right\} \right\}\right\}$
= 1 - V_q \sum_{k,i}{\frac{q_i \frac{\partial f}{\partial k_i} }{\frac{\hbar^2 \vec{k}\cdot\vec{q}}{m} }}.


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

$\sum_\left\{i\right\}\left\{ q_i \frac\left\{\partial f_k\right\}\left\{\partial k_i\right\} \right\} = -\sum_\left\{i\right\}\left\{ q_i \frac\left\{\partial f_k\right\}\left\{\partial \mu\right\} \frac\left\{\partial \epsilon_k\right\}\left\{\partial k_i\right\} \right\} = -\sum_\left\{i\right\}\left\{ q_i k_i \frac\left\{\hbar^2\right\}\left\{m\right\} \frac\left\{\partial f_k\right\}\left\{\partial \mu\right\}\right\}$

here, we used $\epsilon_k = \frac\left\{\hbar^2 k^2\right\}\left\{2m\right\}$ and $\frac\left\{\partial \epsilon_k\right\}\left\{\partial k_i\right\} = \frac\left\{\hbar^2 k_i\right\}\left\{m\right\}$.

Therefore,



\begin{alignat}{2} \epsilon(q,0) & = 1 + V_q \sum_{k,i}{\frac{ q_i k_i \frac{\hbar^2}{m} \frac{\partial f_k}{\partial \mu} }{\frac{\hbar^2 \vec{k}\cdot\vec{q}}{m} }} = 1 + V_q\sum_k{\frac{\partial f_k}{\partial \mu}} = 1 + \frac{4 \pi e^2}{\epsilon q^2} \frac{\partial}{\partial \mu} \frac{1}{L^3} \sum_k{f_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{alignat}

$\kappa$ is 3D screening wave number(3D inverse screening length) defined as $\kappa = \sqrt\left\{ \frac\left\{4\pi e^2\right\}\left\{\epsilon\right\} \frac\left\{\partial n\right\}\left\{\partial \mu\right\} \right\}$.

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

$V_s\left(q,\omega=0\right) \equiv \frac\left\{V_q\right\}\left\{\epsilon\left(q,\omega=0\right)\right\} = \frac \left\{\frac\left\{4 \pi e^2\right\}\left\{\epsilon q^2 L^3\right\} \right\}\left\{ \frac\left\{q^2 + \kappa^2\right\}\left\{q^2\right\} \right\} = \frac\left\{4 \pi e^2\right\}\left\{\epsilon L^3\right\} \frac\left\{1\right\}\left\{q^2 + \kappa^2\right\}$.

And Fourier transformation of this result gives

$V_s\left(r\right) = \sum_q\left\{ \frac\left\{4\pi e^2\right\}\left\{\epsilon L^3 \left(q^2+\kappa^2\right)\right\} e^\left\{i \vec\left\{q\right\} \cdot \vec\left\{r\right\}\right\} \right\} = \frac\left\{e^2\right\}\left\{\epsilon r\right\} e^\left\{-\kappa r\right\}$

known as the Yukawa potential.

For a degenerating gas(T=0), Fermi energy is given by

$E_f = \frac\left\{\hbar^2\right\}\left\{2m\right\}\left(3\pi^2 n\right)^\left\{\frac\left\{2\right\}\left\{3\right\}\right\}$,

So the density is

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

At T=0, $E_f \equiv \mu$, so $\frac\left\{\partial n\right\}\left\{\partial \mu\right\} = \frac\left\{3\right\}\left\{2\right\}\frac\left\{n\right\}\left\{E_f\right\}$.

Inserting this to above 3D screening wave number equation, we get

 $\kappa = \sqrt\left\{ \frac\left\{4\pi e^2\right\}\left\{\epsilon\right\} \frac\left\{\partial n\right\}\left\{\partial \mu\right\} \right\} = \sqrt\left\{ \frac\left\{6\pi e^2 n\right\}\left\{\epsilon E_f\right\} \right\}$.

This is 3D Thomas–Fermi screening wave number.

For reference, Debye-Hückel screening describes the nondegenerate limit case.

The result is $\kappa = \sqrt\left\{ \frac\left\{4\pi e^2 n \beta\right\}\left\{\epsilon\right\} \right\}$, 3D Debye-Hückel screening wave number.

### Two Dimensions

#### Long Wave-length Limit

First, consider the long wavelength limit ($q\to0$).

For denominator of Lindhard formula,

$E_\left\{k-q\right\} - E_k = \frac\left\{\hbar^2\right\}\left\{2m\right\}\left(k^2-2\vec\left\{k\right\}\cdot\vec\left\{q\right\}+q^2\right) - \frac\left\{\hbar^2 k^2\right\}\left\{2m\right\} \simeq -\frac\left\{\hbar^2 \vec\left\{k\right\}\cdot\vec\left\{q\right\}\right\}\left\{m\right\}$,

and for numerator of Lindhard formula,

$f_\left\{k-q\right\} - f_k = f_k - \vec\left\{q\right\}\cdot\nabla_k f_k + \cdots - k_k \simeq - \vec\left\{q\right\}\cdot\nabla_k f_k$.

