Markovian and Non-Markovian Dynamics in the One-Dimensional Transverse-Field XY Model

Abstract

We consider an anisotropic spin-1/2 XY Heisenberg chain in the presence of a transverse magnetic field. Selecting the nearest neighbor pair spins as an open quantum system, the rest of the chain plays the role of the structured environment. In fact, the aforementioned system is used as a quantum probe signifying nontrivial features of the environment with which is interacting. We use a general measure that is based on the trace distance for the degree of non-Markovian behavior in open quantum systems. The witness of non-Markovianity takes on nonzero values whenever there is a flow of information from the environment back to the open system. We have shown that the dynamics of the system with isotropic Heisenberg interaction is Markovian. A dynamical transition into the non-Markovian regime is observed as soon as the anisotropy, $\gamma$, is applied. At the Ising value of the anisotropy $\gamma =1.0$, all the information flows back from the environment to the system. The additional dynamical transition from the non-Markovian to the Markovian is obtained by applying the transverse magnetic field. In addition, we have focused on the time evolution of the Loschmidt-echo return rate function. It is found that a non-analyticity can be seen in the time evolution of the Loschmidt-echo return rate function exactly at the critical points where a dynamical transition from the Markovian to the non-Markovian occurs.

Appendix

Here we try to calculate $X^{+}_{mm'}$ which is related to the fermion operators as

$$\begin{aligned} X^{+}_{mm'}= & {} \langle a^{\dag }_{m}(t) a_{m}(t)\rangle \langle a^{\dag }_{m'}(t) a_{m'}(t)\rangle \nonumber \\&+\langle a^{\dag }_{m}(t) a_{m'}(t)\rangle \langle a_{m}(t)a^{\dag }_{m'}(t)\rangle . \end{aligned}$$

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At first, the method of calculating $\langle a^{\dag }_{m}(t) a_{m}(t)\rangle$ is explained.

$$\begin{aligned} \langle a^{\dag }_{m}(t) a_{m}(t)\rangle= & {} \langle \psi _0|e^{iHt}a^{\dag }_{m}a_{m}e^{-iHt}|\psi _0\rangle \nonumber \\= & {} \frac{1}{\sqrt{N}}\sum _{k}e^{i(k_1-k_2)m}\nonumber \\&\langle \psi _0|e^{iHt}a^{\dag }_{k_1}a_{k_2}e^{-iHt}|\psi _0\rangle \end{aligned}$$

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$$\begin{aligned} e^{iHt}a^{\dag }_{k_1}e^{-iHt}= & {} e^{it\sum _{k}\varepsilon (k) (\beta _{k}^{\dagger }\beta _{k})/\hbar }(cos(k_1)\beta _{k_1}^{\dagger } \nonumber \\&+i sin(k_1)\beta _{-k_1})e^{-it\sum _{k}\varepsilon (k) (\beta _{k}^{\dagger }\beta _{k})/\hbar } \nonumber \\= & {} cos(k_1)e^{it\varepsilon (k_1)}\beta _{k_1}^{\dagger }\nonumber \\&+i sin(k_1)e^{-it\varepsilon (-k_1)}\beta _{-k_1} \end{aligned}$$

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In the same way one can show

$$\begin{aligned} e^{iHt}a_{k_2}e^{-iHt}= & {} cos(k_2)e^{-it\varepsilon (k_2)}\beta _{k_2}^{\dagger }\nonumber \\&-isin(k_2)e^{it\varepsilon (-k_2)}\beta _{-k_2}^{\dagger }. \end{aligned}$$

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The relation 22 is simplified as follows

$$\begin{aligned} \langle a^{\dag }_{m}(t) a_{m}(t)\rangle&=\frac{1}{\sqrt{N}}\sum _{k}e^{i(k_1-k_2)m}\nonumber \\&\quad \langle \psi _0|(cos(k_1)e^{it\varepsilon (k_1)}\beta _{k_1}^{\dagger }\nonumber \\&\quad +i sin(k_1)e^{-it\varepsilon (-k_1)}\beta _{-k_1}) \nonumber \\&\quad (cos(k_2)e^{-it\varepsilon (k_2)}\beta _{k_2}^{\dagger }\nonumber \\&\quad -isin(k_2)e^{it\varepsilon (-k_2)}\beta _{-k_2}^{\dagger })|\psi _0\rangle . \end{aligned}$$

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Using the Bogoliubov operators

$$\begin{aligned} \langle a^{\dag }_{m}(t) a_{m}(t)\rangle= & {} \frac{1}{2 N^2}\sum _{k,k'}(1+e^{i(k+\phi )}+e^{-i(k'+\phi ))}\\&+e^{i(k-k')})\\&(e^{-it(\varepsilon (k)-\varepsilon (k'))}cos^2(k)cos^2(k')\\&+e^{it(\varepsilon (k)+\varepsilon (k'))}sin^2(k)cos^2(k')\\&+e^{-it(\varepsilon (k)+\varepsilon (k'))}sin^2(k')cos^2(k)\\&+ e^{it(\varepsilon (k)-\varepsilon (k'))}sin^2(k)sin^2(k')\\&+\frac{1}{4}sin(2k)sin(2k')(-e^{it(\varepsilon (k)-\varepsilon (k'))}\\&+e^{it(\varepsilon (k)+\varepsilon (k'))}\\&+e^{-it(\varepsilon (k)+\varepsilon (k'))}-e^{-it(\varepsilon (k)-\varepsilon (k'))}))\\&+ \frac{1}{4 N^2}\sum _{k,k'}(sin^2(2k') (1+cos(k+\phi ))\\&(2-e^{2it(\varepsilon (k')}-e^{-2it(\varepsilon (k')}) \end{aligned}$$

Similarly, $\langle a^{\dag }_{m}(t) a_{m+1}(t)\rangle$ can be written as follows

$$\begin{aligned} \langle a^{\dag }_{m}(t) a_{m+1}(t)\rangle= & {} \frac{1}{2 N^2}\sum _{k,k'}(1+e^{i(k+\phi )}+e^{-i(k'+\phi ))}\nonumber \\&+e^{i(k-k')}) e^{-ik}\nonumber \\&(e^{-it(\varepsilon (k)-\varepsilon (k'))}cos^2(k)cos^2(k')\nonumber \\&+e^{it(\varepsilon (k)+\varepsilon (k'))}sin^2(k)cos^2(k')\nonumber \\&+e^{-it(\varepsilon (k)+\varepsilon (k'))}sin^2(k')cos^2(k)\nonumber \\&+e^{it(\varepsilon (k)-\varepsilon (k'))}sin^2(k)sin^2(k')\nonumber \\&+\frac{1}{4}e^{ik'}sin(2k)sin(2k')(-e^{it(\varepsilon (k)-\varepsilon (k'))}\nonumber \\&+e^{it(\varepsilon (k)+\varepsilon (k'))}\nonumber \\&+e^{-it(\varepsilon (k)+\varepsilon (k'))}-e^{-it(\varepsilon (k)-\varepsilon (k'))}))\nonumber \\&+ \frac{1}{4 N^2}\sum _{k,k'}(sin^2(2k')\nonumber \\&(1+cos(k+\phi ))e^{-ik'}(2-e^{2it(\varepsilon (k')}-e^{-2it(\varepsilon (k')}) \end{aligned}$$

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