Non-Markovian Dynamics of Macroscopic Quantum Systems in Interaction with Non-equilibrium Environments

Abstract

We study the dynamics of a macroscopic superconducting qubit coupled to two independent non-stationary reservoirs by using time-dependent perturbation theory. We show that an equilibrium environment surpasses the coherent evolution of the macroscopic qubit completely. When the qubit couples to two different reservoirs, exemplifying a non-equilibrium environment, the short-time dynamics is affected by the interference between two reservoirs, implying the non-additivity of effects of two reservoirs. The non-additivity can be traced back to a non-Markovian effect, even though two reservoirs are independently assumed to be Markovian. Explicitly, the non-equilibrium environment intensifies both coherent and incoherent parts of the evolution. Therefore, the macroscopic qubit would evolve more coherently but at the price of a shorter decoherence time.

Appendix: Calculation of Matrix Elements of \(U_{I}(t)\)

We expand U(t) up to the second order with respect to the interaction Hamiltonian $H_{int}(t)=\sum _{{\mathcal {R}}}H_{S{\mathcal {R}}}(t)$ as

$$\begin{aligned} U(t)\simeq 1-\frac{i}{h}\int _{0}^{t}dt_{1}H_{int}(t_{1})-\frac{1}{h^{2}}\int _{0}^{t}dt_{2}\int _{0}^{t_{2}}dt_{1} H_{int}(t_{2}) H_{int}(t_{1}) \end{aligned}$$

(A.1)

The contribution of two-photon excitations in each reservoir and also of simultaneous single excitaions in both reservoirs are small. Therefore, we first calculate

$$\begin{aligned}&U(t)|0\rangle _{A}|0\rangle _{B}\simeq U_{0A,0B}(t)|0\rangle _{A}|0\rangle _{B}+\sum _{\alpha }\Big \{U_{\alpha A,\alpha B}(t)|\alpha \rangle _{A}|\alpha \rangle _{B}+U_{0A,\alpha B}(t)|0\rangle _{A}|\alpha \rangle _{B}\nonumber \\&\quad +U_{\alpha A,0B}(t)|\alpha \rangle _{A}|0\rangle _{B}\Big \}, \end{aligned}$$

(A.2)

where

$$\begin{aligned} U_{0A,0B}(t)=&1-\frac{i}{2h}\sum _{{\mathcal {R}}}\sum _{\alpha }\int _{0}^{t}dt_{1}f_{\alpha ,{\mathcal {R}}}^{2}(t_{1})\nonumber \\&-\frac{1}{2h}\sum _{{\mathcal {R}}}\sum _{\alpha }\omega _{\alpha ,{\mathcal {R}}} \int _{0}^{t}dt_{2}\int _{0}^{t_{2}}dt_{1}e^{-i(t_{2}-t_{1})\omega _{\alpha ,{\mathcal {R}}}}f_{\alpha ,{\mathcal {R}}}(t_{2})f_{\alpha ,{\mathcal {R}}}(t_{1})\nonumber \\&~-\frac{1}{4h^{2}}\sum _{\alpha }\int _{0}^{t}dt_{2}\int _{0}^{t_{2}}dt_{1}\Big \{f_{\alpha ,A}^{2}(t_{2}),f_{\alpha ,B}^{2}(t_{1})\Big \}\nonumber \\ U_{\alpha A,\alpha B}(t)=&-\frac{1}{2h}\omega _{\alpha ,A}^{1/2}\omega _{\alpha ,B}^{1/2}\int _{0}^{t}dt_{2}\int _{0}^{t_{2}}dt_{1} \Big \{e^{i\omega _{\alpha ,A}t_{2}}f_{\alpha ,A}(t_{2}),e^{i\omega _{\alpha ,B}t_{1}}f_{\alpha ,B}(t_{1})\Big \}\nonumber \\ U_{0A,\alpha B}(t)=&\frac{i}{\sqrt{2h}}\omega _{\alpha ,B}^{1/2}\int _{0}^{t}dt_{1}e^{i\omega _{\alpha ,B}t_{1}}f_{\alpha ,B}(t_{2}) \nonumber \\&+\frac{1}{\sqrt{8h^{3}}}\omega _{\alpha ,B}^{1/2}\int _{0}^{t}dt_{2}\int _{0}^{t_{2}}dt_{1} \Big \{f_{\alpha ,A}^{2}(t_{2}),e^{i\omega _{\alpha ,B}t_{1}}f_{\alpha ,B}(t_{1})\Big \}, \end{aligned}$$

