I'm having a question concerning the Lindhard function. The reference I'm using is the standard text "Quantum Theory of Solids" by Charles Kittel. I'm concerned with Chapter 6, subchapter "Method of the self-consistent field".

The goal is to solve the quantum mechanical eigenvalue problem of the Hamiltonian $H=H_{0}+V(x,t)$ where $V(x,t)=V_{0}(t)+V_{s}$ is a Hartree-Fock-type potential consisting of a time-dependent external potential (e.g. the source of an external electric field) and a screening potential $V_{s}$. $H_{0}$ is a standard kinetic-energy Hamiltonian.

The problem is solved based on quantum statistical mechanics. We assume that the eigenstates of $|m \rangle$ and the eigenstates of $H_{0}$ are plane waves $|k \rangle$.

The overall idea seems to be to find the density matrix time evolution to linear order expressed in terms of the plane wave states.

Without alaborating more on the details of Kittels calculation, the point is that I don't why the "eigenfrequencies" (what eigenfrequencies does he mean?? The diagonal elements of the density matrix when the density matrix is diagonalized) are the zeros of $\epsilon(\omega,q)$. I don't see why this is the case. Furthermore I don't understand why there are two types of eigenfrequencies. Obviously the Plasma frequency produces a zero of $\epsilon(q,w)$ at small $q$ but why is $\omega = \epsilon_{k+q} -\epsilon_{k}$ a zero. To me this gives a $2\omega$ in the the denominator of the dielectric function.

I'm a bit confused. Can someone maybe address my question and summarize briefly the key idea of the whole thing?

This would be really helpful!! :))

I'm looking forward to your responses!!


1 Answer 1


I hope the numbers of the equations are the same in our copies of the book.


The solution of the equation (6.15) has the following form: $$ \delta\rho(t) = \delta\rho \; e^{-i\omega t}. $$ If you substitute this into the equation it will remove the time derivative: $$ i\frac{\partial}{\partial t} \rightarrow \omega. $$ The values of $\omega$ that correspond to the solutions of (6.15) are the eigenfrequencies.

Zeros of dielectric function

The equation (6.15) can be rewritten as follows: $$ \left<\mathbf{k}\right|\delta\rho\left|\mathbf{k}+\mathbf{q}\right> = \frac{4\pi e^2}{q^2} \frac{ f_0(\varepsilon_{\mathbf{k}+\mathbf{q}}) - f_0(\varepsilon_\mathbf{k}) }{ \varepsilon_{\mathbf{k}+\mathbf{q}} - \varepsilon_\mathbf{k} + \omega } \sum_{\mathbf{k}'} \left<\mathbf{k}'\right|\delta\rho\left|\mathbf{k}'+\mathbf{q}\right> $$

If you introduce small dissipation ($\omega + is$) and take a sum over $\mathbf{k}$ for both parts of this equation you will get something close to the expression (6.23) for $\epsilon(\omega,\mathbf{q})$.

Values of the eigenfrequency

Before saying that the following $$ \omega \approx \varepsilon_{\mathbf{k}+\mathbf{q}} - \varepsilon_\mathbf{k} \qquad (a) $$ is a solution, Charles Kittel refers to this article. In the article in equation (7) the denominator looks like this: $$ \varepsilon_{\mathbf{k}+\mathbf{q}} - \varepsilon_\mathbf{k} - \omega; $$ while in Kittel's book we see $$ \varepsilon_{\mathbf{k}+\mathbf{q}} - \varepsilon_\mathbf{k} + \omega. $$ So the eigenfrequency should be $$ \omega \approx \varepsilon_\mathbf{k} - \varepsilon_{\mathbf{k}+\mathbf{q}}. \qquad (b) $$

The difference is in the order of the states in the matrix element which is always the following: $$ \left<\text{final}\right|\rho\left|\text{initial}\right>. $$ In Kittel's book the initial state is denoted as $\mathbf{k}+\mathbf{q}$ while $\mathbf{k}$ is the final state. So the position of the hole (initial position of the electron inside the Fermi sphere) should be $\mathbf{k}+\mathbf{q}$ not $\mathbf{k}$. There is a typo in the corresponding paragraph in the book. The correct sentence is:

The eigenvalues of the first type - $\omega \approx \varepsilon_\mathbf{k} - \varepsilon_{\mathbf{k}+\mathbf{q}}$ - describe the energy necessary for the birth of an electron-hole couple with transition of an electron from the state $\mathbf{k}+\mathbf{q}$ inside the Fermi sphere to the state $\mathbf{k}$ outside.

  • $\begingroup$ Thanks for your answer. I think it's clear to me now. Nonetheless I think I got another questino which I posted here: physics.stackexchange.com/questions/19415/…. $\endgroup$
    – CondMatter
    Commented Jan 12, 2012 at 10:55
  • $\begingroup$ I'd be more than happy if you could adress this one as well:) $\endgroup$
    – CondMatter
    Commented Jan 12, 2012 at 10:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.