Consider bipartite (qubit) systems. The classical mutual information between a pair of binary registers, $$I(X:Y)\equiv H(X)+H(Y)-H(X,Y),$$ is always lesser than $1$ (and non-negative). On the other hand, the quantum mutual information of a bipartite state $\rho$, defined as $$I(\rho) \equiv S(\rho_A) + S(\rho_B) - S(\rho),$$ with $S(\rho)$ von Neumann entropy of $\rho$, and $\rho_A\equiv\operatorname{Tr}_B(\rho)$, $\rho_B\equiv\operatorname{Tr}_A(\rho)$, satisfies $0\le I(\rho)\le 2$.

If $\rho$ is pure, we also know that $I(\rho)=2J(\rho)$, where $J(\rho)$ is the accessible mutual information (the one obtained computing the mutual information via the conditional entropy, maximising over the possible measurement choices). Therefore, a pure $\rho$ is separable (i.e. a product state) iff $I(\rho)=J(\rho) = 0$.

For a more general classical-quantum state, some $\rho=\sum_i p_i \,|i\rangle\!\langle i|\otimes \rho_i$, we have $$I(\rho) = H(\mathbf p)+S\left(\sum_{i}p_{i}\rho_{i}\right)-S(\rho), \\ S(\rho) = \sum_i p_i S(\rho_i) + H(\mathbf p),$$ and thus $$ I(\rho) = S\left(\sum_{i}p_{i}\rho_{i}\right) - \sum_i p_i S(\rho_i) \le S\left(\sum_{i}p_{i}\rho_{i}\right) \le 1, $$ because $S(\sigma)\le1$ for any single-qubit state $\sigma$.

These are all examples of separable states with quantum mutual information $I(\rho)\le 1$. More generally, it's clear that a state can give $I(\rho)>1$ only if it has nonzero discord, so a separable state with $I(\rho)>1$ would have to be discordant. But classical-quantum states are separable and can have nonzero discord (with respect to measurements on the second space), but still always give $I(\rho)\le1$. Other classical examples of discordant separable states that are not classical-quantum are Werner states: $$\rho_z = \frac{1-z}{4}I +z |\Phi^+\rangle\!\langle\Phi^+| =\frac14\begin{pmatrix}1+z & 0&0& 2z \\ 0& 1-z & 0 & 0 \\ 0&0&1-z&0 \\ 2z & 0 & 0 & 1+z\end{pmatrix}, \\ |\Phi^+\rangle\equiv\frac{1}{\sqrt2}(|00\rangle+|11\rangle).$$ These are separable for $z\le1/3$, but as discussed in (Ollivier, Zurek 2001), have nonzero discord. Their quantum mutual information reads $$I(\rho_z) = \frac14\left[ (1+3z) \log_2(1+3z) + 3(1-z)\log_2(1-z) \right],$$ which is smaller than $1$ for $z\le 1/3$.

Is the above a general feature? In other words, do all separable states have $I(\rho)\le1$?


1 Answer 1


If $\rho_{AB}$ is separable then $$ I(A:B) \leq \min\{H(A), H(B)\}. $$

To see this first note that $$ I(A:B) = H(A) + H(B) - H(AB) = H(A) - H(A|B). $$ Now consider the conditional entropy term $H(A|B) := H(AB) - H(B)$. We will show that it is nonnegative for separable states. Let $\rho_{AB} = \sum_x p_x \sigma_x \otimes \tau_x$ be a separable state and define a classical extension $\rho_{ABX} = \sum_x p_x \sigma_x \otimes \tau_x \otimes |x\rangle \langle x |$, where $X$ is a classical register. Note that $\mathrm{Tr}_X[\rho_{ABX}] = \rho_{AB}$. Now by strong subadditivity of the von Neumann entropy we have that $$ H(A|BX) \leq H(A|B). $$ So it suffices to show that $H(A|BX)\geq 0$. Well as $X$ is classical we have $$ H(A|BX) = \sum_x p_x H(A|B,X=x) $$ But notice that conditioned on a particular value of $X=x$ the state $\rho_{ABX}$ becomes a product state. It is not difficult to show that for product states, i.e., $\sigma_x \otimes \tau_x$ we have $$ H(AB) = H(A) + H(B) $$ hence $$ H(A|B,X=x) = H(A|X=x) + H(B|X=x) - H(B|X=x) = H(A|X=x). $$ But then $$ H(A|B) \geq H(A|BX) = \sum_x p_x H(A|X=x) \geq 0 $$ And so, $$ I(A:B) = H(A) - H(A|B) \leq H(A) - \sum_x p_x H(A)_{\sigma_x} \leq H(A). $$ We can do the same argument for $I(A:B) = H(B) - H(B|A)$ so the result holds.

  • $\begingroup$ very nice trick passing through the purification to show that $H(A|B)\ge0$. In hindsight, this reminds me the discussions about state merging, which are all about how the negativity of $H(A|B)$ can be exploited to perform entanglement (if I remember correctly). So also from that point of view, we must have $H(A|B)\ge0$ for separable states $\endgroup$
    – glS
    Nov 1, 2022 at 21:11
  • $\begingroup$ Thanks. Note though that the X system isn't purifying the state, it's just a classical register that keeps track of which term in the convex combination you have. Perhaps one can do something similar with a purification but I don't immediately see it. The upshot of this proof is that negative conditional entropy is a sufficient condition for bipartite entanglement (it's only necessary if the state is pure). $\endgroup$
    – Rammus
    Nov 2, 2022 at 17:19
  • $\begingroup$ ok, I finally had a chance of going through this more carefully. You're right of course. I do wonder why this trick works though. It feels like you're removing the discord contributions from $H(A|B)$ and finding it to still be positive, by making the various states in the separable decomposition "artificially distinguishable" via addition of an additional classical register $\endgroup$
    – glS
    Nov 6, 2022 at 9:45

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