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



\begin{alignat}{2} \epsilon(0,\omega) & \simeq 1 + V_q \sum_{k,i}{ \frac{q_i \frac{\partial f_k}{\partial k_i}}{\hbar \omega_0 - \frac{\hbar^2 \vec{k}\cdot\vec{q}}{m}} }\\ & \simeq 1 + \frac{V_q}{\hbar \omega_0} \sum_{k,i}{q_i \frac{\partial f_k}{\partial k_i}}(1+\frac{\hbar \vec{k}\cdot\vec{q}}{m \omega_0})\\ & \simeq 1 + \frac{V_q}{\hbar \omega_0} \sum_{k,i}{q_i \frac{\partial f_k}{\partial k_i}}\frac{\hbar \vec{k}\cdot\vec{q}}{m \omega_0}\\ & = 1 + \frac{V_q}{\hbar \omega_0} 2 \int d^2 k (\frac{L}{2 \pi})^2 \sum_{i,j}{q_i \frac{\partial f_k}{\partial k_i}}\frac{\hbar k_j q_j}{m \omega_0}\\ & = 1 + \frac{V_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_k}{\partial k_i}}\\ & = 1 + \frac{V_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_k}{\partial k_i}}\\ & = 1 - \frac{V_q L^2}{m \omega_0^2} \sum_{i,j}{ q_i q_j 2 \int \frac{d^2 k}{(2 \pi)^2} k_k \frac{\partial f_j}{\partial k_i}}\\ & = 1 - \frac{V_q L^2}{m \omega_0^2} \sum_{i,j}{ q_i q_j n \delta_{ij}}\\ & = 1 - \frac{2 \pi e^2}{\epsilon q L^2} \frac{L^2}{m \omega_0^2} q^2 n\\ & = 1 - \frac{\omega_{pl}^2(q)}{\omega_0^2}, \end{alignat} where we used $E_k = \hbar \epsilon_k$, $V_q = \frac\left\{2 \pi e^2\right\}\left\{\epsilon q L^2\right\}$ and $\omega_\left\{pl\right\}^2\left(q\right) = \frac\left\{2 \pi e^2 n q\right\}\left\{\epsilon m\right\}$.

#### Static Limit

Second, consider the static limit ($\omega + i\delta \to 0$). The Lindhard formula becomes

$\epsilon\left(q,0\right) = 1 - V_q \sum_k\left\{\frac\left\{f_\left\{k-q\right\}-f_k\right\}\left\{E_\left\{k-q\right\}-E_k\right\}\right\}$.

Inserting above equalities for denominator and numerator to this, we obtain

$\epsilon\left(q,0\right) = 1 - V_q \sum_\left\{k,i\right\}\left\{\frac\left\{-q_i \frac\left\{\partial f\right\}\left\{\partial k_i\right\} \right\}\left\{ -\frac\left\{\hbar^2 \vec\left\{k\right\}\cdot\vec\left\{q\right\}\right\}\left\{m\right\} \right\}\right\}$
= 1 - V_q \sum_{k,i}{\frac{q_i \frac{\partial f}{\partial k_i} }{\frac{\hbar^2 \vec{k}\cdot\vec{q}}{m} }}.


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

$\sum_\left\{i\right\}\left\{ q_i \frac\left\{\partial f_k\right\}\left\{\partial k_i\right\} \right\} = -\sum_\left\{i\right\}\left\{ q_i \frac\left\{\partial f_k\right\}\left\{\partial \mu\right\} \frac\left\{\partial \epsilon_k\right\}\left\{\partial k_i\right\} \right\} = -\sum_\left\{i\right\}\left\{ q_i k_i \frac\left\{\hbar^2\right\}\left\{m\right\} \frac\left\{\partial f_k\right\}\left\{\partial \mu\right\}\right\}$

here, we used $\epsilon_k = \frac\left\{\hbar^2 k^2\right\}\left\{2m\right\}$ and $\frac\left\{\partial \epsilon_k\right\}\left\{\partial k_i\right\} = \frac\left\{\hbar^2 k_i\right\}\left\{m\right\}$.

Therefore,



\begin{alignat}{2} \epsilon(q,0) & = 1 + V_q \sum_{k,i}{\frac{ q_i k_i \frac{\hbar^2}{m} \frac{\partial f_k}{\partial \mu} }{\frac{\hbar^2 \vec{k}\cdot\vec{q}}{m} }} = 1 + V_q\sum_k{\frac{\partial f_k}{\partial \mu}} = 1 + \frac{2 \pi e^2}{\epsilon q L^2} \frac{\partial}{\partial \mu} \sum_k{f_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{alignat}

$\kappa$ is 2D screening wave number(2D inverse screening length) defined as $\kappa = \frac\left\{2\pi e^2\right\}\left\{\epsilon\right\} \frac\left\{\partial n\right\}\left\{\partial \mu\right\}$.

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

$V_s\left(q,\omega=0\right) \equiv \frac\left\{V_q\right\}\left\{\epsilon\left(q,\omega=0\right)\right\} = \frac\left\{2 \pi e^2\right\}\left\{\epsilon q L^2\right\} \frac\left\{q\right\}\left\{q + \kappa\right\} = \frac\left\{2 \pi e^2\right\}\left\{\epsilon L^2\right\} \frac\left\{1\right\}\left\{q + \kappa\right\}$.

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

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

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

So, the 2D screening wave number is

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

Note that this result is independent of n.

### One Dimension

This time, let's consider some generalized case for lowering the dimension. The lower the dimensions is, the weaker the screening effect is. In lower dimension, some of the field lines pass through the barrier material wherein the screening has no effect. For 1 dimensional case, we can guess that the screening effects only on the field lines which are very close to the wire axis.

#### Experiment

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. D. Davis applied the Thomas–Fermi screening to an electron gas confined to a filament and a coaxial cylinder. For K2Pt(CN)4Cl0.32·2.6H20, it was found that the potential within the region between the filament and cylinder varies as $e^\left\{-k_\left\{eff\right\}r\right\}/r$ and its effective screening length is about 10 times that of metallic platinum.

• Electric field screening

## References

• D. Davis Thomas-fermi screening in one dimension, Phys. Rev. B, 7(1), 129, (1973)
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