Mutual Information

Quantum mutual information measures the total correlation between two subsystems of a quantum state. In finite dimensions, for a bipartite density operator $\rho_{AB}$,

$$I(A:B)_\rho = S(\rho_A) + S(\rho_B) - S(\rho_{AB}),$$

where

$$\rho_A=\operatorname{Tr}_B\rho_{AB}, \qquad \rho_B=\operatorname{Tr}_A\rho_{AB},$$

and

$$S(\rho) = -\operatorname{Tr}(\rho\log_2\rho)$$

is the von Neumann entropy.

Helpful background. Subsystem entropy supplies the entropy formula; marginals and correlations explains which information is absent from the local states. All entropies below use base-two logarithms.

Mutual information is a correlation measure, not an entanglement measure for arbitrary mixed states. It responds to both classical and quantum correlations, without defining a unique decomposition into separate contributions. It vanishes exactly for product density operators.

Mutual information of classical distributions

For classical random variables $X$ and $Y$ with joint distribution $p(x,y)$, the classical mutual information is

$$I(X:Y) = H(X)+H(Y)-H(X,Y),$$

where $H$ is Shannon entropy.

Equivalently,

$$I(X:Y) = \sum_{x,y} p(x,y) \log_2 \frac{p(x,y)}{p_X(x)p_Y(y)}.$$

This form shows that mutual information is a relative entropy comparing the joint distribution with the product of its marginals. It is zero exactly when

$$p(x,y)=p_X(x)p_Y(y).$$

Quantum mutual information is the density-operator analogue. It compares $\rho_{AB}$ with $\rho_A\otimes\rho_B$.

Quantum Definition

For a finite-dimensional bipartite state $\rho_{AB}$, define

$$I(A:B)_\rho = S_A+S_B-S_{AB},$$

with

$$S_A=S(\rho_A), \qquad S_B=S(\rho_B), \qquad S_{AB}=S(\rho_{AB}).$$

Multiplying a value in bits by $\ln2$ converts it to nats.

Quantum mutual information satisfies

$$I(A:B)_\rho\ge0,$$

and

$$I(A:B)_\rho=0 \quad \Longleftrightarrow \quad \rho_{AB}=\rho_A\otimes\rho_B.$$

The nonnegativity is equivalent to subadditivity of von Neumann entropy:

$$S_{AB}\le S_A+S_B.$$

Thus mutual information measures exactly how far the joint entropy falls below the sum of the local entropies.

Product states and a shared random bit

For a product density operator, entropy additivity gives $S(\rho_A\otimes\rho_B)=S_A+S_B$, hence $I(A:B)=0$. The factors may themselves be mixed; local uncertainty alone creates no correlation.

In contrast, $\rho_{\rm cc}=\tfrac12|00\rangle\langle00|+\tfrac12|11\rangle\langle11|$ has $S_A=S_B=S_{AB}=1$, so $I(A:B)=1$ bit. It is separable, but the outcomes of local $Z$ measurements share one random bit.

Why a Bell pair has two bits of quantum mutual information

A pure joint state has $S_{AB}=0$ and equal reduced entropies by Schmidt decomposition. Therefore

$$I(A:B)=2S(\rho_A)=2E(\Psi).$$

For a Bell pair the answer is two bits, while its entanglement entropy is one ebit. This does not mean that one local measurement of each qubit returns two shared classical bits. Measuring both in $Z$ produces one shared random bit and loses other quantum correlations.

A nonorthogonal preparation

Consider

$$\rho_{AB}=\tfrac12|0\rangle\langle0|\otimes|0\rangle\langle0| +\tfrac12|1\rangle\langle1|\otimes|+\rangle\langle+|.$$

The two joint vectors are orthogonal because their $A$ labels are orthogonal, so $S_{AB}=1$. Also $S_A=1$, while

$$\rho_B=\begin{pmatrix}3/4&1/4\\1/4&1/4\end{pmatrix}, \qquad \lambda_\pm=\frac{1\pm1/\sqrt2}{2}.$$

Thus $I(A:B)=S_B=h_2(\lambda_+)\simeq0.601$ bits, where $h_2(p)=-p\log_2p-(1-p)\log_2(1-p)$. The state is separable but the conditional states of $B$ are not orthogonal, so the label on $A$ is not perfectly distinguishable by measuring $B$.

For example, a $Z$ measurement of $B$ yields $Y=0$ with probability $3/4$. Its conditional entropy given Alice’s label $X$ is $H(Y|X)=\tfrac12(0)+\tfrac12(1)=1/2$. The measured classical mutual information is $h_2(3/4)-1/2\simeq0.311$ bits, below the quantum value. This is a particular measurement, not a proof that it is optimal.

Relation to Relative Entropy

The classical comparison measure is Relative Entropy. The finite-dimensional quantum relative entropy is

$$D(\rho\Vert\sigma) = \operatorname{Tr} \bigl[ \rho(\log_2\rho-\log_2\sigma) \bigr],$$

when the support of $\rho$ is contained in the support of $\sigma$; otherwise it is $+\infty$.

Quantum mutual information can be written as

$$I(A:B)_\rho = D \bigl( \rho_{AB} \Vert \rho_A\otimes\rho_B \bigr).$$

This identity makes the interpretation precise: mutual information is the relative entropy comparison (not a metric distance) from the joint state to the uncorrelated product state with the same marginals.

