Dynamics of open quantum spin systems: An assessment of the quantum master equation approach

Data of the numerical solution of the time-dependent Schrödinger equation of a system containing one spin-$\frac{1}{2}$ particle interacting with a bath of up to 32 spin-$\frac{1}{2}$ particles is used to construct a Markovian quantum master equation describing the dynamics of the system spin. The procedure of obtaining this quantum master equation, which takes the form of a Bloch equation with time-independent coefficients, accounts for all non-Markovian effects inasmuch the general structure of the quantum master equation allows. Our simulation results show that, with a few rather exotic exceptions, the Bloch-type equation with time-independent coefficients provides a simple and accurate description of the dynamics of a spin-$\frac{1}{2}$ particle in contact with a thermal bath. A calculation of the coefficients that appear in the Redfield master equation in the Markovian limit shows that this perturbatively derived equation quantitatively differs from the numerically estimated Markovian master equation, the results of which agree very well with the solution of the time-dependent Schrödinger equation.

Figure 1

Time evolution of the average of the system spin as obtained by solving the TDSE with a random thermal state at $\beta =2$ as the initial state. The Hamiltonian of the bath is given by Eq. (5) with $K=-\frac{1}{4}$ and $\mathrm{\Delta }=1$ (antiferromagnetic Heisenberg model). The parameters of the system-bath Hamiltonian (3) are $J=\frac{1}{4}$ and $h_{\mathrm{B}}^x=h_{\mathrm{B}}^z=\frac{1}{8}$. The system-bath interaction $\lambda =0.1$. (a) $N_{\mathrm{B}}=13$; (b) $N_{\mathrm{B}}=28$. Lines connecting the data points are guide to the eye.

Figure 2

The absolute values of three of the nine bath-operator correlations Eq. (47) as obtained by solving the TDSE for a bath of $N_{\mathrm{B}}=32$ spins with a random thermal state at $\beta =1$ as the initial state. 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 parameters of the system-bath Hamiltonian $H_{\mathrm{SB}}$ are $J=\frac{1}{4}$ and $h_{\mathrm{B}}^x=h_{\mathrm{B}}^z=\frac{1}{8}$. (a) The bath Hamiltonian $H_{\mathrm{B}}$ is given by Eq. (5) with $K=-\frac{1}{4}$ and $\mathrm{\Delta }=1$ (antiferromagnetic Heisenberg model) and $\lambda =0$; (b) same as (a) except that $\lambda =0.1$; (c) the bath Hamiltonian $H_{\mathrm{B}}$ is given by Eq. (6) with $K=\frac{1}{4}$ and $\lambda =0$; (d) same as (a) except that $\lambda =0.1$.

Figure 3

Comparison between the spin averages as obtained by solving the TDSE (solid lines) and the QMEQ (solid circles) with $e^{\tau \mathbf{A}}$ and $\mathbf{B}$ extracted from the TDSE data. (a) Initial state $|y\rangle |\varphi \rangle$; (b) initial state $|\uparrow \rangle |\varphi \rangle$. The model parameters are $\lambda =0, N_{\mathrm{B}}=13, \beta =0, K=-\frac{1}{4}, \mathrm{\Delta }=1$, and $h_{\mathrm{B}}^x=h_{\mathrm{B}}^z=\frac{1}{8}$. For clarity, the system-spin averages are shown with a time interval of 100. The markers represent the data obtained by least-square fitting to $15000$ numbers generated by the TDSE solver.

Figure 4

Comparison between the spin averages as obtained by solving the TDSE (solid lines) and the QMEQ (solid circles) with $e^{\tau \mathbf{A}}$ and $\mathbf{B}$ extracted from the TDSE data. (a)–(c) Show how the TDSE data (solid line) are being sampled, namely, at times indicated by the $t$ values of the markers, which in the case corresponds to a time steps of 0.2. (d)–(f) The sampled data of the whole interval $[0,1000]$, in this case $15000$ numbers, are used to determine by the least-square procedure described in Sec. 5, the parameters that enter the time evolution of the Markovian master equation (37). The latter is then used to compute the time evolution of the spin components, the data being represented by the markers. For clarity, in the bottom figures, the data are shown with a time interval of 10. The model parameters are $h_{\mathrm{B}}^x=h_{\mathrm{B}}^z=\frac{1}{8}$ and $\lambda =0.1, N_{\mathrm{B}}=13, \beta =0, K=-\frac{1}{4}$, and $\mathrm{\Delta }=1$. (a), (d) initial state $|x\rangle |\varphi \rangle$; (b), (e) initial state $|y\rangle |\varphi \rangle$; (c) initial state $|z\rangle |\varphi \rangle$; (f) the error $e_{\max }\left(t\right)$.

Figure 5

Same as Fig. 4, except that the bath contains $N_{\mathrm{B}}=24$ spins and $\lambda =0.05$. The markers represent the data obtained by least-square fitting to $15000$ numbers generated by the TDSE solver. For clarity, the data are shown with a time interval of 6.

Figure 6

Same as Fig. 5, except that the bath is initially at $\beta =1$.

