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How can we show that when the Lindblad operators satisfy the condition:

$$\sum_{\mu}L_{\mu} L_{\mu}^{\dagger} = \sum_{\mu} L_{\mu}^{\dagger}L_{\mu},\tag{1}$$

the master equation evolution monotonically increases the von Neumann entropy. When measurements are made in the basis in which $\rho$ is diagonal, the von Neumann entropy coincides with the Boltzmann-Shannon entropy.

I have worked with the basis which is going to diagonalise $\rho$ 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.

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I'll show you how to do it by brute force, since this will demonstrate a lot of techniques that will be useful for you if you have to derive something more complicated.

The Lindblad evolution:

$$ \frac{\textrm{d}\rho}{\textrm{d}t} = -\frac{\textrm{i}}{\hbar}[H,\rho] + \sum_{\mu,\nu} h_{\mu,\nu} \left( L_\mu \rho L_\nu^\dagger -\frac{1}{2}\left\{ L^\dagger_\nu L_\mu ,\rho \right\}\right), \tag{1} $$

can be re-written in diagonal form:

$$\tag{2} \frac{\textrm{d}\rho}{\textrm{d}t} = -\frac{\textrm{i}}{\hbar}[H,\rho] + \sum_{\mu} \gamma_{\mu} \left( L_\mu \rho L_\mu^\dagger -\frac{1}{2}\left\{ L^\dagger_\mu L_\mu ,\rho \right\}\right), $$

and in the special case in which we have:

$$\tag{3} \sum_\mu L_\mu L_\mu^\dagger = \sum_\mu L_\mu^\dagger L_\mu, $$

we get:

$$\tag{4} \frac{\textrm{d}\rho}{\textrm{d}t} = -\frac{\textrm{i}}{\hbar}[H,\rho] + \sum_{\mu} \gamma_{\mu} \left( L_\mu \rho L_\mu^\dagger -\frac{1}{2}\left( L^\dagger_\mu L_\mu\rho + \rho L_\mu L^\dagger_\mu \right)\right). $$

Let us now look at the von Neumann entropy:

$$ S = - \textrm{Tr}\left( \rho \ln \rho \right),\tag{5} $$

and it's rate of change:

\begin{align} \frac{\textrm{d}S}{\textrm{d}t} &= - \textrm{Tr}\left( \frac{\textrm{d}\rho}{\textrm{d}t} \ln \rho + \rho \frac{\textrm{d}\ln\rho}{\textrm{d}t}\right)\tag{6}\\ &= - \textrm{Tr}\left( \frac{\textrm{d}\rho}{\textrm{d}t} \ln \rho + \rho \frac{1}{\rho}\frac{\textrm{d}\rho}{\textrm{d}t}\right),\tag{7}\\ &= - \textrm{Tr}\left( \frac{\textrm{d}\rho}{\textrm{d}t} \ln \rho + \frac{\textrm{d}\rho}{\textrm{d}t}\right),\tag{8}\\ &=- \textrm{Tr}\left( \frac{\textrm{d}\rho}{\textrm{d}t} \ln \rho\right) - \textrm{Tr} \left(\frac{\textrm{d}\rho}{\textrm{d}t} \right).\tag{9}\\ \end{align}

Let me start the right-most term for you:

\begin{align} \textrm{Tr} \left(\frac{\textrm{d}\rho}{\textrm{d}t} \right) & = \textrm{Tr} \left( -\frac{\textrm{i}}{\hbar}[H,\rho] + \sum_{\mu} \gamma_{\mu} \left( L_\mu \rho L_\mu^\dagger -\frac{1}{2}\left( L^\dagger_\mu L_\mu\rho + \rho L_\mu L^\dagger_\mu \right)\right)\right)\tag{10}\\ & = \sum_{\mu} \gamma_{\mu}\textrm{Tr} \left( L_\mu \rho L_\mu^\dagger -\frac{1}{2}\left( L^\dagger_\mu L_\mu\rho + \rho L_\mu L^\dagger_\mu \right)\right)\tag{11}\\ & = \sum_{\mu} \gamma_{\mu} \left( \textrm{Tr}\left(L_\mu \rho L_\mu^\dagger\right) -\frac{1}{2}\textrm{Tr}\left( L^\dagger_\mu L_\mu\rho \right) - \frac{1}{2}\textrm{Tr}\left(\rho L_\mu L^\dagger_\mu \right)\right)\tag{12}\\ &=\sum_{\mu} \gamma_{\mu} \left( \textrm{Tr}\left(L_\mu^\dagger L_\mu \rho \right) -\frac{1}{2}\textrm{Tr}\left( L^\dagger_\mu L_\mu\rho \right) - \frac{1}{2}\textrm{Tr}\left( L^\dagger_\mu L_\mu\rho \right)\right)\tag{13}\\ & = 0. \tag{14} \end{align}

I used the cyclic property of the trace operator, and the fact that its linear, in the last set of equations.

We now have:

$$ \frac{\textrm{d}S}{\textrm{d}t} = - \textrm{Tr}\left( \frac{\textrm{d}\rho}{\textrm{d}t} \ln \rho\right) .\tag{15}\\ $$

To do this we have to examine various pieces:

\begin{align} &\textrm{Tr} \left( [H,\rho]\ln \rho \right)\tag{16}\\ =& \textrm{Tr} \left( H\rho \ln \rho - \rho H \ln \rho\right) \tag{17} \\ = &0. \end{align}

That works because we have three operators in the argument of the trace operator, and they are all Hermitian (this wouldn't necessarily work if we had the trace of a product of 4 or more Hermitian operators).

Let us now look at another piece:

\begin{align} &-\frac{1}{2}\textrm{Tr} \left( L^\dagger L \rho \ln \rho \right) -\frac{1}{2}\textrm{Tr} \left( \rho L^\dagger L \ln \rho \right) \tag{19}\\ =& -\textrm{Tr} \left( L^\dagger L \rho \ln \rho \right). \end{align}

The above result comes from using the cyclic property of the trace for the first term and the ability to equate arbitrary permutations for cases in which the argument of the trace operator is the product of three Hermitian arguments (treating $L^\dagger L$ as a single Hermitian operator).

So now we're left with:

$$ \frac{\textrm{d}S}{\textrm{d}t} = -\sum_\mu \gamma_\mu\left( \textrm{Tr}\left( L_\mu \rho L_\mu^\dagger \ln \rho \right) - \textrm{Tr}\left( L_\mu^\dagger L_\mu \rho \ln \rho \right)\right) .\tag{20}\\ $$

That would be 0 if $L_\mu^\dagger = L_\mu^{-1}$ but can you otherwise show that $\frac{\textrm{d}S}{\textrm{d}t}$ never changes sign (i.e. that it's either monotonically increasing or monotonically decreasing)?

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  • $\begingroup$ How can we show with this condition, $\rho \propto I$ is the fixed point of the evolution with the maximum entropy. (This corresponds to the microcanonical distribution.) $\endgroup$ Nov 28 '21 at 13:00
  • $\begingroup$ That would have to be asked as a separate question, since comments are for discussing answers. $\endgroup$ Nov 28 '21 at 16:41

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