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  • Calculation of the displacement charge in a semi-infinite chain coupled to a localized impurity

    In this note we derive the displacement charge of a site which is coupled to a noninteracting tight binding semi-infinite chain. First we derive the Green function for a semi-infinite chain and after we coupled such chain to an impurity with a energy level \(\epsilon_d\). \[H=\sum_{\sigma} \epsilon_0 c^\dagger_{i} c_{i} + t \sum_{n=1}^{n=\infty} \left(c^\dagger_{i}c_{i+1}+h.c\right)\]

    Semi-infinite chain Green function

    We start with the following wavefunction \[|\Psi\rangle =\sum_{\beta=1}^{\infty} \sin(k\beta)|\beta\rangle\] When, he Hamiltonian of the chain \(H\) acts on this wavefunction, then, we get the eigenenergies \[H\Psi\rangle = (\epsilon_0 + 2 t \cos k)|\Psi\rangle\] Here \(k=2\pi/a\) with \(a\) being the lattice spacing. Hereafter, we set \(\epsilon_0=0\). Now we compute the Dyson equation for this semi-infinite chain. We do not consider now the site \(0\). \[G_{11}=g_{11}+g_{11} t_{12} G_{21}\] We consider all hopping amplitude for all sites the same, \(t_{12}=t\). Here, \(g_{11}\) is the bare Green function for an uncoupled site, ie., \[g_{11}=\frac{1}{\omega-\epsilon_0}\] Under these considerations we have \[G_{21}=g_{21} + g_{22} t_{21} G_{11} + g_{22} t_{23} G_{31},\] and similar equations for \(G_{31}\), etc. This sequence of Green functions for \(G_{n1}\) generates an infinite number of equations. Now, it comes the trick. Consider that site \(2\) is in reality another semi-infinite chain coupled to site \(1\). Then, the equation for \(G_{21}\) becomes simply \[G_{12}=g_{ch} t_{21} G_{11}.\] By replacing this expression for \(G_{12}\) in \(G_{11}\) we can close the Dyson equation and to obtain explicitly \(G_{11}\). Therefore \[G_{11}=g_{11}+ g_{11}|t|^2 g_{ch} G_{11}.\] Besides, we notice that \(G_{11}\) corresponds to a semi-infinite chain itself, thus \[g_{ch}=g_{11}+g_{11}|t|^2 g_{ch}^2\] With the notorious result \[g_{ch}=\frac{\omega-\epsilon_0}{2t^2}\pm\frac{1}{2t^2} \sqrt{(\omega-\epsilon_0)^2-4t^2},\] Considering \(|\omega-\epsilon_0|<2t\), then \(g_{ch}\) acquires an imaginary part, then \[g_{ch}=\frac{\omega-\epsilon_0}{2t^2}\pm\frac{i}{t} \sqrt{1-\left(\frac{\omega-\epsilon_0}{2t}\right)^2},\] With a DOS \[\rho_{ch}(\omega)=-\frac{1}{\pi}= \frac{1}{\pi t} \sqrt{1-\left(\frac{\omega-\epsilon_0}{2t}\right)^2},\] This is the surface (boundary) DOS, which never diverges in contrast with the bulk DOS

    Density of states for a semi-infinite chain.

    At the band center the DOS is constant: \(\rho_0=1/\pi t\). There is an alternative expression for the chain DOS by defining \(\cos\phi=\frac{\omega-\epsilon_0}{2t}\). Then \[g^r_{ch}(\omega)=\frac{1}{t}\left(\cos\phi-i\sin\phi\right)=\exp{(-i\phi)}{t}\] In general we get \[G^{r}_{n1}=\exp{-i\phi}{t}\]

