Thermal Entanglement Phase Transition in Coupled Harmonic Oscillators with Arbitrary Time-Dependent Frequencies

TL;DR

This paper derives the thermal state of two coupled harmonic oscillators with time-dependent parameters, analyzes their entanglement properties, and identifies how frequency changes influence thermal entanglement phase transitions.

Contribution

It provides explicit analytical expressions for the thermal state and entanglement measures in a time-dependent coupled oscillator system, highlighting how frequency differences affect critical temperature.

Findings

Critical temperature increases with frequency difference.

Large frequency differences can protect entanglement against temperature.

Analytical formulas for purity, entropy, and mutual information.

Abstract

We derive explicitly the thermal state of the two coupled harmonic oscillator system when the spring and coupling constants are arbitrarily time-dependent. In particular, we focus on the case of sudden change of frequencies. In this case we compute purity function, R\'{e}nyi and von Neumann entropies, and mutual information analytically and examine their temperature-dependence. We also discuss on the thermal entanglement phase transition by making use of the negativity-like quantity. Our calculation shows that the critical temperature $T_c$ increases with increasing the difference between the initial and final frequencies. In this way we can protect the entanglement against the external temperature by introducing large difference of initial and final frequencies.

Equations

$$H_1=\frac{1}{2}{p}^2+\frac{1}{2}{\omega }^2(t)x^2.$$

$$S[x]={\int }_0^t\left[\frac{1}{2}{\dot{x}}^2-\frac{1}{2}{\omega }^2(t)x^2\right].$$

$$K[x^{\prime },x:t]={\int }_{(0,x)}^{(t,x^{\prime })}\mathcal{D}x{e}^{iS[x]}.$$

$$K[{\mathbf{x}}^{\prime },t_2:\mathbf{x},t_1]=\sum _n{\psi }_n\left({\mathbf{x}}^{\prime },t_2\right){\psi }_n^{\ast }\left(\mathbf{x},t_1\right)$$

$${\psi }_n(x,t)=e^{-i{E}_n\tau (t)}{e}^{\frac{i}{2}\left(\frac{\dot{b}}{b}\right)x^2}{\varphi }_n\left(\frac{x}{b}\right)$$

$$E_n=\left(n+\frac{1}{2}\right)\omega (0),\tau (t)={\int }_0^t\frac{ds}{b^2(s)}$$

$${\varphi }_n(x)=\frac{1}{\sqrt{2^{n}n!}}{\left(\frac{\omega (0)}{\pi b^2}\right)}^{1/4}{H}_n\left(\sqrt{\omega (0)}x\right)e^{-\frac{\omega (0)}{2}{x}^2}.$$

$$\ddot{b}+{\omega }^2(t)b=\frac{{\omega }^2(0)}{b^3}$$

$$\displaystyle\omega(t)=\left\{\begin{array}[]{cc}\omega_{0}&\hskip 28.45274ptt=0\\ \omega&\hskip 28.45274ptt>0,\end{array}\right.$$

$$b(t)=\sqrt{\frac{{\omega }^2-{\omega }_0^2}{2{\omega }^2}\cos (2\omega t)+\frac{{\omega }^2+{\omega }_0^2}{2{\omega }^2}}.$$

$$\sum _{n=0}^{\infty }\frac{t^n}{n!}{H}_n(x)H_n(y)=(1-4t^2{)}^{-1/2}\exp \left[\frac{4txy-4t^2(x^2+y^2)}{1-4t^2}\right],$$

$$K[x^{\prime },x:t]=e^{\frac{i}{2}\left(\frac{\dot{b}}{b}\right)x^{\prime 2}}\frac{({\omega }_0{\omega }^{\prime }{)}^{1/4}}{\sqrt{2\pi i\sin \Gamma (t)}}\exp \left[\frac{i}{2\sin \Gamma (t)}\left\{({\omega }_0{x}^2+{\omega }^{\prime }{x}^{\prime 2})\cos \Gamma (t)-2\sqrt{{\omega }_0{\omega }^{\prime }}x{x}^{\prime }\right\}\right]$$

$${\omega }_0=\omega (t=0),{\omega }^{\prime }(t)=\frac{{\omega }_0}{b^2(t)},\Gamma (t)={\int }_0^t{\omega }^{\prime }(s)ds.$$

$$K[x^{\prime },x:t]=\sqrt{\frac{\omega }{2\pi i\sin \omega t}}\exp \left[\frac{i\omega }{2\sin \omega t}\left\{(x^2+x^{\prime 2})\cos \omega t-2x{x}^{\prime }\right\}\right].$$

$$K[x^{\prime },x:t]=e^{-\left(\frac{i({\omega }^2-{\omega }_0^2)}{4\omega b^2}\sin 2\omega t\right)x^{\prime 2}}\sqrt{\frac{{\omega }_0}{2\pi ib\sin \Gamma (t)}}\exp \left[\frac{i{\omega }_0}{2\sin \Gamma (t)}\left\{\left(x^2+\frac{x^{\prime 2}}{b^2}\right)\cos \Gamma (t)-\frac{2x{x}^{\prime }}{b}\right\}\right]$$

