1 Introduction
Quantum systems in heat bath follow the master equations.[1-2] Usually only the weak system-environment coupling regime is considered. In this case the Markovian approximation is used and the system is described by the master equation, in which the decay coefficient does not depend on time.[3-4] For the simplest systems such as harmonic oscillator and two-level system, the master equations can be solved using different methods under the Markovian approximation.
Beyond the Markovian approximation, the non-Markovian dynamics of open systems are investigated in the past few years.[5-20] In this case, the decay coefficient is no longer time-independent, rather it is a function of time $t$.[21-23] Such master equation is well studied for different systems. It seems that there is not an exact solution yet for this kind system.
In this paper we shall investigate the exact solution of harmonic oscillator and two-level system for the non-Markovian process. The exact solutions are achieved by making use of the hyper-operator techniques[24] and Wei-Norman theorem[25] for the harmonic oscillator system and two-level system. Harmonic system is solved exactly for any time-dependent decay coefficient and two-level system is determined by the second-order differential equation. We hope that this approach can help us to understand the decay process of quantum systems beyond the Markovian approximation.
This paper is organized as follows. In Sec. 2, we investigate the exact solution of the time evolution of the harmonic oscillator in a heat bath beyond the Markovian approximation. The time evolution operator of the two-level system is exactly found in Sec. 3. We conclude in Sec. 4.
2 Harmonic oscillator in heat bath
Consider a Harmonic oscillator in heat bath
$H=\hbar\omega a^{\dagger}a+\sum_{i}\hbar\omega_{i}b_{i}^{\dagger}b_{i}+\sum_{i}\lambda_{i}(a^{\dagger}b_{i}+ab_{i}^{\dagger})\,,$
where $a, a^{\dagger}$ are annihilation and creation operators of harmonic oscillator and $b_{i}, b_{i}^{\dagger}$ are annihilation and creation operators of heat bath.
Then its master equation is known as[1]
$\frac{d\rho(t)}{d t}=-\frac{i}{\hbar}[H_{0},\rho(t)]\\\ +\;\Gamma(t)\Bigl\{ a\rho a^{\dagger}-\frac{1}{2}a^{\dagger}a\rho-\frac{1}{2}\rho a^{\dagger}a\Bigr\} , $
where $\Gamma(t)$ is time-dependent in general. Under Markovian approximation, $\Gamma(t)$ is time independent. Changing to interaction picture, we have
$\frac{\mbox{ d}\rho_{I}}{\mbox{d}t}= \Gamma(t) \Bigl[a\rho_{I}a^{\dagger}-\frac{1}{2}a_{I}^{\dagger}a_{I}\rho_{I}-\frac{1}{2}\rho_{I}a_{I}^{\dagger}a_{I} \Bigr]\\ =\Gamma(t)\Bigl[{\cal L}_{a}R_{a^{\dagger}}-\frac{1}{2}{\cal L}_{a^{\dagger}a}-\frac{1}{2} {\cal R}_{a^{\dagger}a}\Bigr]\rho_{I}\,. $
Let $V$ be the Hilbert space of oscillator and $\mbox{End}(V)$ be the linear space of all linear operators on $V$. Furthermore, all operators on $\mbox{End}(V)$ constitute a linear space $\mbox{End(End}(V))$ and those operators are called hyper-operator. The operators ${\cal L}_{a}$ and ${\cal R}_{a}$ etc. introduced in Eq. (2) are hyper-operators acting on $\rho_I \in \mbox{End}V$, by multiplying $a$ on the left hand side and right hand side, respectively
${\cal L}_{a} A=aA\,, \quad {\cal R}_{a}A = A a\,, \quad \forall A\in \mbox{End}(V)\,, $
which are called the left-multiplication and right-multiplication hyper-operators, respectively.
In terms of the hyper-operators, the master equation (2) is rewritten as
$\frac{\mbox{d}\rho_{I}}{\mbox{d}t}={\cal H}\rho_{I}\,,$
where
${\cal H}=\Gamma(t)\Bigl[{\cal L}_{a}R_{a^{\dagger}}-\frac{1}{2}{\cal L}_{a^{\dagger}a} -\frac{1}{2}{\cal R}_{a^{\dagger}a}\Bigr]. $
Note that, although Eq. (4) looks like the Schrödinger equation, it is not usual Schrödinger equation. In fact, ${\cal H}$ is a hyper-operator and $\rho \in \mbox{End}(V)$. Therefore, Eq. (4) can be understood as the Schrödinger equation in $\mbox{End}{V}$ and ${\cal H}$ is the Hamiltonian acting on $\mbox{End}{(V)}$ rather than $V$.
