Temporal nonlocality of a two-level system interacting with a dephasing environment

Abstract

Temporal nonlocality of a two-level system interacting with a thermal reservoir is studied in terms of the Leggett–Garg inequality and the quantum witness, where the dephasing coupling and the Ohmic-like spectral density are assumed for an system-reservoir interaction. When the whole system is initially in the thermal equilibrium state, which is a quantum-classical correlated state, violation of the Leggett–Garg inequality and the quantum witness are investigated in detail. It is shown how the time evolution of the temporal nonlocality depends on the system-reservoir coupling strength, the reservoir temperature and the Ohmicity parameter. The results are compared with those obtained for a non-correlated initial state, where the two-level system and the thermal reservoir are initially in their own thermal equilibrium states. The comparison reveals that ignoring the initial correlation yields an additional phase shift in two-time correlation functions of the two-level system. The effect of the system-reservoir initial correlation becomes more significant as the Ohmicity parameter is smaller.

Appendices

Derivation of $\langle V(t_{j},t_{k})\rangle$ and $\langle V(t_{j},t_{k})\sigma _{z}\rangle$ in Sects. 3 and 4

In this “Appendix”, we calculate the average values $\langle V(t_{j},t_{k})\rangle$ and $\langle V(t_{j},t_{k})\sigma _{z}\rangle$ which are used to examine the Leggett–Garg inequality and the quantum witness in Sects. 3 and 4. First, substituting Eq. (2) into (9) and using the commutation relation of bosonic annihilation and creation operators, we can rewrite the operator $V(t_{j},t_{k})$ into

$$\begin{aligned} V(t_{j},t_{k})= & {} \mathrm {T}_{\rightarrow } \exp \left( i\sum _{\ell }\int _{t_{k}}^{t_{j}}\text {d}t\, [g_{\ell }a_{\ell }^{\dagger }\text {e}^{i\omega _{\ell }(t-t_{k})} +g_{\ell }^{*}a_{\ell }\text {e}^{-i\omega _{\ell }(t-t_{k})}]\right) \nonumber \\&\times \, \mathrm {T}_{\leftarrow } \exp \left( i\sum _{\ell }\int _{t_{k}}^{t_{j}}\text {d}t\, [g_{\ell }a_{\ell }^{\dagger }\text {e}^{i\omega _{\ell }(t-t_{k})} +g_{\ell }^{*}a_{\ell }\text {e}^{-i\omega _{\ell }(t-t_{k})}]\right) \nonumber \\= & {} \exp \left[ i\sum _{\ell }(C_{\ell }(t_{j}-t_{k})a_{\ell }^{\dagger } +C_{\ell }^{*}(t_{j}-t_{k})a_{\ell })\right] , \end{aligned}$$

(45)

where $C_{k}(t_{j}-t_{k}){=}(2g_{k}/i\omega _{k})(\text {e}^{i\omega _{k}(t_{j}-t_{k})}-1)$ and $\mathrm {T}_{\leftarrow }$ ($\mathrm {T}_{\rightarrow }$) stands for the time-ordering (the ant-time-order) operation. Then, using Eq. (3), we obtain the average value of Eq. (45),

