When the evolution of the system is not unitary, one can describe this evolution by using the Master equation, wich contains the quantum jump operators (called also the Lindblad operators).

The expression of the Lindblad operators clearly depends on the rate $K$ of transition and the operator $O$, which assures this transition:

$L_u= \sqrt{[K]}\hat{O}$

I found this expression of the Lindblad operator always the same, but in the case of an cavity with loss operator $K$ wich interacts with a bath of $n$ number of photons, the expression is changed to:

$L_u= \sqrt{[K(n+1)]}\hat{O}$


$L_u^{+}= \sqrt{[K(n)]}\hat{O}$

How can one explain the $(n+1)$ and $(n)$ in the last expressions?


1 Answer 1


If $\hat{O}$ causes a transition $|0\rangle \rightarrow |1\rangle$, a photon from the bath is used up. So if you know have $n$ photons, when the transition happened there were $n+1$.

$n$ is the number of photons present in the bath at any one time. The rate $K$ depends on the number of photons.


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