Marginals and Correlations

Marginals are reduced descriptions of subsystems. Correlations are features of the joint state that are not fixed by those marginal descriptions alone.

For a bipartite state $\rho_{AB}$, the marginal states are

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

They determine all local measurement statistics on $A$ and $B$ separately. They do not determine the joint statistics of measurements on $A$ and $B$ together.

Marginal States

The word “marginal” is borrowed from probability theory. A joint probability distribution $p(a,b)$ has marginals

$$p_A(a) = \sum_b p(a,b), \qquad p_B(b) = \sum_a p(a,b).$$

In quantum mechanics, the density-operator analogue is the reduced state:

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

For local measurement effects $\{E_a\}$ on $A$ and $\{F_b\}$ on $B$, the joint probabilities are

$$p(a,b) = \operatorname{Tr}_{AB} \bigl[ \rho_{AB}(E_a\otimes F_b) \bigr].$$

The marginals are

$$p_A(a) = \operatorname{Tr}_A(\rho_A E_a), \qquad p_B(b) = \operatorname{Tr}_B(\rho_B F_b).$$

Thus $\rho_A$ and $\rho_B$ are enough for local statistics, while $\rho_{AB}$ is needed for joint statistics.

Joint States Contain More Than Marginals

The same pair of marginals can come from many different joint states. For example,

$$\lvert\Phi^+\rangle = \frac{1}{\sqrt2} \bigl( \lvert00\rangle+\lvert11\rangle \bigr)$$

and

$$\rho_{\mathrm{cc}} = \frac12 \lvert00\rangle\langle00\rvert + \frac12 \lvert11\rangle\langle11\rvert$$

both satisfy

$$\rho_A = \rho_B = \frac12 I.$$

They nevertheless have different joint states and different correlations. The first is pure and entangled. The second is mixed, separable, and classically correlated.

This is the quantum version of a familiar classical fact: knowing the marginal distributions of two random variables does not determine their joint distribution.

Product States and Zero Correlation

A bipartite density operator is a product state if

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

In a product state, every product-observable expectation factorizes:

$$\begin{aligned} \langle M_A\otimes N_B\rangle &= \operatorname{Tr}_{AB} \bigl[ (\rho_A\otimes\rho_B)(M_A\otimes N_B) \bigr]\\ &= \operatorname{Tr}_A(\rho_A M_A) \operatorname{Tr}_B(\rho_B N_B)\\ &= \langle M_A\rangle\langle N_B\rangle. \end{aligned}$$

Equivalently, all connected correlations vanish:

$$\langle M_A\otimes N_B\rangle - \langle M_A\otimes I_B\rangle \langle I_A\otimes N_B\rangle = 0$$

for all local observables $M_A$ and $N_B$.

Conversely, in finite dimensions, if this factorization holds for all local operators $M_A$ and $N_B$, then $\rho_{AB}=\rho_A\otimes\rho_B$. Product operators span the operator space on $\mathcal H_A\otimes\mathcal H_B$, so all matrix elements of $\rho_{AB}$ are fixed by those product expectations.

Correlation Functions

A correlation function is an expectation value involving observables on multiple subsystems. For two subsystems,

$$C_{MN} = \langle M_A\otimes N_B\rangle = \operatorname{Tr}_{AB} \bigl[ \rho_{AB}(M_A\otimes N_B) \bigr].$$

The connected correlation is

$$C^{\mathrm{conn}}_{MN} = \langle M_A\otimes N_B\rangle - \langle M_A\rangle \langle N_B\rangle.$$

Connected correlations remove the part already explained by the one-subsystem averages. They are the quantum analogue of covariance.

Nonzero connected correlation means the joint state is not a product state. It does not, by itself, mean the state is entangled. Separable mixed states can have nonzero connected correlations.

Classical and Bell Examples

Use Pauli operators with eigenvalues $\pm1$. For the classically correlated state

$$\rho_{\mathrm{cc}} = \frac12 \lvert00\rangle\langle00\rvert + \frac12 \lvert11\rangle\langle11\rvert,$$

the local means vanish:

$$\langle Z\otimes I\rangle = \langle I\otimes Z\rangle = 0.$$

But the joint $Z$ correlation is

$$\langle Z\otimes Z\rangle = 1.$$

Thus

$$C^{\mathrm{conn}}_{ZZ} = 1.$$

This correlation is classical: the state is a separable mixture of matching computational-basis preparations.

For the Bell state $\lvert\Phi^+\rangle$,

$$\langle Z\otimes Z\rangle = 1, \qquad \langle X\otimes X\rangle = 1, \qquad \langle Y\otimes Y\rangle = -1.$$

The Bell state has correlations in complementary bases because its joint density operator contains coherence terms absent from $\rho_{\mathrm{cc}}$.

Correlation Matrix for Two Qubits

For two qubits, a useful summary is the Pauli correlation matrix

$$T_{ij} = \operatorname{Tr} \bigl[ \rho_{AB}(\sigma_i\otimes\sigma_j) \bigr], \qquad i,j\in\{x,y,z\}.$$

The local Bloch vectors are

$$r_i = \operatorname{Tr} \bigl[ \rho_{AB}(\sigma_i\otimes I) \bigr], \qquad s_j = \operatorname{Tr} \bigl[ \rho_{AB}(I\otimes\sigma_j) \bigr].$$

The connected Pauli correlation matrix is

$$T^{\mathrm{conn}}_{ij} = T_{ij}-r_i s_j.$$

This matrix is not an entanglement measure by itself, but it is a compact diagnostic for examples, tomography, Bell inequalities, and two-qubit model calculations.

Mutual Information Preview

The quantum mutual information is

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

where

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

is the von Neumann entropy.