$$\begin{aligned} \langle V(t_{j},t_{k})\rangle= & {} \frac{\text {e}^{{-}\beta \hbar \omega /2}}{\text {e}^{\beta \hbar \omega /2}{+}\text {e}^{{-}\beta \hbar \omega /2}} \mathrm {Tr}[V(t_{j},t_{k})\rho _{+}] {+}\frac{\text {e}^{\beta \hbar \omega /2}}{\text {e}^{\beta \hbar \omega /2}+\text {e}^{-\beta \hbar \omega /2}} \mathrm {Tr}[V(t_{j},t_{k})\rho _{-}] \nonumber \\= & {} \frac{\text {e}^{-\beta \hbar \omega /2}}{\text {e}^{\beta \hbar \omega /2}+\text {e}^{-\beta \hbar \omega /2}} \nonumber \\&\times \exp \left[ -i\sum _{\ell }\left( \frac{g_{\ell }^{*}}{\omega _{\ell }} C_{\ell }(t_{j}-t_{k}) +\frac{g_{\ell }}{\omega _{\ell }}C_{\ell }^{*}(t_{j}-t_{k})\right) \right. \nonumber \\&\quad \left. -\,\frac{1}{2}\sum _{\ell }\vert C_{\ell }(t_{j}-t_{k}) \vert ^{2}(2{\bar{n}}_{\ell }+1)\right] \nonumber \\&+\,\frac{\text {e}^{\beta \hbar \omega /2}}{\text {e}^{\beta \hbar \omega /2}+\text {e}^{-\beta \hbar \omega /2}} \nonumber \\&\times \exp \left[ i\sum _{\ell }\left( \frac{g_{\ell }^{*}}{\omega _{\ell }} C_{\ell }(t_{j}-t_{k}) +\frac{g_{\ell }}{\omega _{\ell }}C_{\ell }^{*}(t_{j}-t_{k})\right) \right. \nonumber \\&\quad \left. -\,\frac{1}{2}\sum _{\ell }\vert C_{\ell }(t_{j}-t_{k})\vert ^{2} (2{\bar{n}}_{\ell }+1)\right] . \end{aligned}$$

(46)

The real and imaginary parts in the exponentials are calculated to be

$$\begin{aligned} \frac{1}{2}\sum _{\ell }\vert C_{\ell }(t_{j}-t_{k}) \vert ^{2}(2{\bar{n}}_{\ell }+1)= & {} \sum _{\ell }\frac{4\vert g_{\ell }\vert ^{2}}{\omega _{\ell }^{2}} (2{\bar{n}}_{\ell }+1)[1-\cos \omega _{\ell }(t_{j}-t_{k})] \nonumber \\= & {} \int _{0}^{\infty }\text {d}w\,J(w)(2{\bar{n}}(w)+1) \frac{1-\cos w(t_{j}-t_{k})}{w^{2}} \nonumber \\= & {} \gamma (t_{j}-t_{k}), \end{aligned}$$

(47)

and

$$\begin{aligned} \sum _{\ell }\left( \frac{g_{\ell }^{*}}{\omega _{\ell }}C_{\ell }(t_{j}-t_{k}) +\frac{g_{\ell }}{\omega _{\ell }}C_{\ell }^{*}(t_{j}-t_{k})\right)= & {} \sum _{\ell }\frac{4\vert g_{\ell }\vert ^{2}}{\omega _{\ell }^{2}} \sin \omega _{\ell }(t_{j}-t_{k}) \nonumber \\= & {} \int _{0}^{\infty }\text {d}w\,J(w) \frac{\sin w(t_{j}-t_{k})}{w^{2}}\nonumber \\= & {} \phi (t_{j}-t_{k}). \end{aligned}$$

(48)

Then, Substituting these equations into Eq. (46), we obtain the average value,

$$\begin{aligned} \langle V(t_{j},t_{k})\rangle =\left[ \cos \phi (t_{j},t_{k}) +i\tanh (\beta \hbar \omega /2)\sin \phi (t_{j}-t_{k})\right] \text {e}^{-\gamma (t_{j}-t_{k})}, \end{aligned}$$

(49)

