Rosa edited section_Density_of_states_In__.tex  over 8 years ago

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\end{mutline}  Using the fact that $\sin k(N+1)=\sin kN \cos k +\cos kN \sin k$, then the whole expression is simplified to  \begin{multline}  \sin k(N+1) \cos kN \sin k + \sin k N (\sin kN-\cos k\sin k(N+1)) = \sin^ k \end{mutline} Therefore the final result reads \begin{equation} |c_1|^2 = D^2\frac{\sin^2 q \sin ^2 k}{\sin^k -\frac{V_g}{t}\sin kN \sin k(N+1)} \end{equation} Finally the density of states coincides with the one in \cite{Park_2013}. Using the fact that we are using $k$ and $q$ wavevectors in units of $a$, the lattice spacing, we have \begin{equation} \rho(E)=\frac{\sin q a\sin^ka}{(\pi a t)(\left (\sin^2(ka)+\frac{eV_g}{t}\sin k(N+1)a \right) \sin kNa} \end{equation}