Consider the Lie algebra generated by hyper-operators
${\cal A}\equiv{-\frac{1}{2}{\cal L}_{a^{\dagger}a}-\frac{1}{2}{\cal R}_{a^{\dagger}a}},$
${\cal B}\equiv{\cal L}_{a}{\cal R}_{a^{\dagger}}.$
It is easy to see that
$[{\cal A},{\cal B}]=-\frac{1}{2}\left[{\cal L}_{N}+{\cal R}_{N},{\cal L}_{a}{\cal R}_{a^{\dagger}}\right]\\ =\frac{1}{2}\left(\left[{\cal L}_{a},{\cal L}_{N}\right]{\cal R}_{a^{\dagger}}+{\cal L}_{a} \left[{\cal R}_{a^{\dagger}},{\cal R}_{N}\right]\right)\\ =\frac{1}{2}\left({\cal L}_{\left[a,N\right]}{\cal R}_{a^{\dagger}}+{\cal L}_{a}{\cal R}_{\left[N,a^{\dagger}\right]}\right)\\ =\frac{1}{2}\left({\cal L}_{a}{\cal R}_{a^{\dagger}}+{\cal L}_{a}{\cal R}_{a^{\dagger}}\right)={\cal L}_{a}{\cal R}_{a^{\dagger}}\\ ={\cal B}\,. $
So the hyper-operators ${\cal A}$ and ${\cal B}$ generate a 2-dimensional solvable Lie algebra with commutation relations
$$[{\cal A},{\cal A}]=0,\quad[{\cal B},{\cal B}]=0,\quad[{\cal A},{\cal B}]={\cal B}\,.$$
According to Wei-Norman theorem, the time evolution operator $U(t)$ can be written in the form
$$U(t)=\exp\left[f(t){\cal A}\right]\cdot\exp\left[g(t){\cal B}\right],$$
where $f(t)$ and $g(t)$ are time-dependent parameters to be determined.
It is easy to evaluate that
$\frac{\mbox{d}U}{\mbox{d}t}U^{-1}=\dot{f}{\cal A}+\dot{g}\exp[f(t){\cal A}]{\cal B}\exp[-f(t){\cal A}]\\ =\dot{f}{\cal A}+\dot{g}\Bigl({\cal B}+f[{\cal A,B}]+\frac{1}{2!}f^{2}[{\cal A},[{\cal A,B]]+\cdots}\Bigr)\\ =\dot{f}{\cal A}+\dot{g}{\cal B}\exp{f(t)}={\cal H}\,. $
In comparison with Eq. (5), we find
$$\Gamma(t)=\dot{f},\quad \dot{g}\exp(f)=\Gamma(t).$$
One can easily find the coefficient $f(t)$ from the first equation. Combining the first and the second equations, we can obtain $\dot{g}=\dot{f}\exp\left(-f\right)$ and the coefficient $g\left(t\right)$ can be expressed by
$g(t)=1-\exp\left(-f\left(t\right)\right)$
satisfying the initial conditions $f\left(0\right)=g\left(t\right)=0.$
For the Markovian approximation, where $\Gamma(t)=\Gamma$ is a constant, from Eq. (5), we can directly find that
$f(t)=\Gamma t$
$g(t)=1-\exp(-\Gamma t)$
and the time evolution operator is
$U(t)=\exp(\Gamma{\cal A}t)\exp\left[\left(1-e^{-\Gamma t}\right){\cal B}\right]\,,$
which is a hyper-operator and density matrix at time $t$ is then obtained as
$$\rho(t)=U(t)\rho(0)\,.$$
We then obtained the exact solution of harmonic oscillator in the heat bath.
3 Two-level Atom in Heat Bath
Consider a two-level atom in an environment consisting infinite number of harmonic oscillators. The atom can be described by a reduced density operator $\rho$ satisfying the master equation
$\dot{\rho} = r(N+1)\sigma_{-}\rho\sigma_{+}+rN\sigma_{+}\rho\sigma_{-}\\ -\frac{r}{2}(N+1)\sigma_{+}\sigma_{-}\rho-\frac{r}{2}N\sigma_{-}\sigma_{+}\rho\\ -\frac{r}{2}(N+1)\rho\sigma_{+}\sigma_{-}-\frac{r}{2}N\rho\sigma_{-}\sigma_{+}\,, \label{matrix-eq-atom} $
where $\sigma_+$, $\sigma_-$ are Pauli matrices of two-level atom, $r$ is the time-dependent decay rate and $N$ is the mean number of the bath quanta at the temperature of the heat bath.