It also explains why

$$I(A:B)_\rho\ge0,$$

because quantum relative entropy is nonnegative, and why equality occurs only for

$$\rho_{AB} = \rho_A\otimes\rho_B.$$

To derive the identity, use $\log(\rho_A\otimes\rho_B)=\log\rho_A\otimes I+I\otimes\log\rho_B$ on the support and take the partial traces. The support of $\rho_{AB}$ is contained in $\operatorname{supp}\rho_A\otimes\operatorname{supp}\rho_B$, so this comparison is well defined. In infinite dimensions, use a suitable relative-entropy definition; subtracting infinite subsystem entropies as $\infty-\infty$ is not valid.

Local processing loses correlations

For local trace-preserving quantum channels $\Lambda_A$ and $\Lambda_B$, relative-entropy data processing gives

$$I(A':B')_{(\Lambda_A\otimes\Lambda_B)(\rho)}\le I(A:B)_\rho.$$

Local measurements are such channels when their classical records are kept, so measured classical mutual information cannot exceed the quantum value. Dephasing one member of a Bell pair in $Z$ turns it into $\rho_{\rm cc}$ and lowers mutual information from two bits to one.

The assumption is independent local channels without communication. LOCC includes communication and can create classical correlations: Alice may generate a random bit and send a copy to Bob, changing an initially uncorrelated pair into $\rho_{\rm cc}$. Thus mutual information is not an LOCC entanglement monotone. Postselected individual branches also require separate treatment.

Conditional Entropy Identities

Define the quantum conditional entropy

$$S(A\vert B) = S_{AB}-S_B.$$

Then

$$I(A:B) = S_A-S(A\vert B) = S_B-S(B\vert A).$$

Unlike classical conditional entropy, quantum conditional entropy can be negative. For a Bell state,

$$S(A\vert B) = 0-1 = -1$$

bit. This negativity is a quantum-information signal tied to entanglement, but mutual information itself still measures total correlation rather than only entanglement.

The Araki–Lieb inequality $S_{AB}\ge|S_A-S_B|$ also gives

$$0\le I(A:B)\le2\min(S_A,S_B)\le2\log_2\min(d_A,d_B).$$

A maximally entangled pure state reaches the dimensional bound. Negative conditional entropy certifies entanglement, but many entangled mixed states have nonnegative conditional entropy.

Many-Body and QFT Preview

In many-body physics, $I(A:B)$ is used to quantify total correlation between spatial regions, lattice blocks, sites, modes, or subsystems. It includes relationships not visible in one chosen two-point function. Connected Correlation Functions develops the complementary operator-cumulant hierarchy and explains why one selected connected correlator is neither a complete total-correlation measure nor an entanglement witness.

In thermal states, mutual information can be nonzero because of thermal and classical correlations even when entanglement is absent or small. In ground states, it can help compare correlation structure across regions and system sizes.

In QFT, individual entanglement entropies of spatial regions are often regulator-dependent. Mutual information between separated regions is often better behaved because some local boundary divergences cancel, though adjacent regions and continuum limits still require care. The finite-dimensional formula above is the clean starting point, not the full field-theoretic analysis.

For the broader many-body setting where mutual information is compared with entanglement entropy, area laws, thermalization, and localization, see Entanglement in Many-Body Physics. For the continuum role of mutual information beside modular Hamiltonians and relative entropy, see Entanglement in QFT Preview.

References

• H. Araki and E. H. Lieb, “Entropy inequalities”, Communications in Mathematical Physics 18, 160–170 (1970).
• T. M. Cover and J. A. Thomas, Elements of Information Theory, 2nd ed., Wiley (2006).
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press (2010).
• J. Watrous, The Theory of Quantum Information, Cambridge University Press (2018), Ch. 5.
• M. M. Wilde, Quantum Information Theory, 2nd ed., Cambridge University Press (2017).

Exercises

1. Show that a product state has zero mutual information.

Solution

If $\rho_{AB}=\rho_A\otimes\rho_B$, then

$$S_{AB} = S(\rho_A\otimes\rho_B) = S_A+S_B.$$

Therefore

$$I(A:B) = S_A+S_B-S_{AB} = 0.$$

2. A uniformly random bit $X$ is copied to $Y$, then flipped independently with probability $\epsilon$. Compute their mutual information.

Solution

$Y$ is uniform, so $H(Y)=1$. Given $X$, only the independent flip remains uncertain, so $H(Y|X)=h_2(\epsilon)$. Therefore $I(X:Y)=1-h_2(\epsilon)$. It is one bit at $\epsilon=0$ and at $\epsilon=1$ (perfect anticorrelation), and zero at $\epsilon=1/2$. Correlation need not mean identical values.

3. Partially dephase $|\Phi^+\rangle$ to

$$\rho_c=\tfrac12\bigl(|00\rangle\langle00|+|11\rangle\langle11|+c|00\rangle\langle11|+c|11\rangle\langle00|\bigr),\quad0\le c\le1.$$

Find its mutual information and the mutual information of local $Z$ outcomes.

Solution

Both marginals are $I/2$. The nonzero joint eigenvalues are $(1+c)/2$ and $(1-c)/2$, so $I(A:B)=2-h_2((1+c)/2)$. The $Z$ outcomes remain perfectly correlated and uniform for every $c$, giving one classical bit. That measurement does not reveal the loss of off-diagonal coherence.

4. For a pure bipartite state, show that $I(A:B)=2S_A$.

Solution

If $\rho_{AB}$ is pure, then $S_{AB}=0$. The reduced states of a pure bipartite state have the same nonzero eigenvalues, so $S_A=S_B$. Thus

$$I(A:B) = S_A+S_B-S_{AB} = S_A+S_A-0 = 2S_A.$$

5. Why can a separable state have nonzero mutual information?

Solution

Separable means the state can be written as a classical mixture of product states. The shared classical label in the mixture can correlate the subsystems. Mutual information counts total correlation, including classical correlation, so it can be nonzero even when the state is not entangled.