Let us compute the following quantity:

\[\mathbb{E}\left [S\left (\boldsymbol \rho \| \boldsymbol \sigma\right )\right ]\]

this could in principle be computed as

\[\mathbb{E}\left [S\left (\boldsymbol \rho \| \boldsymbol \sigma\right )\right ] = \frac{1}{|G|} \sum \limits_{g \in G} S\left (\boldsymbol \rho \| \boldsymbol \sigma\left (\mathbf{L}_g\right )\right )\]

Let us define:

\[\boldsymbol \sigma\left (\mathbf{X}\right ) = \frac{e^{-\beta \mathbf{X}}}{\mathrm{Tr} \left \lbrack e^{-\beta \mathbf{X}} \right \rbrack}\]

\mathbf{I}f Jensen inequality holds here (and it should by the transfer rule for positive definite matrices) then we have:

\[\mathbb{E}\left \lbrack\boldsymbol \sigma\left (\mathbf{L}\right )\right \rbrack \geq \boldsymbol \sigma \left (\mathbb{E}\left \lbrack \mathbf{L}\right \rbrack\right )\]

where the inequality is considered on the spectral norm, i.e. the largest eigenvalue.

Expand it and exploit the positivity of relative entropy:

\[\mathrm{Tr} \left \lbrack \boldsymbol \rho \log \boldsymbol \rho \right \rbrack \geq \frac{1}{|G|} \sum \limits_{g \in G} \mathrm{Tr} \left \lbrack\boldsymbol \rho \log\left( \boldsymbol \sigma\left (\mathbf{L}_g\right ) \right) \right \rbrack\]

the right hand side can be written as:

\[\frac{1}{|G|} \sum \limits_{g \in G} \mathrm{Tr} \left \lbrack\boldsymbol \rho \log\left( \boldsymbol \sigma\left (\mathbf{L}_g\right ) \right) \right \rbrack = \mathrm{Tr} \left \lbrack \boldsymbol \rho \mathbb{E}\left \lbrack \log \boldsymbol \sigma\left (\mathbf{L}\right ) \right \rbrack \right \rbrack\]

because \(\boldsymbol \rho\) is a constant matrix and trace and expectations commute.

Now we would like to find some bound on the term \(\mathbb{E}\left \lbrack \log \boldsymbol \sigma\left (\mathbf{L}\right ) \right \rbrack\). For positive definite matrices is simple to show that:

\[\mathbb{E} \left \lbrack \log \left (\boldsymbol \sigma\left (\mathbf{L}\right )\right ) \right \rbrack = \mathbb{E} \left \lbrack \log \frac{e^{-\beta \mathbf{L}}}{\mathrm{Tr} \left \lbrack e^{-\beta \mathbf{L}} \right \rbrack} \right \rbrack = \mathbb{E} \left \lbrack -\beta \mathbf{L} - \mathbf{I}\log\left ( \mathrm{Tr} e^{-\beta \mathbf{L}} \right ) \right \rbrack =\] \[= -\beta \mathbb{E}\left \lbrack \mathbf{L} \right \rbrack - \mathbb{E}\left \lbrack \mathbf{I} \log\left ( \mathrm{Tr} e^{-\beta \mathbf{L}} \right )\right \rbrack\]

Now we would like to find an expression for \(\mathbb{E}\left \lbrack \mathbf{I} \log\left ( \mathrm{Tr}\left \lbrack e^{-\beta \mathbf{L}} \right \rbrack \right )\right \rbrack\) that highlights the presence of the term \(\mathbb{E}[\mathbf{L}]\), so we could find some inequality for the relative entropy that we can use. By the Peierls-Bogoliubov inequality, the quantity \(\log\left ( \mathrm{Tr}\left \lbrack e^{-\beta \mathbf{L}}\right \rbrack \right )\) is convex, so we can use the Jensen inequality to get:

