# How to find the Heisenberg picture from the master equation without Lindblad structure?

My question is closely related to this post and this one too. I understand that with a Lindbladian type master equation, it is possible to find the differential equation for an observable. However, my master equation looks like this: $$\frac{d}{dt}\rho(t) = -\frac{i}{\hbar} [H,\rho(t)]-f(t)[q,[q,\rho(t)]]+g(t)[q,[p,\rho(t)]].$$ Firstly, does the argument from the answer in the first linked post hold here? I'm not too sure if the caveats raised in the comments apply here.

Secondly, if we describe the above as $$\frac{d}{dt}\rho(t) = \mathcal{L}[\rho(t)]$$, then how do I find $$\mathcal{L}^{\dagger}$$, so as to find the Heisenberg picture? Of course I'd know how to find $$(\mathcal{L}[\rho(t)])^{\dagger}$$, but I'm not clear on how to get the adjoint of the superoperator itself.

## 1 Answer

I have been doing similar calculations with non-Lindbladian dissipators, except with time independent coefficients. In that case it seems that by simply using the cyclical nature of the trace is perfectly reasonable. That is $\operatorname{tr}(A[q,[p,\rho]]) = \operatorname{tr}([[A,q],p]\rho)$, so $[[A,q],p]$ would be the adjoint form for the final term in your master equation.

• Doesn't this reasoning somehow assume the only thing of interest in all of the equations is the trace of the operators? Mar 30 at 11:05