$$\Gamma (t)={\tan }^{-1}\left(\frac{{\omega }_0}{\omega }\tan \omega t\right)=\frac{1}{2i}\ln \frac{\omega +i{\omega }_0\tan \omega t}{\omega -i{\omega }_0\tan \omega t}.$$

$${\rho }_T[x^{\prime },x:\beta ]=\frac{1}{\mathcal{Z}(\beta )}G[x^{\prime },x:\beta ]$$

$$b(\beta )=\sqrt{\frac{{\omega }^2-{\omega }_0^2}{2{\omega }^2}\cosh (2\omega \beta )+\frac{{\omega }^2+{\omega }_0^2}{2{\omega }^2}},\Gamma (\beta )=-i{\Gamma }_0(\beta )$$

$${\Gamma }_0(\beta )=\frac{1}{2}\ln \left(\frac{\omega +{\omega }_0\tanh \omega \beta }{\omega -{\omega }_0\tanh \omega \beta }\right).$$

$$\mathcal{Z}(\beta )=\sqrt{\frac{{\omega }_0}{2\pi b\sinh {\Gamma }_0}}\sqrt{\frac{\pi }{a_{0,-}}}$$

$$a_{0,\pm }(\beta )=A_0(\beta )+\frac{{\omega }_0}{2\sinh {\Gamma }_0}\left[\left(1+\frac{1}{b^2}\right)\cosh {\Gamma }_0\pm \frac{2}{b}\right]$$

$$A_0(\beta )=\frac{{\omega }^2-{\omega }_0^2}{4\omega b^2}\sinh (2\omega \beta ).$$

$${\rho }_0[x^{\prime },x:\beta ]=\sqrt{\frac{a_{0,-}}{\pi }}{e}^{-A_0(\beta )x^{\prime 2}}\exp \left[-\frac{{\omega }_0}{2\sinh {\Gamma }_0}\left\{\left(x^2+\frac{x^{\prime 2}}{b^2}\right)\cosh {\Gamma }_0-\frac{2x{x}^{\prime }}{b}\right\}\right].$$

$$P_0(\beta )\equiv \int dxd{x}^{\prime }{\rho }_0[x^{\prime },x:\beta ]{\rho }_0[x,x^{\prime }:\beta ]=\sqrt{\frac{a_{0,-}}{a_{0,+}}}.$$

$$\int dx{\rho }_0[x^{\prime },x:\beta ]f_n(x)={\lambda }_n(\beta )f_n(x^{\prime }).$$

$$\int dx\left(A{e}^{-a_1{x}^2-a_2{x}^{\prime 2}+2bx{x}^{\prime }}\right)f_n(x)={\lambda }_n{f}_n(x^{\prime })$$

$$f_n(x)={\mathcal{C}}_n^{-1}{H}_n(\sqrt{{ϵ}_0}x)e^{-\frac{{\alpha }_0}{2}{x}^2}$$

$${\lambda }_n=A\sqrt{\frac{2\pi }{(a_1+a_2)+{ϵ}_0}}{\left[\frac{(a_1+a_2)-{ϵ}_0}{(a_1+a_2)+{ϵ}_0}\right]}^{n/2}$$

$${\int }_{-\infty }^{\infty }{e}^{-(x-y{)}^2}{H}_m(cx)H_n(cx)dx$$

$$\displaystyle=\sqrt{\pi}\sum_{k=0}^{\min(m,n)}2^{k}k!\left(\begin{array}[]{c}m\\ k\end{array}\right)\left(\begin{array}[]{c}n\\ k\end{array}\right)(1-c^{2})^{\frac{m+n}{2}-k}H_{m+n-2k}\left(\frac{cy}{\sqrt{1-c^{2}}}\right)$$

$${\mathcal{C}}_n^2=\frac{1}{\sqrt{{\alpha }_0}}\sum _{k=0}^n{2}^{2n-k}{\left(\frac{{ϵ}_0}{{\alpha }_0}-1\right)}^{n-k}\frac{{\Gamma }^2(n+1)\Gamma \left(n-k+\frac{1}{2}\right)}{\Gamma (k+1){\Gamma }^2(n-k+1)}.$$

$${\mathcal{C}}_n^{-1}=\frac{1}{\sqrt{2^{n}n!}}{\left(\frac{{ϵ}_0}{\pi }\right)}^{1/4}.$$

Full text

I Introduction

Entanglementschrodinger-35 ; text ; horodecki09 is a key physical resource in quantum information processing. For example, it plays crucial role in quantum teleportationteleportation , superdense codingsuperdense , quantum cloningclon , quantum cryptographycryptography ; cryptography2 , quantum metrologymetro17 , and quantum computerqcomputer ; qcreview . In particular, physical realization of quantum cryptography and quantum computer seems to be accomplished in the near future111see Ref. white and web page https://www.computing.co.uk/ctg/news/3065541/european-union-reveals-test-projects-for-first-tranche-of-eur1bn-quantum-computing-fund..