\[\mathbb{E}\left \lbrack \mathbf{I} \log\left ( \mathrm{Tr}\left [ e^{-\beta \mathbf{L}}\right ] \right )\right \rbrack \geq \mathbf{I} \log \mathbb{E}\left \lbrack \mathrm{Tr} \left \lbrack e^{-\beta \mathbf{L}}\right \rbrack \right \rbrack = \mathbf{I} \log \mathrm{Tr} \left \lbrack \mathbb{E}\left \lbrack e^{-\beta \mathbf{L}} \right \rbrack \right \rbrack\]

but also again by Jensen for convex function:

\[\mathbb{E}\left \lbrack e^{-\beta \mathbf{L}}\right \rbrack \geq e^{-\beta \mathbb{E}\left [\mathbf{L}\right ]}\]

and because the log-trace is a monotone growing function we can say:

\[\log \mathrm{Tr} \left \lbrack \mathbb{E}\left \lbrack e^{-\beta \mathbf{L}} \right \rbrack\right \rbrack \geq \log \mathrm{Tr} \left \lbrack e^{-\beta \mathbb{E}\left [\mathbf{L}\right ]} \right \rbrack\]

We now know that \(\log e^{-\beta \mathbb{E}\left [\mathbf{L}\right ]} = -\beta \mathbb{E}\left [\mathbf{L}\right ]\) and start to substitute it back til the expression for the relative entropy:

We have:

\[\mathbb{E}\left \lbrack \mathbf{I} \log\left ( \mathrm{Tr} e^{-\beta \mathbf{L}} \right )\right \rbrack \geq \log \mathbb{E}\left \lbrack \left ( \mathrm{Tr} e^{-\beta \mathbf{L}} \right )\right \rbrack = \log \mathrm{Tr} \left \lbrack \mathbb{E}\left \lbrack e^{-\beta \mathbf{L}} \right \rbrack \right \rbrack \geq \log \mathrm{Tr} \left \lbrack e^{-\beta \mathbb{E}\left [\mathbf{L}\right ]} \right \rbrack\]

Multiply by \(\boldsymbol \rho\) and take the trace on both sides, to get:

\[\mathrm{Tr} \left \lbrack \boldsymbol \rho \mathbb{E}\left \lbrack \mathbf{I} \log\left ( \mathrm{Tr} e^{-\beta \mathbf{L}} \right )\right \rbrack \right \rbrack \geq \mathrm{Tr} \left \lbrack \boldsymbol \rho \log \mathrm{Tr} \left \lbrack e^{-\beta \mathbb{E}\left [\mathbf{L}\right ]} \right \rbrack \right \rbrack\]

Now change sign (and also direction of the inequality) and then add the term \(\mathrm{Tr} \left [\boldsymbol \rho \log \boldsymbol \rho\right ]\) to both sides, to get:

\[\mathrm{Tr} \left \lbrack \boldsymbol \rho \log \boldsymbol \rho \right \rbrack - \mathrm{Tr} \left \lbrack \boldsymbol \rho \mathbb{E}\left \lbrack \log\left ( \mathrm{Tr} e^{-\beta \mathbf{L}} \right )\right \rbrack \right \rbrack \leq \mathrm{Tr} \left \lbrack \boldsymbol \rho \log \boldsymbol \rho \right \rbrack - \mathrm{Tr} \left \lbrack \boldsymbol \rho \mathbf{I} \log \mathrm{Tr} \left \lbrack e^{-\beta \mathbb{E}\left [\mathbf{L}\right ]} \right \rbrack \right \rbrack\]

in other words we get:

\[\mathbb{E}\left \lbrack S(\boldsymbol \rho \| \boldsymbol \sigma(\mathbf{L})) \right \rbrack \leq S(\boldsymbol \rho \| \boldsymbol \sigma(\mathbb{E}[\mathbf{L}]))\]

Maybe there is an error: we know that the relative entropy is jointly convex so we must have (and it’s verified numerically):

\[\mathbb{E}\left \lbrack S(\boldsymbol \rho \| \boldsymbol \sigma(\mathbf{L})) \right \rbrack \geq S(\boldsymbol \rho \| \mathbb{E}[\boldsymbol \sigma(\mathbf{L})])\]

Inequalities, traces, densities, matrices

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