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\end{equation}  We now determine the displaced charge at a distance $r$ from the impurity. This quantity is defined as  \begin{equation}  \Delta_{11}=G_{11}-G^{0}_{11} \Delta G_{11}=G_{11}-G^{0}_{11}  \end{equation}  where $G^{0}_{11}$ is the Green function in the absence of impurity. We write down the two Dyson equations for $G_{11}$ and $G_{01}$. These are  \begin{eqnarray} 
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\begin{equation}  G_{11}\approx \frac{1}{g_{11}^{-1}-|V_{01}|^2 g_{00}}  \end{equation}  Once we know $G_{11}(\omega)$ then we can easily write the displaced charge   \begin{equation}  \Delta G_{11}=G_{11}-G^{0}_{11}= g_{11}|V|^2 g_{d}G_{11}=\frac{g_{11} |V|^2 g_d g_{11}}{1-g_{11}|V|^2 g_d}  \end{equation}  On the other hand $\frac{G_{00}-g_d}{G_{00}}=g_d |V|^2 g_{11}$, and then  \begin{equation}  \Delta_{11}=\frac{g_{11} |V|^2 g_d g_{11}}{1-\frac{G_{00}- g_d}{G_{00}}}  \end{equation}  Let us now find $\Delta G_{22}=G_{22}-G^{0}_{22}$ using the same procedure, then  \begin{equation}  G_{22}=g_{22} + G_{21} t_{12} g_{22}  \end{equation}  We compute $G_{21}$ and $g_{21}$  \begin{equation}  G_{21}=g_{21} + g_{22} t_{21} G_{11}  \end{equation}  A remark for $g_{21}$ is in order here. This Green function is the bare Green function for $V=0$ but $t\neq 0$. Then  \begin{equation}  g_{21}= g_{22} t_{21} g_{11}  \end{equation}  We replace back $G_{21}$, and $g_{21}$ into $G_{22}$  \begin{equation}  G_{22}=g_{22} + |t|^2 g_{22} (G_{11}-g_{11})= g_{22} + |t|^2 g_{22} \Delta G_{11}  \end{equation}  Then, the result, using the relation of $\Delta G_{11}$ as a function of the impurity site Green function $G_{00}$ reads  \begin{equation}  \Delta G_{22}=|t|^2 g_{22}^2 \Delta G_{11} = t^2 g^4_{ch} |V|^2 G_{00}  \end{equation}  This expression can be generalized to the following result for an arbitrary site $r$  \begin{equation}  \Delta G_{rr}=\left(\frac{1}{t}(|t| g_{ch})^r\right)^r |V|^2 G_{00}  \end{equation}  We define now in a more precise way the notion of displaced charge as  \begin{equation}  \Delta n(r)=-\frac{2}{\pi}\Im m \int_{-\infty}^{\infty} d\omega f(\omega, T) \Delta G_{rr}  \end{equation}  From the chain Green function we have $tg_{ch}=\exp{i\phi}$, using this result in $\Delta n(r)$ we have  \begin{equation}  \Delta n(r) =\frac{2}{\pi}\Im m \int_{-\infty}^{\infty} d\omega f(\omega, T) \exp{-2ir\phi(\omega)} \frac{|V^2|}{t^2} G_{00}  \end{equation}  In the "non-interacting limit" we have the following integral to solve  \begin{equation}  \Delta n(r) =\frac{2}{\pi}\Im m \int_{-\infty}^{\infty} d\omega f(\omega, T) \exp{-2ir\phi(\omega)} \frac{\Gamma/2t}{\omega-\epsilon_d+i\Gamma/2}   \end{equation}