# Superoperator cannot increase relative entropy

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?

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