Two baths

In the previous model, we had only one heat bath. After enough time has passed, i.e. in steady state, the central level reaches equilibrium with the bath, i.e. it aquires at \(t\rightarrow\infty\) the same temperature and chemical potential \(T_L,\mu_L\) as the bath¹. We know this from statistical physics and we obtained this from the calculation. This entire model is poses an equilibrium problem.

If such a problem always leads to a steady-state, should we even care about non-equilibrium calculations? The answer is yes. We can have a steady-state situation (at infinite time), where the system is time-independent (stationary), but still in a state of non-equilibrium.

We will study this with the model of having two separate heat-baths with different temperatures and/or chemical potentials. This is a system where we’ll find a genuine non-equilibrium state, even in the steady state.

The Hamiltonian used for one bath was² \[H_0=\sum_{p\neq 0} \epsilon_p c_p^\dagger c_p +\epsilon_0 c^\dagger_0 c_0\] with the perturbation \[V=\sum_{p\neq 0}V_p\pqty{c_p^\dagger c_0 + c^\dagger_0 c_p}\,.\] Formally, the extension to a system of two baths is easy. We allow the indices \(p\) to be negative and define negative indices to denote the levels of the left bath, whereas the positive indices belong to the levels of the right bath: \[\begin{aligned} P(L)&=\qty{p\in\mathbb{Z}:p<0}\\ P(R)&=\qty{p\in\mathbb{Z}:p>0} \end{aligned}\] Up to this point, it is formally not much different from assigning all levels to one big bath with levels indexed by \(p\neq 0\). However, it’s important that the bath is infinite. This is required for a small region to aquire the thermodynamic properties of the bath. If we have two infinite baths with different thermodynamic properties, we’ll not reach equilibrium, as we are going to see.

We can take the result from the model with one bath as a starting point. The Dyson equation in index notation is: \[\underline G_{00}=\underline g_{00}+\underline g_{00} \underline{\tilde\Sigma}\underline G_{00}\] We also had used this (kind of) self-energy term \[\underline{\tilde\Sigma}=\sum_{p\neq 0}\underline V_{0p} \underline g_{pp}\underline V_{p0}=\sum_{p>0}\dots+\sum_{p<0}\dots\] \[=\underline{\tilde\Sigma}_L+\underline{\tilde\Sigma}_R\] which we are now splitting into two parts, one for each bath.

Retarded part

For the retarded Green’s function part we have \[G^r_{00}=\pqty{\omega-\epsilon_0 + i0^+ -\tilde\Sigma^r_L-\tilde\Sigma^r_R}^{-1}\] The retarded “self-energy” term \[\tilde{\Sigma}_\alpha^r(\omega)=R_\alpha(\omega)+i\Gamma_\alpha(\omega), \;\alpha\in\qty{L,R}\] was split into a real part and an imaginary part, where \(R_\alpha(\omega), \Gamma_\alpha(\omega)\) are terms that we have already evaluated: \[\begin{aligned} R_\alpha(\omega)&=V^2\sum_{p\in P(\alpha)}\Re g^r_{pp}(\omega)\\ \Gamma_\alpha(\omega)&=\pi V^2\rho_\alpha(\omega)\\ \end{aligned}\] where \(\rho_\alpha(\omega)\) is the density of states for the respective bath. In total we get: \[\begin{aligned} G_{00}^r&=\pqty{\omega-E(\omega)+i\Gamma(\omega)}^{-1}\\ E(\omega)&=\epsilon_0+\sum_\alpha R_\alpha(\omega)\\ &=\epsilon_0+R_L(\omega)+R_R(\omega)\\ \Gamma(\omega)&=\sum_\alpha \Gamma_\alpha(\omega)\\ &=\Gamma_L(\omega)+\Gamma_R(\omega) \end{aligned}\]

Keldysh component

For the retarded part, we could also consider the two baths as a single bath with a total D.O.S. of \(\rho=\rho_L+\rho_R\). However, we’ll have to keep both D.O.S. separate. The reason for this becomes obvious when we start looking at the Keldysh component:

\[\begin{aligned} G_{00}^K&=G_{00}^r\pqty{\sum_p V_{0p}^2 g_{pp}^K}G_{00}^a\\ &=|G_{00}^r(\omega)|^2\sum_\alpha\pqty{-2i\Gamma_\alpha(\omega)S_\alpha(\omega)} \end{aligned}\] Here we don’t have the total D.O.S. \(\rho_L+\rho_R\) because these two terms are weighted differently via \[S_\alpha(\omega)=1-2f_F(\omega,\mu_\alpha,T_\alpha)\,.\] This is the point where the fact that we have two thermodynamically different baths enters.