which yields Eq. (19). In the same way, we can derive the average value,

$$\begin{aligned} \langle V(t_{j},t_{k})\sigma _{z}\rangle= & {} \frac{\text {e}^{-\beta \hbar \omega /2}}{\text {e}^{\beta \hbar \omega /2}{+}\text {e}^{-\beta \hbar \omega /2}} \mathrm {Tr}[V(t_{j},t_{k})\rho _{+}] -\frac{\text {e}^{\beta \hbar \omega /2}}{\text {e}^{\beta \hbar \omega /2}+\text {e}^{-\beta \hbar \omega /2}} \mathrm {Tr}[V(t_{j},t_{k})\rho _{-}] \nonumber \\= & {} \frac{\text {e}^{-\beta \hbar \omega /2}}{\text {e}^{\beta \hbar \omega /2}+\text {e}^{-\beta \hbar \omega /2}} \nonumber \\&\times \exp \left[ -i\sum _{\ell }\left( \frac{g_{\ell }^{*}}{\omega _{\ell }} C_{\ell }(t_{j}-t_{k}) +\frac{g_{\ell }}{\omega _{\ell }}C_{\ell }^{*}(t_{j}-t_{k})\right) \right. \nonumber \\&\left. -\,\frac{1}{2}\sum _{\ell }\vert C_{\ell }(t_{j}-t_{k}) \vert ^{2}(2{\bar{n}}_{\ell }+1)\right] \nonumber \\&-\,\frac{\text {e}^{\beta \hbar \omega /2}}{\text {e}^{\beta \hbar \omega /2}+\text {e}^{-\beta \hbar \omega /2}} \nonumber \\&\times \exp \left[ i\sum _{\ell }\left( \frac{g_{\ell }^{*}}{\omega _{\ell }} C_{\ell }(t_{j}-t_{k}) +\frac{g_{\ell }}{\omega _{\ell }}C_{\ell }^{*}(t_{j}-t_{k})\right) \right. \nonumber \\&\left. -\,\frac{1}{2}\sum _{\ell }\vert C_{\ell }(t_{j}-t_{k}) \vert ^{2}(2{\bar{n}}_{\ell }+1)\right] \nonumber \\= & {} -[\tanh (\beta \hbar \omega /2)\cos \phi (t_{j}-t_{k}) +i\sin \phi (t_{j}-t_{k})]\text {e}^{-\gamma (t_{j}-t_{k})}, \end{aligned}$$

(50)

from which we can obtain Eq. (36).

Derivation of the parameters $\phi (t)$, $\gamma (t)$ and ${\dot{\gamma }}(t)$

When the thermal reservoir has the Ohmic-like spectral density (22), we derive Eqs. (23)–(26) and (28). First, substituting Eq. (22) into (20) and (21), we obtain $\gamma (t)=\gamma _{0}(t)+\gamma _{1}(t)$ and ${\dot{\gamma }}(t)={\dot{\gamma }}_{0}(t)+{\dot{\gamma }}_{1}(t)$ with

$$\begin{aligned} \gamma _{0}(t)= & {} \lambda \int _{0}^{\infty }\text {d}x\,\text {e}^{-x}x^{s-2}(1-\cos ax), \end{aligned}$$

(51)

$$\begin{aligned} \gamma _{1}(t)= & {} \lambda \int _{0}^{\infty }\text {d}x\, \text {e}^{-x}x^{s-2}(1-\cos ax)(\coth bx/2-1), \end{aligned}$$

(52)

$$\begin{aligned} {\dot{\gamma }}_{0}(t)= & {} \lambda \varOmega \int _{0}^{\infty }\text {d}x\, \text {e}^{-x}x^{s-1}\sin ax, \end{aligned}$$

(53)

$$\begin{aligned} {\dot{\gamma }}_{1}(t)= & {} \lambda \varOmega \int _{0}^{\infty }\text {d}x\, \text {e}^{-x}x^{s-1}\sin ax(\coth bx/2-1), \end{aligned}$$

(54)

and

$$\begin{aligned} \phi (t)=\lambda \int _{0}^{\infty }\text {d}x\,\text {e}^{-x}x^{s-2}\sin ax. \end{aligned}$$

(55)

In these equation, we set $a=\varOmega t$ and $b=\hbar \varOmega /kT$ for the sake of simplicity. To calculate the integral on the right-hand side of Eq. (51), we first differentiate $\gamma _{0}(t)$ with respect to the parameter a and then perform the integration with respect to x first and then a. Hence after straightforward calculation, we find that

$$\begin{aligned} \gamma _{0}(t)=\frac{1}{2}\lambda \ln (1+a^2), \end{aligned}$$

(56)

for $s=1$ and

$$\begin{aligned} \gamma _{0}(t)=\lambda \varGamma (s-1) \left[ 1-\frac{\text {Re}(1+ia)^{s-1}}{(1+a^2)^{s-1}}\right] , \end{aligned}$$

(57)

for $s\ne 1$. In this equation, $\varGamma (z)$ is the gamma function. Furthermore, we note that the relations $\cos (u\tan ^{-1}v)=\mathrm {Re}(1+iv)^{u}/(1+v^2)^{u/2}$ and $\sin (u\tan ^{-1}v)=\mathrm {Im}(1+iv)^{u}/(1+v^2)^{u/2}$ hold for real parameters u and v. Then we finally obtain $\gamma _{0}(t)=\frac{1}{2}\lambda \ln [1+(\varOmega t)^{2}]$ for $s=1$ and $\gamma _{0}(t)=\lambda \varGamma (s-1)f_{s}(\varOmega t)$ for $s\ne 1$, where $f_{s}(z)$ is given by Eq. (27).

