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

Question

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?

Answer 1

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.

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^{$^1$ 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.}