I was reading Tommy Ohlsson's book on Relativistic Quantum Mechanics where he goes to a little digression on electrons and holes in Graphene. He claims that electrons and holes in Graphene can be considered as relativistic massless fermions and hence can be described by Weyl equations. Does anyone have any idea whether this description is really viable? If yes, why?
[Physics] Why are the electrons in graphene massless
condensed-mattergraphene
Related Solutions
According to this article: http://physics.aps.org/articles/v5/24:
The statement that in graphene the "conduction electrons are massless" is because the energy levels (bands) are proportional to their momenta.
So the $E = \sqrt{p^2+m^2}$ relation of a free electron becomes $E\propto p$ in graphene.
Massless particles travel all at the same speed because of the $E\propto p$ relation but this characteristic velocity in graphene is far below c though, only 0.3% of the speed of light.
The reason that the relation $E\propto p$ leads to a characteristic speed is due to the quantum mechanical wave character. $E$ is proportional to the phase changes in time, $p$ is proportional to the phase changes in space and therefor $p/E$ is proportional to the velocity. In the case that $E\propto p$ there is a characteristic velocity $v$ independent of the energy level.
The most striking aspect of graphene is that its electronic energy levels, or “bands,” produce conduction electrons whose energies are directly proportional to their momentum. This is the energy-momentum relationship exhibited by photons, which are massless particles of light. Electrons and other particles of matter normally have energies that depend on the square of their momentum.
When the bands are plotted in three dimensions, the photonlike energy-momentum relationship appears as an inverted cone, called a Dirac cone. This unusual relationship causes conduction electrons to behave as though they were massless, like photons, so that all of them travel at roughly the same speed (about 0.3 percent of the speed of light). This uniformity leads to a conductivity greater than copper.
Hans
The group velocity $v_g$ of a wave packet (that's the speed of the maximum of the wave packet) is given by $v_g=\frac{\partial\omega}{\partial k}$. In this case, $\frac{\partial\omega}{\partial k}=\frac 1 \hbar\frac{\partial E}{\partial k}$, which easily evaluates to $v_g=\frac{3ta}{2}=:v_f$ for $k=0$. That's actually the definition of $v_f$: it is the group velocity at $k=K$ ($K$ is the point in the Graphene bandstructure where the Dirac cone occurs - note that it is a vector because $k$ has an $x$ and a $y$ component), because $E(K)=E_f$.
The effective mass from solid state physics is indeed infinite. If one talks about "zero effective mass Dirac fermions" in Graphene, this comes from the massless Dirac equation which has the same dispersion relation. The solid state physics effective mass doesn't work here, because the dispersion relation needs to be parabolic (not linear with a cusp), there are two papers about that on arXiv here and here.
Best Answer
Yes, low energy electrons and holes in graphene can be thought of as massless because of the linear dispersion of the band structure near the K points. This is an analogy to the relativistic energy dispersion $E^2=p^2c^2+m^2c^4$, which becomes linear in momentum for $m=0$. This linear dispersion has been confirmed countless times via optical and electrical means, and it is the source of many of the bizarre relativistic-like properties that made graphene famous (e.g. Klein tunneling and the half-integer quantum Hall effect).