Next, to calculate the integral on the right-hand side of Eq. (52), expanding $\coth bx/2$ into a power series of $\text {e}^{-bx}$, we obtain $\gamma _{1}(t)=2\lambda \sum _{n=1}^{\infty }f_{n}(t)$ with

$$\begin{aligned} f_{n}(t)=\int _{0}^{\infty }\text {d}x\, \text {e}^{-(1+nb)x}x^{s-2}(1-\cos ax), \end{aligned}$$

(58)

which can be calculated in the same way used for deriving $\gamma _{0}(t)$. In the case of $s=1$, we obtain

$$\begin{aligned} f_{n}(t)=\frac{1}{2}\left[ \ln (1+bn-ia)+\ln (1+bn+ia)\right] -\ln (1+bn), \end{aligned}$$

(59)

which yields

$$\begin{aligned} \gamma _{1}(t)= & {} -2\lambda \sum _{n=1}^{\infty } \left[ \ln (1+bn)-\frac{1}{2}\ln (1+bn-ia) -\frac{1}{2}\ln (1+bn+ia)\right] \nonumber \\= & {} 2\lambda \left[ \ln \varGamma \left( 1+\frac{kT}{\hbar \varOmega }\right) -\frac{1}{2}\ln \left| \varGamma \left( 1+\frac{kT}{\hbar \varOmega }+i\frac{kT}{\hbar }t\right) \right| ^{2}\right] , \end{aligned}$$

(60)

where we have used [54]

$$\begin{aligned} \varGamma (z)=\lim _{n\rightarrow \infty } \frac{n!n^{z}}{\prod _{k=0}^{n}(k+z)}. \end{aligned}$$

(61)

Thus we find that $\gamma (t)=\gamma _{0}(t)+\gamma _{1}(t)$ is equal to Eq. (23) for $s=1$. On the other hand, when $s\ne 1$, $f_{n}(t)$ is calculated to be

$$\begin{aligned} f_{n}(t)= & {} \varGamma (s-1)\left[ \frac{1}{(1+bn)^{s-1}}-\frac{1}{(1+bn)^{s-1}} \frac{\mathrm {Re}\left( 1+\frac{ia}{1+bn}\right) ^{s-1}}{\left[ 1+\left( \frac{a}{1+bn}\right) ^{2}\right] ^{s-1}}\right] \nonumber \\= & {} \frac{\varGamma (s-1)}{(1+bn)^{s-1}}\left[ 1-\frac{\cos \left[ (s-1)\tan ^{-1}\left( \frac{a}{1+bn}\right) \right] }{\left[ 1+\left( \frac{a}{1+bn}\right) ^{2}\right] ^{(s-1)/2}}\right] . \end{aligned}$$

(62)

Then we obtain

$$\begin{aligned} \gamma _{1}(t)&=2\lambda \varGamma (s-1)\sum _{n=1}^{\infty } \frac{1}{(1+bn)^{s-1}}\left\{ 1-\frac{\cos \left[ (s-1)\tan ^{-1}\left( \frac{a}{1+bn}\right) \right] }{\left[ 1+\left( \frac{a}{1+bn}\right) ^{2}\right] ^{(s-1)/2}}\right\} \nonumber \\&=2\lambda \left( \frac{kT}{\hbar \varOmega }\right) ^{s-1}\varGamma (s-1) \sum _{n=1}^{\infty } \frac{1}{(n+kT/\hbar \varOmega )^{s-1}} f_{s}\left( \frac{\frac{k_{\mathrm {B}}T}{\hbar }t}{n+\frac{k_{\mathrm {B}}T}{\hbar \varOmega }}\right) , \end{aligned}$$

(63)

which shows that $\gamma (t)=\gamma _{0}(t)+\gamma _{1}(t)$ is equal to Eq. (25) for $s\ne 1$.