Although entanglement is highly useful property of quantum state, it is normally fragile when quantum system interacts with its surroundings. Interaction with the environments makes the given quantum system undergo decoherencezurek03 and as a result, it loses its quantum properties. Thus, decoherence significantly changes the quantum entanglement. Sometimes entanglement exhibits an exponential decay in time by successive halves. Sometimes, however, entanglement sudden death (ESD) occurs when the entangled multipartite quantum system is embedded in Markovian environmentsmarkovian ; yu05-1 ; yu06-1 ; yu09-1 ; almeida07 ; park-16 . This means that the entanglement is completely disentangled at finite times.

Most typical surrounding is external temperature. At finite temperature quantum mechanics the external temperature is introduced via imaginary time at zero temperature quantum mechanics. Thus, the exponential decay or ESD-like phenomenon can occur in external temperature. If external temperature induces the ESD-like phenomenon in temperature, this means there exists a critical temperature $T_{c}$, below or above which the entanglement of a system is nonzero or completely zero. We will call this phenomenon thermal entanglement phase transition (TEPT) between nonzero entanglement phase and zero entanglement phase. The TEPT and the critical temperature $T_{c}$ were exploredpark-19 recently by making use of concurrenceform2 ; form3 in anisotropic Heisenberg $XYZ$ spin model with Dzyaloshinskii-Moriya interactiondzya58 ; mori60 .

The purpose of this paper is to study on the TEPT phenomenon in continuous variable system. Most simple continuous variable system seems to be two-coupled harmonic oscillator system. In this reason we will choose this system to explore the TEPT when the spring constant $k_{0}$ and coupling constant $J$ are arbitrarily time-dependent. Another reason we choose this system is because of the fact that the thermal state of this system is Gaussian. It is known that the Peres-Horodecki positive partial transposition (PPT) criterionperes96 ; horodecki96 ; horodecki97 provides a necessary and sufficient condition for separability of Gaussian continuous variable statesduan-2000 ; simon-2000 . Thus, the temperature-dependence of entanglement can be roughly deduced by considering the negativity-like quantityvidal01 . What we are interested in is to examine how the arbitrarily time-dependent parameters affect the critical temperature. In particular, we focus in this paper on the sudden quenched model, where the system parameters abruptly change at $t=0$.

The paper is organized as follows. In section II we derive the thermal state of single harmonic oscillator system when the frequency is arbitrarily time-dependent. We focus on the case of sudden quenched model (SQM). For SQM we derive the purity function and von Neumann entropy of the thermal state analytically. In section III we derive explicitly the thermal state of two coupled harmonic oscillator system when the spring constant $k_{0}$ and coupling constant $J$ are arbitrarily time-dependent. In section IV we compute the purity function, Rényi and von Neumann entropies, and mutual information analytically for the thermal state of two coupled harmonic oscillator system in the case of SQM. It is shown that the thermal state is less mixed with increasing the difference between initial and final frequencies at the given external temperature. The mutual information shows that the common information parties $A$ and $B$ share does not completely vanish even in the infinity temperature limit. In section V the TEPT is discussed for the case of SQM by making use of the negativity-like quantity. It is shown that the critical temperature $T_{c}$ increases with increasing the frequency difference. Thus, using SQM with large difference of initial and final frequencies it seems to be possible to protect entanglement against external temperature. In section VI a brief conclusion is given. In appendix A the eigenvalue equation for the thermal state of the coupled harmonic oscillator system is explicitly solved.

II Thermal State for single harmonic oscillator with arbitrary time-dependent frequency

Let us consider a single harmonic oscillator with time-dependent frequency, whose Hamiltonian is

[TABLE]

Then, the action functional of this system is given by

[TABLE]

Usually Kernel for any quantum system can be derived by computing the path-integralfeynman

[TABLE]

Although the path-integral with constant frequency can be computedfeynman ; kleinert , it does not seem to be simple matter to compute the path-integral explicitly when $\omega$ is arbitrary time-dependent. However, it is possible to derive the Kernel without computing the path-integral if one uses the Schrödinger description of Kernel

[TABLE]

where $n$ represents all possible quantum numbers and $\psi_{n}\left({\bm{x}},t\right)$ is linearly-independent solution of time-dependent Schrödinger equation (TDSE).

The TDSE of our system was exactly solved in Ref. lewis68 ; lohe09 . The linearly independent solutions $\psi_{n}(x,t)\hskip 2.84544pt(n=0,1,\cdots)$ are expressed in a form

[TABLE]

where

[TABLE]

In Eq. (6) $H_{n}(z)$ is $n^{th}$-order Hermite polynomial and $b(t)$ satisfies the Ermakov equation

[TABLE]

with $b(0)=1$ and $\dot{b}(0)=0$.

Solutions of the Ermakov equation were discussed in Ref. pinney50 . If $\omega(t)$ is time-independent, $b(t)$ is simply one. If $\omega(t)$ is instantly changed as

[TABLE]

then $b(t)$ becomes

[TABLE]

For more general time-dependent case the Ermakov equation should be solved numerically.

