# What can be said about the non-negativity of the relative entropy of $S(\rho_{AB}||\rho_{B})$?

Taking $$\rho_{AB}=\rho_{A}\otimes \rho_{B}$$, where $$S(\rho_{A})$$ and $$S(\rho_{B})$$ aren't 0, it's easy to see that

$$S(\rho_{AB}||I \otimes \rho_{B})=-S(\rho_{A})-S(\rho_{B})+S(\rho_{B})=-S(\rho_{A}).$$

Now clearly this is less than 0 due to the non-negativtiy of the von-neumann entropy. However, the non-negativty of the relative entropy is given in every textbook, with the caveat that for $$S(\rho||\sigma) \ge0$$ provided the support of $$\rho\subseteq \sigma$$, which by the violation of the inequality, can't be the case in this example. Although I don't see how, as if I take each marginal to be maximally mixed, the above still holds, yet in that case the support of $$\rho \subseteq \sigma$$. Perhaps I am missing something here?

However, if $$\rho_{AB}$$ were a maximally entangled state, it would be non-negative. So given cases like this, wherein the support isn't a subset of the second argument, are there other ways to show it will hold? Possibly something to do with $$S(\rho_{AB}||N(\rho_{AB}))$$ wherein $$N$$ is some CPTP channel?

## Source of the problem

The purported contradiction arises due to the use of incorrect assumptions for Klein equality

$$S(\rho||\sigma) \ge 0.$$

The inequality does not require any particular relationship$$^1$$ between the support of $$\rho$$ and the support of $$\sigma$$. However, it does require that $$\rho$$ and $$\sigma$$ be states, i.e. unit trace positive semidefinite operators. See e.g. theorem 11.7 on page 511 in Nielsen & Chuang or the Wikipedia link above (where the assumption that $$\sigma$$ is a state is used to declare that $$r_i$$ form a probability distribution).

The problem is that $$I\otimes\rho_B$$ is not a state and so we cannot conclude that $$S(\rho_{AB}||I\otimes\rho_B) \ge 0$$.

## Relative entropy of $$\rho_{AB}$$ with respect to $$\rho_B$$

Recall that $$S(\rho||\sigma)$$ is defined as

$$S(\rho||\sigma) = \mathrm{tr}(\rho\log\rho) - \mathrm{tr}(\rho\log\sigma).$$

Note that $$\log\sigma$$ is an operator acting on the same Hilbert space as $$\sigma$$. Therefore, for $$S(\rho||\sigma)$$ to be defined $$\rho$$ and $$\sigma$$ must act on the same Hilbert space. Consequently, $$S(\rho_{AB}||\rho_B)$$ is not defined.

## Maximally mixed state

If you meant to use the maximally mixed state $$\frac{I}{N}$$ in place of $$I$$ then the calculation becomes

\begin{align} S\left(\rho_{AB}\big|\big|\frac{I}{N}\otimes\rho_B\right) &= -S(\rho_A)-S(\rho_B) - \mathrm{tr}\left[\rho_{AB}\log\left(\frac{I}{N}\otimes\rho_B\right)\right] \\ &= -S(\rho_A)-S(\rho_B)-\mathrm{tr}[\rho_{AB}(-\log N + I\otimes\log\rho_B)] \\ &= -S(\rho_A)-S(\rho_B)+\log N + S(\rho_B) \\ &= \log N - S(\rho_A) \end{align}

which is non-negative.

$$^1$$ Note that the definition of $$S(\rho||\sigma)$$ does depend on the relationship between the kernel of $$\sigma$$ and the support of $$\rho$$. See page 511 in Nielsen & Chuang.

• Aha, all right so in the case where they aren't states, I can still use the relative entropy to derive expressions for relations of other entropic values, but I can't use the inequality when doing so to make determinations about the non-negativity. Aug 30 '21 at 10:29