    Impurity coupled to the semi-infinite chain

    Now we consider that the \(0\) state is coupled to the semi-infinite chain with \(V\) as the tunneling amplitude. The Dyson equation for the impurity site Green function reads \[G_{00}=g_d+g_{11}+V_{10}G_{10},\] The bare dot Green function is \[g_{d}=\frac{1}{\omega-\epsilon_d}\] Now we take \(V_{10}=V\). Then, the chain Green function \(G_{10}\) coupled to the impurity reads \[G_{10}=g_{10}+g_{10}V_{01}G_{00}=g_{ch}VG_{10},\] With \(g_{10}=0\) and \(V_{01}=V_{10}=V\). Replacing back \(G_{10}\) in the Dyson equation for \(G_{00}\) we have \[G_{00}=\frac{1}{g_d^{-1}-|V|^2 g_{ch}}\] For energies around the band center we can neglect the energy dependence of \(g_{ch}\approx \frac{-i}{t}\), then \[G_{00}\approx \frac{1}{g_d^{-1}+i|V|^2/t}=\frac{1}{g_d^{-1}+i\Gamma/2}\] with \(\Gamma=2|V|^2/t\). The extension to two leads is straightforward \[G_{00}\approx \frac{1}{g_d^{-1}-\Sigma_{0}}\] with \[\Sigma_{0}=-i(|V_{01}|^2 g_{ch}^L+ |V_{01}|^2 g_{ch}^R)=-i(\Gamma_L+\Gamma_R)\] We now determine the displaced charge at a distance \(r\) from the impurity. This quantity is defined as \[\Delta G_{11}=G_{11}-G^{0}_{11}\] 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 \[\begin{aligned} G_{11}&=&g_{11}+g_{11}V_{10}G_{01}, \\ G_{01}&=&g_{01}+g_{00}V_{01}G_{11}\end{aligned}\] where \(g_{00}=g_{d}\), and \(g_{11}=g_{ch}\). We obtain \[G_{11}\approx \frac{1}{g_{11}^{-1}-|V_{01}|^2 g_{00}}\] Once we know \(G_{11}(\omega)\) then we can easily write the displaced charge \[\Delta G_{11}=G_{11}-G^{0}_{11}= g_{11}|V|^2 g_{d}G_{11}=\frac{g_{11} |V|^2 g_d g_{11}}{1-g_{11}|V|^2 g_d}\] On the other hand \(\frac{G_{00}-g_d}{G_{00}}=g_d |V|^2 g_{11}\), and then \[\Delta_{11}=\frac{g_{11} |V|^2 g_d g_{11}}{1-\frac{G_{00}- g_d}{G_{00}}}\] Let us now find \(\Delta G_{22}=G_{22}-G^{0}_{22}\) using the same procedure, then \[G_{22}=g_{22} + G_{21} t_{12} g_{22}\] We compute \(G_{21}\) and \(g_{21}\) \[G_{21}=g_{21} + g_{22} t_{21} G_{11}\] 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 \[g_{21}= g_{22} t_{21} g_{11}\] We replace back \(G_{21}\), and \(g_{21}\) into \(G_{22}\) \[G_{22}=g_{22} + |t|^2 g_{22} (G_{11}-g_{11})= g_{22} + |t|^2 g_{22} \Delta G_{11}\] Then, the result, using the relation of \(\Delta G_{11}\) as a function of the impurity site Green function \(G_{00}\) reads \[\Delta G_{22}=|t|^2 g_{22}^2 \Delta G_{11} = t^2 g^4_{ch} |V|^2 G_{00}\] This expression can be generalized to the following result for an arbitrary site \(r\) \[\Delta G_{rr}=\left(\frac{1}{t}(|t| g_{ch})^r\right)^r |V|^2 G_{00}\] We define now in a more precise way the notion of displaced charge as \[\Delta n(r)=-\frac{2}{\pi}\Im m \int_{-\infty}^{\infty} d\omega f(\omega, T) \Delta G_{rr}\] From the chain Green function we have \(tg_{ch}=\exp{i\phi}\), using this result in \(\Delta n(r)\) we have \[\Delta n(r) =\frac{2}{\pi}\Im m \int_{-\infty}^{\infty} d\omega f(\omega, T) \exp{[-2ir\phi(\omega)]} \frac{|V^2|}{t^2} G_{00}\] In the “non-interacting limit” we have the following integral to solve \[\Delta n(r) =\frac{2}{\pi}\Im m \int_{-\infty}^{\infty} d\omega f(\omega, T) \exp{[-2ir\phi(\omega)]} \frac{\Gamma/2t}{\omega-\epsilon_d+i\Gamma/2}\]