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\section{Density of states} In this calculation we reproduce the result for the density of states for a semi-infinite chain of sites under the influence of a spatial dependent potential \cite{Park_2013}. The Hamiltonian for the chain is  \begin{equation}  H_{W}=\sum_{j=1}^{j=N} \epsilon_0 c_{j}^\dagger c_{j} + \sum_{j=N+1}^{j=\infty} V_g c_{j}^\dagger c_{j} + \sum_{j=1}^{j=\infty} t \left(c_j^\dagger c_j+1+h.c\right) 
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\end{equation}  and by simplifying the denominator we get  \begin{equation}  |\sin k(N+1)-\sin kN e^{iq}}|^2 e^{iq}|^2  = \sin^2 k(N+1)+\sin k N^2 -2\cos q \sin kN\sin K(N+1) \end{equation}  with 2\cos $2\cos  q =2\cos k-(V_g/t), k-(V_g/t)$,  then \begin{multline}  \sin^2 k(N+1)+\sin k N^2 -2\cos q \sin kN\sin K(N+1) &=& \\  =  \sin^2 k(N+1)+\sin k N^2 -\cos k \sin kN\sin K(N+1) +\frac{V_g}{t} \sin kN\sin K(N+1) \end{multline}  Now we simplify the following term  \begin{equation}  \sin^2 \sin ^2  k(N+1)+\sin k N^2 -2\cos k \sin kN\sin K(N+1) =\sin k(N+1) \left[ \sin k(N+1)-2 \cos k \sin kN \right]+\sin k N^2 \end{equation}  then we split the 2\cos k $2\cos k$  in two contributions (we do not consider the last term $\sin k N^2$ for these simplifications) \begin{equation}  \sin k(N+1) \left[ \sin k(N+1)- \cos k \sin kN \right] =\sin k(N+1) \left[ \sin kN \cos k +\cos kN \sin k- \cos k \sin kN \right]   \end{equation} 
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\sin^2 k(N+1)+\sin k N^2 -\cos k \sin kN\sin K(N+1) =\sin k(N+1) (\cos kN \sin k- \cos k \sin kN ) + \sin k N^2 =  \\  \sin k(N+1) \cos kN \sin k + \sin k N (\sin kN-\cos k\sin k(N+1))  \end{mutline} \end{multline}  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} \end{multline}  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)} 
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\end{equation}