# Is quantum mutual information an entanglement measure?

For a bipartite system, the quantum mutual information is defined via the Von Neumann entropy as follows: $$I(A:B)=S(A)+S(B)-S(AB).$$

It's always positive. Is it an entanglement measurement?

Also how can you find out if a bipartite system is separable by the use of mutual information?

• are you asking specifically about "entanglement measures" (note the term as a very specific definition, see eg arxiv.org/abs/quant-ph/0504163) or more generally what the (quantum) mutual information (QMI) tells you about entanglement?
– glS
Sep 6 at 13:46
• Squashed entanglement, which is defined via the conditional mutual information, is an entanglement monotone. Sep 6 at 17:49

Coherent information is the measure of quantum correlations. Positivity of coherent information indicates that quantum correlations are present. But this is not an iff relation.

Coherent information is denoted as

$$I(A\rangle B) = H(B) - H(AB) = -H(A|B)\,.$$

If $$I(A\rangle B)$$ quantity is positive then there is quantum entanglement between system $$A$$ and $$B$$.

The relation between mutual information and coherent information is as follows:

$$I(A;B) = H(A) + I(A\rangle B) = H(B) + I(B\rangle A)\,.$$

For example, consider $$\rho_{AB} = | \Phi_+ \rangle \langle \Phi_+|\,.$$

Where $$| \Phi_+ \rangle$$ is a bell state. You can see that $$H(AB) =H(\rho_{AB})= 0$$ $$H(B)= H(\text{Tr}_A(\rho_{AB}) )= 1$$

Hence; $$H(A|B) = -1$$

Thus, entanglement is present between system $$A$$ and $$B$$.

TL;DR: Negative conditional entropy $$H(A|B)$$, i.e. Positive Coherent Information $$I(A\rangle B)= -H(A|B)$$ is a hallmark of quantum correlations - that is, of entanglement!

• there might be some typo in the sentence relating coherent information and entanglement. For example, a maximally entangled two-qubit state has $I(A\rangle B)=1>0$. It's true that $H(B)>H(AB)$ (ie $I(A\rangle B)>0$) implies entanglement, but $H(B)< H(AB)$ doesn't imply separability. Consider eg a balanced mixture of a maximally entangled state and $I/4$. This is entangled (as you can see eg via PPT) but still $H(B)<H(AB)$.
– glS
Sep 6 at 1:56