1. Introduction
Quantum entanglement plays a vital role in quantum information technologies [1] such as quantum communication, quantum teleportation, quantum cryptography, quantum computation, and so on. A major challenge in the realization of these quantum based technologies is the decoherence and dissipation induced by the inevitable system-environment couplings. In particular, it has been shown that an entangled quantum system may get disentangled within a finite time via spontaneous emissions even when the environment is a vacuum, which is known as entanglement sudden death [2, 3]. However, on the other hand, due to the indirect interactions provided by the common bath, in certain circumstances, two initially separable atoms may get entangled [4-13]. In particular, it has been shown that, for a two-atom system immersed in a thermal bath, entanglement generation happens for atoms with an appropriate separation in a thermal bath at sufficiently small temperatures, while entanglement sudden death is a general feature if the interatomic separation is nonvanishing [11]. That is, steady-state entanglement is possible only when the two atoms are placed extremely close to each other [13].
The studies also show that the entanglement dynamics of a quantum system is crucially dependent on changes of the environment to which it is coupled. It is well-known that vacuum fluctuations can be modified in the presence of boundaries, which leads to novel effects such as the Casimir [14] and Casimir-Polder effects [15], modifications of the spontaneous transition rates [16-22], position-dependent resonance interactions [23, 24], electromagnetic shielding in quantum metrology [25], and so on. Then, a question naturally arises as to what happens to the entanglement dynamics of a quantum system if the environment is altered by the presence of reflecting mirrors. Recently, the entanglement dynamics for a two-atom system in the presence of a single reflecting mirror has been investigated, which shows that the presence of a mirror plays an important role in controlling the entanglement dynamics [26-28]. Remarkably, it has been shown that, when both of the atoms are placed extremely close the mirror, the two-atom system can behave as if it were isolated from the environment, and the initial entanglement will be preserved no matter how long they are separated [28].
A further question is what happens to the entanglement dynamics if two parallel reflecting mirrors are present. In fact, it has long been known that, the spontaneous emission rate of an excited atom can be completely turned off if its transition dipole moment is parallel to the reflecting mirrors, when the distance between the two mirrors is smaller than half of the transition wavelength [16-19], no matter where the atom is placed between the mirrors. Therefore, novel features which are different from those in a free space and in the presence of a single mirror are expected in the entanglement dynamics of a two-atom system, since it is closely related to the spontaneous transition processes. Here let us note that the entanglement dynamics of a two-atom system in a common cavity has been extensively studied, see, e.g. [29-34]. Usually, it is assumed that the atoms interact with a single-mode cavity field [30-32], or a reservoir with some kinds of spectral density, such as the Lorentzian spectral density [33]. In the present paper, we assume that the environment the atoms immersed in is a thermal bath of quantum scalar fields with all frequencies in the presence of two parallel reflecting mirrors. The two-point function of the quantum fields in the coordinate space is worked out with the method of images, which facilitates the investigation of how the entanglement dynamics is dependent on the geometric configurations of the two-atom system with respect to the mirrors. In particular, we focus on whether entanglement can be generated at the neighborhood of the initial time for a separable two-atom system, and whether entanglement can persist in the steady state. Hereafter, natural units with ℏ = c = kB = 1 are applied.
2. The basic formalism
In this paper, we study the entanglement generation for two atoms immersed in a thermal bath of quantum scalar fields in the presence of two parallel reflecting mirrors. The Hamiltonian of the total system is
Here HA is the Hamiltonian of the two-atom system. We assume that there is no direct interaction between the two atoms, so the Hamiltonian can be written as
where the superscript $\alpha$ ($\alpha$ = 1, 2) labels the operator of the $\alpha$ th atom, ${\sigma }_{i}^{(1)}={\sigma }_{i}\otimes {\sigma }_{0}$, ${\sigma }_{i}^{(2)}={\sigma }_{0}\otimes {\sigma }_{i}$, with σi (i = 1, 2, 3) being the Pauli matrices, σ0 the 2 × 2 unit matrix, and ω the energy level spacing of the atoms. HF is the Hamiltonian of the fluctuating massless scalar fields. HI is the Hamiltonian describing the interaction between the atoms and the field, which takes the form in analogy to the electric dipole interaction as [35]
where $\mu$ is a dimensionless coupling constant which is supposed to be small, Φ is the field operator, τ is the proper time, and xi (i = 1, 2) labels the position of the i th atom.
$\begin{eqnarray}H={H}_{A}+{H}_{F}+{H}_{I}.\end{eqnarray}$
Here HA is the Hamiltonian of the two-atom system. We assume that there is no direct interaction between the two atoms, so the Hamiltonian can be written as
$\begin{eqnarray}{H}_{{\rm{A}}}=\displaystyle \frac{\omega }{2}{\sigma }_{3}^{(1)}+\displaystyle \frac{\omega }{2}{\sigma }_{3}^{(2)},\end{eqnarray}$
where the superscript $\alpha$ ($\alpha$ = 1, 2) labels the operator of the $\alpha$ th atom, ${\sigma }_{i}^{(1)}={\sigma }_{i}\otimes {\sigma }_{0}$, ${\sigma }_{i}^{(2)}={\sigma }_{0}\otimes {\sigma }_{i}$, with σi (i = 1, 2, 3) being the Pauli matrices, σ0 the 2 × 2 unit matrix, and ω the energy level spacing of the atoms. HF is the Hamiltonian of the fluctuating massless scalar fields. HI is the Hamiltonian describing the interaction between the atoms and the field, which takes the form in analogy to the electric dipole interaction as [35]
$\begin{eqnarray}{H}_{I}=\mu [{\sigma }_{2}^{(1)}{\rm{\Phi }}(\tau ,{{\boldsymbol{x}}}_{1})+{\sigma }_{2}^{(2)}{\rm{\Phi }}(\tau ,{{\boldsymbol{x}}}_{2})],\end{eqnarray}$
where $\mu$ is a dimensionless coupling constant which is supposed to be small, Φ is the field operator, τ is the proper time, and xi (i = 1, 2) labels the position of the i th atom.
In the weak-coupling limit, the evolution of the reduced density matrix of the two-atom system satisfies the Gorini−Kossakowski−Lindblad−Sudarshan master equation [36, 37]
where i is the imaginary unit,
and
The coefficients ${C}_{{ij}}^{(\alpha \xi )}$ and ${H}_{{ij}}^{(\alpha \xi )}$ are determined by the Fourier and Hilbert transforms of the field correlation function
which are given as follows,
Here P denotes the principal value, and ${\rm{\Delta }}\tau =\tau -\tau ^{\prime} $. Then, the coefficients ${C}_{{ij}}^{(\alpha \xi )}$ can be explicitly written as
where
Replacing the Fourier transform ${{ \mathcal G }}^{(\alpha \xi )}(\omega )$ with the Hilbert transform ${{ \mathcal K }}^{(\alpha \xi )}(\omega )$, one obtains ${H}_{{ij}}^{(\alpha \xi )}$.