Remember that in the equilibrium case, we had \[G_{00}^K(\omega)=\pqty{G_{00}^r(\omega)-G_{00}^a(\omega)}S_L(\omega)\,.\] This relation is always valid in equilibrium. If this relation doesn’t hold true, that is a good indicator that the system is out of equilibrium. And this fact is connected to the fluctuation-dissipation theorem of statistical physics. Also remember that the retarded and advanced terms are complex conjugate to each other: \[G_{00}^r(\omega)=\pqty{G_{00}^a(\omega)}^*\]

The question is, what do we have in this case? Using³ \[G_{00}^r(\omega)-G_{00}^a(\omega)=-2i\Gamma(\omega)|G_{00}^r|^2\,,\] we find a very similar expression to the one from equilibrium: \[\begin{aligned} G_{00}^K(\omega)&=(G_{00}^r(\omega)-G_{00}^a(\omega))S_{Av}(\omega)\\ S_{Av}(\omega)&=\frac {\Gamma_L(\omega)S_L(\omega)+\Gamma_R(\omega)S_R(\omega)} {\Gamma_L(\omega)+\Gamma_R(\omega)} \end{aligned}\] where the subscript \(Av\) means average. We see that this average is dependent on \(\omega\), so it can’t be attributed to any average. So it is not like a simple average temperature or chemical potential. Following the advice from above, we can already see from the structure of the equation for \(G_{00}^K(\omega)\) that this system is not in equilibrium.

Current

In this model with two baths, the first thing we may be interested in is of course the current flowing from one reservoir to the other. There may also be a heat current if we have a difference in temperature. However, we start with the charge current here. That is the most interesting quantity in this case.

In general, the current is a time derivative of the charge in a certain region⁴: \[\hat I=-\dv{\hat Q}{t}\] That is the generic definition⁵. We want specifically the current flowing away from the left bath: \[\begin{aligned} \hat I_L&=-\dv{\hat Q_L}{t}=-e\sum_{p<0}\dv{t} c^\dagger_p c_p\\ &=-e\sum_{p<0}i\comm{H}{c^\dagger_p c_p}\\ &=-ie\sum_{p<0}V_{p}\pqty{c^\dagger_0 c_p-c_p^\dagger c_0} \end{aligned}\] The quantity \(e\) is the charge of an electron.

Exercise 7.1: prove that \(\forall p\neq 0,\; \comm{H}{c^\dagger_p c_p}=V_{p}\pqty{c^\dagger_0 c_p-c_p^\dagger c_0}\)

The expression \(\pqty{c^\dagger_0 c_p-c_p^\dagger c_0}\) is rather intuitive. The term exchanges electrons between the center level and one lead level. As we’ve seen some lectures ago, getting the expectation value for an operator pair like \(c_p^\dagger c_0\) is exaclty what the Keldysh part of the Green’s function does: \[\ev{c^\dagger_p c_0}=-\frac{i}{2} G^K_{0p}(t=0)\]

\[\begin{aligned} I_L&=\ev{\hat I_L}\\ &=\frac{e}{2}\sum_{p<0}V_p\pqty{G_{0p}^K(t=0)+\mathrm{c.c.}}\\ &=e\sum_{p<0}V_p \Re G_{0p}^K(t=0)\\ \end{aligned}\] with \[G_{0p}^K(t=0)=\int\frac{\dd{\omega}}{2\pi}G_{0p}^K(\omega),\,\] which is an integral over all frequencies. This is essentially a Fourier transform, but without a phase factor since \(e^{-i\omega t}=1\) due to \(t=0\).

