Relaxation, thermalization, and Markovian dynamics of two spins coupled to a spin bath

It is shown that by fitting a Markovian quantum master equation to the numerical solution of the time-dependent Schrödinger equation of a system of two spin-1/2 particles interacting with a bath of up to 34 spin-1/2 particles, the former can describe the dynamics of the two-spin system rather well. The fitting procedure that yields this Markovian quantum master equation accounts for all non-Markovian effects in as much the general structure of this equation allows and yields a description that is incompatible with the Lindblad equation.

Figure 1

Time dependence of the energy (open circles, a–c) and entropy (open circles, d–f) of the two-spin system in contact with a spin bath at moderate temperature, as obtained from the solution of the TDSE for three different initial states $|\psi \rangle \otimes |\mathrm{\Phi }(\beta =1)\rangle$ where $\left|\mathrm{\Phi }\right(\beta =1)\rangle$ denotes a thermal random-state Eq. (21) of the bath only. The system Hamiltonian is given by Eq. (2) with $J_{\mathrm{S}}=1/4$ (antiferromagnetic Heisenberg model). The Hamiltonian of the system-bath interaction and spin-glass bath are given by Eqs. (4) and (7), respectively. The number of bath spins is $N_{\mathrm{B}}=20,K=1/2,h_{\mathrm{B}}^x=h_{\mathrm{B}}^z=0$. The system-bath interaction strength is $\lambda =0.5$. (a) $|\psi \rangle =|S\rangle$, where $|S\rangle$ is the singlet state as given by Eq. (3); (b) $|\psi \rangle =|\uparrow \downarrow \rangle$; (c) $|\psi \rangle =|\uparrow \uparrow \rangle$. Stars in (a–c): ground-state energy of the isolated two-spin system; crosses in (a–c): thermal energy of the isolated two-spin system at $\beta =1$; stars in (d–f): maximum entropy of the isolated two-spin system; crosses in (d–f): entropy of the isolated two-spin system at $\beta =1$.

Figure 2

Time dependence of the energy (open circles, a–c) and entropy (open circles, d–f) of the two-spin system in contact with a spin bath at low temperature, as obtained from the solution of the TDSE with the initial state $|\psi \rangle \otimes |\mathrm{\Phi }(\beta =5)\rangle$. The system Hamiltonian is given by Eq. (2) with $J_{\mathrm{S}}=1/4$ (antiferromagnetic Heisenberg model). The Hamiltonian of the system-bath interaction and spin bath are given by Eqs. (4) and (6), respectively. The number of bath spins is $N_{\mathrm{B}}=20,K=1,h_{\mathrm{B}}^x=h_{\mathrm{B}}^z=1/4$. The system-bath interaction strength is $\lambda =0.25$. (a) $|\psi \rangle =|S\rangle$, where $|S\rangle$ denotes the singlet state given by Eq. (3); (b) $|\psi \rangle =|\uparrow \downarrow \rangle$; (c) $|\psi \rangle =|R\rangle \otimes |R\rangle$, where $|R\rangle$ denotes a random superposition of spin-up and spin-down states. Stars in (a–c): ground-state energy of the isolated two-spin system; crosses in (a–c): thermal energy of the isolated two-spin system at $\beta =5$; stars in (d–f): maximum entropy of the isolated two-spin system; crosses in (d–f): entropy of the isolated two-spin system at $\beta =5$.

Figure 3

Finite-size scaling of the system energy in the stationary state as a function of the number of bath spins as obtained from the solution of the TDSE with the initial state $|\uparrow \uparrow \rangle \otimes |\mathrm{\Phi }(\beta =1)\rangle$. Solid lines represent the function $a+b{N}_{\mathrm{B}}^{-c}$ with $a$ and $b$ determined by the best fit of the data. The Hamiltonian of the system-bath interaction is given by Eq. (4), respectively. (a) Integrable spin-bath Hamiltonian Eq. (6) with $K_n^x=K_n^y=K_n^z=K=1/2$ and $h_{\mathrm{B}}^x=h_{\mathrm{B}}^z=0$, system-bath interaction $\lambda =1/2$, and $N_{\mathrm{B}}=14,16,18,20,22,24,28,32,34$; (b) spin-bath Hamiltonian Eq. (7) with all $K$'s random in the interval $[-1/2,+1/2],h_{\mathrm{B}}^x=h_{\mathrm{B}}^z=0$, system-bath interaction $\lambda =1/2$, and $N_{\mathrm{B}}=14,16,18,20,22,24,28,32$.