$\begin{eqnarray}\displaystyle \frac{\partial \rho (\tau )}{\partial \tau }=-{\rm{i}}\,[{H}_{\mathrm{eff}},\rho (\tau )]+{ \mathcal L }[\rho (\tau )],\end{eqnarray}$
where i is the imaginary unit,
$\begin{eqnarray}{H}_{\mathrm{eff}}={H}_{S}-\displaystyle \frac{{\rm{i}}}{2}\displaystyle \sum _{\alpha ,\xi =1}^{2}\displaystyle \sum _{i,j=1}^{3}{H}_{{ij}}^{(\alpha \xi )}{\sigma }_{i}^{(\alpha )}{\sigma }_{j}^{(\xi )},\end{eqnarray}$
and
$\begin{eqnarray}\begin{array}{l}{ \mathcal L }[\rho ]=\displaystyle \frac{1}{2}\displaystyle \sum _{\alpha ,\xi =1}^{2}\displaystyle \sum _{i,j=1}^{3}{C}_{{ij}}^{(\alpha \xi )}[2{\sigma }_{j}^{(\xi )}\rho {\sigma }_{i}^{(\alpha )}\\ \quad -\,{\sigma }_{i}^{(\alpha )}{\sigma }_{j}^{(\xi )}\rho -\rho {\sigma }_{i}^{(\alpha )}{\sigma }_{j}^{(\xi )}].\end{array}\end{eqnarray}$
The coefficients ${C}_{{ij}}^{(\alpha \xi )}$ and ${H}_{{ij}}^{(\alpha \xi )}$ are determined by the Fourier and Hilbert transforms of the field correlation function
$\begin{eqnarray}{G}^{(\alpha \xi )}(\tau -\tau ^{\prime} )=\langle {\rm{\Phi }}(\tau ,{{\boldsymbol{x}}}_{\alpha }){\rm{\Phi }}(\tau ^{\prime} ,{{\boldsymbol{x}}}_{\xi }){\rangle }_{\beta },\end{eqnarray}$
which are given as follows,
$\begin{eqnarray}{{ \mathcal G }}^{(\alpha \xi )}(\omega )={\int }_{-\infty }^{\infty }{\rm{d}}t\,{{\rm{e}}}^{i\omega {\rm{\Delta }}\tau }{G}^{(\alpha \xi )}({\rm{\Delta }}\tau ),\end{eqnarray}$
$\begin{eqnarray}{{ \mathcal K }}^{(\alpha \xi )}(\lambda )=\displaystyle \frac{P}{\pi {\rm{i}}}{\int }_{-\infty }^{\infty }{\rm{d}}\omega \displaystyle \frac{{{ \mathcal G }}^{(\alpha \xi )}(\omega )}{\omega -\lambda }.\end{eqnarray}$
Here P denotes the principal value, and ${\rm{\Delta }}\tau =\tau -\tau ^{\prime} $. Then, the coefficients ${C}_{{ij}}^{(\alpha \xi )}$ can be explicitly written as
$\begin{eqnarray}{C}_{{ij}}^{(\alpha \xi )}={A}^{(\alpha \xi )}{\delta }_{{ij}}-{\rm{i}}\,{B}^{(\alpha \xi )}{\epsilon }_{{ijk}}{\delta }_{3k}-{A}^{(\alpha \xi )}{\delta }_{3i}{\delta }_{3j},\end{eqnarray}$
where
$\begin{eqnarray}\begin{array}{l}{A}^{(\alpha \xi )}=\displaystyle \frac{{\mu }^{2}}{4}[{{ \mathcal G }}^{(\alpha \xi )}(\omega )+{{ \mathcal G }}^{(\alpha \xi )}(-\omega )],\\ {B}^{(\alpha \xi )}=\displaystyle \frac{{\mu }^{2}}{4}[{{ \mathcal G }}^{(\alpha \xi )}(\omega )-{{ \mathcal G }}^{(\alpha \xi )}(-\omega )].\end{array}\end{eqnarray}$
Replacing the Fourier transform ${{ \mathcal G }}^{(\alpha \xi )}(\omega )$ with the Hilbert transform ${{ \mathcal K }}^{(\alpha \xi )}(\omega )$, one obtains ${H}_{{ij}}^{(\alpha \xi )}$.
For convenience, we employ the coupled basis {∣G⟩ = ∣00⟩, ∣E⟩ = ∣11⟩, ∣A⟩ = $\tfrac{1}{\sqrt{2}}(| 10\rangle -| 01\rangle ),| S\rangle $ = $\tfrac{1}{\sqrt{2}}(| 10\rangle \,+| 01\rangle )\}$ to solve the master equation of the two-atom system. The density matrix of the two-atom system can be written as ρ = ∑I,JρIJ∣I⟩⟨J∣, where ρIJ = ⟨I∣ρ∣J⟩, I, J ∈ {G, E, A, S}. First, we rewrite the coefficients equation (10 ) as
$\begin{eqnarray}\begin{array}{rcl}{C}_{{ij}}^{(11)} & = & {A}_{1}{\delta }_{{ij}}-{\rm{i}}\,{B}_{1}{\epsilon }_{{ijk}}{\delta }_{3k}-{A}_{1}{\delta }_{3i}{\delta }_{3j},\\ {C}_{{ij}}^{(22)} & = & {A}_{2}{\delta }_{{ij}}-{\rm{i}}\,{B}_{2}{\epsilon }_{{ijk}}{\delta }_{3k}-{A}_{2}{\delta }_{3i}{\delta }_{3j},\\ {C}_{{ij}}^{(12)} & = & {C}_{{ij}}^{(21)}={A}_{3}{\delta }_{{ij}}-{\rm{i}}\,{B}_{3}{\epsilon }_{{ijk}}{\delta }_{3k}-{A}_{3}{\delta }_{3i}{\delta }_{3j}.\end{array}\end{eqnarray}$
For brevity, we have defined A1 = A(11), A2 = A(22), A3 = A(12) = A(21), and B1 = B(11), B2 = B(22), B3 = B(12) = B(21). Then, a set of equations, which are decoupled from other matrix elements, can be obtained as follows
Note that ρGG + ρEE + ρAA + ρSS = 1 is satisfied.
$\begin{eqnarray}\begin{array}{l}{\dot{\rho }}_{\mathrm{GG}}=-2({A}_{1}-{B}_{1}+{A}_{2}-{B}_{2}){\rho }_{\mathrm{GG}}\\ \quad +\,({A}_{1}+{B}_{1}+{A}_{2}+{B}_{2}-2{A}_{3}-2{B}_{3}){\rho }_{\mathrm{AA}}\\ \quad +\,({A}_{1}+{B}_{1}+{A}_{2}+{B}_{2}+2{A}_{3}+2{B}_{3}){\rho }_{\mathrm{SS}}\\ \quad +\,({A}_{1}+{B}_{1}-{A}_{2}-{B}_{2})({\rho }_{\mathrm{AS}}+{\rho }_{\mathrm{SA}}),\\ {\dot{\rho }}_{\mathrm{AA}}=-2({A}_{1}+{A}_{2}-2{A}_{3}){\rho }_{\mathrm{AA}}\\ \quad +\,({A}_{1}-{B}_{1}+{A}_{2}-{B}_{2}-2{A}_{3}+2{B}_{3}){\rho }_{\mathrm{GG}}\\ \quad +\,({A}_{1}+{B}_{1}+{A}_{2}+{B}_{2}-2{A}_{3}-2{B}_{3}){\rho }_{\mathrm{EE}}\\ \quad +\,(-{B}_{1}+{B}_{2})({\rho }_{\mathrm{AS}}+{\rho }_{\mathrm{SA}}),\\ {\dot{\rho }}_{\mathrm{SS}}=-2({A}_{1}+{A}_{2}+2{A}_{3}){\rho }_{\mathrm{SS}}\\ \quad +\,({A}_{1}-{B}_{1}+{A}_{2}-{B}_{2}+2{A}_{3}-2{B}_{3}){\rho }_{\mathrm{GG}}\\ \quad +\,({A}_{1}+{B}_{1}+{A}_{2}+{B}_{2}+2{A}_{3}+2{B}_{3}){\rho }_{\mathrm{EE}}\\ \quad +\,(-{B}_{1}+{B}_{2})({\rho }_{\mathrm{AS}}+{\rho }_{\mathrm{SA}}),\\ {\dot{\rho }}_{\mathrm{EE}}=-2({A}_{1}+{B}_{1}+{A}_{2}+{B}_{2}){\rho }_{\mathrm{EE}}\\ \quad +\,({A}_{1}-{B}_{1}+{A}_{2}-{B}_{2}-2{A}_{3}+2{B}_{3}){\rho }_{\mathrm{AA}}\\ \quad +\,({A}_{1}-{B}_{1}+{A}_{2}-{B}_{2}+2{A}_{3}-2{B}_{3}){\rho }_{\mathrm{SS}}\\ \quad +\,(-{A}_{1}+{B}_{1}+{A}_{2}-{B}_{2})({\rho }_{\mathrm{AS}}+{\rho }_{\mathrm{SA}}),\\ {\dot{\rho }}_{\mathrm{AS}}=-2({A}_{1}+{A}_{2}){\rho }_{\mathrm{AS}}\\ \quad +\,({A}_{1}-{B}_{1}-{A}_{2}+{B}_{2}){\rho }_{\mathrm{GG}}\\ \quad +\,(-{A}_{1}-{B}_{1}+{A}_{2}+{B}_{2}){\rho }_{\mathrm{EE}}\\ \quad +\,(-{B}_{1}+{B}_{2})({\rho }_{\mathrm{SS}}+{\rho }_{\mathrm{AA}}),\\ {\dot{\rho }}_{\mathrm{SA}}=-2({A}_{1}+{A}_{2}){\rho }_{\mathrm{SA}}\\ \quad +\,({A}_{1}-{B}_{1}-{A}_{2}+{B}_{2}){\rho }_{\mathrm{GG}}\\ \quad +\,(-{A}_{1}-{B}_{1}+{A}_{2}+{B}_{2}){\rho }_{\mathrm{EE}}\\ \quad +\,(-{B}_{1}+{B}_{2})({\rho }_{\mathrm{SS}}+{\rho }_{\mathrm{AA}}),\\ {\dot{\rho }}_{\mathrm{GE}}=-2({A}_{1}+{A}_{2}){\rho }_{\mathrm{GE}},\\ {\dot{\rho }}_{\mathrm{EG}}=-2({A}_{1}+{A}_{2}){\rho }_{\mathrm{EG}}.\end{array}\end{eqnarray}$
Note that ρGG + ρEE + ρAA + ρSS = 1 is satisfied.
