Transport of correlated electrons through disordered chains: A perspective on entanglement, conductance, and disorder averaging

We investigate electron transport in disordered Hubbard chains contacted to macroscopic leads, via the nonequilibrium Green's function technique. We observe a crossover of currents and conductances at finite bias which depends on the relative strength of disorder and interactions. We provide a proof that the coherent potential approximation, a widely used method for treating disorder averages, fulfills particle conservation at finite bias with or without electron correlations. Finally, our results hint that the observed trends in conductance due to interactions and disorder also appear as signatures in the single-site entanglement entropy.

Figure 1

The effect of interactions and disorder on the current for a bias $b_L=0.5$. (a) Heat map for the current. (b) Cuts along a fixed disorder strength (horizontal cuts in the heat map) as a function of $U$. (c) Cuts along a fixed interaction strength (vertical cuts) as a function of $W$. In (b) and (c), the lines are a guide for the eye. The legend applies to both (b) and (c).

Figure 2

Averaged differential conductance $\sigma =dI/db$ as a function of the chain size $L$ for box disorder and for bias $b_L=0$ (dashed curves) and $b_L=0.5$ (solid curves). The conductance is in units of the quantum of conductance.

Figure 3

Steady-state currents for eight-site chains with binary disorder $(A_{50}{B}_{50}$ alloy) as a function of disorder and interaction strength. The applied bias $b_L=0.5$. Both results for the exact distribution of currents (histograms) and CPA current averages (circles) are shown, discussed in the main text.

Figure 4

Cumulative distribution of the single-site entanglement entropy ${\mathcal{E}}_k$ as a function of the interaction strength for a ten-site chain with $W=2$ and bias $b_L=0.5$. The black solid curves at the base of the entanglement histograms correspond to ${\mathcal{E}}_k$ for a noninteracting system, i.e., $\langle {\widehat{X}}_{k\uparrow }^{\mu }{\widehat{X}}_{k\downarrow }^{\nu }\rangle =\langle {\widehat{X}}_{k\uparrow }^{\mu }\rangle \langle {\widehat{X}}_{k\downarrow }^{\nu }\rangle$, where $n_k\in [0,2]$ and ${\mathcal{E}}_k\in [0,2]$.