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A system Hamiltonian is given by

$$ H=\hbar\omega_{1}\hat{a}^{\dagger}\hat{a}+\Sigma_{i=1}^{N}\left(\hbar\omega_{se}\hat{\sigma}_{ss}^{i}+\hbar\omega_{ge}\hat{\sigma}_{ee}^{i}\right)-\hbar\Sigma_{i=1}^{N}\left[\hat{a}ge^{i\omega_{1}z_{i}/c}\hat{\sigma}_{eg}^{i}+\Omega e^{-i\omega_{2}(t-z_{i}/c)}\hat{\sigma}_{es}^{i}\right]+\textrm{H.c.} $$

After transformation, we got the effective rotating frame Hamiltonian,

$$\tilde{H}=\hbar\Delta\hat{\sigma}_{ee}-\left(\hbar\Omega\hat{\sigma}_{es}+\hbar g\hat{\varepsilon}\hat{\sigma}_{eg}+\textrm{H.c.}\right)$$

And from this result, how can I derive the Heisenberg Equations of Motion?

Note: The Hamiltonian is given in this paper, Photon Storage in $\Lambda$-type Optically Dense Atomic Media. I. Cavity Mode (Physical Review A 76, 033804 (2007))

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closed as unclear what you're asking by ACuriousMind, Kyle Kanos, Ryan Unger, Sebastian Riese, Martin Oct 27 '15 at 15:59

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ $d_t A = \frac i \hbar [H, A] + \partial_t A$ $\endgroup$ – Sebastian Riese Jul 21 '15 at 11:34
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    $\begingroup$ It's not clear to me what you're asking. The Heisenberg equations of motion do not need to be derived for any particular system, they are always the same (just the Hamiltonian differs). What are you trying to do? $\endgroup$ – ACuriousMind Oct 25 '15 at 14:29
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The Heisenberg equation of motion for a general system operator $\hat{A}$ (that has no explicit dependence upon $t$) where the full system is specified by the Hamiltonian $\hat{H}$ is \begin{equation} \frac{d \hat{A}}{dt} = \frac{i}{\hbar} [\hat{H}, \hat{A}] . \end{equation} For example (from the paper you referenced), let's take the Hamiltonian $\hat{\tilde{H}} = \hbar \Delta \hat{\sigma}_{ee} - (\hbar \Omega(t) \hat{\sigma}_{es} + \hbar g \hat{\mathcal{E}} \hat{\sigma}_{eg} + \text{H.c.})$ and calculate the Heisenberg equation of motion for $\hat{\mathcal{E}}$: \begin{equation} \frac{d\hat{\mathcal{E}}}{dt} = \frac{i}{\hbar} [\hbar \Delta \hat{\sigma}_{ee} - (\hbar \Omega(t) \hat{\sigma}_{es} + \hbar g \hat{\mathcal{E}} \hat{\sigma}_{eg} + \text{H.c.}), \hat{\mathcal{E}}] = - i g \hat{\sigma}_{ge} [\hat{\mathcal{E}}^\dagger, \hat{\mathcal{E}}] = i g \hat{\sigma}_{ge} . \end{equation} Note that, in the paper you see the extra terms $- \kappa \hat{\mathcal{E}}$ and $\sqrt{2 \kappa} \hat{\mathcal{E}}_{\text{in}}$ in the equation of motion for $\hat{\mathcal{E}}$; these terms are due to a coupling with a bath that has been traced out.

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