(A.3)

to find

$$\begin{aligned} |\chi _{n}(t)\rangle =&|0\rangle _{A}|0\rangle _{B}\langle n|U_{0A,0B}(t)|L\rangle +\sum _{\alpha }e^{-i\omega _{\alpha ,B}t}|0\rangle _{A}|\alpha \rangle _{B}\langle n|U_{0A,\alpha B}(t)|L\rangle \nonumber \\&+\sum _{\alpha }e^{-i\omega _{\alpha ,A}t}|\alpha \rangle _{A}|0\rangle _{B}\langle n|U_{\alpha A,0B}(t)|L\rangle \nonumber \\&+\sum _{\alpha }e^{-i(\omega _{\alpha ,A}+\omega _{\alpha ,B})t}|\alpha \rangle _{A}|\alpha \rangle _{B}\langle n|U_{\alpha A,\alpha B}(t)|L\rangle , \end{aligned}$$

(A.4)

where we defined $f_{\alpha ,{\mathcal {R}}}(t)=\omega _{\alpha ,{\mathcal {R}}}f_{\alpha ,{\mathcal {R}}}(x(t))$, $x(t)=e^{iH_{S}t/h}xe^{-iH_{S}t/h}$ and $\{g(t),h(t')\}=g(t)h(t')+g(t')h(t)$. Note that $U_{\alpha A,0B}(t)$ can be obtained by interconverting $A\leftrightarrow B$ of $U_{0A,\alpha B}(t)$ in (A.3).

The problem is now reduced to the evaluation of matrix elements of interaction Hamiltonian in (A.4). The potential U(x) being an even function, the energy eigenstates $|n\rangle$ have definite parity. Since $U_{0A,0B}(t)$ and $U_{\alpha A,\alpha B}(t)$ are even functions and $U_{0A,\alpha B}(t)$ and $U_{\alpha A,0B}(t)$ are odd functions, the following selection rules are identified

$$\begin{aligned} \langle m| U_{0A,0B}(t)|n\rangle&=\langle m| U_{\alpha A,\alpha B}(t)|n\rangle =0, \nonumber \\ \langle n| U_{0A,\alpha B}(t)|n\rangle&=\langle n| U_{\alpha A,0B}(t)|n\rangle =0. \end{aligned}$$

(A.5)

The diagonal matrix elements of $U_{0A,0B}(t)$ are evaluated as

$$\begin{aligned} \langle n|U_{0A,0B}(t)|n\rangle&=1-\frac{it}{h}\sum _{{\mathcal {R}}}\Bigg \{\delta E_{n,{\mathcal {R}}}^{(1)}-\frac{1}{\pi }\sum _{m}x_{mn}^{2}\int _{0}^{\infty }d\omega _{{\mathcal {R}}} \frac{J(\omega _{{\mathcal {R}}})}{\omega _{{\mathcal {R}}}+\Omega _{mn}}\Bigg \} \nonumber \\&-\frac{1}{\pi h}\sum _{{\mathcal {R}}}\sum _{m}x_{mn}^{2}\int _{0}^{\infty }d\omega _{{\mathcal {R}}}J(\omega _{{\mathcal {R}}}) \frac{1-e^{\imath t(\omega _{{\mathcal {R}}}+\Omega _{mn)}}}{(\omega _{{\mathcal {R}}}+\Omega _{mn})^{2}}-\frac{t^{2}}{h^{2}}\delta E_{n,A}^{(1)}\delta E_{n,B}^{(1)}, \end{aligned}$$