Inserting Eq. (5) into Eq. (4) and using

[TABLE]

Kernel for this system becomes

[TABLE]

where

[TABLE]

For time-independent case $b(t)=1$, $\omega_{0}=\omega^{\prime}=\omega$, and $\Gamma(t)=\omega t$. Then, the Kernel in Eq. (13) reduces to usual well-known harmonic oscillator Kernel

[TABLE]

It is remarkable to note that the $x\leftrightarrow x^{\prime}$ symmetry in Eq. (15) is broken in Eq. (13) due to the time-dependence of frequency. In fact, it is manifest due to the fact that the system parameters at $t=0$ are different from those at $t>0$.

From now on in this section we consider only the case of SQM given in Eq. (10). In this case the Kernel becomes

[TABLE]

where $b(t)$ is given in Eq. (11) and $\Gamma(t)$ becomes

[TABLE]

In quantum mechanics the inverse temperature $\beta=1/k_{B}T$ is introduced as a Euclidean time $\beta=it$ (see Ch. $10$ of Ref.feynman ), where $k_{B}$ is a Boltzmann constant. Then, the thermal density matrix is defined as

[TABLE]

where $\beta=1/k_{B}T$, $G[x^{\prime},x:\beta]=K[x^{\prime},x:-i\beta]$, and ${\cal Z}(\beta)=\mbox{tr}G[x^{\prime},x:\beta]$ is a partition function. Throughout this paper we use $k_{B}=1$ for convenience. For SQM case $b(t)$ and $\Gamma(t)$ are changed into $b(\beta)$ and $\Gamma(\beta)$, whose explicit expressions are

[TABLE]

where222In fact, one can show that $b(\beta)$ in Eq. (19) is a solution of $\frac{d^{2}b}{d\beta^{2}}-\omega^{2}b=-\frac{\omega_{0}^{2}}{b^{3}}$.

[TABLE]

Then, the partition function of this system becomes

[TABLE]

where

[TABLE]

with

[TABLE]

Using the partition function one can derive the thermal density matrix in a form

[TABLE]

The thermal density matrix is in general mixed state. In order to explore how much it is mixed we first compute the purity function $P_{0}(\beta)=\mbox{tr}\left(\rho_{0}\right)^{2}$. If it is one, this means that $\rho_{0}$ is pure state. If it is zero, this means $\rho_{0}$ is completely mixed state. If $0<P_{0}(\beta)<1$, this means that $\rho_{0}$ is partially mixed state. It is not difficult to show that the purity function of this system is

[TABLE]

Another quantity we want to compute is a von Neumann entropy $S[\rho_{0}]$ of $\rho_{0}$. If $\rho_{0}$ is pure state, $S[\rho_{0}]$ is zero. If its mixedness increases, $S[\rho_{0}]$ also increases from zero and eventually goes to infinity for completely mixed state in this continuum case. In order to compute the von Neumann entropy we should solve the eigenvalue equation

[TABLE]

One can show that the eigenvalue equation

[TABLE]

can be solved, and the eigenfunction and corresponding eigenvalue are

[TABLE]

where $\epsilon_{0}=\sqrt{(a_{1}+a_{2})^{2}-4b^{2}}$ and $\alpha_{0}=\epsilon_{0}-(a_{1}-a_{2})$. By making use of integral formulaintegral

[TABLE]

and various properties of Gamma functionhandbook , the normalization constant ${\cal C}_{n}$ can be written in a form

[TABLE]

If $a_{1}=a_{2}$, $\alpha_{0}=\epsilon_{0}$, which makes nonzero in $k$-summation of Eq. (35) only when $k=n$. Then, ${\cal C}_{n}$ becomes usual harmonic oscillator normalization constant

[TABLE]

Using Eqs. (27) and (28) the eigenvalue of Eq. (26) becomes

[TABLE]

where

[TABLE]

Thus, the spectral decomposition of $\rho_{0}$ can be written as

[TABLE]

where $f_{n}(x:\beta)$ is given by Eq. (28) with $\epsilon_{0}=\sqrt{a_{0,+}a_{0,-}}$ and $\alpha_{0}=\epsilon_{0}+A_{0}(\beta)-\frac{\omega_{0}\cosh\Gamma_{0}}{2\sinh\Gamma_{0}}\left(1-\frac{1}{b^{2}}\right).$ Eq. (37) implies $\sum_{n}\lambda_{n}(\beta)=1$, which is consistent with $\mathrm{tr}\rho_{0}=1$. Then the von Neumann entropy of $\rho_{0}$ becomes

[TABLE]

For constant frequency, i.e. $\omega=\omega_{0}$, $A_{0}=0$, $a_{0,+}=\omega\coth\frac{\omega\beta}{2}$, $a_{0,-}=\omega\tanh\frac{\omega\beta}{2}$, and $\xi_{0}=e^{-\omega\beta}$. For the case of SQM $A_{0}$ and $a_{0,\pm}$ become larger than those in constant frequency case in the entire range of temperature. As a result, $P_{0}(\beta)$ and $\xi_{0}$ become larger and smaller compared to the constant frequency case. Since $-\ln(1-x)-\frac{x}{1-x}\ln x$ is monotonically increasing function in the range $0\leq x\leq 1$, this fact decreases the von Neumann entropy.