3. The condition for entanglement generation
In this section, we investigate whether entanglement can be created between two atoms at the neighborhood of the initial time with the help of the partial transposition criterion [38, 39], i.e. a two-atom state ρ(τ) is entangled if and only if the partial transposition of ρ(τ) does not preserve its positivity. We assume that initially the two atoms are in their ground and excited states respectively, i.e. ρ(0) = ∣1⟩⟨1∣ ⨂ ∣0⟩⟨0∣, which is separable. In this case, based on the partial transposition criterion, the two atoms can be entangled at the neighbourhood of τ = 0 if and only if the following inequality is satisfied [12, 13],
where the superscript T denotes matrix transposition, and ui = vi = {1, − i, 0}. Substituting the explicit form of C($\alpha$ξ) into equation (14 ), one obtains
$\begin{eqnarray}\langle u| {C}^{(11)}| u\rangle \langle v| {\left({C}^{(22)}\right)}^{{\rm{T}}}| v\rangle \lt {\left|\langle u| \mathrm{Re}({C}^{(12)})| v\rangle \right|}^{2},\end{eqnarray}$
where the superscript T denotes matrix transposition, and ui = vi = {1, − i, 0}. Substituting the explicit form of C($\alpha$ξ) into equation (
$\begin{eqnarray}{A}_{3}^{2}+{B}_{1}{B}_{2}\gt {A}_{1}{A}_{2}.\end{eqnarray}$
In this paper, we study the entanglement generation for two atoms coupled with massless scalar fields in a thermal bath at temperature 1/$\beta$ in the presence of two parallel reflecting mirrors. Without loss of generality, the two mirrors are assumed to be located at z = 0 and z = L respectively. We assume that the two atoms are aligned vertically to the mirrors, which are placed at x1 = (0, 0, d1) and x2 = (0, 0, d1 + d) respectively, where d1 is the distance between atom 1 and mirror 1, and d is the interatomic separation, see figure 1. For convenience, we rewrite the distance between the two mirrors as
where $\lambda=\tfrac{2\pi }{\omega }$ is the atomic transition wavelength, $k\in {\mathbb{Z}}$ and 0 ≤ θ < 2π. The coefficients Ai and Bi are given inappendix , and the explicit expressions are dependent on whether the distance between the two mirrors L is an integer times ƛ/2, i.e. whether θ in equation (16 ) is zero or not. Since the expressions are complicated in a general case, first let us consider some special cases in the following.
$\begin{eqnarray}L=\displaystyle \frac{\lambda}{2}\left(k+\displaystyle \frac{\theta }{2\pi }\right),\end{eqnarray}$
where $\lambda=\tfrac{2\pi }{\omega }$ is the atomic transition wavelength, $k\in {\mathbb{Z}}$ and 0 ≤ θ < 2π. The coefficients Ai and Bi are given in
Figure 1. Two atoms separated from each other by a distance d are aligned vertically to two parallel mirrors separated by a distance L. |
Case 1. When the transition wavelength is larger than twice the distance between the two mirrors, i.e. ƛ > 2L, it can be found with the help of equation (32 ) that, A1 = A2 = A3 = 0, and B1 = B2 = B3 = 0, so the condition (15 ) cannot be satisfied, and thus entanglement can not be created. In fact, in this case, the spontaneous emission is turned off [16-19], so the atoms are locked up in their initial state, and it is natural that entanglement cannot be created in this case. Note that this conclusion does not rely on how the atoms are aligned with respect to the boundaries. Actually, from the definition equation (7 ), it is clear that the two-point functions G(11)(Δτ) and G(22)(Δτ) (and thus A1, A2, B1, B2) are independent of the relative positions of the two atoms and the alignment with respect to the boundaries, so they are always zero when ƛ > 2L. Then, according to the Cauchy−Schwarz inequality, ∣G(12)(Δτ)∣2 ≤ G(11)(Δτ)G(22)(Δτ) = 0, so G(12)(Δτ) (and thus A3, B3) are also zero when ƛ > 2L.
A similar case is when one or both of the two atoms are placed extremely close to the mirrors, i.e. when d1 → 0 or d1 + d → L, it can be found from equations (32 ) and (33 ) that ${A}_{1}{A}_{2}={B}_{1}{B}_{2}={A}_{3}^{2}=0$. Therefore, the condition (15 ) is not satisfied, and entanglement cannot be created. Actually, when an atom is placed infinitely close to a reflecting mirror, it behaves as if it were isolated from the environment, so there is no environment-induced entanglement. Also, it does not rely on how the atoms are aligned with respect to the boundary.
Case 2. When the transition wavelength is larger than one but smaller than or equal to twice the distance between the two mirrors, i.e. L < ƛ ≤ 2L, it is straightforward to verify that ${A}_{3}^{2}={A}_{1}{A}_{2}$. Then, as long as none of the two atoms is placed infinitely close to the mirrors, i.e. d1 ≠ 0 and d + d1 ≠ L, then B1B2 > 0, so the condition (15 ) can be satisfied, and entanglement can be created at the initial time, no matter how high the environmental temperature is.
