The quantum Jarzynski inequality for superconducting optical cavities

Abstract

The quantum version of Jarzynski equality was theoretically investigated for the work operator in the interaction picture. Since the Hamiltonian is time-dependent, it does not commute with itself at different times, then we have solved analytically its dynamics with lie-algebraic techniques. The physical object is a superconducting optical cavity. It was shown that the quantum dynamics of a cavity does not obey Jarzynski’s equality in the small temperature regime $(\beta \omega _0 \hbar \gg 1)$, if we consider a quantum work operator defined in a protocol interaction picture. The case of higher temperatures $(\beta \omega _0 \hbar \ll 1)$ we are in Jarzynski regime and Jarzynski equality can be used to obtain statistical information about work done on the cavity. Since, for all used parameters, the state is a thermal state and its dynamics is of the harmonic oscillator, then this quantum regime is related to the protocol action. The protocol can be implemented by injecting a field in the cavity, thus it creates quantum resources.

Graphic Abstract

Appendices

Appendix I: Time evolution operator

The Hamiltonian is given by

$$\begin{aligned} H\left( t\right) =\hbar \omega _{0} \left( a^{\dag }a+\frac{1}{2}\right) +\sqrt{ \hbar }\alpha \left( t\right) \left( a^{\dag }+a\right) , \end{aligned}$$

(20)

where $\alpha \left( t\right) =$ $\sqrt{\frac{1}{2\omega _{0} }}lvt$. The group $\{1,a^{\dag }a,a^{\dag },a\}$ form a closed algebra, thus we can write the time evolution operator as

$$\begin{aligned} U\left( t\right) =e^{\zeta \left( t\right) a^{\dag }}e^{-i\Omega \left( t\right) a^{\dag }a}e^{\xi \left( t\right) a}e^{\Gamma \left( t\right) }. \end{aligned}$$

(21)

Since it must obeys

$$\begin{aligned} \frac{d}{dt}U=-\frac{i}{\hbar }H\left( t\right) U\left( t\right) , \end{aligned}$$

(22)

then we have

$$\begin{aligned} \frac{d}{dt}U= & {} \left( \dot{\zeta }a^{\dag }+\dot{\Gamma } \right) U-i\dot{\Omega }e^{\zeta \left( t\right) a^{\dag }}a^{\dag }ae^{-i\Omega \left( t\right) a^{\dag }a}e^{\xi \left( t\right) a}e^{\Gamma \left( t\right) }\nonumber \\&+\dot{\xi }e^{\zeta \left( t\right) a^{\dag }}e^{-i\Omega \left( t\right) a^{\dag }a}ae^{\xi \left( t\right) a}e^{\Gamma \left( t\right) }. \end{aligned}$$

(23)

After some algebra we can write (23) as

$$\begin{aligned} \frac{d}{dt}U=\left[ \left( \dot{\zeta }+i\dot{\Omega }\zeta \right) a^{\dag }-i\dot{\Omega }a^{\dag }a+\dot{\xi } e^{i\Omega }a+\dot{\Gamma }-{\dot{\xi \zeta }}e^{i\Omega }\right] U.\nonumber \\ \end{aligned}$$

(24)

Comparing (22) with (24) we find

$$\begin{aligned} -i\dot{\Omega }=-i\omega _{0} \Longrightarrow \Omega \left( t\right) =\omega _{0} t, \end{aligned}$$

(25)

where we have assume that $\Omega \left( 0\right) =0$. Also we have

$$\begin{aligned} \dot{\zeta }+i\omega _{0} \zeta= & {} -\frac{i}{\sqrt{\hbar }}\alpha \left( t\right) , \end{aligned}$$

(26)

$$\begin{aligned} \dot{\xi }e^{i\omega _{0} t}= & {} -\frac{i}{\sqrt{\hbar }}\alpha \left( t\right) , \end{aligned}$$

(27)

$$\begin{aligned} \dot{\Gamma }-\dot{\xi }\zeta e^{i\omega _{0} t}= & {} 0. \end{aligned}$$

(28)

