I am confused about what quantum mutual information gives us.

Does it give all kinds of quantum correlations? Or does it give all kinds of quantum and classical correlations?

If it consists of classical correlations too, then how can you calculate only quantum correlations? Is there a measure which only contains information pertaining to quantum correlations?


1 Answer 1


Definition and Intuition

Quantum mutual information is defined as:

$$ S(A:B) = S(A) + S(B) - S(A,B) $$

Here, $ S(A) $ is the Von Neumann entropy of the quantum state described by density matrix $ A $, i.e.,

$$ S(A) = -\text{tr}(A\log(A)), $$

and $S(A,B)$ is the joint entropy of two quantum states, defined by

$$ S(A,B) = -\text{tr}(\rho^{AB}\log(\rho^{AB})), $$

with $ \rho^{AB} $ being the density matrix of the entire system.

However, from this definition (in my opinion) all the implications may not seem intuitive at first glance. For something more intuitive, let's first consider the classical mutual information using Shannon entropy. In classical information theory, the mutual information of two random variables $ X $ and $ Y $ reflects the amount of information they share. It's calculated as:

$$ H(X :Y ) ≡ H(X) + H(Y ) − H(X, Y ). $$

This equation accounts for the overlap in information content between $ X $ and $ Y $ (if you want to learn more about classical information theory, Shannon's orignal paper is a nice place to start!).

Quantum Differences

We find that certain properties of Shannon entropy do not hold for Von Neumann entropy. For example, in classical scenarios, we have $H(X) \leq H(X, Y ) $, aligning with our intuition that uncertainty about $ X $ cannot exceed that about the joint state of $ X $ and $ Y $. This intuition isn't always correct in quantum contexts.

Consider a two-qubit system $ AB $ in an entangled state $ (|00\rangle + |11\rangle)/\sqrt{2} $. This system, being in a pure state, has $ S(A, B) = 0 $, but the entropy of system $ A $ alone is one, indicating that the quantity $ S(A|B)$ can be negative (which, clearly, is impossible in the classical case).


  1. For Pure States: When $ \rho^{AB} $ is pure ( $ S(\rho^{AB})=0 )$, the mutual information is twice the entanglement entropy of the state:

    $$ S(A:B) = 2S(\rho^{A}) $$

  2. Quantum vs. Classical Correlations: A positive quantum mutual information does not necessarily indicate entanglement. For instance, consider a classical mixture of separable states. This will always have zero entanglement, such as

    $$ \rho^{AB} = \frac{1}{2}(|00\rangle \langle 00| + |11\rangle \langle 11|). $$

    We can see this state isn't entangled as $\text{tr}(\rho_A^2 )= 1$ (if you want to know why this means the state is not entangled, see this answer). However, the mutual information is non-zero: $$ S(A:B) = \log 2, $$

    denoting the presence of classical correlations rather than quantum entanglement.


Quantum mutual information provides a measure of correlations in quantum systems, encompassing both classical and quantum aspects, and its properties deviate from classical intuition.

  • $\begingroup$ $\operatorname{tr}(\rho_A^2)=1$ isn't enough to say that a state is or isn't entangled. The maximally entangled state $\frac{1}{\sqrt2}(|00\rangle+|11\rangle)$ also gives the same value and clearly is entangled. Here $\rho^{AB}$ is obviously separable simply by definition of separability: it'a convex mixture of product states $\endgroup$
    – glS
    Jan 30 at 16:38
  • $\begingroup$ also, the rest of that paragraph is confusing. You start arguing that "positive QMI doesn't necessarily indicate entanglement", then show a separable state and write "however its QMI is non-zero", which seems disconnected. If anything, you have to show an entangled state that has non-zero QMI here? But regardless, nonzero QMI doesn't say anything about classicality or lack thereof, so that discussion seems moot $\endgroup$
    – glS
    Jan 30 at 16:44

Your Answer

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

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