The time-derivative ${\dot{\gamma }}(t)$ can be calculated directly. In fact, we can derive

$$\begin{aligned} {\dot{\gamma }}(t)= & {} \frac{\lambda \varOmega }{2i}\int _{0}^{\infty }\text {d}x\, [\text {e}^{-(1-ia)x}-\text {e}^{-(1+ia)x}]x^{s-1}\nonumber \\&+\,2\lambda \varOmega \sum _{n=1}^{\infty }\frac{1}{2i} \int _{0}^{\infty }\text {d}x\,[\text {e}^{-(1+bn-ia)x}-\text {e}^{-(1+bn+ia)x}]x^{s-1} \nonumber \\= & {} \lambda \varOmega \varGamma (s) \frac{\text {Im}(1+ia)^{s}}{(1+a^2)^{s}} +2\lambda \varOmega \varGamma (s)\sum _{n=1}^{\infty } \frac{1}{(1+bn)^{s}} \frac{\text {Im}(1+ia/(1+bn))^{s}}{[1+(a/(1+bn))^{2}]^{s}} \nonumber \\= & {} \lambda \varOmega \varGamma (s)g_{s}(\varOmega t) +2\lambda \varOmega \varGamma (s) \left( \frac{k_{\mathrm {B}}T}{\hbar \varOmega }\right) ^{s} \nonumber \\&\times \sum _{n=1}^{\infty }\frac{1}{(n+k_{\mathrm {B}}T/\hbar \varOmega )^{s}} g_{s}\left( \frac{\frac{k_{\mathrm {B}}T}{\hbar }t}{n+\frac{k_{\mathrm {B}}T}{\hbar \varOmega }}\right) , \end{aligned}$$

(64)

where $g_{s}(z)$ is given by Eq. (29). Thus we have derived Eq. (28). Finally, in the same way for deriving the parameter $\gamma _{0}(t)$, the phase parameter $\phi (t)$ is calculated to be

$$\begin{aligned} \phi (t)= & {} \frac{1}{2}\lambda \varGamma (s) \int _{0}^{a}\text {d}a' \left[ \frac{1}{(1-ia')^{s}}+\frac{1}{(1+ia')^{s}}\right] \nonumber \\= & {} \left\{ \begin{array}{ll} \displaystyle {\lambda \tan ^{-1}\varOmega t} &{}\quad (s=1) \\ \displaystyle { \lambda \varGamma (s-1)\frac{\sin [(s-1)\tan ^{-1}\varOmega t]}{[1+(\varOmega t)^{2}]^{(s-1)/2}}} &{}\quad (s\ne 1), \end{array}\right. \end{aligned}$$

(65)

which is equal to Eqs. (24) and (26).

Derivation of the correlation functions in Sect. 5

We calculate $\langle V(t_{j},t_{0})V^{\dagger }(t_{k},t_{0})\rangle _{R}$, $\langle V^{\dagger }(t_{j},t_{0})V(t_{k},t_{0})\rangle _{R}$, $\langle V(t_{k},t_{0})V^{\dagger }(t_{j},t_{0})\rangle _{R}$ and $\langle V^{\dagger }(t_{k},t_{0})V(t_{j},t_{0})\rangle _{R}$ used to examine the violation of the Leggett–Garg inequality and the quantum witness for the non-correlated initial state in Sect. 5. First we calculate the correlation function $\langle V(t_{j},t_{0})V^{\dagger }(t_{k},t_{0})\rangle _{R}$. From Eq. (45), we obtain