By setting $\zeta \left( t\right) =z\left( t\right) e^{-i\omega _{0} t}$ and using (26) we obtain

$$\begin{aligned} \dot{\zeta }=\dot{z}e^{-i\omega _{0} t}-i\omega _{0} \zeta \left( t\right) \end{aligned}$$

(29)

and

$$\begin{aligned} \dot{z}=-\frac{i}{\sqrt{\hbar }}\alpha \left( t\right) e^{i\omega _{0} t}. \end{aligned}$$

(30)

Now, from (27), we get

$$\begin{aligned} \dot{\xi }=-\frac{i}{\sqrt{\hbar }}\alpha \left( t\right) e^{-i\omega _{0} t} \end{aligned}$$

(31)

We note that $\alpha \in \mathfrak {R}$, then

$$\begin{aligned} i\dot{\xi }=\left( i\dot{z}\right) ^{*}\Longrightarrow \dot{\xi }=-\dot{z}^{*}\Longrightarrow \xi =-z^{*} \end{aligned}$$

(32)

thus

$$\begin{aligned} \zeta \left( t\right) =-\xi ^{*}\left( t\right) e^{-i\omega _{0} t} \end{aligned}$$

(33)

and

$$\begin{aligned} \dot{\xi }=-\frac{i}{\sqrt{\hbar }}\alpha \left( t\right) e^{-i\omega _{0} t} \end{aligned}$$

(34)

Considering (34) and the initial condition as initial $\xi \left( 0\right) =0$ then we obtain

$$\begin{aligned} \xi \left( t\right) =-i\int \nolimits _{0}^{t}\frac{1}{\sqrt{\hbar }}\alpha \left( t^{\prime }\right) e^{-i\omega _{0} t^{\prime }}dt^{\prime } \end{aligned}$$

(35)

If we set $\alpha =\frac{lvt}{\sqrt{2\omega _{0} }}$ then

$$\begin{aligned} \xi \left( t\right)= & {} -i\int \nolimits _{0}^{t}\frac{lvt^{\prime }}{\sqrt{ 2\hbar \omega _{0} }}e^{-i\omega _{0} t^{\prime }}dt^{\prime }\nonumber \\= & {} \frac{lv}{\sqrt{2\hbar \omega _{0} ^{3}}}\left[ te^{-i\omega _{0} t}-\frac{i}{\omega _{0} }\left( e^{-i\omega _{0} t}-1\right) \right] . \end{aligned}$$

(36)

Now we use the solution obtained in (36) into (33) and we obtain

$$\begin{aligned} \zeta \left( t\right) =-\frac{lv}{\sqrt{2\hbar \omega _{0} ^{3}}}\left[ t+\frac{i }{\omega _{0} }\left( 1-e^{-i\omega _{0} t}\right) \right] \end{aligned}$$

(37)

It is easy to show that U(t) is unitary, we observe that

$$\begin{aligned} U^{\dag }\left( t\right) U\left( t\right)= & {} e^{\xi ^{*}\left( t\right) a^{\dag }}e^{i\Omega \left( t\right) a^{\dag }a}e^{\zeta ^{*}\left( t\right) a}e^{\Gamma ^{*}\left( t\right) }e^{\zeta \left( t\right) a^{\dag }}\nonumber \\&e^{-i\Omega \left( t\right) a^{\dag }a}e^{\xi \left( t\right) a}e^{\Gamma \left( t\right) } \end{aligned}$$

(38)

As $\Omega \left( t\right) =\omega _{0} t$, then

$$\begin{aligned} U^{\dag }\left( t\right) U\left( t\right)= & {} e^{\Gamma ^{*}\left( t\right) +\Gamma \left( t\right) +\left| \zeta \left( t\right) \right| ^{2}}e^{\left[ \left( \xi ^{*}\left( t\right) +\zeta \left( t\right) e^{i\omega _{0} t}\right) a^{\dag }\right] }\nonumber \\&e^{\left[ \left( \zeta ^{*}\left( t\right) e^{-i\omega _{0} t}+\xi \left( t\right) \right) a\right] }. \end{aligned}$$

