Spontaneous emission and thermalization of cold bosons in optical lattices

We study the thermalization of excitations generated by spontaneous emission events for cold bosons in an optical lattice. Computing the dynamics described by the many-body master equation, we characterize equilibration time scales in different parameter regimes. For simple observables, we find regimes in which the system relaxes rapidly to values in agreement with a thermal distribution, and others where thermalization does not occur on typical experimental time scales. Because spontaneous emissions lead effectively to a local quantum quench, this behavior is strongly dependent on the low-energy spectrum of the Hamiltonian, and undergoes a qualitative change at the Mott insulator–superfluid transition point. These results have important implications for the understanding of thermalization after localized quenches in isolated quantum gases, as well as the characterization of heating in experiments.

Figure 1

(a) Absorption and spontaneous emission of a lattice photon leads effectively to localization of single atoms. Tunneling and interactions between atoms then redistribute the energy added to the system. (b) Localization of an atom in space corresponds initially to a distribution of the atom over the whole Brillouin zone (the tails of the quasimomentum distribution are lifted). Subsequent unitary evolution leads to a broadened quasimomentum distribution, i.e., the $n_{q=0}$ peak and the tails decrease, while the small quasimomentum components increase (t-DMRG, $U=2J$, $N=48$ particles on $M=48$ sites, $d_l=6$, $D=512$).

Figure 2

Time evolution after a single spontaneous emission. (a), (b) For a superfluid initial state ($U=2J$), the kinetic energy relaxes to the equilibrium value obtained from a Monte Carlo calculation $E_{\mathrm{eq}}^{\mathrm{kin}}$. For MI states ($U=4J$), the energy relaxes, but not to $E_{\mathrm{eq}}^{\mathrm{kin}}$. The zero value of kinetic energy for this plot is the ground-state kinetic energy $E_{\mathrm{gs}}^{\mathrm{kin}}$. (c) The difference in the infinite time value of the kinetic energy (obtained from an extrapolation of an exponential fit) to the equilibrium energy. For MI states with $U/J\gtrsim 3.37$, the difference increases rapidly for $M=24,48,96$ sites. (d) The decay rate extracted from the exponential fit as a function of $U$. (e) Comparison of the time-evolved quasimomentum distribution at $t=10/J$ (dots) to the equilibrium distribution from a QMC calculation. (f), (g) Differences between the two distributions as a function of time for the $qa=0$ peak and for a large quasimomentum of $qa=(40/48)\pi$. In the superfluid [$U=2J$ in (f)], the components for large momenta relax rapidly to thermal values, for $qa\sim 0$, the relaxation time scale is much longer. In the MI [$U=5J$ in (g)], the same is true, but for large momenta there is a discrepancy to the thermal value. (t-DMRG, $d_l=6$, $D=256,512$; error bars represent fitting errors and statistical errors from QMC.)

Figure 3

Expectation values of the kinetic energy of the lowest 1000 eigenstates as a function of the energy in a system with $M=10$ and $N=10$ (exact diagonalization). The gray line in the upper plot shows the equilibrium kinetic energy $E_{\mathrm{eq}}^{\mathrm{kin}}$ for increasing temperatures as a function of the mean energy of the underlying Boltzmann distributions. In the SF, the eigenvalue expectations are distributed around $E_{\mathrm{eq}}^{\mathrm{kin}}$, but are far from these values in the MI. The lower parts show the occupation probabilities for eigenstates after a single spontaneous emission.

Figure 4

Quantum trajectory simulations for heating with spontaneous emission rate $\gamma$, which is switched on for a time $t=1/J$, as illustrated above (a). $M=N=48,$ and the standard error of the mean is given as the shaded area. (a) The increase in kinetic energy during the heating and the subsequent relaxation for $\gamma =0.02,0.04,0.06$. For superfluid initial states, the kinetic energy relaxes to the equilibrium value (QMC calculations, dashed lines). For a Mott insulating initial state, on the same time scale, the energy does not thermalize. This can be seen in (b), where we plot the difference between the actual kinetic energy and the equilibrium energy (t-DMRG results, $D=256$, $d_l=6$, 500 trajectories).