The properties of the system depend on the property of the bath (only the left one, for simplicity) via the density of states.
\[\sum_p \mathcal{V}_p^2 g_{pp}^r(\omega)= V^2 R_L(\omega)-i\pi V^2\rho_L(\omega)\] \[=R_L(\omega)-i\Gamma_L(\omega)\]
\(\mathcal{V}_p\) is the coupling to each bath level. The left and right bath may couple differently.
Error: \(V^2\) is missing here before \(R_L\).
\[G_{00}^r(\omega)=\pqty{\omega-\bqty{\Delta+R_L(\omega)} +i\Gamma(\omega)}^{-1}\] with \(E(\omega):=\Delta+R_L(\omega)\).
Consider the case of a central impurity, which couples to a series of bath levels \(\epsilon_p\) that become continous in the end, via coupling terms \(\mathcal{V}_p\).
\[G_{00}^K(\omega)=-2i\Gamma_L|G_{00}^r(\omega)|^2 S(\omega)\] \[=\pqty{G_{00}^r(\omega)-G_{00}^a(\omega)}S(\omega)\]
Exercise 6.1: prove the relation above.
\[=-2i\pi A_{00}(\omega)S_L(\omega)\]
\[S(\omega)=1-2f_F(\omega,\mu_L,T_L)\]
Next we have
\[\ev{c_0^\dagger c_0}=\frac{1}{2}-\frac{i}{2} \int\frac{\dd{\omega}}{2\pi}G_{00}^K(\omega)\] \[=\frac{1}{2}-\pi\int\frac{\dd{\omega}}{2\pi}A_{00}(\omega)S_L(\omega)\] \[=\frac{1}{2}-\frac{1}{2}\int\dd{\omega}A_{00}(\omega) \bqty{1-\color{red}2\color{black}f_L(\omega)}\] \[=\int\dd{\omega}A_{00}(\omega)f_L(\omega)\]
That result we also had in the previous course. We used \(\int\dd{\omega}A_{00}(\omega)=1\) and \[G_{00}^r(\omega)-G_{00}^a(\omega)=-2i\pi A_{00}(\omega)\] with \[A_{00}(\omega)=-\frac{1}{\pi}\Im G_{00}^{r}(\omega)\] where \(A_{00}(\omega)\) is the spectral function. See the previous Green’s function course for more information on this.
We can think of a spectral function as something like this:
It gives me the distribution of states on the site (00) as a function of \(\omega\). At zero temperature, the Fermi function will be an inverse step function. The occupation \(\ev{c_0^\dagger c_0}\) is represented by the blue area in the screenshot above. This is expected as we are still in equilibrium here. Now we do the calculation in that case.
\[-\Im G_{00}^r(\omega)=\frac{\Gamma_L(\omega)}{ \bqty{\omega-E_L(\omega)}^2+\Gamma_L(\omega)^2}\]
That is the situation where we can do analytic calculations, as we can neglect the omega dependence of \(\rho(\omega)\), where the density of the bath is a constant. That is the wide-band limit, i.e. the bath has a wide band width. Just one thing we have to be mathematically careful about: If \(\rho_L\) becomes a constant, the integral over it diverges, so we’ll have to take the limit of a small \(V\) instead. In the limit, the bandwidth goes to infinity. That gives a quantity \(\Gamma\) independent of \(\omega\) and the real part \(R_L(\omega)\) can be taken to zero under certain conditions, e.g. when the distribution of energy is symmetric above and below. We won’t go too much into details here.
Then at \(T=0\), the occupation is simply given by \[\ev{c_0^\dagger c_0}=\int_{-\infty}^{\mu} \frac{\Gamma}{(\omega-\Delta)^2+\Gamma^2}\dd{\omega}\,.\]
Exercise 6.2: Write down an expression for the full Green’s function \(\underline G_{pp}\). You’ll have to start from Dyson’s equation in index notation. E.g. previously we had the following case:
In two equations we got an expression for \(\underline G_{00}\) by iterating it. What we do is the following:
Remember Dysons equation: \[\underline G=\underline g+\underline g\underline V\underline G\] \[=\underline g+\underline G\underline V\underline g\] The second line is the right Dyson equation, which we’ll also need here. The point is to write it in indices: \[\underline G_{pp'}=\underline g_{pp'}+ \underline g_{pp_1}\underline V_{p_1p_2}\underline G_{p_2p'}\] where \(\underline g_{pp'}\) is diagonal in \(p,p'\). \(V_{p_1p_2}\) is non-zero only if either \(p_1\) or \(p_2\) is zero.
So again, we’ll need two equations. The first is for \(\underline G_{pp'}\). We see already that if we put \(p=p'\) here, the \(p_2\) in the last equation turns out to be zero and so we get an equation for \(G_{0p'}\).
Hint:
\(G_{pp}=\dots\), \(G_{0p}=\dots\). That contains \(G_{00}=\dots\). Once we have done this
Evaluate at \(\ev{c_p^\dagger c_p}\) for the wide-band limit.
Exercise 6.3:
We have the same situation as before, with is a level (\(c_0\)) coupled to a bath. We add another level (\(f\)) with the same energy \(\Delta\) and a hopping \(T\) (not to be confused with the temperature). The central Hamiltonian is:
\[H_C=\Delta(c_0^\dagger c_0+f^\dagger f)+ T(c_0^\dagger f + f^\dagger c_0)\]
The Hamiltonian for the bath is the same as before. \(H_{\textrm{Bath}}+\)
The trick is the following: First we consider the original situation, as we have calculated it already, up to a certain time \[t_1 \gg t_0, T=0\,.\] Now we consider the \(T\) as a perturbation and write down a recursive expression for \(G_{ff}\).
Optional: Evaluate the expectation value \(\ev{f^\dagger f}\).
Exercise 6.4:
Here we have a semi-infinite (tight-binding) chain. We have nearest-neighbor hopping \(\mathcal{V}\).
\[\sum_{n=0}^\infty \mathcal{V} (c^\dagger_n c_{n+1}+c^\dagger_{n+1}c_n)\]
The trick to solve this is this: Consider the last hopping \(\mathcal{V}_0\) as a perturbation, even though it has the same value as the other ones. Write an expression for \(G_{00}\) in terms of \(g_{11}\). \(g_{11}\) is the Green’s function of the isolated chain, isolated without the last site. But due to the infinite system, the Green’s function must be the same as the coupled one. Realize that \(g_{11}=G_{00}\). This gives a recursive equation. Compute the spectral function of the model.
It is enough to do the first and then one of the other three exercises.