The mutual information measures total correlation, both classical and quantum. It satisfies

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

and

$$I(A:B)=0 \qquad \text{if and only if} \qquad \rho_{AB}=\rho_A\otimes\rho_B.$$

It is not the same as entanglement entropy for mixed states. For a pure bipartite state, it equals twice the subsystem entropy:

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

For a classically correlated separable state, it can be nonzero even though there is no entanglement.

The reduced-state entropy itself is developed in Subsystem Entropy, including examples where the same local entropy comes from entanglement, classical correlation, or uncorrelated local noise. The full entropy-based total-correlation measure is developed in Mutual Information.

Many-Body and QFT Language

In many-body physics, one often studies local or few-body observables at sites, modes, or spatial regions. A two-point function has the schematic form

$$\langle O_i O_j\rangle,$$

and the connected two-point function is

$$\langle O_i O_j\rangle_{\mathrm c} = \langle O_i O_j\rangle - \langle O_i\rangle\langle O_j\rangle.$$

This is the same idea as $C^{\mathrm{conn}}_{MN}$, expressed for observables labeled by position, site, mode, or field operator. Higher $n$-point functions and their connected parts organize more detailed information about the state.

In QFT and statistical mechanics, correlation functions often become the primary objects: they diagnose phases, response, clustering, critical behavior, and particle content. The finite-dimensional reduced-state language here is the clean entry point; Connected Correlation Functions develops the ordered cumulant hierarchy and clustering boundary without duplicating the reduced-state reconstruction given here.

Common Mistakes

• Assuming the marginals $\rho_A$ and $\rho_B$ determine the joint state $\rho_{AB}$.
• Treating one nonzero connected correlation as proof of entanglement.
• Testing only one pair of observables and concluding that a state is product.
• Forgetting that separable mixed states can have strong correlations.
• Confusing correlation with causal influence or signaling.
• Treating mutual information as an entanglement measure for arbitrary mixed states.
• Ignoring the subsystem split before discussing marginals or correlations.

Cross-Links

• Reduced Density Operators
• Local Measurement Statistics
• Partial Trace
• Subsystem Entropy
• Product States
• Classical Correlation versus Entanglement
• Bell States
• Operators on Composite Systems
• Entanglement Entropy
• Mutual Information

References

• A. Peres, Quantum Theory: Concepts and Methods, Kluwer, 1995.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press, 2018.
• R. Horodecki, P. Horodecki, M. Horodecki, and K. Horodecki, “Quantum Entanglement,” Reviews of Modern Physics 81, 865-942, 2009.
• A. Altland and B. Simons, Condensed Matter Field Theory, 2nd ed., Cambridge University Press, 2010.
• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, McGraw-Hill, 1971.

Exercises

1. Show that product density operators have factorized product-observable expectations.

Solution

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

$$\begin{aligned} \langle M_A\otimes N_B\rangle &= \operatorname{Tr}_{AB} \bigl[ (\rho_A\otimes\rho_B)(M_A\otimes N_B) \bigr]\\ &= \operatorname{Tr}_A(\rho_A M_A) \operatorname{Tr}_B(\rho_B N_B)\\ &= \langle M_A\rangle\langle N_B\rangle. \end{aligned}$$

2. Compute the $Z$ connected correlation for $\rho_{\mathrm{cc}}$.

Solution

For

$$\rho_{\mathrm{cc}} = \frac12 \lvert00\rangle\langle00\rvert + \frac12 \lvert11\rangle\langle11\rvert,$$

the local means vanish:

$$\langle Z\otimes I\rangle = \langle I\otimes Z\rangle = 0.$$

The joint expectation is

$$\langle Z\otimes Z\rangle = \frac12(+1)(+1) + \frac12(-1)(-1) = 1.$$

Therefore

$$C^{\mathrm{conn}}_{ZZ} = 1-0\cdot0 = 1.$$

3. Compare $\langle X\otimes X\rangle$ for $\rho_{\mathrm{cc}}$ and $\lvert\Phi^+\rangle$.

Solution

For $\rho_{\mathrm{cc}}$, the operator $X\otimes X$ maps $\lvert00\rangle$ to $\lvert11\rangle$ and $\lvert11\rangle$ to $\lvert00\rangle$, so diagonal matrix elements in $\lvert00\rangle$ and $\lvert11\rangle$ vanish. Hence

$$\langle X\otimes X\rangle_{\rho_{\mathrm{cc}}} = 0.$$

For

$$\lvert\Phi^+\rangle = \frac{1}{\sqrt2} \bigl( \lvert00\rangle+\lvert11\rangle \bigr),$$

we have

$$(X\otimes X)\lvert\Phi^+\rangle = \lvert\Phi^+\rangle,$$

so

$$\langle X\otimes X\rangle_{\Phi^+} = 1.$$

4. Give two different joint states with the same marginals.

Solution

The Bell state $\lvert\Phi^+\rangle$ and the classically correlated state $\rho_{\mathrm{cc}}$ both have

$$\rho_A = \rho_B = \frac12 I.$$

They are different joint states: $\lvert\Phi^+\rangle$ is pure and entangled, while $\rho_{\mathrm{cc}}$ is mixed and separable. Their joint measurements differ, for example in the $X\otimes X$ expectation value.

5. Why does vanishing connected correlation for one observable pair not prove that a state is product?

Solution

Product structure requires factorization for all local observables, not just one chosen pair. A state may have zero covariance for $M_A$ and $N_B$ while still being correlated for other observables. To conclude $\rho_{AB}=\rho_A\otimes\rho_B$ in finite dimensions, product-observable expectations must factorize for an operator basis on each subsystem, or equivalently for all local operators.