# Can separable states have quantum mutual information larger than one?

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 $$I(\rho)\le1$$?

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.
• 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
• 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