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Final Form of Master Equation

Using these definitions, we can now write
$\displaystyle \frac{d}{dt}\rho(t)$ $\displaystyle =$ $\displaystyle -i[\Omega a^{\dagger}a,{\rho}(t)] %%\nonumber\\
-\frac{1}{2}\Big...
...2 i \Delta_+)a^{\dagger} a
+ (\gamma + 2i \Delta) a a^{\dagger}\right]{\rho}(t)$ (46)
  $\displaystyle +$ $\displaystyle {\rho}(t)
\left[ (\gamma - 2i \Delta) a a^{\dagger} +
(\gamma_+ -...
...mber\\
-2\gamma_+
a{\rho}(t)a^{\dagger}
-2\gamma a^{\dagger}{\rho}(t)a
\Big\}.$  

We write $ 2i\Delta aa^{\dagger} = 2i\Delta (a^{\dagger}a+1)$ and obtain
$\displaystyle \frac{d}{dt}\rho(t)$ $\displaystyle =$ $\displaystyle -i[(\Omega +\Delta_++\Delta) a^{\dagger}a,{\rho}(t)]$ (47)
  $\displaystyle -$ $\displaystyle \frac{1}{2}\gamma_+ \Big\{ a^{\dagger} a {\rho}(t)
+ {\rho}(t) a^...
...a^{\dagger}{\rho}(t)
+ {\rho}(t) a a^{\dagger} -2 a^{\dagger}{\rho}(t)a
\Big\}.$  

This can be further re-arranged into
$\displaystyle \fbox{$ \begin{array}{rcl} \displaystyle \frac{d}{dt}\rho(t) &=& ...
... a a^{\dagger} - a \rho a^{\dagger} -a^{\dagger} \rho a \Big\},
\end{array}$\ }$     (48)

where
$\displaystyle \bar{\Omega}$ $\displaystyle \equiv$ $\displaystyle \Omega + P\int_{0}^{\infty}d\omega\frac{\rho(\omega)}{\Omega-\omega},
\quad \kappa\equiv \pi \rho(\Omega).$ (49)

Remarks


next up previous contents index
Next: Expectation Values (RWA Model) Up: Master Equation II: the Previous: Rates and Energy Shift   Contents   Index
Tobias Brandes 2004-02-18