The temperature dependence of the purity function and von Neumann entropy is plotted in Fig. 1(a) and Fig. 1(b) when $\omega=3$ (black line), $5$ (red line), $7$ (blue line) with $\omega_{0}=3$. All figures show that $\rho_{0}$ becomes more and more mixed with increasing the external temperature. Both figures also show that $\rho_{0}$ becomes less mixed with increasing $|\omega-\omega_{0}|$ at the given temperature. Thus, we can use SQM model to protect the purity against the external temperature.

III Thermal State for two coupled harmonic oscillators with arbitrary time-dependent frequencies

In this section we will derive the thermal state for two coupled harmonic oscillator system, whose Hamiltonian is

[TABLE]

We choose the potential $V(x_{1},x_{2})$ as a quadratic function with arbitrary time-dependent spring and coupling parameters. The explicit expression of the potential is chosen in a form

[TABLE]

Then, the action functional of this system is

[TABLE]

It is easy to show that the potential is diagonalized by introducing $y_{1}=\frac{1}{\sqrt{2}}(x_{1}+x_{2})$ and $y_{2}=\frac{1}{\sqrt{2}}(x_{1}-x_{2})$ . In terms of new canonical variables the action becomes that of two non-interacting harmonic oscillators in a form

[TABLE]

where $\omega_{1}(t)=\sqrt{k_{0}(t)}$ and $\omega_{2}(t)=\sqrt{k_{0}(t)+2J(t)}$. Thus, the Kernel for this system is

[TABLE]

where $\omega_{j,0}=\omega_{j}(t=0)$, $\omega_{j}^{\prime}=\frac{\omega_{j,0}}{b_{j}^{2}(t)}$, and $\Gamma_{j}(t)=\int_{0}^{t}\omega_{j}^{\prime}(s)ds$. Of course, $b_{1}(t)$ and $b_{2}(t)$ satisfy the Ermakov equation

[TABLE]

with $\dot{b}_{j}(0)=0$ and $b_{j}(0)=1$. Then the thermal density matrix of this system is given by

[TABLE]

where $\beta=1/k_{B}T$ and ${\cal Z}(\beta)=\mbox{tr}K[x_{1}^{\prime},x_{2}^{\prime}:x_{1},x_{2}:-i\beta]$.

In this paper we will examine only the case of SQM. More general time-dependent cases will be explored elsewhere. If spring and coupling constants are abruptly changed as

[TABLE]

$\omega_{1}$ and $\omega_{2}$ become

[TABLE]

Then, the thermal density matrix of this system is given by

[TABLE]

where333The subscript $E$ in $\Gamma_{E,j}$ stands for “Euclidean”. This subscript is attached to stress the point that the inverse temperature $\beta$ is introduced as a Euclidean time.

[TABLE]

with $j=1,2$. For the limit of $\omega_{j,i}=\omega_{j,f}\equiv\omega_{j}$, we have $A_{j}=0$, $b_{j}=1$, $\Gamma_{E,j}=\omega_{j}\beta$, $a_{j,+}=\omega_{j}\coth(\omega_{j}\beta/2)$, and $a_{j,-}=\omega_{j}\tanh(\omega_{j}\beta/2)$. In terms of $x_{j}$-coordinates the thermal state reduces to

[TABLE]

where

[TABLE]

It is worthwhile noting that $\alpha_{j}$ satisfy

[TABLE]

Using Eq. (62) it is straightforward to show $\mbox{tr}\rho_{T}=1$. In next section we compute several quantum information quantities analytically, which measure how much $\rho_{T}$ is mixed.

IV Various Quantities of Thermal State: Case of SQM

In this section we will compute purity function, Rényi and von Neumann entropies, and mutual information of $\rho_{T}[x_{1}^{\prime},x_{2}^{\prime}:x_{1},x_{2}:\beta]$ given in Eq. (58) or equivalently Eq. (60). As a by-product we derive the spectral decomposition of $\rho_{T}[x_{1}^{\prime},x_{2}^{\prime}:x_{1},x_{2}:\beta]$.

IV.1 Purity function

The purity function is defined as

[TABLE]

If $P(\beta)=1$ or [math], this means that $\rho_{T}$ is pure or completely mixed state. Direct calculation shows

[TABLE]

For the case of constant frequencies $\omega_{j,i}=\omega_{j,f}\equiv\omega_{j}$ it reduces to

[TABLE]

The temperature-dependence of the purity function is plotted in Fig. 2 (a) when $k_{0,f}=J_{f}=6$ (red line) and $k_{0,f}=J_{f}=9$ (blue line). The $k_{0,i}$ and $J_{i}$ are fixed as $k_{0,i}=J_{i}=3$. The black dashed line corresponds to constant frequencies $k_{0}=J=3$. As expected $\rho_{T}$ becomes more and more mixed with increasing temperature. Fig. 2(a) also show that $\rho_{T}$ is less mixed when $|k_{0,f}-k_{0,i}|$ and $|J_{f}-J_{i}|$ increase.

