# Deriving coupled equations of motion using optical master equation for atom-plasmon system

I am learning the interaction between surface plasmons (which are bosons that act much like photons) and atoms.

I came across the following scenario describing the interaction between a two level atom and a single surface plasmon mode in a metal nano-particle under the classical driving field $E_i = E_0e^{-i\omega t} + c.c$ , with the following Hamiltonian and master equation:

(Let $\textbf{a}$ and $\textbf{a}^\dagger$ be the annihilation and creation operators of the surface plasmon mode, $\sigma$ and $\sigma^\dagger$ be the lowering and raising operators of the atom and $\omega_{sp}$ and $\omega_x$ be the plasmon and atomic resonance frequencies respectively. $\rho$ is the density matrix of the combined system. Coupling g and nanoparticle property $\chi$ are to be determined)

\begin{align}H_0 &= \hbar\omega_{sp}\textbf{a}^\dagger\textbf{a} + \hbar\omega_{x}\sigma^\dagger\sigma\\H_\textrm{int} &= i\hbar g\left(\textbf{a}^\dagger \sigma-\textbf{a} \sigma^\dagger\right)\\ H_\textrm{drive} &=-~E_o\left(\chi \textbf{a}^\dagger+\chi^* \textbf{a}\right)-\mu E_0 \left(\sigma ^\dagger+\sigma\right)\\ H_S &=H_0+H_\textrm{int}+H_\textrm{drive}\end{align}

The Markovian interaction with reservoirs determining the decay rates $\gamma_x$ and $\gamma_{sp}$ for the atom and surface plasmon mode respectively are given by the following Liouvillian terms;

\begin{align}\mathscr{L}_{sp} &= \frac{\gamma_{sp}}{2} \left(2 \textbf{a}\rho\textbf{a}^\dagger - \textbf{a}^\dagger\textbf{a}\rho - \rho\textbf{a}^\dagger\textbf{a}\right)\\ \mathscr{L}_{x} &= \frac{\gamma_{x}}{2} \left(2 \sigma\rho\sigma^\dagger - \sigma^\dagger\sigma\rho - \rho\sigma^\dagger \sigma\right)\end{align}

The master equation for the density operator reads: $$\dot{\rho}=\frac{\mathrm i}{\hbar}[\rho,H_s]+\mathscr{L}_{sp}+\mathscr{L}_{x}$$

My questions:

1) Assuming all the above equations are in Schrödinger picture (please correct me if I'm wrong) what assumptions and properties should I use to arrive at an equation of motion of the following form (where $\langle\sigma\rangle = \textrm{Tr}[\sigma\rho]$)? Even the slightest guidance or reference would be a big help. $$\frac{\mathrm d}{\mathrm dt} \langle\sigma\rangle = -\left[\mathrm i(\omega_x-\omega)+\frac{\gamma_x}{2}\right]\langle \sigma \rangle - g \langle \textbf{a} \rangle + 2g \langle \textbf{a} \sigma^\dagger\sigma \rangle + \frac{\mathrm i\mu E_0}{\hbar}\langle\left(1-2\langle \sigma^\dagger\sigma\rangle\right)$$

2) It seemed to me that $H_\textrm{drive}$ is due to the interaction of the atom and plasmon as separate dipoles with the classical driving field. In this case, should $E_0$ in $H_\textrm{drive}$ be replaced by $E_i$?

• Have you tried simply using the master eq., as in $Tr[\sigma{\dot \rho}] = Tr\left[\sigma\left(\frac{i}{\hbar}[\rho, H_s] + {\mathcal L}_{sp} + {\mathcal L}_x\right)\right]$? Looks like all the terms in your EOM would be accounted for. As for question (2), I think your Hamiltonian was constructed using $E_i$ and a rotating frame approximation that removed the time-dependent exponentials. – udrv Oct 28 '16 at 2:04
• Thanks a lot for your reply. I assumed the calculations are done in Schrodinger picture looking at the master equation. Please let me know if it's correct to assume so. Assuming $\sigma$ is not time varying, I tried the master equation approach suggested. I am not clear of how to get an $\omega$ term in the result. Also, are there any other property specific to a and $\sigma$ that you could suggest to be used in the simplification other than the cyclic property of trace and commutator relations? (Can we use $\frac{d}{dt}\langle\sigma\rangle=Tr[\sigma\dot{\rho}]$ for time varying $\sigma$ too?) – Toshi_H Oct 28 '16 at 5:58
• Welcome. The calculations seem to be in the Schroedinger picture, although if a rotating frame transformation was previously applied, you may have an interaction picture with the $\omega$ time-dependence transferred onto the $\sigma$-s. In this case you must account for it in the time-derivative of the average: $\frac{d}{dt}\langle \sigma\rangle = Tr[\sigma{\dot\rho} + {\dot\sigma} \rho]$ (so no, can't use simply $\frac{d}{dt}\langle \sigma\rangle = Tr[\sigma{\dot\rho}]$ when $\sigma$ is time-dependent). The rest is indeed a matter of cyclic permutations under trace and commutation relations. – udrv Oct 28 '16 at 15:50