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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 $[\mathcal{H}_I, \mathcal{L}]$.

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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.

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  • $\begingroup$ 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. $\endgroup$ – Andris Erglis Apr 1 at 9:34

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