(A.6)

The quantity embraced by the bracket on the right-hand side of (A.6) coincides with $\delta E_{n,{\mathcal {R}}}$ in (11). Thus, the following expression is valid up to the second order

$$\begin{aligned}&\langle n|U_{0A,0B}(t)|n\rangle \simeq e^{-it/h\sum _{{\mathcal {R}}}\delta E_{n,{\mathcal {R}}}}\Bigg \{1-\frac{1}{\pi h}\sum _{{\mathcal {R}}}\sum _{m}x_{mn}^{2}\int _{0}^{\infty }d\omega _{{\mathcal {R}}}J(\omega _{{\mathcal {R}}})\nonumber \\&\quad \times \Big (\frac{2\sin ^{2}\{(\omega _{{\mathcal {R}}}+\Omega _{mn})t/2\}}{(\omega _{{\mathcal {R}}}+\Omega _{mn})^{2}}-i\frac{\sin \{(\omega _{{\mathcal {R}}} +\Omega _{mn})t\}}{(\omega _{{\mathcal {R}}}+\Omega _{mn})^{2}}-\frac{t^{2}}{h^{2}}\delta E_{n,A}^{(1)}\delta E_{n,B}^{(1)}\Big )\Bigg \}. \end{aligned}$$

(A.7)

We assume that the reservoirs’ cut-off strengths are much higher than the system’s characteristic strength (Markov approximation), i.e. $\Lambda _{{\mathcal {R}}}\gg \Omega$, so that at times much higher than $\Omega ^{-1}$, the first term of the integral in (A.7) can be approximated by a delta function $\delta (\omega _{{\mathcal {R}}}+\Omega _{mn})$. Obviously, the result of the corresponding integral would be $J(\omega _{{\mathcal {R}}}+\Omega _{mn})$, which is zero for $\Omega _{mn}\ge 0$. The elements of $U_{0A,0B}(t)$ are then reduced to

$$\begin{aligned} \langle 1|U_{0A,0B}(t)|1\rangle&\simeq e^{-it/h\sum _{{\mathcal {R}}}\delta E_{1,{\mathcal {R}}}}\Bigg \{1+\frac{i}{\pi h}\sum _{{\mathcal {R}}}\int _{0}^{\infty }d\omega _{{\mathcal {R}}}J(\omega _{{\mathcal {R}}})\frac{\sin \{(\omega _{{\mathcal {R}}} +\Delta )t\}}{(\omega _{{\mathcal {R}}}+\Delta )^{2}}\nonumber \\&\quad -\frac{t^{2}}{h^{2}}\delta E_{1A}^{(1)}\delta E_{1B}^{(1)}\Bigg \},\nonumber \\ \langle 2|U_{0A,0B}(t)|2\rangle&\simeq e^{-it/h\sum _{{\mathcal {R}}}\delta E_{2,{\mathcal {R}}}}\Bigg \{1-\sum _{{\mathcal {R}}}\frac{\Gamma _{2,{\mathcal {R}}}t}{2}\nonumber \\&\quad +\frac{i}{\pi h}\sum _{{\mathcal {R}}}\int _{0}^{\infty }d\omega _{{\mathcal {R}}}J(\omega _{{\mathcal {R}}})\frac{\sin \{(\omega _{{\mathcal {R}}} -\Delta )t\}}{(\omega _{{\mathcal {R}}}-\Delta )^{2}}-\frac{t^{2}}{h^{2}}\delta E_{2A}^{(1)}\delta E_{2B}^{(1)}\Bigg \}. \end{aligned}$$

(A.8)

