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In graphene, we have (in the low energy limit) the linear energy-momentum dispersion relation: $E=\hbar v_{\rm{F}}|k|$. This expression arises from a tight-binding model, in fact $E =\frac{3\hbar ta}{2}|k|$ where $t$ is the nearest-neighbor hopping energy and $a$ the interatomic distance. But how does one know that the Fermi velocity is given by $v_{\rm{F}} = \frac{3ta}{2}$? Normally one would use $m_{\rm{eff}}^{-1} =\frac{1}{\hbar^{2}} \frac{\partial^{2}E}{\partial k^{2}}$ and $v_{\rm{F}} = \frac{\hbar k_{\rm{F}}}{m_{\rm{eff}}}$, but in this case $m_{\rm{eff}} = \infty$.

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up vote 4 down vote accepted

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.

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Are you sure that the definition of $v_{F}$ isn't the group velocity at $k=k_{F}$? – Nick Thompson Jan 25 '13 at 0:45
You're right, $k=0$ is wrong, but $k_F$ is only a valid number for the isotropic case (i.e. free electron gas). – Rafael Reiter Jan 25 '13 at 8:57

In any case, for undoped graphene, the Fermi level of electrons and holes are symmetric so the Fermi Energy lies at the Dirac Point, so these two definitions would be equivalent.

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