Except for the situations discussed in Case 1, the spontaneous emission rates of the atoms are non-zero, i.e. A1A2 ≠ 0 and B1B2 ≠ 0. Then, the condition (15 ) can be rewritten as
Here let us note that the criterion for entanglement generation can be written as equation (17 ) only if A1A2 ≠ 0. As an example, in Case 1, although A1 = A2 = A3 = 0, and B1 = B2 = B3 = 0, the limits of $\tfrac{{A}_{3}^{2}}{{A}_{1}{A}_{2}}$ and $\tfrac{{B}_{1}{B}_{2}}{{A}_{1}{A}_{2}}$ exist, and $\tfrac{{A}_{3}^{2}}{{A}_{1}{A}_{2}}+\tfrac{{B}_{1}{B}_{2}}{{A}_{1}{A}_{2}}\gt 1$ is possible. However, it is obvious from equation (13 ) that the two-atom system is lock up in its initial state, and entanglement generation is impossible. From equations (32 ) and (33 ), it can be found that $\tfrac{{B}_{1}{B}_{2}}{{A}_{1}{A}_{2}}$ depends on the environmental temperature only, which takes the form
and decreases monotonically as temperature increases, while $\tfrac{{A}_{3}^{2}}{{A}_{1}{A}_{2}}$ is temperature-independent, and it is dependent on the distance between the two mirrors and the positions of the two atoms. When the environment is a vacuum, i.e. $\beta$ → ∞, $\tfrac{{B}_{1}{B}_{2}}{{A}_{1}{A}_{2}}=1$, then entanglement can be generated unless $\tfrac{{A}_{3}^{2}}{{A}_{1}{A}_{2}}$ is zero. Also, if the separation between the two atoms is vanishing, i.e. d → 0, then $\tfrac{{A}_{3}^{2}}{{A}_{1}{A}_{2}}$ becomes 1, and the inequality (17 ) is always satisfied and entanglement can be created, no matter where the atoms are placed, as long as the environment temperature is not infinite. When the distance between the two mirrors L → ∞ , but the distance d1 between atom 1 and mirror 1 and the interatomic separation d are finite, the coefficients Ai and Bi are given in equation (34 ) in appendix , which reduce to those in the case of a single mirror, see equations (37)-(42) in [27] in the zero-acceleration limit. Furthermore, if d1 also approaches infinity, the coefficients are given in equation (35 ), which reduce to those in a free space [13], as expected.
$\begin{eqnarray}\displaystyle \frac{{A}_{3}^{2}}{{A}_{1}{A}_{2}}+\displaystyle \frac{{B}_{1}{B}_{2}}{{A}_{1}{A}_{2}}\gt 1.\end{eqnarray}$
Here let us note that the criterion for entanglement generation can be written as equation (
$\begin{eqnarray}\displaystyle \frac{{B}_{1}{B}_{2}}{{A}_{1}{A}_{2}}={\left(\displaystyle \frac{1-{{\rm{e}}}^{-\beta \omega }}{1+{{\rm{e}}}^{-\beta \omega }}\right)}^{2},\end{eqnarray}$
and decreases monotonically as temperature increases, while $\tfrac{{A}_{3}^{2}}{{A}_{1}{A}_{2}}$ is temperature-independent, and it is dependent on the distance between the two mirrors and the positions of the two atoms. When the environment is a vacuum, i.e. $\beta$ → ∞, $\tfrac{{B}_{1}{B}_{2}}{{A}_{1}{A}_{2}}=1$, then entanglement can be generated unless $\tfrac{{A}_{3}^{2}}{{A}_{1}{A}_{2}}$ is zero. Also, if the separation between the two atoms is vanishing, i.e. d → 0, then $\tfrac{{A}_{3}^{2}}{{A}_{1}{A}_{2}}$ becomes 1, and the inequality (
Since in a general case, the ratio $\tfrac{{A}_{3}^{2}}{{A}_{1}{A}_{2}}$ is complicated, in the following, we study its behavior numerically. First, we fix the interatomic separation d, and see how the ratio $\tfrac{{A}_{3}^{2}}{{A}_{1}{A}_{2}}$ varies with the distance between atom 1 and the first mirror d1. Note that since the atoms are aligned vertically to the mirrors, the range of d1 is restricted to [0, L−d]. As shown in figure 2, the ratio $\tfrac{{A}_{3}^{2}}{{A}_{1}{A}_{2}}$ oscillates as d1 increases, and it is always positive when the interatomic separation d is small. That is, there is always a critical temperature below which entanglement can be generated no matter where the vertically aligned two-atom system is placed. As d becomes larger, there are zeros of the ratio $\tfrac{{A}_{3}^{2}}{{A}_{1}{A}_{2}}$, i.e. there are positions at which the two-atom system cannot get entangled at the neighborhood of the initial time, even if the environment is a vacuum. When d becomes close to the distance between the two mirrors L (i.e. when the two atoms are close to the two mirrors respectively), $\tfrac{{A}_{3}^{2}}{{A}_{1}{A}_{2}}$ is almost unchanged as d1 varies, and it is always positive, i.e. entanglement can always be generated below a critical temperature.
Figure 2. The ratio $\tfrac{{A}_{3}^{2}}{{A}_{1}{A}_{2}}$ as a function of ωd1, with ωL = 10, and ωd = 1 (a), ωd = 3 (b), ωd = 8 (c). The red real, green dashed, and blue dotdashed lines correspond to the cases in the presence of two parallel mirrors, one mirror and the free space, respectively. |
Second, we fix the distance between atom 1 and the first mirror d1, and show how the ratio $\tfrac{{A}_{3}^{2}}{{A}_{1}{A}_{2}}$ varies with the interatomic separation d in figure 3. It is shown that the ratio $\tfrac{{A}_{3}^{2}}{{A}_{1}{A}_{2}}$ oscillates with a damping amplitude as d increases. Thus, as temperature increases, the range of d within which entanglement can be created becomes narrower. When d1 approaches L, i.e. when both the two atoms are placed close to one of the mirrors, the ratio $\tfrac{{A}_{3}^{2}}{{A}_{1}{A}_{2}}$ is close to unity and remains nearly unchanged, so entanglement can always be generated below some critical temperature.
Figure 3. The ratio $\tfrac{{A}_{3}^{2}}{{A}_{1}{A}_{2}}$ as a function of ωd, with ωL = 10, and ωd1 = 1 (a), ωd1 = 3 (b), ωd1 = 8 (c). The red real, green dashed, and blue dotdashed lines correspond to the cases in the presence of two parallel mirrors, one mirror and the free space, respectively. |
In figures 2 and 3, we have also compared the ratio $\tfrac{{A}_{3}^{2}}{{A}_{1}{A}_{2}}$ in the current case with those in the cases of a single mirror and the free space, which in general can be both larger and smaller than those in the other two cases. However, when the atoms are placed close to one of (i.e. d1 is large, see figure 3 (c)) or both of (i.e. d is large, see figure 2 (c)) the mirrors, the ratio $\tfrac{{A}_{3}^{2}}{{A}_{1}{A}_{2}}$ in the two-mirror case is always larger than those in the other two cases. That is, the range of temperature within which entanglement can be created is wider in the two-mirror case compared with the cases of a free space and one mirror.
4. The steady-state entanglement
In the previous section, we reveal that in certain circumstance entanglement can be generated in the presence of two mirrors. Another problem of interest is whether the entanglement can be protected and the system is still entangled in the steady state. Here, we characterize the degree of entanglement with the help of concurrence [40], which ranges from 0, for separable states, to 1, for maximally entangled states. In this paper, for simplicity, the initial density matrix is chosen as of the X shape, i.e. the only nonzero elements are those along the diagonal and anti-diagonal elements in the basis ∣00⟩, ∣01⟩, ∣10⟩, ∣11⟩. The X form will be maintained during evolution [41], and the concurrence can be calculated to be in the following form [9]
where
Moreover, as will be shown that ρGE(∞) is zero, so K2(∞) is negative, and $C[\rho (\infty )]=\max \{0,{K}_{1}(\infty )\}$.
$\begin{eqnarray}C[\rho (\tau )]=\max \{0,{K}_{1}(\tau ),{K}_{2}(\tau )\},\end{eqnarray}$
where
$\begin{eqnarray}\begin{array}{l}{K}_{1}(\tau )=\sqrt{{[{\rho }_{\mathrm{AA}}(\tau )-{\rho }_{\mathrm{SS}}(\tau )]}^{2}-{[{\rho }_{\mathrm{AS}}(\tau )-{\rho }_{\mathrm{SA}}(\tau )]}^{2}}\\ \quad -2\sqrt{{\rho }_{\mathrm{GG}}(\tau ){\rho }_{\mathrm{EE}}(\tau )},\\ {K}_{2}(\tau )=2| {\rho }_{\mathrm{GE}}(\tau )| \\ \quad -\sqrt{{[{\rho }_{\mathrm{AA}}(\tau )+{\rho }_{\mathrm{SS}}(\tau )]}^{2}-{[{\rho }_{\mathrm{AS}}(\tau )+{\rho }_{\mathrm{SA}}(\tau )]}^{2}}.\end{array}\end{eqnarray}$
Moreover, as will be shown that ρGE(∞) is zero, so K2(∞) is negative, and $C[\rho (\infty )]=\max \{0,{K}_{1}(\infty )\}$.
