2
$\begingroup$

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?

$\endgroup$

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ Aug 30, 2021 at 10:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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