Plasmons are quantized oscillations of the charge density in solids, and are found in basically all conducting materials in nature.

For light in vacuum, the dispersion relation is required for propagation at $c$, and you have $\omega=ck$ as a strict relation. For acoustic phonons, you similarly have a linear dispersion relation via Goldstones theorem (with the usual subtleties).

The plasmon however, is essentially always gapped/massive, and I would assume then that Goldstone's theorem and its extensions would have nothing to say about the plasmon dispersion relation.

It is still possible to estimate the dispersion relation within the self-consistent field approximation/RPA (basically valid in the large electron density limit). In this approximation you generally get a dispersion relation for the plasmon of the following form (Section 15 of Quantum Theory of Many-Particle Systems by Fetter and Walecka):

$$\omega(q) = \omega_0\left(1+\frac{9}{10}\left(\frac{q}{q_{\mathrm{TF}}}\right)^2 + O(q^4)\right)=\omega_0\left(1+a q^2 + O(q^4)\right)$$

Here $q_\mathrm{TF}$ is the Thomas-Fermi wavelength, which is simply $q_\mathrm{TF} = \sqrt{\frac{6\pi n e^2}{E_f}}$. Where $n$ is the electron density, and $E_f$ is the Fermi energy, which in 3D is proportional to $n^{2/3}$. The plasmon frequency is given by $\omega_0=\sqrt{\frac{4\pi n e^2}{m}}$

We can rewrite this expression in terms of the Fermi energy and plasma frequency as well

$$\omega(q)-\omega_0 \propto \frac{E_F}{\omega_0} q^2 \propto n^{1/6}q^2$$

Although the calculation for this dispersion term is very simple and straightforward, I am struggling to understand why I would expect

  1. To lowest order, $q^2$ dispersion for the plasmon and
  2. A coefficient of the form $a=\frac{\omega_0}{q_{\mathrm{TF}}^2}$

There are basically 3 contributions to calculating $1=V\,\mathrm{Re}\left[\Pi\right]$, which gives the plasmon energy: the Coulomb interaction ($e^2/q^{2}$ in 3D), the phase space over which $\Pi$ is integrated (a mix of $\omega$ and $q$), and how the energy of electrons disperse $q^2$. Which ones are giving what contributions?

Naively if I think of the $q^2$ from coming due to purely kinetic energy contributions, you would expect the coefficient $a\propto n$, but instead we have $a\propto n^{1/6}$. Additionally, the classical solution (parallel plate capacitor essentially) predicts no dispersion, so the effect should be due to the existence of the Fermi surface and its dispersion.


1 Answer 1


So what you do, is you search for the long living, resonant, quasi-particle excitations of your system. Their dispersion is given by $\epsilon(q, \omega(q))=1-v_{q}\chi(q, \omega(q))=0$, where $\epsilon(q, \omega)$ is the dielectric function and $\chi(q, \omega)$ is the polarization buble and $v_{q}$ is the Fourier image of the bare interaction. You search for the solution to this equation near $q=0$, that is because the damping of the quasi-particle excitations is the least in that region (in RPA it is actually zero).You get the quadratic dispersion because you assume that your bare interaction is the long-ranged Coulomb one, i.e. $v_{q}=\frac{4\pi}{q^{2}}$. If instead you've started with a short range interaction $v(x)\propto\delta(x)$, you would get a linear dispersion. Just try it. Note that the polarisation bubble (in RPA, the Lindhard function) would not change. In classical theory you do not get any dispersion, that just corresponds to long wavelength limit of the quantum result $q\rightarrow0$, then $\omega\rightarrow\omega_{pl}$.


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.