In order to investigate the steady-state entanglement, first we need to work out the steady state of the two-atom system, which can be found by taking the time derivatives in equation (13 ) to be zero. Since only five of the first six equations in equation (13 ) are independent, we solve a system of six linear equations composed of five of the first six equations in equation (13 ) by taking their right-hand sides to be zero, and ρGG + ρEE + ρAA + ρSS = 1.
When the determinant of coefficient matrix is nonzero, i.e.
the nonzero density matrix elements of the steady state can be solved as
where $\eta ={A}_{i}/{B}_{i}=\coth \left(\tfrac{\beta \omega }{2}\right)$ is introduced for brevity. That is, the steady state depends on the temperature of the thermal bath only in this case. Since ρAA(∞) = ρSS(∞) and ρAS(∞) = ρSA(∞) = 0, it is obvious from equation (20 ) that K1(∞) is negative, so the concurrence C[ρ(∞)] is zero when the determinant Δ is nonzero, i.e. the steady state of the two-atom system cannot be entangled in this case.
$\begin{eqnarray}\begin{array}{l}{\rm{\Delta }}=256\left({B}_{1}+{B}_{2}\right){\eta }^{3}\\ \qquad \times \,\left\{{B}_{3}^{2}\left[2{B}_{1}{B}_{2}\left(2{\eta }^{2}-1\right)+{B}_{1}^{2}+{B}_{2}^{2}\right]\right.\\ \qquad \left.-\,{B}_{1}{B}_{2}\left({B}_{1}+{B}_{2}\right){}^{2}{\eta }^{2}\right\}\ne 0,\end{array}\end{eqnarray}$
the nonzero density matrix elements of the steady state can be solved as
$\begin{eqnarray}\begin{array}{rcl}{\rho }_{\mathrm{GG}} & = & \displaystyle \frac{{(1+\eta )}^{2}}{4{\eta }^{2}},\quad {\rho }_{\mathrm{EE}}=\displaystyle \frac{{(\eta -1)}^{2}}{4{\eta }^{2}},\\ \quad {\rho }_{\mathrm{AA}} & = & {\rho }_{\mathrm{SS}}=\displaystyle \frac{{\eta }^{2}-1}{4{\eta }^{2}},\end{array}\end{eqnarray}$
where $\eta ={A}_{i}/{B}_{i}=\coth \left(\tfrac{\beta \omega }{2}\right)$ is introduced for brevity. That is, the steady state depends on the temperature of the thermal bath only in this case. Since ρAA(∞) = ρSS(∞) and ρAS(∞) = ρSA(∞) = 0, it is obvious from equation (
On the other hand, when the determinant Δ = 0, the steady state is also related to the initial state, and might be entangled. Explicit cases will be shown in the following in which Δ = 0 can be achieved.
Case 1. When the interatomic separation d → 0, we have A1 = A2 = A3, and B1 = B2 = B3. Then it can be found that the nonzero density matrix elements are,
and
When ${\rho }_{\mathrm{AA}}(0)\gt \tfrac{3({\eta }^{2}-1)}{2(3{\eta }^{2}-1)}$, K1(∞) > 0, and the steady state is entangled. In particular, when the environment is a vacuum, i.e. $\eta$ = 1, the steady state is entangled as long as ρAA(0) > 0. This agrees with the result in [13, 26]. In fact, a vanishing interatomic separation is necessary for atoms in a free space to achieve steady-state entanglement.
$\begin{eqnarray}\begin{array}{rcl}{\rho }_{\mathrm{SS}}(\infty ) & = & -\displaystyle \frac{({\rho }_{\mathrm{AA}}(0)-1)({\eta }^{2}-1)}{3{\eta }^{2}+1},\\ {\rho }_{\mathrm{AA}}(\infty ) & = & {\rho }_{\mathrm{AA}}(0),\\ {\rho }_{\mathrm{EE}}(\infty ) & = & -\displaystyle \frac{({\rho }_{\mathrm{AA}}(0)-1){\left(\eta -1\right)}^{2}}{3{\eta }^{2}+1},\\ {\rho }_{\mathrm{GG}}(\infty ) & = & \displaystyle \frac{({\rho }_{\mathrm{AA}}(0)-1){\left(\eta +1\right)}^{2}}{3{\eta }^{2}+1},\end{array}\end{eqnarray}$
and
$\begin{eqnarray}{K}_{1}(\infty )=\displaystyle \frac{| (4{\rho }_{\mathrm{AA}}(0)-1){\eta }^{2}+1| +2({\rho }_{\mathrm{AA}}(0)-1)({\eta }^{2}-1)}{3{\eta }^{2}+1}.\end{eqnarray}$
When ${\rho }_{\mathrm{AA}}(0)\gt \tfrac{3({\eta }^{2}-1)}{2(3{\eta }^{2}-1)}$, K1(∞) > 0, and the steady state is entangled. In particular, when the environment is a vacuum, i.e. $\eta$ = 1, the steady state is entangled as long as ρAA(0) > 0. This agrees with the result in [13, 26]. In fact, a vanishing interatomic separation is necessary for atoms in a free space to achieve steady-state entanglement.
Case 2. When one of the atoms is placed extremely close to a mirror, e.g. atom 1 is placed extremely close to mirror 1, i.e. d1 → 0, then A1 = A3 = 0, and B1 = B3 = 0. The nonzero density matrix elements can be calculated as
Since ρAA(∞) = ρSS(∞) and ρAS(∞) = ρSA(∞) = 0, K1(∞) is negative, thus the steady state cannot be entangled.
$\begin{eqnarray}\begin{array}{rcl}{\rho }_{\mathrm{SS}}(\infty ) & = & {\rho }_{\mathrm{AA}}(\infty )=\displaystyle \frac{\eta +{\rho }_{\mathrm{AA}}(0)+{\rho }_{\mathrm{SS}}(0)+2{\rho }_{\mathrm{EE}}(0)+{\rho }_{\mathrm{AS}}(0)+{\rho }_{\mathrm{SA}}(0)-1}{4\eta },\\ {\rho }_{\mathrm{AS}}(\infty ) & = & {\rho }_{\mathrm{SA}}(\infty )=\displaystyle \frac{1+\eta \,[\,{\rho }_{{AA}}(0)+{\rho }_{\mathrm{SS}}(0)+2{\rho }_{\mathrm{EE}}(0)+{\rho }_{\mathrm{AS}}(0)+{\rho }_{\mathrm{SA}}(0)-1\,]}{4\eta },\\ {\rho }_{\mathrm{EE}}(\infty ) & = & \displaystyle \frac{(\eta -1)[\,{\rho }_{\mathrm{AA}}(0)+{\rho }_{\mathrm{SS}}(0)+2{\rho }_{\mathrm{EE}}(0)+{\rho }_{\mathrm{AS}}(0)+{\rho }_{\mathrm{SA}}(0)]}{4\eta },\\ {\rho }_{\mathrm{GG}}(\infty ) & = & \displaystyle \frac{(\eta +1)[\,{\rho }_{\mathrm{AA}}(0)+{\rho }_{\mathrm{SS}}(0)+2{\rho }_{\mathrm{EE}}(0)+{\rho }_{\mathrm{AS}}(0)+{\rho }_{\mathrm{SA}}(0)-2\,]}{4\eta }.\end{array}\end{eqnarray}$
Since ρAA(∞) = ρSS(∞) and ρAS(∞) = ρSA(∞) = 0, K1(∞) is negative, thus the steady state cannot be entangled.
