Quantum Mutual Information

Mutual information measures how much two systems know about each other. Classically it counts the shared bits between two random variables; quantumly it captures the total correlation — classical and quantum together — held in a joint state.

Definition

For a bipartite state $\rho_{AB}$ with reduced states $\rho_A = \operatorname{Tr}_B\rho_{AB}$ and $\rho_B = \operatorname{Tr}_A\rho_{AB}$, the quantum mutual information is

This mirrors the classical identity $I(X;Y) = H(X) + H(Y) - H(X,Y)$, with Shannon entropies replaced by von Neumann entropies. It quantifies how much the entropy of the parts exceeds the entropy of the whole: if the whole carries less uncertainty than its pieces taken separately, the deficit is shared information.

Key properties

• Non-negativity. $I(A\!:\!B) \ge 0$ always, a direct consequence of the subadditivity of von Neumann entropy, $S(AB) \le S(A) + S(B)$ (covered in the next lesson).
• Zero iff uncorrelated. $I(A\!:\!B) = 0$ exactly when $\rho_{AB} = \rho_A \otimes \rho_B$, a product state with no correlations of any kind.
• Symmetry. $I(A\!:\!B) = I(B\!:\!A)$.
• Upper bound. $I(A\!:\!B) \le 2\min\{\log_2 d_A,\ \log_2 d_B\}$ — twice the classical ceiling, the extra factor of two being room for quantum correlations.

Pure states: maximal correlation

If $\rho_{AB} = |\psi\rangle\langle\psi|$ is pure, then $S(\rho_{AB}) = 0$, and (as the Schmidt decomposition will show) the two reduced states share the same spectrum, so $S(\rho_A) = S(\rho_B)$. The mutual information becomes

exactly twice the entanglement entropy. A pure product state has $S(\rho_A) = 0$ and hence $I = 0$; a maximally entangled state has $S(\rho_A) = \log_2 d_A$ and hence the maximal $I = 2\log_2 d_A$.

The Bell state, computed

Take $|\Phi^+\rangle = \tfrac{1}{\sqrt2}(|00\rangle + |11\rangle)$. Its global density matrix $|\Phi^+\rangle\langle\Phi^+|$ is pure, so $S(\rho_{AB}) = 0$. Tracing out $B$,

the maximally mixed qubit with $S(\rho_A) = 1$ bit; likewise $S(\rho_B) = 1$. Therefore

saturating the bound $2\log_2 2$. The two qubits are perfectly correlated, yet the whole has zero entropy — the hallmark of pure-state entanglement.

Try it

Compute the quantum mutual information $I(A\!:\!B)$, in bits, of the Bell state $|\Phi^+\rangle = \tfrac{1}{\sqrt2}(|00\rangle + |11\rangle)$ using $I = S(A) + S(B) - S(AB)$. Return the value.

Sign in on the full site to ask questions and join the discussion.