(39)

On the other hand we have

$$\begin{aligned} \xi ^{*}\left( t\right) +\zeta \left( t\right) e^{i\omega _{0} t}= & {} \xi ^{*}\left( t\right) +\left( -\xi ^{*}\left( t\right) e^{-i\omega _{0} t}\right) e^{i\omega _{0} t}\nonumber \\= & {} 0 \end{aligned}$$

(40)

and

$$\begin{aligned} \zeta ^{*}\left( t\right) e^{-i\omega _{0} t}+\xi \left( t\right)= & {} \left( -\xi ^{*}\left( t\right) e^{-i\omega _{0} t}\right) ^{*}e^{-i\omega _{0} t}+\xi \left( t\right) \nonumber \\= & {} 0 \end{aligned}$$

(41)

then

$$\begin{aligned} \left| \zeta \left( t\right) \right| ^{2}=\left| \xi \left( t\right) \right| ^{2}. \end{aligned}$$

(42)

Also note that $\dot{\Gamma }+\dot{\xi }\xi ^{*}=0$ and $\dot{\Gamma ^{*}}+\dot{\xi ^{*}}\xi =0$ then

$$\begin{aligned} \dot{\Gamma }+{\dot{\Gamma ^{*}}}+\dot{\xi }\xi ^{*}+{\dot{\xi ^{*}}}\xi =0 \end{aligned}$$

(43)

Then we obtain

$$\begin{aligned} \frac{d}{dt}\left( \Gamma +\Gamma ^{*}+\left| \xi \right| ^{2}\right) =0. \end{aligned}$$

(44)

Then we observe that $\Gamma +\Gamma ^{*}+\left| \xi \right| ^{2}$ is constant and $\Gamma \left( 0\right) =\xi \left( 0\right) =0,$ and $\Gamma +\Gamma ^{*}+\left| \xi \right| ^{2}=0$. Finally we conclude that U(t) is unitary.

Appendix II: Work operator

The work operator can be defined [26] as

$$\begin{aligned} \Delta E=U^{\dag }(t)H\left( t\right) U(t)-H_{0}. \end{aligned}$$

(45)

We need to evaluate the quantity $\widetilde{\Delta E}$, where

$$\begin{aligned} \widetilde{\Delta E}\left( \beta ,t\right) =\left\langle \beta \left| U^{\dag }(t)H\left( t\right) U(t)-H_{0}\right| \beta \right\rangle , \end{aligned}$$

(46)

then

$$\begin{aligned} \widetilde{\Delta E}\left( \beta ,t\right)= & {} e^{{\overline{\rho }}}\left\langle \beta \left| e^{i\omega _{0} ta^{\dag }a}e^{\zeta ^{*}a}\left[ \hbar \omega _{0} \left( a^{\dag }a+\frac{1}{2}\right) \right. \right. \right. \nonumber \\&\left. \left. \left. +\sqrt{\hbar }\alpha \left( a^{\dag }+a\right) \right] e^{\zeta a^{\dag }}e^{-i\omega _{0} ta^{\dag }a}\right| \beta \right\rangle +\nonumber \\&-\hbar \omega _{0} \left( \beta ^{*}\beta +\frac{1}{2}\right) , \end{aligned}$$

(47)

where $e^{{\overline{\rho }}}=e^{\Gamma ^{*}}e^{\Gamma }e^{\xi ^{*}\beta ^{*}}e^{\xi \beta }$. The action of the operator $exp({i\theta a^{\dag }a})$ on a coherent state can be easily calculated as

$$\begin{aligned} e^{i\theta a^{\dag }a}|\beta \rangle= & {} e^{-\frac{\left| \beta \right| ^{2}}{2}}\sum \frac{\beta ^{n}}{\sqrt{n!}}e^{i\theta n}|\beta \rangle \\= & {} e^{-\frac{\left| \beta \right| ^{2}}{2}}\sum \frac{\left( \beta e^{i\theta }\right) ^{n}}{ \sqrt{n!}}|\beta \rangle =|\beta e^{i\theta }\rangle . \end{aligned}$$