Case 3. When both of the two atoms are placed extremely close to the mirrors, then A1 = A2 = A3 = 0, and B1 = B2 = B3 = 0. From equation (13 ), it is obvious that the two-atom system does not evolve. So the initial entanglement can persist in the steady state. Note that this condition can also be satisfied when only one mirror is present [28].
Case 4. A similar case to Case 3 is when the transition wavelength is smaller than twice the distance between the two mirrors, i.e. ƛ < 2L. In this case, A1 = A2 = A3 = 0, and B1 = B2 = B3 = 0 are also satisfied, so there exists steady-state entanglement.
Case 5. When the transition wavelength is larger than one but smaller than or equal to twice the distance between the two mirrors, i.e. L < ƛ ≤ 2L, then ${A}_{3}^{2}={A}_{1}{A}_{2}$ and ${B}_{3}^{2}={B}_{1}{B}_{2}$, and the result depends on whether the environment is a vacuum or not.
If the environment is a vacuum, $\eta$ = 1, it can be calculated that
which can be positive depending on the initial state and the value of B1 and B2. As an example, if both the two atoms are initially in the excited state, ρEE(0) = 1, there exists steady-state entanglement when B1 ≠ B2, i.e. the distance between atom 1 and mirror 1 must be different from that between atom 2 and mirror 2. When B1 = B2, equation (26 ) can be simplified as K1(∞) = ρAA(0). That is, it is required that the initial state satisfies ρAA(0) ≠ 0 to obtain steady-state entanglement.
$\begin{eqnarray}\begin{array}{l}{K}_{1}(\infty )=\displaystyle \frac{\sqrt{{B}_{1}{B}_{2}}({B}_{1}+{B}_{2})+2{B}_{1}{B}_{2}}{{\left({B}_{1}+{B}_{2}\right)}^{2}}{\rho }_{\mathrm{AA}}(0)\\ \quad +\,\displaystyle \frac{\sqrt{{B}_{1}{B}_{2}}({B}_{1}+{B}_{2})-2{B}_{1}{B}_{2}}{{\left({B}_{1}+{B}_{2}\right)}^{2}}{\rho }_{\mathrm{SS}}(0)\\ \quad +\,\displaystyle \frac{2\sqrt{{B}_{1}{B}_{2}}{\left({B}_{1}-{B}_{2}\right)}^{2}}{{\left({B}_{1}+{B}_{2}\right)}^{3}}{\rho }_{\mathrm{EE}}(0)\\ \quad +\,\displaystyle \frac{\sqrt{{B}_{1}{B}_{2}}({B}_{2}-{B}_{1})}{{\left({B}_{1}+{B}_{2}\right)}^{2}}[{\rho }_{\mathrm{AS}}(0)+{\rho }_{\mathrm{SA}}(0)],\end{array}\end{eqnarray}$
which can be positive depending on the initial state and the value of B1 and B2. As an example, if both the two atoms are initially in the excited state, ρEE(0) = 1, there exists steady-state entanglement when B1 ≠ B2, i.e. the distance between atom 1 and mirror 1 must be different from that between atom 2 and mirror 2. When B1 = B2, equation (
If the environment is not a vacuum, i.e. $\eta$ ≠ 1, it can be shown that B1 = B2 = B3 is required to have Δ = 0 and ${B}_{3}^{2}={B}_{1}{B}_{2}$ satisfied simultaneously, i.e. the distance between atom 1 and mirror 1 must be the same as that between atom 2 and mirror 2. It is obvious that the result should be the same as that has been discussed in Case 1, and the steady state can be entangled.
5. Discussion and summary
We have investigated, in the framework of open quantum systems, the entanglement generation for a two-atom system immersed in a thermal bath in the presence of two parallel reflecting plane mirrors. We have shown that the presence of mirrors plays an important role in entanglement generation, in the sense that the ranges of temperature and interatomic separation within which entanglement can be generated are significantly changed compared with those in a free space. Remarkably, when the atomic transition wavelength is larger than twice the distance between the two mirrors, the atoms behave as if they were isolated from the environment, and the initial entanglement can persist in the steady state but no entanglement can be created if they are initially separable, no matter how the atoms are placed with respect to the mirrors and to each other. This is in stark contrast to the fact that in a free space, steady-state entanglement is possible only when the two atoms are placed extremely close to each other, while in the presence of one mirror, it is possible when the two atoms placed extremely close to the mirror.
It should be pointed out that we have only considered the case in which the two atoms are aligned vertically to the mirrors. The reason is twofold. First, for the parallel alignment case, terms such as $\tfrac{\sin (\omega \sqrt{{d}^{2}+{\left(2{nL}\right)}^{2}})}{\omega \sqrt{{d}^{2}+{\left(2{nL}\right)}^{2}}}$ will appear in the infinite summation in the two-point functions. This makes a mere analytical calculation of the two-point functions impossible, let alone the coefficients Ai and Bi which determine the entanglement generation. So, a study of such case would require numerical computations from at very early stages. Second, more importantly, the main features of the entanglement generation, which have been summarized above, do not depend on how the atoms are aligned with respect to the mirrors.
Nevertheless, there are some unique features of entanglement generation in the case of vertical alignment. The range of temperature within which entanglement can be created is wider in the two-mirror case compared with the cases of a free space and one mirror, when the atoms are placed close to one or both of the mirrors, and entanglement can both be generated and be maintained in the asymptotic regime, when the transition wavelength is larger than one but smaller than or equal to twice the distance between the two mirrors. It can be checked numerically that there are no such features in the case of parallel alignment.
Appendix. Calculations of the coefficients
In this appendix, we show the details of the calculations of the coefficients Ai and Bi of a two-atom system immersed in a thermal bath of quantum scalar fields in the presence of two perfectly reflecting plane mirrors.
The field correlation function can be obtained with the help of the method of images [42], which takes the following form,
where 1/$\beta$ is the temperature of the thermal bath.
$\begin{eqnarray}\begin{array}{l}G(x(\tau ),x(\tau ^{\prime} ))=-\displaystyle \frac{1}{4{\pi }^{2}}\\ \,\times \,\displaystyle \sum _{m=-\infty }^{\infty }\displaystyle \sum _{n=-\infty }^{\infty }\left[\displaystyle \frac{1}{{\left({\rm{\Delta }}\tau -{\rm{i}}m\beta -{\rm{i}}\epsilon \right)}^{2}-{\left({\rm{\Delta }}x\right)}^{2}-{\left({\rm{\Delta }}y\right)}^{2}-{\left({\rm{\Delta }}z-2{nL}\right)}^{2}}\right.\\ \quad \left.-\,\displaystyle \frac{1}{{\left({\rm{\Delta }}\tau -{\rm{i}}m\beta -{\rm{i}}\epsilon \right)}^{2}-{\left({\rm{\Delta }}x\right)}^{2}-{\left({\rm{\Delta }}y\right)}^{2}-{\left(z+z^{\prime} -2{nL}\right)}^{2}}\right],\end{array}\,\end{eqnarray}$
where 1/$\beta$ is the temperature of the thermal bath.