IV.2 Rényi and von Neumann entropies

In order to solve the Rényi and von Neumann entropies of $\rho_{T}$ we should solve the eigenvalue equation

[TABLE]

Eq. (66) is solved in appendix A and the eigenvalue $p_{mn}(\beta)$ is

[TABLE]

where

[TABLE]

In terms of $\xi_{1}$ and $\xi_{2}$ the purity function $P(\beta)$ in Eq. (64) can be written as

[TABLE]

Then, the Rényi and von Neumann entropies of $\rho_{T}$ reduce to

[TABLE]

where

[TABLE]

with $j=1,2$.

One can show also that the normalized eigenfunction $u_{mn}(x_{1},x_{2}:\beta)$ is

[TABLE]

where

[TABLE]

For the case of constant frequencies $\alpha_{1}-\alpha_{2}=\alpha_{3}-\alpha_{4}=0$, which results in $\mu_{j}=\frac{\epsilon_{j}}{2}$. In this case the sum in ${\cal C}_{1.m}^{2}$ or ${\cal C}_{2,n}^{2}$ is nonzero only when $k=m$ or $k=n$, and this fact yields well-known quantities ${\cal C}_{1,m}^{-1}=\frac{1}{\sqrt{2^{m}m!}}\left(\frac{\epsilon_{1}}{\pi}\right)^{1/4}$ and ${\cal C}_{2,n}^{-1}=\frac{1}{\sqrt{2^{n}n!}}\left(\frac{\epsilon_{2}}{\pi}\right)^{1/4}$. Thus, the spectral decomposition of $\rho_{T}$ can be written as

[TABLE]

The temperature-dependence of the von Neumann entropy is plotted in Fig. 2 (b) when $k_{0,f}=J_{f}=6$ (red line) and $k_{0,f}=J_{f}=9$ (blue line). The $k_{0,i}$ and $J_{i}$ are fixed as $k_{0,i}=J_{i}=3$. The black dashed line corresponds to constant frequencies $k_{0}=J=3$. As expected $\rho_{T}$ becomes more and more mixed with increasing temperature. Fig. 2(b) also show that $\rho_{T}$ is less entangled when $|k_{0,f}-k_{0,i}|$ and $|J_{f}-J_{i}|$ increase as purity function exhibits.

IV.3 mutual information

From $\rho_{T}$ in Eq. (60) one can derive the substates $\rho_{T,A}=\mbox{tr}_{B}\rho_{T}$ and $\rho_{T,B}=\mbox{tr}_{A}\rho_{T}$ by performing partial trace appropriately. Then, the substates become

[TABLE]

where

[TABLE]

It is not difficult to show that the eigenvalues of $\rho_{T,A}$ or $\rho_{T,B}$ are $(1-\zeta)\zeta^{n}$, where

[TABLE]

with $\nu=\sqrt{(B_{1}+B_{2})^{2}-4B_{3}^{2}}$. Using the eigenvalues the Rényi and von Neumann entropies of $\rho_{T,A}$ and $\rho_{T,B}$ can be obtained as

[TABLE]

Therefore, the mutual information of $\rho_{T}$ is given by

[TABLE]

The temperature-dependence of the mutual information is plotted in Fig. 2 (c) when $k_{0,f}=J_{f}=6$ (red line) and $k_{0,f}=J_{f}=9$ (blue line). The $k_{0,i}$ and $J_{i}$ are fixed as $k_{0,i}=J_{i}=3$. The black dashed line corresponds to constant frequencies $k_{0}=J=3$. Like other quantities mutual information also decreases with increasing temperature. However, it does not completely vanish at $T=\infty$. Fig. 2 (c) shows that the mutual information seems to approach $0.144$ at the large temperature limit. This implies that the common information parties $A$ and $B$ share does not completely vanish even in the infinity temperature limit.

V Thermal Entanglement Phase Transition: Case of SQM

Since the thermal state $\rho_{T}$ given in Eq. (60) is mixed state, its entanglement is in general defined via the convex-roof methodbenn96 ; uhlmann99-1 ;

[TABLE]

where minimum is taken over all possible pure state decompositions, i.e. $\rho_{T}=\sum_{j}p_{j}\lvert\psi_{j}\rangle\langle\psi_{j}\lvert$, with $0\leq p_{j}\leq 1$ and $\sum_{j}p_{j}=1$. The decomposition which yields minimum value is called the optimal decomposition. However, it seems to be highly difficult problem to derive the optimal decomposition in the continuous variable system.

