# How to get Heisenberg equation of motion? [closed]

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))

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

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• $d_t A = \frac i \hbar [H, A] + \partial_t A$ – Sebastian Riese Jul 21 '15 at 11:34
• 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? – ACuriousMind Oct 25 '15 at 14:29

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 $$\frac{d \hat{A}}{dt} = \frac{i}{\hbar} [\hat{H}, \hat{A}] .$$ 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 $\hat{\mathcal{E}}$: $$\frac{d\hat{\mathcal{E}}}{dt} = \frac{i}{\hbar} [\hbar \Delta \hat{\sigma}_{ee} - (\hbar \Omega(t) \hat{\sigma}_{es} + \hbar g \hat{\mathcal{E}} \hat{\sigma}_{eg} + \text{H.c.}), \hat{\mathcal{E}}] = - i g \hat{\sigma}_{ge} [\hat{\mathcal{E}}^\dagger, \hat{\mathcal{E}}] = i g \hat{\sigma}_{ge} .$$ 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.