# Why does the velocity of 2D plasmon diverge at small q?

## Main Question or Discussion Point

Now plasma represents collective wave-like motions of charged particles. In 3D, their frequency is well known to be almost a constant, $\omega^{3D}_p \approx \sqrt{4\pi ne^2/m}$ with n=charge density, m=particle mass. However, in 2D, one can show that it becomes $\omega^{2D}_p \sim \sqrt{q}$, where q= wave number. It gets dispersion. The group velocity is then $v^{2D}_g \sim 1/\sqrt{q}$, which diverges as q→0. How could this be possible ? How could $v^{2D}_g$ exceed the speed of light ? This sounds ridiculous !

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DrDu
Could you provide a reference? There are no truely 2d systems in nature, so this may depend on details of the system.

DrDu
In that derivation, they assumed that the charges only interact via Coulombic forces. This is strictly true in the 3D case, as the charge density is longitudinal and the Coulomb field is the only longitudinal electric field. In the 2D case this is no longer true. You can calculate the field of a time dependent charge distribution locatet in the xy-plane and travelling in the x-direction: $\rho(t,r)=\rho_0 \delta(z) \exp(iqx-i\omega t)$ and you will find that it's electric field is not purely longitudinal. Hence the interaction is not purely Coulombic. I suppose this will change the dispersion relation, especially for small q.

Thanks for your suggestion. But that kind of dispersion had been experimentally verified: see e.g. Phys. Rev. Lett. 36: 145(1976).

DrDu
I must say that I don't understand this paper: In eq (5) they express div E in terms of first the potential and then again in terms of the charge density, but $\nabla\cdot E=\rho$ by Maxwells equation, so there must be something fishy.

DrDu
Ok, I suppose he means the nabla in reduced density. Using the ansatz for the charge distribution given above, I find for the electric potential $E=(\kappa/(iq) e_x+\mathrm{sgn}(z)e_z)\rho/2 \exp(-\kappa |z|) \exp(iqx-i\omega t)$, where $\kappa=\sqrt(|\omega^2-q^2|)$.
Hence the driving force in equation 5 does not depend linearly on q but on kappa. For large values of q, kappa goes like q but for small values like omega. Hence for small values of q there should the dispersion relation changes.

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DrDu
there might be some small errors, but the expression for E is a correct solution of the wave eqation for z not equal zero while forming the divergence yields the correct charge density. Forming the 2d-divergence in eq. 5 ammounts to multiplication of the x component with iq which yields kappa.
Maybe you could post your solution?

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I found the electric potential to be $V(\vec{r},z) = \frac{\rho_0}{2}\kappa^{-1}e^{i(\vec{q}\vec{r}-\omega t)}e^{-\kappa|z|}$, where $\vec{r}$ denotes the in-plane components. The electric field strength in the x-direction is $E_x = \frac{\rho_0}{2}(-iq_x\kappa^{-1})e^{i(\vec{q}\vec{r}-\omega t)}e^{-\kappa|z|}$. Here I keep the speed of light so that $\kappa = \sqrt{|q^2-\omega^2/c^2|}$. Now the plasmon dispersion is determined by this equation, $\omega^4|q^2-\omega^2/c^2| = \frac{2\pi n_0 e^2}{m}q^4$. Indeed the initial problem is solved !

I also found that, if both scalar and vector potentials are included, the plasmon dispersion is determined by an even simple equation: $\left(2\pi n_0 e^2/m\right)^2 |q^2-\omega^2/c^2| = \omega^4$. Obviously, the dispersion becomes linear for small $q$.

This is what looks like:

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DrDu
You can't derive a retarded field from only a potential. Namely your potential gives $\nabla^2 V=\omega^2 V\ne 0$ for $z \ne 0$, however, it should be the charge density which vanishes for nonzero z.

The retarded potential I gave is a solution to this equation: $(\partial^2_{\vec{x}}-\partial^2_t/c^2) V(\vec{x},t)= -\rho_0e^{i(\vec{q}\vec{r}-\omega t)}\delta(z)$.

DrDu
But the electric field is cannot be expressed in terms of a retarded potential alone, you will also have to specify the retarded magnetic vector potential.

That is exactly what I did. I have worked out the vector potential as well. The final result was shown in the figure attached in previous post.

DrDu
So what is the z-component of the electric field you found?

The z-component is obtained as $E_z = -\partial_z V(\vec{x},t) - c^{-1}\partial_tA_z(\vec{x},t)$, where $A_z$ is the z-component of the vector potential. The final expressions are a little complicated, because now the vector potential depends on the current density, which further depends on the conductivity of the system.

DrDu
Current density is completely determined by continuity.

Dear DrDu,

I'm much grateful for your comments which have led me to a good understanding of my problem. As this understanding is significant in improving one of my recent manuscripts, which I'll submit for publication, I'm thinking formally acknowledging you or adding you as a coauthor. So, if possible, I hope to discuss more by private emails, instead here. I definitely have to know you real profile before I can do this.

Thanks.

hiyok

DrDu
I just want to tie some loose ends: due to continuity $\nabla \cdot j=\dot{\rho}$ we find $j=\omega \rho/q e_x$ and thus $A=V\omega/q e_x$ (don't try to pin me down on signs). Hence $E_z$ can be derived from the potential you gave, but there is an extra term in $E_x$ due to the vector potential. A superficial check showed that the final expression for $E_x$ coincides with the one I derived earlier.
Now I agree with you. I thought your were referring to the part due to $V$ only. Putting all pieces together, we agree.