We can define the following hyper-operators
$H=\frac{1}{2}\left({\cal L}_{\sigma_{+}\sigma_{-}}+{\cal R}_{\sigma_{+}\sigma_{-}}-1\right)\,,$
$E={\cal L}_{\sigma_{+}}{\cal R}_{\sigma_{-}}\,,$
$F={\cal L}_{\sigma_{-}}{\cal R}_{\sigma_{+}}\,.$
It is easy to prove that those operators satisfy the following communication relations
$\left[E,F\right]=2H\,,\quad\left[H,E\right]=E\,,\quad\left[H,F\right]=-F\,.$
Therefore, Eq. (14) can be rewritten as
$\dot{\rho}= \Bigl[\beta E+\alpha F+(\beta-\alpha)H-\frac{1}{2}(\beta+\alpha)\Bigr]\rho\,,$
where $\alpha$ and $\beta$ are
$\alpha=r(N+1)\,,\quad\beta=rN\,.$
Time evolution operator satisfies the following equation
$$\frac{d U(t)}{d t}=\left[a(t)H+b(t)E+c(t)F\right]U(t),$$
and the time evolution of the density matrix
$\rho\left(t\right)=U(t)\rho(0)$
with $a(t)=\beta-\alpha$, $b(t)=\beta$, and $c(t)=\alpha$.
Write $U(t)$ in the following form
$$U(t)=\exp\left[h(t)H\right]\cdot\exp\left[g(t)E\right]\cdot\exp\left[f(t)F\right]\,,$$
and our aim to find the coefficients $a(t), b(t)$, and $c(t)$. We have
$\frac{d U}{d t}=\dot{h}He^{hH}e^{gE}e^{fF}+e^{hH}\dot{g}Ee^{gE}e^{fF}\\ +e^{hH}e^{gE}\dot{f}Fe^{fF}\,.$
Multiplication of $U^{-1}(t)$ on right side of above equation gives
$\frac{d U}{d t}U^{-1}=\dot{h}H+\dot{g}e^{hH}Ee^{-hH} +\dot{f}e^{hH}e^{gE}Fe^{-gE}e^{-hH}\\ =\dot{h}H+\dot{g}e^{\scriptsize{\rm ad}(hH)}E+\dot{f}e^{\scriptsize{\rm ad}(hH)}e^{\scriptsize{\rm ad}(e E)}F\,, $
where ad$(x)$ is the adjoint operator defined as
$$\mbox{ad}\left(x\right)y=[x,y]\,.$$
It is easy to find that
$$ e^{\scriptsize{\rm ad}(hH)}E=\sum_{n=0}^{\infty}\frac{h^{n}}{n!}[H,[H,\cdots[H,E]\cdots]]\\ \hphantom{ e^{\scriptsize{\rm ad}(hH)}E }=\sum_{n=0}^{\infty}\frac{h^{n}}{n!}E=e^{h}E\,,\\ e^{\scriptsize{\rm ad}(gE)}F= F+g[E,F]+\frac{1}{2!}g^{2}[E,[E,F]]\\ \hphantom{ e^{\scriptsize{\rm ad}(gE)}F= }+\frac{1}{3!}g^{3}[E,[E,[E,F]]]+\cdots\\ \hphantom{ e^{\scriptsize{\rm ad}(gE)}F }= F+2gH-g^{2}E\,,\\ e^{\scriptsize{\rm ad}(hH)}e^{\scriptsize{\rm ad}(gE)}F=e^{\scriptsize{\rm ad}(hH)}(F+2gH-g^{2}E)\\ \hphantom{e^{\scriptsize{\rm ad}(hH)}e^{\scriptsize{\rm ad}(gE)}F }=e^{-h}F+2gH-g^{2}e^{h}E\,. $$
So
$\frac{d U}{d t}U^{-1}= \dot{h}H+\dot{g}e^{h}E+\dot{f}\left(e^{-h}F+2gH-g^{2}e^{h}E\right)\\ =(\dot{h}+2g\dot{f})H+(\dot{g}e^{h}-\dot{f}g^{2}e^{h})E \\ \quad +\dot{f}e^{-h}F\,. $
On the other hand
$\frac{d U}{d t}U^{-1}=HUU^{-1}=H\\=a(t)H+b(t)E+c(t)F\,.$
Comparing the coefficients, we find
$a=\dot{h}+2\dot{f}g,$
$b=\dot{g}e^{h}-\dot{f}g^{2}e^{h}\,,$
$c=\dot{f}e^{-h}\,,$
or in the following more elegant form
$\left[\begin{array}{c} a\\ b\\ c \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 2g\\ 0 & e^{h} & -g^{2}e^{h}\\ 0 & 0 & e^{-h} \end{array}\right]\left[\begin{array}{c} \dot{h}\\ \dot{g}\\ \dot{f} \end{array}\right].$