Allowing for the trajectories of the two atoms, ${x}_{1}^{\mu }=(\tau ,0,0,{d}_{1})$ and ${x}_{2}^{\mu }=(\tau ,0,0,{d}_{1}+d)$, the correlation functions can be written as
The Fourier transforms of the correlation functions above can be obtained as
In order to solve the infinite sum over n, we employ the following formulae [43],
with k being a positive integer. For convenience, we rewrite the distance between the two mirrors as
where $k\in {\mathbb{Z}}$ and 0 ≤ θ < 2π. When θ ≠ 0, the coefficients can be calculated as
where Γ0 = $\mu$2ω/2π being the spontaneous emission rate. When θ = 0, we have
$\begin{eqnarray}\begin{array}{l}{G}^{(11)}(x(\tau ),x(\tau ^{\prime} ))=-\displaystyle \frac{1}{4{\pi }^{2}}\\ \quad \times \,\displaystyle \sum _{m=-\infty }^{\infty }\displaystyle \sum _{n=-\infty }^{\infty }\left[\displaystyle \frac{1}{{\left({\rm{\Delta }}\tau -{\rm{i}}m\beta -{\rm{i}}\epsilon \right)}^{2}-{\left(-2{nL}\right)}^{2}}\right.\\ \quad \left.-\,\displaystyle \frac{1}{{\left({\rm{\Delta }}\tau -{\rm{i}}m\beta -{\rm{i}}\epsilon \right)}^{2}-{\left(2{d}_{1}-2{nL}\right)}^{2}}\right],\\ {G}^{(22)}(x(\tau ),x(\tau ^{\prime} ))=-\displaystyle \frac{1}{4{\pi }^{2}}\\ \quad \times \,\displaystyle \sum _{m=-\infty }^{\infty }\displaystyle \sum _{n=-\infty }^{\infty }\left[\displaystyle \frac{1}{{\left({\rm{\Delta }}\tau -{\rm{i}}m\beta -{\rm{i}}\epsilon \right)}^{2}-{\left(-2{nL}\right)}^{2}}\right.\\ \quad \left.-\,\displaystyle \frac{1}{{\left({\rm{\Delta }}\tau -{\rm{i}}m\beta -{\rm{i}}\epsilon \right)}^{2}-{\left(2{d}_{1}+2d-2{nL}\right)}^{2}}\right],\\ {G}^{(12)}(x(\tau ),x(\tau ^{\prime} ))=-\displaystyle \frac{1}{4{\pi }^{2}}\\ \quad \times \,\displaystyle \sum _{m=-\infty }^{\infty }\displaystyle \sum _{n=-\infty }^{\infty }\left[\displaystyle \frac{1}{{\left({\rm{\Delta }}\tau -{\rm{i}}m\beta -{\rm{i}}\epsilon \right)}^{2}-{\left(-d-2{nL}\right)}^{2}}\right.\\ \quad \left.-\,\displaystyle \frac{1}{{\left({\rm{\Delta }}\tau -{\rm{i}}m\beta -{\rm{i}}\epsilon \right)}^{2}-{\left(2{d}_{1}+d-2{nL}\right)}^{2}}\right],\\ {G}^{(21)}(x(\tau ),x(\tau ^{\prime} ))=-\displaystyle \frac{1}{4{\pi }^{2}}\\ \quad \times \,\displaystyle \sum _{m=-\infty }^{\infty }\displaystyle \sum _{n=-\infty }^{\infty }\left[\displaystyle \frac{1}{{\left({\rm{\Delta }}\tau -{\rm{i}}m\beta -{\rm{i}}\epsilon \right)}^{2}-{\left(d-2{nL}\right)}^{2}}\right.\\ \quad \left.-\,\displaystyle \frac{1}{{\left({\rm{\Delta }}\tau -{\rm{i}}m\beta -{\rm{i}}\epsilon \right)}^{2}-{\left(2{d}_{1}+d-2{nL}\right)}^{2}}\right].\end{array}\end{eqnarray}$
The Fourier transforms of the correlation functions above can be obtained as
$\begin{eqnarray}\begin{array}{l}{{ \mathcal G }}^{(11)}(\omega )=\displaystyle \frac{\omega }{2\pi }\displaystyle \frac{1}{1-{{\rm{e}}}^{-\beta \omega }}\\ \quad \times \,\displaystyle \sum _{n=-\infty }^{\infty }\left\{\displaystyle \frac{\sin (2{nL}\omega )}{2{nL}\omega }-\displaystyle \frac{\sin [(2{d}_{1}-2{nL})\omega ]}{(2{d}_{1}-2{nL})\omega }\right\},\\ {{ \mathcal G }}^{(22)}(\omega )=\displaystyle \frac{\omega }{2\pi }\displaystyle \frac{1}{1-{{\rm{e}}}^{-\beta \omega }}\\ \quad \times \,\displaystyle \sum _{n=-\infty }^{\infty }\left\{\displaystyle \frac{\sin (2{nL}\omega )}{2{nL}\omega }-\displaystyle \frac{\sin [(2{d}_{1}+2d-2{nL})\omega ]}{(2{d}_{1}+2d-2{nL})\omega }\right\},\\ {{ \mathcal G }}^{(21)}(\omega )=\displaystyle \frac{\omega }{2\pi }\displaystyle \frac{1}{1-{{\rm{e}}}^{-\beta \omega }}\\ \quad \times \,\displaystyle \sum _{n=-\infty }^{\infty }\left\{\displaystyle \frac{\sin [(d-2{nL})\omega ]}{(d-2{nL})\omega }-\displaystyle \frac{\sin [(2{d}_{1}+d-2{nL})\omega ]}{(2{d}_{1}+d-2{nL})\omega }\right\},\\ {{ \mathcal G }}^{(12)}(\omega )=\displaystyle \frac{\omega }{2\pi }\displaystyle \frac{1}{1-{{\rm{e}}}^{-\beta \omega }}\displaystyle \sum _{n=-\infty }^{\infty }\\ \quad \times \,\left\{\displaystyle \frac{\sin [(-d-2{nL})\omega ]}{(-d-2{nL})\omega }-\displaystyle \frac{\sin [(2{d}_{1}+d-2{nL})\omega ]}{(2{d}_{1}+d-2{nL})\omega }\right\}.\end{array}\end{eqnarray}$
In order to solve the infinite sum over n, we employ the following formulae [43],
$\begin{eqnarray}\begin{array}{l}\displaystyle \sum _{n=1}^{\infty }\displaystyle \frac{\cos ({nx})}{{n}^{2}-{\gamma }^{2}}=\displaystyle \frac{1}{2{\gamma }^{2}}-\displaystyle \frac{\pi \cos \{\gamma [(2k+1)\pi -x]\}}{2\gamma \sin (\gamma \pi )},\\ \quad (\gamma \notin {\mathbb{Z}},2k\pi \leqslant x\leqslant (2k+2)\pi )\\ \displaystyle \sum _{n=1}^{\infty }\displaystyle \frac{n\sin ({nx})}{{n}^{2}-{\gamma }^{2}}=\displaystyle \frac{\pi \sin \{\gamma [(2k+1)\pi -x]\}}{2\sin (\gamma \pi )},\\ \quad (\gamma \notin {\mathbb{Z}},2k\pi \lt x\lt (2k+2)\pi )\\ \displaystyle \sum _{n=1}^{\infty }\displaystyle \frac{\sin ({ny})}{n}=\displaystyle \frac{\pi -y}{2},\\ \quad (0\lt y\lt 2\pi )\end{array}\end{eqnarray}$
with k being a positive integer. For convenience, we rewrite the distance between the two mirrors as
$\begin{eqnarray}L=\displaystyle \frac{\pi }{\omega }\left(k+\displaystyle \frac{\theta }{2\pi }\right),\end{eqnarray}$
where $k\in {\mathbb{Z}}$ and 0 ≤ θ < 2π. When θ ≠ 0, the coefficients can be calculated as
