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Note: Cross-posted on Physics SE.

So I have to show that a superoperator $\$$ cannot increase relative entropy using the monotonicity of relative entropy:

$$S(\rho_A || \sigma_A) \leq S(\rho_{AB} || \sigma_{AB}).$$

What I have to prove:

$$S(\$\rho|| \$ \sigma) \leq S(\rho || \sigma).$$

Now the hint is that I should use the unitary representation of the superoperator $\$$. I know that we can represent $ \$ \rho = \sum_i M_i \rho M_i^{\dagger} $ with $\sum_i M_i M_i^{\dagger} = I$. Now I am able to write out $S(\$\rho|| \$ \sigma_A)$ in this notation, but that doesn't bring me any further.

Does anyone have any idea how to show this in the way that the questions hints to? I already read the original paper of Lindblad but this doesn't help me (he does it another special way). Any clues or how to do this?

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I'm not an expert with this sort of thing (i.e. there may be imperfections in this argument), but hopefully this will set you in the right direction...

Consider $\rho_{AB}=\rho_A\otimes |0\rangle\langle 0|$ and $\sigma_{AB}=\sigma_A\otimes |0\rangle\langle 0|$. It must be that $S(\rho_A\|\sigma_A)=S(\rho_{AB}\|\sigma_{AB})$.

Now, your superoperator can be described by a unitary $U$ over a larger space: $$ \$\rho=\text{Tr}_B\left(U(\rho\otimes|0\rangle\langle 0|)U^\dagger\right) $$ So, let $$ \tilde\rho_{AB}=U\rho_{AB}U^\dagger\qquad\tilde\sigma_{AB}=U\sigma_{AB}U^\dagger. $$ Since it's unitary, $$ S(\rho_A\|\sigma_A)=S(\rho_{AB}\|\sigma_{AB})=S(\tilde \rho_{AB}\|\tilde\sigma_{AB}), $$ but from your original statement (I assume, you've not quite stated it precisely enough) $$ S(\tilde \rho_{AB}\|\tilde\sigma_{AB})\geq S(\$\rho_A\|\$\sigma_A) $$

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  • $\begingroup$ This is exactly how I did it! Thanks for the confirmation. $\endgroup$ – CFRedDemon May 7 at 18:53

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