# Effective Mass and Fermi Velocity of Electrons in Graphene:

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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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 $E(K)=E_f$.
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