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I am reading a paper https://journals.aps.org/prb/pdf/10.1103/PhysRevB.96.224302. In this paper the initial state of the system and environment is given as \begin{equation} |\Psi(0)\rangle=|\phi_{s}(0)\rangle\otimes|\Phi_{env}(0)\rangle \end{equation} The system state is given by $\phi_{s}(0)=c_{g}|0\rangle+c_{e}|1\rangle$. The total Hamiltonian is given by

\begin{equation} H=H_{env}+H_{s}+H_{s,env} \end{equation} The system Hamiltonian is given by $H_{s}=\omega_{e}|e\rangle\langle e|$ and the $H_{s,env}=-\delta|e\rangle\langle e|\hat{O}$, here $\hat{O}$ is some local operator. Given that $[H_{s},H_{s,env}]=0$, The authors write the evolved state as

\begin{equation} |\Psi(t)\rangle=c_{g}|g\rangle\otimes\exp[-iH_{env}t]|\Phi(0)\rangle+\exp[-i\omega_{e}t]c_{e}|e\rangle \otimes\exp[-iH^{(\delta)}t]|\Phi_{env}\rangle. \end{equation} Where $H^{(\delta)}$ is the perturbed Hamiltonian. I want to reproduce the expression for $|\Psi(t)\rangle$. Do I need to go to the interaction picture?

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$\left|\Psi(t)\right>$ is just a state in the Schrödinger picture (by the way, I think that by the initial system state you meant $\left|\phi_s(0)\right> = c_g\left|g\right> + c_e\left|e\right>$).

To obtain $\left|\Psi(t)\right>$ just apply the time evolution operator (supposing that the Hamiltonian does not depend on time). Because $[H_s,H_{s,env}] = 0$ both $\left|g\right>$ and $\left|e\right>$ must be eigenstates of $H_{s,env}$ (moreover note that $H_{s,env}$ you gave should act only on the system part in order to be well defined). This means that $\left<e\right|\hat{O} \left|g\right>= 0$. Then: \begin{eqnarray} \left|\Psi(t)\right> = \exp{(-itH)}\left|\Psi(0)\right> = \left(\exp{(-itH_s-itH_{s,env})}\left|\phi_s(0)\right>\right)\otimes \left(\exp{(-itH_{env})}\left|\Phi_{env}(0)\right>\right) = \\ = \left(\exp{(-it0-it0)}c_g \left|g\right> + \exp{(-it\omega_e+it\delta_eo_e)}c_e \left|e\right>\right)\otimes\left(\exp(-itH_{env})\left|\Phi_{env}(0)\right>\right)= \\ = c_g \left|g\right>\otimes\exp(-itH_{env})\left|\Phi_{env}(0)\right> + \exp{(-it\omega_e)}c_e \left|e\right> \otimes\left(\exp(-itH_{\delta}\left|\Phi_{env}(0)\right>\right) \end{eqnarray} where in the second line I wrote $o_e = \left<e\right|\hat{O} \left|e\right>$ and in the third line I wrote $H_\delta = H_{env} -\delta_e o_e 1$.

Usually, $H_{s,env}$ acts on the environment part as well (because the one you wrote is just a perturbation of the Hamiltonian of the system, and does not include any interaction with the environment). The more realistic case is: $$H_{s,env} = -\delta \left(\left|e\right>\left< e\right|\hat{O}_s\ \otimes \hat{O}_{env}\right) $$ which I suspect is somehow what the authors meant. In this case the only difference is that $$H_\delta = H_{env} -\delta_e o_e \hat{O}_{env}$$ where again $o_e = \left<e\right|\hat{O}_s \left|e\right> $

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  • $\begingroup$ In $H_{s,env}=-\delta|e\rangle\langle e|_{s}\hat{O}_{env}$, the system and environment parts are written in subscript. $\endgroup$
    – user123
    Commented Jan 25, 2018 at 10:42
  • $\begingroup$ Is there any assumption in decomposing the $\exp[-iHt]$ into $\exp[-itH_{s}-itH_{s,env}]\otimes \exp[-itH_{env}]$? $\endgroup$
    – user123
    Commented Jan 25, 2018 at 10:45
  • $\begingroup$ The assumption is that $H_{s,env}$ does not act on the environment, which is the case for $H_{s,env}$ you wrote. In the more general case I wrote after, that decomposition is not valid, but you could write it the same way with $H_{s,env}$ replaced only by the left part $-\delta \left| e\right>\left<e\right| \hat{O}_s$ and adding a second $H_{s,env}$ on the second part of the tensor product with the right part $-\delta \hat{O}_{env}$. This can be done because $H_{s,env}$ is written as a tensor product. For other cases where $H_{s,env}$ cannot be written this way it is not valid. $\endgroup$
    – amrgoloh
    Commented Jan 25, 2018 at 11:21

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