Show that if the Lindblad satisfy $\sum_\mu L_\mu L_\mu^\dagger=\sum_\mu L_\mu^\dagger L_\mu$ then $\rho\propto I$ is a fixed point of an evolution

Question

How can we show that the Lindblad condition: $$\sum_{\mu}L_{\mu} L_{\mu}^{\dagger} = \sum_{\mu} L_{\mu}^{\dagger}L_{\mu},\tag{1}$$ implies that $\rho \propto I$ is the fixed point of the evolution with the maximum entropy (this corresponds to the microcanonical distribution)?

I have worked with the basis which is going to diagonalize $\rho$ and also I have taken the necessary condition where von Neumann entropy has been increased monotonically but how to proceed the next step I am not getting.

This is a follow-up to: Show that if the Lindblad satisfy $\sum_\mu L_\mu L_\mu^\dagger=\sum_\mu L_\mu^\dagger L_\mu$ then the von Neumann entropy increases monotonically.

Answer 1

Go back to the Lindblad master equation: $$ \frac{d\rho}{dt}=i[H,\rho]+\sum_nL_n^\dagger\rho L_n-\frac12\sum_nL_nL_n^\dagger \rho-\frac12\sum_n\rho L_nL_n^\dagger. $$ The statement that the maximally mixed state is a fixed point is equivalent to saying that if $\rho=I/d$ then $\frac{d\rho}{dt}=0$. So, that's check that. \begin{align*} d\frac{d\rho}{dt}&=i[H,I]+\sum_nL_n^\dagger L_n-\frac12\sum_nL_nL_n^\dagger-\frac12\sum_n L_nL_n^\dagger. \\ &=\sum_nL_n^\dagger L_n-\sum_nL_nL_n^\dagger \\ &=0 \end{align*} where the last line follows from the assumption about the form of the operators.