Figure 4

(a) The absolute values of three of the nine bath-operator correlations Eq. (23) as obtained by solving the TDSE for a bath of $N_{\mathrm{B}}=20$ spins with the initial state $|\uparrow \downarrow \rangle \otimes |\mathrm{\Phi }(\beta =5)\rangle$. The bath-operator correlations that have absolute values that are too small to be seen on the scale of the plot have been omitted. The Hamiltonian of the bath is given by Eq. (6) with $K=1/2$ and $h_{\mathrm{B}}^x=h_{\mathrm{B}}^z=0$; (b) corresponding two-spin averages; (c) $F\left(t\right)$ as obtained from the 20 lowest eigenvalues of the bath Hamiltonian.

Figure 5

A single-spin average (a, d) and a two-spin correlation (b, e) as obtained by solving the TDSE (solid lines) and the QMEQ (circles) with $e^{\tau \mathbf{A}}$ and $\mathbf{B}$ obtained by least-square fitting to the TDSE data. The other 13 expectation values show similar agreement and are therefore not shown. Also shown is the RMSE error (c, f) defined by Eq. (27). The system Hamiltonian is given by Eq. (2) with $J_{\mathrm{S}}=1/4$ (antiferromagnetic Heisenberg model). The Hamiltonian of the system-bath interaction and spin bath are given by Eqs. (4) and (6), respectively. The number of bath spins is $N_{\mathrm{B}}=28,K=1,h_{\mathrm{B}}^x=h_{\mathrm{B}}^z=1/4$. The system-bath interaction strength is $\lambda =0.25$. The initial state is $|\uparrow \downarrow \rangle \otimes |\mathrm{\Phi }\left(\beta \right)\rangle$. (a, b, c) $\beta =1$; (d, e, f) $\beta =5$.

Figure 6

A single-spin average (a) and two-spin correlations (b, c) as obtained by solving the TDSE (solid lines) and the QMEQ (circles) with $e^{\tau \mathbf{A}}$ and $\mathbf{B}$ obtained by least-square fitting to the TDSE data. The other expectation values show similar agreement and are therefore not shown. The system Hamiltonian is given by Eq. (2) with $J_{\mathrm{S}}=1/4$ (antiferromagnetic Heisenberg model). The Hamiltonian of the system-bath interaction and spin bath are given by Eq. (4) and the spin-glass model Eq. (7), respectively. The number of bath spins is $N_{\mathrm{B}}=32,K=1/2$, and $h_{\mathrm{B}}^x=h_{\mathrm{B}}^z=0$. The system-bath interaction strength is $\lambda =0.5$. The initial state is $|\uparrow \uparrow \rangle \otimes |\mathrm{\Phi }(\beta =1)\rangle$.

Figure 7

The time evolution of the system energy $E_S$ with the model parameters $a=0.5,b=0.01$, and $\lambda =0.01$. The number of bath oscillators is $N_{\mathrm{B}}=511$. The initial state of the system is $T_S=0$. The initial state of the bath is drawn randomly from the canonical distribution with $\beta =1$. The straight line denotes the energy per oscillator of the total system.

Figure 8

The histogram ${\rho }_S\left(E_S\right)$ at different times and two different initial states of the system, $T_S=0$ (a) and $T_S=1.1$ (b). The model parameters are the same as in Fig. 7. The histogram represents an ensemble of ${10}^4$ random realizations of initial thermal states of the bath at $\beta =1$. The straight line denotes the expected distribution $exp(-\beta E_S)$ with $\beta =1$.