$$\begin{aligned} V(t_{j},t_{0})V^{\dagger }(t_{k},t_{0})= & {} \exp \left[ i\sum _{\ell } [C_{\ell }(t_{j}-t_{0})a_{\ell }^{\dagger } +C_{\ell }^{*}(t_{j}-t_{0})a_{\ell }]\right] \nonumber \\&\times \exp \left[ -i\sum _{\ell } [C_{\ell }(t_{k}-t_{0})a_{\ell }^{\dagger } +C_{\ell }^{*}(t_{k}-t_{0})a_{\ell }]\right] \nonumber \\= & {} \exp \left[ i\sum _{\ell } [(C_{\ell }(t_{j}-t_{0})-C_{\ell }(t_{k}-t_{0}))a_{\ell }^{\dagger }\right. \nonumber \\&\left. +\,(C_{\ell }^{*}(t_{j}-t_{0})-C_{\ell }^{*}(t_{k}-t_{0}))a_{\ell }]\right] \exp (\varPhi _{jk}), \end{aligned}$$

(66)

with

$$\begin{aligned} \varPhi _{jk}= & {} \frac{1}{2}\sum _{\ell } [-C_{\ell }(t_{j}-t_{0})C_{\ell }^{*}(t_{k}-t_{0}) +C_{\ell }^{*}(t_{j}-t_{0})C_{\ell }(t_{k}-t_{0})] \nonumber \\= & {} -i\phi (t_{j}-t_{k})-i\delta _{jk}, \end{aligned}$$

(67)

where $\phi (t_{j}-t_{k})$ is given by Eq. (21) or (48) and $\delta _{jk}$ is given by Eq. (40). Since the thermal reservoir is in the equilibrium state $\rho _{R}=\text {e}^{-H_{R}/k_{\mathrm {B}}T}/\mathrm {Tr}\text {e}^{-H_{R}/k_{\mathrm {B}}T}$, the average value of the exponential including the annihilation and creation operators in Eq. (66) is calculated to be

$$\begin{aligned}&\left\langle \exp \left[ i\sum _{\ell } [(C_{\ell }(t_{j}-t_{0})-C_{\ell }(t_{k}-t_{0}))a_{\ell }^{\dagger } +(C_{\ell }^{*}(t_{j}-t_{0})-C_{\ell }^{*}(t_{k}-t_{0}))a_{\ell }]\right] \right\rangle _{R} \nonumber \\&\quad =\exp \left[ -\frac{1}{2}\sum _{\ell } \vert C_{\ell }(t_{j}-t_{0})-C_{\ell }(t_{k}-t_{0})\vert ^{2} (2{\bar{n}}_{\ell }+1)\right] =\text {e}^{-\gamma (t_{j}-t_{k})}. \end{aligned}$$

(68)

Hence we obtain the correlation function,

$$\begin{aligned} \langle V(t_{j},t_{0})V^{\dagger }(t_{k},t_{0})\rangle _{R} =\text {e}^{-\gamma (t_{j}-t_{k})-i\phi (t_{j}-t_{k})-i\delta _{jk}}. \end{aligned}$$

(69)

In the same way, we can derive

$$\begin{aligned}&\langle V^{\dagger }(t_{j},t_{0})V(t_{k},t_{0})\rangle _{R} \nonumber \\&\quad =\left\langle \exp \left[ -i\sum _{\ell } [(C_{\ell }(t_{j}-t_{0})-C_{\ell }(t_{k}-t_{0}))a_{\ell }^{\dagger }\right. \right. \nonumber \\&\qquad \left. \left. +\,(C_{\ell }^{*}(t_{j}-t_{0})-C_{\ell }^{*}(t_{k}-t_{0}))a_{\ell }]\right] \right\rangle _{R} \exp (\varPhi _{jk}) \nonumber \\&\quad =\text {e}^{-\gamma (t_{j}-t_{k})-i\phi (t_{j}-t_{k})-i\delta _{jk}}. \end{aligned}$$

(70)

Taking complex conjugate of Eqs. (69) and (69), we obtain

$$\begin{aligned} \langle V(t_{k},t_{0})V^{\dagger }(t_{j},t_{0})\rangle _{R} =\langle V^{\dagger }(t_{k},t_{0})V(t_{j},t_{0})\rangle _{R} =\text {e}^{-\gamma (t_{j}-t_{k})+i\phi (t_{j}-t_{k})+i\delta _{jk}}. \end{aligned}$$

(71)

Therefore we have derived the correlation functions given by Eqs. (39) and (43).