Because of this difficulty, we will consider the negativity-like quantityvidal01 of $\rho_{T}$. Let $\sigma_{T}$ be a partial transpose of $\rho_{T}$, i.e.,

[TABLE]

Then, the negativity-like quantity ${\cal N}(\rho_{T})$ is defined as

[TABLE]

where $\Lambda_{mn}$ is eigenvalue of $\sigma_{T}$, i.e.,

[TABLE]

One may wonder why the negativity-like quantity is introduced, because the PPT is known as necessary and sufficient criterion of separability for only $2\times 2$ qubit-qubit and $2\times 3$ qubit-qudit statesperes96 ; horodecki96 ; horodecki97 . However, as Ref. duan-2000 ; simon-2000 have shown, PPT also provides a necessary and sufficient criterion of the separability for Gaussian continuous variable quantum states. Furthermore, the distillation protocols to maximally entangled state have been already suggested in Ref. duan-00-1 ; giedke-2000 in the Gaussian states. Thus, our negativity-like quantity is valid at least to determine whether the given Gaussian state is entangled or not. Since ${\cal N}(\rho_{T})$ is proportional to ${\cal E}(\rho_{T})$, ${\cal N}(\rho_{T})=0$ at the critical temperature $T=T_{c}$ of the TEPT if the external temperature induces the ESD phenomenon. Thus, if the eigenvalue equation (83) is solved, it is possible to compute $T_{c}$.

As we will show in the following, however, it seems to be very difficult to solve Eq. (83) directly. In order to solve Eq. (83) we define

[TABLE]

Then, Eq. (83) can be written as

[TABLE]

If one changes the variables as $y_{1}=\frac{1}{\sqrt{2}}(x_{1}+x_{2})$ and $y_{2}=\frac{1}{\sqrt{2}}(-x_{1}+x_{2})$, Eq. (85) reduces to

[TABLE]

where

[TABLE]

The difficulty arises because of the fact that $A$ is not symmetric matrix if $\alpha_{3}\neq\alpha_{4}$. Due to this fact it seems to be impossible to factorize Eq. (95) into two single-party eigenvalue equations as we did in appendix A.

However, Eq. (95) can be solved for the case of constant frequencies, i.e., $\omega_{1,i}=\omega_{1,f}\equiv\omega_{1}$ and $\omega_{2,i}=\omega_{2,f}\equiv\omega_{2}$, because in this case $\alpha_{3}$ is exactly equals to $\alpha_{4}$. Furthermore, in this case we get

[TABLE]

Since $\alpha_{3}=\alpha_{4}$ in this case, Eq. (95) is factorized into the following two single-party eigenvalue equations:

[TABLE]

Then, the total eigenvalue $\Lambda_{mn}$ and the normalized eigenfunction $f_{mn}(x_{1},x_{2}:\beta)$ are expressed as

[TABLE]

where $g_{1,m}(y_{1}:\beta)$ and $g_{2,n}(y_{2}:\beta)$ are normalized eigenfunctions of Eq. (100). Solving Eq. (100) it is straightforward to show that the normalized eigenfunctions are

[TABLE]

where

[TABLE]

One can also show that the eigenvalue $\Lambda_{mn}$ is

[TABLE]

where

[TABLE]

One can compute $\pm 1-\zeta_{1}$ and $\pm 1-\zeta_{2}$ explicitly, which result in $-1<\zeta_{1},\zeta_{2}\leq 1$ for arbitrary temperature. Thus, it is easy to show $\sum_{m,n}\Lambda_{mn}(\beta)=1$ as expected. Eq. (82) and Eq. (104) make ${\cal N}(\beta)$ to be

[TABLE]

The $T$-dependence of ${\cal N}(\beta)$ is plotted in Fig. 3 for (a) positive and (b) negative $J$ with fixing $k_{0}=1$. Both figures show ${\cal N}(\beta)$ is zero at $T\geq T_{c}$. Similar results were obtained for general bosonic harmonic lattice systemsperes96 ; plenio . Since ${\cal N}(\beta)$ is proportional to entanglement of $\rho_{T}$, this fact implies that $\rho_{T}$ is entangled (or separable) state at $T<T_{c}$ (or $T\geq T_{c}$). The critical temperature temperature $T_{c}$ increases with increasing $|J|$.

From Eq. (106) it is evident that $\rho_{T}$ is separable when $\zeta_{1}\geq 0$ and $\zeta_{2}\geq 0$. Eq. (105) implies that this separability criteria can be rewritten in a form

[TABLE]

where $x=\omega_{1}\beta/2$ and $y=\omega_{2}\beta/2$. If $J\geq 0$, first equation of Eq. (107) is automatically satisfied. Hence, the second equation plays a role as a genuine separability criterion. If $J<0$, first equation is true criterion. It is worthwhile noting that two equations in Eq. (107) can be transformed into each other by interchanging $x$ and $y$. This fact implies that the region in $x$-$y$ plane, where the separable states reside, is symmetric with respect to $y=x$.

The shaded region in Fig. 4(a) is a region where the separable states of $\rho_{T}$ reside in $x$-$y$ plane. As expected, the region is symmetric with respect to $y=x$. It is shown that most separable states are accumulated in $0\leq x,y\leq 1$. The boundary of the region contains an information about the critical temperature $T_{c}$. The black dashed line in the region is $y=x\coth x$. Since this is very close to upper boundary, this can be used to compute $T_{c}$ approximately.