From here on, we assume that all Green’s functions are specified in frequency space, unless explicitly stated otherwise: \[G_{pq}^X \mapsto G_{pq}^X(\omega)\] Now we have to provide an expression for \(G_{0p}^K\). So far we only have \(G_{00}^K\). But we can get it via another step from the Dyson equation: \[\underline G_{0p}=\underline G_{00}\underline V_{p} \underline g_{pp}\] Remember that \((AB)^K=A^r B^K + A^K B^a\) and that \(\underline V_{p}\) is diagonal in Keldysh space. \[G_{0p}^K=\pqty{G_{00}^r g_{pp}^K+G_{00}^K g_{pp}^a}V_{p}\] As we only look at the left bath for now, we assume \(p<0\). \[=V_{p}\bqty{ G_{00}^r(g_{pp}^r-g_{pp}^a)S_L(\omega)+ (G_{00}^r-G_{00}^a)S_{Av}(\omega)g_{pp}^a }\] We need \(\sum_{p<0}V_{p}G_{0p}^K\) and use the fact that \[\sum_{p<0}V_p^2g_{pp}^r=R_L-i\Gamma_L\] as we have done before. Putting everything together, we get:

\[\sum_{p<0}V_p G_{0p}^K\] \[=G_{00}^r\pqty{-2i\Gamma_L}S_L(\omega)+ (G_{00}^r-G_{00}^a)(R_L+i\Gamma_L)S_{Av}(\omega)\]

Also useful here is the fact that \(g_{pp}^a=(g_{pp}^r)^*\). Now since we really only need the real part, we can already simplify a few things. For example, \(G_{00}^r-G_{00}^a\) is purely imaginary, so \(R_L\) disappears here. The argument in the integral for \(I_L\) reads: \[\frac{e}{2\pi}\sum_p\Re V_p G_{0p}^K\] \[=\frac{e}{\pi}\pqty{\Gamma_L S_L\Im G_{00}^r-\Gamma_L S_{Av}\Im G_{00}^r}\] \[=\frac{e}{\pi}\Gamma_L G_{00}^r(S_L-S_{Av})\equiv\mathcal{J}\] The caligraphic \(\mathcal{J}\) is the term under the integral for \(I_L\): \[I_L=\int_{-\infty}^\infty\dd{\omega}\mathcal{J}(\omega)\,.\]

Next we calculate the difference \[\begin{aligned} S_L-S_{Av}&=\frac{(\Gamma_L+\Gamma_R)S_L-S_L\Gamma_L-S_R\Gamma_R} {\Gamma_R+\Gamma_L}\\ &=\frac{\Gamma_R}{\Gamma_R+\Gamma_L}\pqty{S_L-S_R}\,. \end{aligned}\] In total we get \[\mathcal{J}=\frac{e}{\pi}\frac{\Gamma_L\Gamma_R}{\Gamma} (S_L-S_R)\Im G_{00}^r\] which looks a lot more symmetric then the previous expressions. In fact, the term is anti-symmetric as it changes sign if we swap the bath indices \(L,R\). Using \[S_\alpha=1-2f_{F\alpha},\quad\Im G_{00}^r=-\Gamma|G_{00}^r|^2\] yields \[\mathcal{J}=\frac{2e}{\pi}\Gamma_L\Gamma_R|G_{00}^r|^2(f_{FL}-f_{FR})\] \[I_L=\int_{-\infty}^\infty\dd{\omega}\mathcal{J}(\omega)\]

Some notes on this result:

1. Obviously \(I_R\) is obtained by exchanging \(L\mapsto R\), so we have \(I_L=-I_R\) as expected from current conservation.
2. The integrand is proportional to the difference in occupation \(f_{FL}-f_{FR}\) times the joint density of states \(\rho_L\rho_R\propto\Gamma_L\Gamma_R\). The rest, which is \(\propto |G_{00}^r|^2\), describes a transmission probability.

The figure below shows (schematically) the Fermi functions of the left and right bath.

We only find current in the region where \(\mu_L\) and \(\mu_R\) are different. That region is called the transport window. For \(T\rightarrow 0\), the difference within the transport window is exactly \(f_{FL}-f_{FR}=1\).

These results are quite intuitive. which is in particular interesting when we introduce correlations in the center region. Then the transport coefficient depends on interactions between the electrons that region, which yields some interesting results.

We arrived at an interesting part of this lecture, where we have found our first non-equilibrium effect. This is something, we could not have done with pure equilibrium calculations.