How can I show that the Liouvillian superoperator $\mathcal{L}$ satisfies $$ \langle A,\mathcal{L}B\rangle=\langle \mathcal{L}^\dagger A,B\rangle $$ where $\langle A,B\rangle =\mathrm{Tr}(A^\dagger B)$ is the Hilber-Schmidt product, and $A$ and $B$ are two operators. In other words, how can I prove that $$ \mathrm{Tr}(A^\dagger \mathcal{L} B)=\mathrm{Tr}((\mathcal{L}^\dagger A^\dagger)B) $$
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sry, the one in the body – Turbotanten May 04 '23 at 05:52
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7Isn’t this how the Hermitian adjoint of an operator is defined? – Ghoster May 04 '23 at 06:27
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What Ghoster said + see https://physics.stackexchange.com/questions/280398/lindblad-equation-for-heisenberg-operators – Wolpertinger May 17 '23 at 13:34