Then we can write the work operator as

$$\begin{aligned} \widetilde{\Delta E}\left( \beta ,t\right)= & {} e^{\Gamma ^{*}}e^{\Gamma }e^{\xi ^{*}\beta ^{*}}e^{\xi \beta }\left\langle \beta e^{-i\omega _{0} t}\left| e^{\zeta ^{*}a}\left[ \hbar \omega _{0} \left( a^{\dag }a+\frac{1 }{2}\right) \right. \right. \right. \nonumber \\&\left. \left. \left. +\sqrt{\hbar }\alpha \left( a^{\dag }+a\right) \right] e^{\zeta a^{\dag }}\right| \beta e^{-i\omega _{0} t}\right\rangle \nonumber \\&-\hbar \omega _{0} \left( \beta ^{*}\beta +\frac{1}{2}\right) . \end{aligned}$$

(48)

Now we use the BCH formula and we write the products as

$$\begin{aligned} e^{\zeta ^{*}a}a^{\dag }e^{\zeta a^{\dag }}= & {} e^{\left| \zeta \right| ^{2}}e^{\zeta a^{\dag }}\left( a^{\dag }+\zeta ^{*}\right) e^{\zeta ^{*}a}, \end{aligned}$$

(49)

$$\begin{aligned} e^{\zeta ^{*}a}ae^{\zeta a^{\dag }}= & {} e^{\left| \zeta \right| ^{2}}e^{\zeta a^{\dag }}e^{\zeta ^{*}a}\left( a+\zeta \right) \nonumber \\= & {} e^{\left| \zeta \right| ^{2}}e^{\zeta a^{\dag }}\left( a+\zeta \right) e^{\zeta ^{*}a}, \end{aligned}$$

(50)

$$\begin{aligned} e^{\zeta ^{*}a}a^{\dag }ae^{\zeta a^{\dag }}= & {} e^{\zeta a^{\dag }}\left( a^{\dag }+\zeta ^{*}\right) \left( a+\zeta \right) e^{\zeta ^{*}a}e^{\left| \zeta \right| ^{2}}. \end{aligned}$$

(51)

Now inserting (49), (50) and (51) into (48) we have

$$\begin{aligned}&\widetilde{\Delta E}\left( \beta ,t\right) \nonumber \\&\quad =\exp \left( \Gamma ^{*}+\Gamma +\xi ^{*}\beta ^{*}+\xi \beta +\left| \zeta \right| ^{2}\right) \nonumber \\&\qquad \hbar \omega _{0} e^{\zeta \beta ^{*}e^{i\omega _{0} t}}e^{\zeta ^{*}\beta e^{-i\omega _{0} t}}\nonumber \\&\qquad \left[ \left( \beta ^{*}e^{i\omega _{0} t}+\zeta ^{*}\right) \left( \beta e^{-i\omega _{0} t}+\zeta \right) +\frac{1}{2}\right] \nonumber \\&\qquad +\exp \left( \Gamma ^{*}+\Gamma +\xi ^{*}\beta ^{*}+\xi \beta +\left| \zeta \right| ^{2}\right) \nonumber \\&\qquad \sqrt{\hbar }\alpha e^{\zeta \beta ^{*}e^{i\omega _{0} t}}\left[ \left( \beta ^{*}e^{i\omega _{0} t}+\zeta ^{*}\right) \right. \nonumber \\&\qquad \left. +\left( \beta e^{-i\omega _{0} t}+\zeta \right) \right] e^{\zeta ^{*}\beta e^{-i\omega _{0} t}} \nonumber \\&\qquad -\hbar \omega _{0} \left( \beta ^{*}\beta +\frac{1}{2}\right) . \end{aligned}$$

(52)

As we observe before, we have that $\xi ^{*}+\zeta e^{i\omega _{0} t}=0$, then

$$\begin{aligned} \xi ^{*}\beta ^{*}+\zeta \beta ^{*}e^{i\omega _{0} t}=0 \end{aligned}$$