$\begin{eqnarray}\begin{array}{l}{A}_{1}=\displaystyle \frac{{{\rm{\Gamma }}}_{0}}{8}\displaystyle \frac{\pi }{\omega L}\coth \left(\displaystyle \frac{\beta \omega }{2}\right)\\ \,\times \,\left\{1+2k-\displaystyle \frac{\sin [(2k+1)\tfrac{\pi {d}_{1}}{L}]}{\sin (\tfrac{\pi {d}_{1}}{L})}\right\},\\ {B}_{1}=\displaystyle \frac{{{\rm{\Gamma }}}_{0}}{8}\displaystyle \frac{\pi }{\omega L}\\ \qquad \times \,\left\{1+2k-\displaystyle \frac{\sin [(2k+1)\tfrac{\pi {d}_{1}}{L}]}{\sin (\tfrac{\pi {d}_{1}}{L})}\right\},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{A}_{2}=\displaystyle \frac{{{\rm{\Gamma }}}_{0}}{8}\displaystyle \frac{\pi }{\omega L}\coth \left(\displaystyle \frac{\beta \omega }{2}\right)\\ \,\times \,\left\{1+2k-\displaystyle \frac{\sin [(2k+1)\tfrac{\pi ({d}_{1}+d)}{L}]}{\sin [\tfrac{\pi ({d}_{1}+d)}{L}]}\right\},\\ {B}_{2}=\displaystyle \frac{{{\rm{\Gamma }}}_{0}}{8}\displaystyle \frac{\pi }{\omega L}\\ \,\times \,\left\{1+2k-\displaystyle \frac{\sin [(2k+1)\tfrac{\pi ({d}_{1}+d)}{L}]}{\sin [\tfrac{\pi ({d}_{1}+d)}{L}]}\right\},\\ {A}_{3}=\displaystyle \frac{{{\rm{\Gamma }}}_{0}}{8}\displaystyle \frac{\pi }{\omega L}\coth \left(\displaystyle \frac{\beta \omega }{2}\right)\\ \left\{\displaystyle \frac{\sin [(2k+1)\tfrac{\pi d}{2L}]}{\sin (\tfrac{\pi d}{2L})}-\displaystyle \frac{\sin [(2k+1)\tfrac{\pi (2{d}_{1}+d)}{2L}]}{\sin [\tfrac{\pi (2{d}_{1}+d)}{2L}]}\right\},\\ {B}_{3}=\displaystyle \frac{{{\rm{\Gamma }}}_{0}}{8}\displaystyle \frac{\pi }{\omega L}\\ \,\times \,\left\{\displaystyle \frac{\sin [(2k+1)\tfrac{\pi d}{2L}]}{\sin (\tfrac{\pi d}{2L})}-\displaystyle \frac{\sin [(2k+1)\tfrac{\pi (2{d}_{1}+d)}{2L}]}{\sin [\tfrac{\pi (2{d}_{1}+d)}{2L}]}\right\},\end{array}\end{eqnarray}$
where Γ0 = $\mu$2ω/2π being the spontaneous emission rate. When θ = 0, we have
$\begin{eqnarray}\begin{array}{l}{A}_{1}=\displaystyle \frac{{{\rm{\Gamma }}}_{0}}{4}\coth \left(\displaystyle \frac{\beta \omega }{2}\right)\\ \,\times \,\left\{1-\displaystyle \frac{\sin (2\omega {d}_{1})}{2k}\cot \left(\displaystyle \frac{\omega {d}_{1}}{k}\right)\right\},\\ {B}_{1}=\displaystyle \frac{{{\rm{\Gamma }}}_{0}}{4}\\ \qquad \times \,\left\{1-\displaystyle \frac{\sin (2\omega {d}_{1})}{2k}\cot \left(\displaystyle \frac{\omega {d}_{1}}{k}\right)\right\},\\ {A}_{2}=\displaystyle \frac{{{\rm{\Gamma }}}_{0}}{4}\coth \left(\displaystyle \frac{\beta \omega }{2}\right)\\ \qquad \times \,\left\{1-\displaystyle \frac{\sin [2\omega ({d}_{1}+d)]}{2k}\cot \left[\displaystyle \frac{\omega ({d}_{1}+d)}{k}\right]\right\},\\ {B}_{2}=\displaystyle \frac{{{\rm{\Gamma }}}_{0}}{4}\\ \qquad \times \,\left\{1-\displaystyle \frac{\sin [2\omega ({d}_{1}+d)]}{2k}\cot \left[\displaystyle \frac{\omega ({d}_{1}+d)}{k}\right]\right\},\\ {A}_{3}=\displaystyle \frac{{{\rm{\Gamma }}}_{0}}{4}\coth \left(\displaystyle \frac{\beta \omega }{2}\right)\left\{\displaystyle \frac{\sin (\omega d)}{2k}\cot \left(\displaystyle \frac{\omega d}{2k}\right)\right.\\ \qquad \left.-\,\displaystyle \frac{\sin [\omega (2{d}_{1}+d)]}{2k}\cot \left[\displaystyle \frac{\omega (2{d}_{1}+d)}{2k}\right]\right\},\\ {B}_{3}=\displaystyle \frac{{{\rm{\Gamma }}}_{0}}{4}\left\{\displaystyle \frac{\sin (\omega d)}{2k}\cot \left(\displaystyle \frac{\omega d}{2k}\right)\right.\\ \qquad \left.-\,\displaystyle \frac{\sin [\omega (2{d}_{1}+d)]}{2k}\cot \left[\displaystyle \frac{\omega (2{d}_{1}+d)}{2k}\right]\right\}.\end{array}\end{eqnarray}$
When the distance between the two mirrors L → ∞ , but the distance between atom 1 and the first mirror d1 and the interatomic separation d are finite, then
which reduces to the result in the case of a single mirror, see equations (37)-(42) in [27] in the limit a → 0. Furthermore, if d1 also approaches infinity, then
i.e. the result reduces to that in a free space [13].
$\begin{eqnarray}\begin{array}{rcl}{A}_{1} & = & \displaystyle \frac{{{\rm{\Gamma }}}_{0}}{4}\coth \left(\displaystyle \frac{\beta \omega }{2}\right)\left[1-\displaystyle \frac{\sin (2\omega {d}_{1})}{2\omega {d}_{1}}\right],\\ {A}_{2} & = & \displaystyle \frac{{{\rm{\Gamma }}}_{0}}{4}\coth \left(\displaystyle \frac{\beta \omega }{2}\right)\left[1-\displaystyle \frac{\sin (2\omega {d}_{1}+2\omega d)}{2\omega {d}_{1}+2\omega d}\right],\\ {A}_{3} & = & \displaystyle \frac{{{\rm{\Gamma }}}_{0}}{4}\coth \left(\displaystyle \frac{\beta \omega }{2}\right)\left[\displaystyle \frac{\sin (\omega d)}{\omega d}-\displaystyle \frac{\sin (2\omega {d}_{1}+\omega d)}{2\omega {d}_{1}+\omega d}\right],\\ {B}_{1} & = & \displaystyle \frac{{{\rm{\Gamma }}}_{0}}{4}\left[1-\displaystyle \frac{\sin (2\omega {d}_{1})}{2\omega {d}_{1}}\right],\\ {B}_{2} & = & \displaystyle \frac{{{\rm{\Gamma }}}_{0}}{4}\left[1-\displaystyle \frac{\sin (2\omega {d}_{1}+2\omega d)}{2\omega {d}_{1}+2\omega d}\right],\\ {B}_{3} & = & \displaystyle \frac{{{\rm{\Gamma }}}_{0}}{4}\left[\displaystyle \frac{\sin (\omega d)}{\omega d}-\displaystyle \frac{\sin (2\omega {d}_{1}+\omega d)}{2\omega {d}_{1}+\omega d}\right],\end{array}\end{eqnarray}$
which reduces to the result in the case of a single mirror, see equations (37)-(42) in [27] in the limit a → 0. Furthermore, if d1 also approaches infinity, then
$\begin{eqnarray}\begin{array}{rcl}{A}_{1} & = & {A}_{2}=\displaystyle \frac{{{\rm{\Gamma }}}_{0}}{4}\coth \left(\displaystyle \frac{\beta \omega }{2}\right),\\ {A}_{3} & = & \displaystyle \frac{{{\rm{\Gamma }}}_{0}}{4}\displaystyle \frac{\sin (\omega d)}{\omega d}\coth \left(\displaystyle \frac{\beta \omega }{2}\right),\\ {B}_{1} & = & {B}_{2}=\displaystyle \frac{{{\rm{\Gamma }}}_{0}}{4},\\ {B}_{3} & = & \displaystyle \frac{{{\rm{\Gamma }}}_{0}}{4}\displaystyle \frac{\sin (\omega d)}{\omega d},\end{array}\end{eqnarray}$
i.e. the result reduces to that in a free space [13].