Let the upper boundary of Fig. 4(a) be expressed by $y_{c}=x_{c}g(x_{c})$, where $x_{c}$ and $y_{c}$ are $x$ and $y$ at $T=T_{c}$. Then the low boundary should be $x_{c}=y_{c}g(y_{c})$. The function $g(z)$ can be derived numerically by using Eq. (107) after changing the inequality into equality. Then, $T_{c}$ can be computed by

[TABLE]

where $\omega_{min}=\min(\omega_{1},\omega_{2})$ and $\omega_{max}=\max(\omega_{1},\omega_{2})$. If one uses $g(z)\approx\coth z$, the critical temperature is approximately

[TABLE]

In Fig. 4(b) the $J$-dependence of $T_{c}$ is plotted when $k_{0}=1$. The black solid line and red dashed line correspond to Eq. (108) and Eq. (109) respectively. It is shown that $T_{c}$ increases with increasing $|J|$ as expected from Fig. 3.

As we commented earlier, for the case of SQM it seems to be highly difficult problem to solve the eigenvalue equation (83) directly. However, we can conjecture the eigenvalue $\Lambda_{mn}(\beta)$ without deriving the eigenfunction $f_{mn}[x_{1},x_{2}:\beta]$ as follows. Since $\sum_{m,n}\Lambda_{mn}(\beta)=1$, $\Lambda_{mn}(\beta)$ might be represented as Eq. (104). If this is right, we can compute $\zeta_{1}$ and $\zeta_{2}$ by making use of the Rényi entropy. If the eigenvalue is represented as Eq. (104), the Rényi entropy of $\sigma_{T}$ can be written as

[TABLE]

Putting $\alpha=2$ or $3$ in Eq. (110), it is possible to derive

[TABLE]

where

[TABLE]

with

[TABLE]

Solving Eq. (111), we get

[TABLE]

Thus, $\zeta_{1}$ and $\zeta_{2}$ for the case of SQM become

[TABLE]

At the case of constant frequency $\beta_{1}$ and $\beta_{2}$ reduce to

[TABLE]

Using Eq. (116) and after tedious calculation, one can show that $\zeta_{1}$ and $\zeta_{2}$ in Eq. (115) exactly coincide with those in Eq. (105) when $\omega_{1,i}=\omega_{1,f}=\omega_{1}$ and $\omega_{2,i}=\omega_{2,f}=\omega_{2}$. Then, the negativity-like quantity can be written in a form

[TABLE]

The temperature dependence of ${\cal N}(\beta)/{\cal N}(\infty)$ is plotted in Fig. 5. In Fig. 5(a) we choose $k_{0,f}=1$ (black dashed line), $k_{0,f}=20$ (red line), and $k_{0,f}=40$ (blue line) when $k_{0,i}=1$ and $J_{i}=J_{f}=5$. As this figure exhibits, the critical temperature $T_{c}$ increases with increasing $|k_{0,f}-k_{0,i}|$. In Fig. 5(b) we choose $J_{f}=5$ (black dashed line), $J_{f}=25$ (red line), and $J_{f}=45$ (blue line) when $k_{0,i}=k_{0,f}=1$ and $J_{i}=5$. This figure also shows that $T_{c}$ increases with increasing $|J_{f}-J_{i}|$.

VI Conclusions

In this paper we derive explicitly the thermal state of the two coupled harmonic oscillator system when the spring and coupling constants are arbitrarily time-dependent. In particular, we focus on the SQM model (see Eq. (52) and Eq. (57)). In this model we compute purity function, Rényi and von Neumann entropies, and mutual information analytically and examine their temperature-dependence. We also discuss on the TEPT by making use of the negativity-like quantity. Our calculation shows that the critical temperature $T_{c}$ increases with increasing the difference between the initial and final frequencies. In this way we can use the SQM model to protect the entanglement against the external temperature by introducing a large difference of frequencies, i.e. $|\omega_{f}-\omega_{i}|\gg 1$.

There are several issues related to our paper. Since the SQM model we consider involves a discontinuity at $t=0$, it is unrealistic in some sense. In order to escape this fact we can introduce the time-dependence of frequencies as a form $\omega=\omega_{i}+(\omega_{f}-\omega_{i})\sin\Omega t$. Then, we have to solve the Ermakov equation numerically. In this case the critical temperature $T_{c}$ might be dependent on $\Omega$ and $|\omega_{f}-\omega_{i}|$. Then, it may be possible to protect the entanglement in the thermal bath by adjusting $\Omega$ and $|\omega_{f}-\omega_{i}|$ appropriately.

In this paper we introduce the negativity-like quantity to examine the thermal entanglement, because we do not know how to derive the optimal decomposition of Eq. (80). Recently, the upper and lower bounds of entanglement of formation (EoF) are examined for arbitrary two-mode Gaussian stateralph19 . It seems to be of interest to examine the TEPT with EoF.