If the coupling between the system and each reservoir is considered to be weak, we have $\Gamma _{2,{\mathcal {R}}}\ll \Omega$. At the temporal domain $\Omega ^{-1}\ll t\ll \Gamma _{2,{\mathcal {R}}}^{-1}$, we finally obtain

$$\begin{aligned} \langle 1|U_{0A,0B}(t)|1\rangle&\approx \prod _{{\mathcal {R}}}\exp \Bigg \{-\frac{i}{h}\Big (t\delta E_{1,{\mathcal {R}}}-\frac{1}{\pi }\int _{0}^{\infty }d\omega _{{\mathcal {R}}}J(\omega _{{\mathcal {R}}})\frac{\sin \{(\omega _{{\mathcal {R}}} +\Delta )t\}}{(\omega _{{\mathcal {R}}}+\Delta )^{2}}\Big )\nonumber \\&\quad -\frac{t^{2}}{h^{2}}\delta E_{1A}^{(1)}\delta E_{1B}^{(1)}\Bigg \},\nonumber \\ \langle 2|U_{0A,0B (t)}|2\rangle&\approx \prod _{{\mathcal {R}}}\exp \Bigg \{-\frac{\Gamma _{2,{\mathcal {R}}}t}{2}-\frac{i}{h}\Big (t\delta E_{2,{\mathcal {R}}}\nonumber \\&\quad -\frac{1}{\pi }\int _{0}^{\infty }d\omega _{{\mathcal {R}}}J(\omega _{{\mathcal {R}}})\frac{\sin \{(\omega _{{\mathcal {R}}} -\Delta )t\}}{(\omega _{{\mathcal {R}}}-\Delta )^{2}}\Big )-\frac{t^{2}}{h^{2}}\delta E_{2A}^{(1)}\delta E_{2B}^{(1)}\Bigg \}. \end{aligned}$$

(A.9)

The elements of $U_{\alpha A,\alpha B}(t)$ are obtained as

$$\begin{aligned}&\langle n|U_{\alpha A,\alpha B}(t)|n\rangle \nonumber \\&\quad =-\frac{\pi }{4h}\gamma _{\alpha ,A}\omega _{\alpha ,A}^{3/2}\gamma _{\alpha ,B}\omega _{\alpha ,B}^{3/2}\nonumber \\&\qquad \times \Bigg \{\frac{1-e^{-i(\omega _{\alpha ,A}+\omega _{\alpha ,B})t}}{(\omega _{\alpha ,A}+\omega _{\alpha ,B})(\omega _{\alpha ,B}+(-1)^{n+1}\Delta )}+\frac{e^{-i(\omega _{\alpha ,A}+(-1)^{n+1}\Delta )t}-1}{(\omega _{\alpha ,A}+(-1)^{n+1}\Delta )(\omega _{\alpha ,B}+(-1)^{n+1}\Delta )}\nonumber \\&\qquad +\frac{1-e^{-i(\omega _{\alpha ,A}+\omega _{\alpha ,B})t}}{(\omega _{\alpha ,A}+\omega _{\alpha ,B})(\omega _{\alpha ,A}+(-1)^{n+1}\Delta )}+\frac{e^{-i(\omega _{\alpha ,B}+(-1)^{n+1}\Delta )t}-1}{(\omega _{\alpha ,A}+(-1)^{n+1}\Delta )(\omega _{\alpha ,B}+(-1)^{n+1}\Delta )}\Bigg \}. \end{aligned}$$

(A.10)

The elements of $U_{0A,\alpha B}(t)$ are evaluated as

$$\begin{aligned} \langle m|U_{0A,\alpha B}(t)|n\rangle= & {} \sqrt{\frac{\pi }{h}}\Big (i+\frac{t}{h}\delta E_{n,A}^{(1)}\Big )\gamma _{\alpha ,B}^{2}\omega _{\alpha ,B}^{3}\frac{\sin \big \{(\omega _{\alpha ,B}+(-1)^{m}\Delta )t/2\big \}}{\omega _{\alpha ,B}+(-1)^{m}\Delta }\nonumber \\&\times e^{i(\omega _{\alpha ,B}+(-1)^{m}\Delta )t/2}. \end{aligned}$$

(A.11)

The elements of $U_{\alpha A,0B}(t)$ are calculated by interconverting $A\leftrightarrow B$ in $\langle m|U_{0A,\alpha B}(t)|n\rangle$ in (A.11).