Figure 7

Simulation data for a bath with $N_{\mathrm{B}}=28$ spins and system-bath interaction $\lambda =0.1$. The model parameters are $h_{\mathrm{B}}^x=h_{\mathrm{B}}^z=\frac{1}{8}, K=-\frac{1}{4}$, and $\mathrm{\Delta }=1$ eps. Solid lines: TDSE data; solid circles: QMEQ data. Top row: $\langle {\sigma }^x\left(t\right)\rangle$ as obtained by starting from the initial state $|x\rangle |\varphi \rangle$, (a)–(c) corresponding to $\beta =0,1,2$, respectively. Bottom row: $\langle {\sigma }^y\left(t\right)\rangle$ as obtained by starting from the initial state $|y\rangle |\varphi \rangle$, (d)–(f) corresponding to $\beta =0,1,2$, respectively. The TDSE simulations yield $\langle \left|{\sigma }_0^x(t=200)\right|\rangle =0.044, \langle \left|{\sigma }_0^x(t=200)\right|\rangle =0.475$, and $\langle \left|{\sigma }_0^x(t=200)\right|\rangle =0.756$ for $\beta =0,1,2$, respectively, whereas for the system in equilibrium we have $\langle {\sigma }_0^x\rangle =0,0.462,0.762$ for $\beta =0,1,2$, respectively. For clarity, the data are shown with a time interval of 0.6. The TDSE solver provided 3000 numbers as input to the least-square procedure.

Figure 8

Simulation data for a bath with $N_{\mathrm{B}}=32$ spins prepared at $\beta =1$ and system-bath interaction $\lambda =0.1$. The model parameters are $h_{\mathrm{B}}^x=h_{\mathrm{B}}^z=\frac{1}{8}, K=-\frac{1}{4}$, and $\mathrm{\Delta }=1$. (a), (b) Show TDSE data (solid lines) and QMEQ data (solid circles). (a) Initial state $|x\rangle |\varphi \rangle$; (b) initial state $|y\rangle |\varphi \rangle$; (c) the error $e_{\max }\left(t\right)$. The data obtained with the initial state $|\uparrow \rangle |\varphi \rangle$ are very similar as the data obtained with the initial state $|y\rangle |\varphi \rangle$ and are therefore not shown. For clarity, the data are shown with a time interval of 0.4. The TDSE solver provided 3000 numbers as input to the least-square procedure.

Figure 9

Same as Fig. 8 except that $\beta =0$ and $\lambda =0.2$.

Figure 10

Simulation data for a bath with $N_{\mathrm{B}}=32$ spins prepared at $\beta =0$ and system-bath interaction $\lambda =0.2$. The system Hamiltonian is given by Eq. (2). The system-bath interaction is given by Eq. (3) with $J_n^x=J_n^y=J_n^z=\frac{1}{4}$. The bath Hamiltonian is given by Eq. (5) with $K=-\frac{1}{4}, \mathrm{\Delta }=1$, and $h_n^x=h_n^z=0$. The full Hamiltonian is given by Eq. (56). (a), (b) Show TDSE data (solid lines) and QMEQ data (solid circles). (a) Initial state $|x\rangle |\varphi \rangle$; (b) initial state $|y\rangle |\varphi \rangle$; (c) the error $e_{\max }\left(t\right)$. The data obtained with the initial state $|\uparrow \rangle |\varphi \rangle$ are very similar as the data obtained with the initial state $|y\rangle |\varphi \rangle$ and are therefore not shown. With this choice of parameters of bath and system-bath Hamiltonians, the system does not relax to its thermal equilibrium state ${lim}_{t\to \infty }\langle {\sigma }_0^x\left(t\right)\rangle ={lim}_{t\to \infty }\langle {\sigma }_0^y\left(t\right)\rangle ={lim}_{t\to \infty }\langle {\sigma }_0^z\left(t\right)\rangle =0$. For clarity, the data are shown with a time interval of 0.4. The TDSE solver provided 3000 numbers as input to the least-square procedure.

Figure 11

Simulation data for a bath with $N_{\mathrm{B}}=32$ spins prepared at $\beta =0$ and system-bath interaction $\lambda =0.2$. The system Hamiltonian is given by Eq. (2). The system-bath interaction is given by Eq. (3) with $J_n^x=J_n^y=0$ and $J_n^z$ uniformly random between $-\frac{1}{4}$ and $\frac{1}{4}$, in which case the interaction of the system and bath spins is through the coupling of the $z$ components of the spins only. The bath Hamiltonian is given by Eq. (6) with $K_n^x=K_n^y=h_n^x=h_n^z=0$ and $K_n^z$ uniformly random between $-1$ and 1. The full Hamiltonian is given by Eq. (57). (a), (b) Show TDSE data (solid lines) and QMEQ data (solid circles). (a) Initial state $|x\rangle |\varphi \rangle$; (b) initial state $|y\rangle |\varphi \rangle$; (c) the error $e_{\max }\left(t\right)$. The data obtained with the initial state $|\uparrow \rangle |\varphi \rangle$ are very similar as the data obtained with the initial state $|y\rangle |\varphi \rangle$ and are therefore not shown. With this choice of bath and system-bath Hamiltonians, the system does not relax to its thermal equilibrium state ${lim}_{t\to \infty }\langle {\sigma }_0^x\left(t\right)\rangle ={lim}_{t\to \infty }\langle {\sigma }_0^y\left(t\right)\rangle ={lim}_{t\to \infty }\langle {\sigma }_0^z\left(t\right)\rangle =0$. For clarity, the data are shown with a time interval of 0.4. The TDSE solver provided 3000 numbers as input to the least-square procedure.

Figure 12

Same as Fig. 8 except that $\beta =1$ and $\lambda =1$.