We finally find the relationship between the coefficients in $H(t)$ and that in $U(t)$
$\dot{h}=a-2ge^{h}c\,,$
$\dot{g}=be^{-h}+g^{2}e^{h}c\,,$
$\dot{f}=e^{h}c\,.$
Equations (29)$-$(31) are nonlinear differential equations and in general they are difficult to be solved. Fortunately, we can change it to the Reccati equation for $\dot{h}$
$\ddot{h}=\frac{1}{2}(\dot{h})^{2}+\Bigl(\frac{\dot{c}}{c}\Bigr)\dot{h} \\ +\Bigl[\dot{a}-2bc-\frac{1}{2}a^{2}-a\Bigl(\frac{\dot{c}}{c}\Bigr)\Bigr]. $
Letting
$u=\frac{1}{2}\dot{h}\,,\quad p(t)=-\frac{\dot{c}}{c}\,,\\ q(t)=\frac{1}{2}\Bigl[2bc+\frac{1}{2}a^{2}+a\Bigl(\frac{\dot{c}}{c}\Bigr)-\dot{a}\Bigr], $
we obtain the Riccati equation satisfied by $u=({1}/{2})\dot{h}$ and $\dot{u}-u^{2}+p(t)u+q(t)=0.$ Letting $u=-v^{\prime}/v$, above equation is further changed to the second-order differential equation
$v^{\prime\prime}-p(x)v^{\prime}+q(x)v=0, \label{final-eq}$
which have been well studied in mathematics. Therefore, solving the master equation of two-level atom reduces to find solutions to Eq. (34).
4 Conclusion
In this paper we investigated solutions to the non-Markovian master equation using Wei-Norman theorem and hyper-operator technique. For harmonic oscillator system, the time evolution operator is exactly derived, and for two-level system the time evolution operator is expressed as linear combination of hyper-operator and the coefficient in the evolution operator can be determined by solving a second-order differential equation.
Appendix A Hyper-operator
Let ${\cal H}$ be a Hilbert space with dimension $N<\infty$. Then all linear operators on ${\cal H}$ span an $N^{2}$-dimensional space End(${\cal H}$). Furthermore, we can consider all linear operators on End(${\cal H}$) which span an $N^{2}\times N^{2}$ linear space End(End(${\cal H}$)). We call End(End(${\cal H}$)) the Hyper-space and any elements in End(End(${\cal H}$)) a hyper-operator.
In the hyperspace End(End(${\cal H}$)), one can define the left-multiplication hyper-operator and right-multiplication hyper-operator
${\cal L}_A X = AX\,, \quad {\cal R}_A X = XA\,,$
for any $X\in\mbox{End}({\cal H)}$.
Then one can easily prove that
${\cal L}_{A}{\cal L}_{B}={\cal L}_{AB},\quad{\cal R}_{A}{\cal R}_{B}={\cal R}_{BA},\quad[{\cal L}_{A},{\cal R}_{B}]=0\,.$
Suppose that $\left\{ |n\rangle\ |\ n=1,2,\ldots,n\right\} $ is a basis of ${\cal H}$, then it is known that $\{|m\rangle\langle n|\ |\ m,n=1,2,\ldots,N\}$ is a basis of End$({\cal H)}$. Furthermore,
${\cal L}_{|m\rangle\langle n|} {\cal R}_{|p\rangle\langle q|}, \quad{}m,n,p,q = 1,2,\ldots,N\,,$
is a basis of End(End(${\cal H}$)). For mathematical details see Ref. [24].