Exact Bethe Ansatz Spectrum of a Tight-Binding Chain with Dephasing Noise

We construct an exact map between a tight-binding model on any bipartite lattice in the presence of dephasing noise and a Hubbard model with imaginary interaction strength. In one dimension, the exact many-body Liouvillian spectrum can be obtained by application of the Bethe ansatz method. We find that both the nonequilibrium steady state and the leading decay modes describing the relaxation at late times are related to the $\eta$-pairing symmetry of the Hubbard model. We show that there is a remarkable relation between the time evolution of an arbitrary $k$-point correlation function in the dissipative system and $k$-particle states of the corresponding Hubbard model.

Figure 1

Spectrum of the tight-binding model with dephasing for $L=6$, $2\gamma =1.75$. Vertical dashed lines indicate multiples of $-2\gamma$. As a result of the $\mathbb{P}\mathbb{T}$-symmetry [25] of the Liouvillian, the spectrum exhibits a $D_2$ point group symmetry. (Left inset) Schematic representation of the $1k\mathrm{\Lambda }$-string solution of the Bethe equations, for the imaginary $u$-Hubbard model (the solid symbols), and for the usual Hubbard chain with a real $u$ (the empty symbols). (Right inset) Close-up of the slowest decaying modes. The numbers indicate the degeneracies, which are consistent with the translation from lower to higher magnetization sectors via the $\eta$ symmetry. For example, $10=5\times 2$ is obtained by counting the number of $\eta$-pairing descendant states in sectors with higher numbers of particles (there are $L-1=5$), while the factor of 2 is due to a degeneracy in the sector $N=2$, $M=1$ of the $\eta$-pairing lowest weight states, which is a consequence of the parity symmetry of the Hamiltonian (5). The nondegenerate state shown has $N=2M=6$ and is a singlet with respect to both parity and the $\eta$ pairing.