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A e^{ik (j+1)} +B e^{-ik(j+1)} + A e^{ik (j+1)} +B e^{-ik(j+1)} = \epsilon A e^{ikj)} +B e^{-ikj}  \end{equation}  Then the solution for the energy is $(E-\epsilon_0)/t=2\cos k$, with $E=\epsilon_0+2\cos k$  We determine now the A $A$  and B $B$  coefficients in $c_j$. For such purpose we just employ the Schrödinger equation for the site $1$: \begin{equation}  c_2=[(E-\epsilon_0)/t] c_1 = \rightarrow  A e^{2ik}+Be^{-2ik}= A e^{ik} + t B e^{-i k} (A e^{ik}+Be^{-ik}) \end{equation}  Then we infer that $A=-B$, and therefore $c_j= 2Ai\sin k j$ with $E=\epsilon+2t\cos k$ k$.   For the problem that we want to solve we have to consider a semi-infinite chain running from $j=N+1$ to $j=\infty$ in which the sites have an energy $\epsilon_0-V_g$ and a   dispersion relation $E=\epsilon_0-V_g+2t\cos q$. Then, we consider another finite chain where sites have an energy $\epsilon_0$ for $j=1$ to $j=N$. In order to obtain (local) the density of states at site 1 we need to solve the matching problem at the boundary  for $j=N$ and $j=N+1$, then we have  \begin{eqnarray}  c_{N-1}+c_{N+1}&=& \frac{E-\epsilon_0}{t} c_N \\ \nonumber  c_{N}+c_{N+2}= \frac{E-(\epsilon_0-V_g)}{t} c_{N+1}  \end{eqnarray}  We employ the previous solutions for the two chains  \begin{equation}  c_N=A \sin kN, \quad c_{N-1}=A\sin k(N-1), \quad c_{N-1}= C e^{i (N-1)q} + D e^{-i(N-1)q}  \end{equation}  We obtain the following result after matching:  \begin{equation}  \frac{A}{D}=\frac{1}{2i}\frac{-e^{i q N }\sin q}{\sin{k(N+1)}-\sin KN e^{i q} }  \end{equation}  Using this result we can compute the local density of states at site 1 as follows  \begin{equation}  \rho(E)=\sum_k |c_1|^2 \delta (E-\epsilon_k)  \end{equation}  therefore we get  \begin{equation}  |c_1|^2 =|D|^2\frac{ \sin^ q \sin ^k}{\sin k(N+1)-\sin kN e^{iq}}  \end{equation}  and by simplifying the denominator we get  \begin{equation}  |\sin k(N+1)-\sin kN e^{iq}}|^2 = |\sin k(N+1) -\sin k N\cos q -i\sin kN\sin q |^2=  \end{equation}