# Superoperator Commutation Relations

Let's say we are given a generic Master equation:

$$\dot{\rho} = -\frac{i}{\hbar}[H_0 + H_I, \rho] - \mathcal{L}\rho$$

where $$H_0$$ is unperturbed Hamiltonian and $$H_I$$ is the interaction Hamiltonian, and $$\mathcal{L}$$ is the Lindblad operator. We can write this in an alternative way by using superoperators:

$$\dot{\rho} = -(\mathcal{H}_0 \rho + \mathcal{H}_I \rho) - \mathcal{L}\rho$$

where $$\mathcal{H}_0=\frac{i}{\hbar}[H_0 , \cdot]$$ and $$\mathcal{H}_I=\frac{i}{\hbar}[H_I , \cdot]$$.

What my question is - if I want to calculate commutation relations of these superoperators alone, how one could proceed? For example, I would like to calculate $$[\mathcal{H}_0, \mathcal{H}_I]$$, $$[\mathcal{H}_0, \mathcal{L}]$$ and $$[\mathcal{H}_I, \mathcal{L}]$$.

One option is to just use the Jacobi identity and act on some state $$\rho$$:

$$[{\cal A},{\cal B}]\rho = [A,[B,\rho]]-[B,[A,\rho]]=[A,[B,\rho]+[B,[\rho,A]]$$

then by Jacobi we have:

$$[{\cal A},{\cal B}]\rho = -[\rho,[A,B]] = [[A,B],\rho]$$

so $$[{\cal A},{\cal B}] = [[A,B],\cdot]$$. This answers your question about $${\cal H}_0, {\cal H}_I$$. I don't think any answer will be forthcoming about $${\cal L}$$ which is not of this form.

If you assume Markovianity and so $${\cal L}$$ is in Lindblad form then you might be able to make some progress in terms of commutators/anti-commutators of $$H$$ and the jump operators $$L$$ entering the Lindbladian.

• Thanks for the answer! About Lindblad terms - in my case they indeed contain jump operators derived using Markov approximation. Actually my problem is to find this guy $e^{(\mathcal{H}_0+\mathcal{L})t} \mathcal{H}_I e^{-(\mathcal{H}_0+\mathcal{L})t}$ which is in the interaction picture. I will then try to evaluate that Lindbladian commutator. – Andris Erglis Apr 1 at 9:34