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?


1 Answer 1


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) $$ S(\tilde \rho_{AB}\|\tilde\sigma_{AB})\geq S(\$\rho_A\|\$\sigma_A) $$

  • $\begingroup$ This is exactly how I did it! Thanks for the confirmation. $\endgroup$
    – CFRedDemon
    May 7, 2019 at 18:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.