(53)

In a similar way we have $\xi +\zeta ^{*}e^{-i\omega _{0} t}=0$, thus

$$\begin{aligned} \xi \beta +\zeta ^{*}\beta e^{-i\omega _{0} t}=0. \end{aligned}$$

(54)

We have that , $\Gamma ^{*}+\Gamma +\left| \zeta \right| ^{2}=0$, then, after some algebra, we find

$$\begin{aligned} \widetilde{\Delta E}\left( t\right)= & {} \hbar \omega _{0} \left( a^{\dag }\zeta e^{i\omega _{0} t}+a\zeta ^{*}e^{-i\omega _{0} t}+\left| \zeta \right| ^{2}\right) \nonumber \\&+\sqrt{\hbar }\alpha \left( a^{\dag }e^{i\omega _{0} t}+ae^{-i\omega _{0} t}+\zeta ^{*}+\zeta \right) \end{aligned}$$

(55)

Using the result (37) we obtain

$$\begin{aligned} \widetilde{\Delta E}= & {} \frac{lv}{\omega _{0} }x\sin \left( \omega _{0} t\right) -\frac{lv }{\omega _{0} ^{2}}p\cos \left( \omega _{0} t\right) \nonumber \\&-\frac{lv}{\omega _{0} ^{2}}p+\frac{ l^{2}v^{2}}{\omega _{0} ^{4}}\left( 1-\cos \left( \omega _{0} t\right) \right) -\frac{ l^{2}v^{2}t^{2}}{2\omega _{0} ^{2}} \end{aligned}$$

(56)

Now we can verify the quantum analog for Jarzynski’s equality, the mean of the work operator is

$$\begin{aligned}&\left\langle \exp \left( -\beta \widetilde{\Delta E}\right) \right\rangle \nonumber \\&=Tr\left\{ \begin{array}{c} {\hat{\rho }}_{0}\exp \left[ -\frac{\beta lv}{\omega _{0} }{\hat{x}}\sin \left( \omega _{0} t\right) +\frac{\beta lv}{\omega _{0} ^{2}}{\hat{p}}\cos \left( \omega _{0} t\right) +\frac{\beta lv}{\omega _{0} ^{2}}{\hat{p}}\right] \\ \times \exp \left[ -\frac{\beta l^{2}v^{2}}{\omega _{0} ^{4}}\left( 1-\cos \left( \omega _{0} t\right) \right) +\frac{\beta l^{2}v^{2}t^{2}}{2\omega _{0} ^{2}}\right] \end{array} \right\} \nonumber \\ \end{aligned}$$

(57)

and we have

$$\begin{aligned} \rho _{0}(x,x^{\prime })= & {} \sqrt{\frac{\omega _{0} }{\pi }\tanh \left( \frac{\beta \omega _{0} }{2}\right) }\nonumber \\&\exp \left[ -\frac{\omega _{0} }{4}\left( \left( x+x^{\prime }\right) ^{2}\tanh \left( \frac{\beta \omega _{0} }{2}\right) \right. \right. \nonumber \\&\left. \left. +\left( x-x^{\prime }\right) ^{2}\coth \left( \frac{\beta \omega _{0} }{2}\right) \right) \right] . \end{aligned}$$

(58)

Then after some algebraic manipulations we have

$$\begin{aligned} \left\langle \exp \left( -\beta \widetilde{\Delta E}\right) \right\rangle= & {} \exp \left\{ \frac{\beta ^{2}l^{2}v^{2}\left[ 1-\cos \left( \omega _{0} t\right) \right] }{2\omega _{0} ^{3}\tanh \left( \frac{\beta \omega _{0} }{2}\right) }+\frac{ \beta l^{2}v^{2}t^{2}}{2\omega _{0} ^{2}}\right. \nonumber \\&\left. -\frac{\beta l^{2}v^{2}}{\omega _{0} ^{4}} \left[ 1-\cos \left( \omega _{0} t\right) \right] \right\} . \end{aligned}$$

(59)