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 generally do not determine the joint statistics of measurements on $A$ and $B$ together.

Helpful background. Reduced density operators explains local states; local measurement statistics distinguishes paired outcomes from separate marginals.

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

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

For bounded local observables (or unbounded observables with the required moments), every product-observable expectation in a product state 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.

There is a useful exception to the usual nonuniqueness of joint states: if either marginal is pure, the joint state must be a product. For example, $\rho_A=|0\rangle\langle0|$ forces $\rho_{AB}=|0\rangle\langle0|\otimes\rho_B$, even if $\rho_B$ is mixed. A pure subsystem cannot remain correlated with another subsystem. Exercise 4 proves this using positivity.

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.

A connected correlation removes local bias

Consider the separable family

$$\rho_p=p|00\rangle\langle00|+(1-p)|11\rangle\langle11|, \qquad 0\le p\le1.$$

Every run gives matching $Z$ outcomes, so $\langle Z\otimes Z\rangle=1$ for every $p$. But the local mean is $2p-1$, and therefore

$$C^{\mathrm{conn}}_{ZZ}=1-(2p-1)^2=4p(1-p).$$

At $p=1/2$ the matching outcomes are individually unpredictable and the connected correlation is $1$. At $p=0$ or $1$, the outcomes are already fixed locally; the state is product and the connected correlation vanishes despite the raw correlation remaining $1$.

Thus a large raw product expectation can reflect local bias. A nonzero connected value establishes correlation, but the explicit separable preparation above shows that it does not by itself establish entanglement.

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

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

These quantities reconstruct the full two-qubit state:

$$\rho_{AB}=\frac14\left[I\otimes I +\sum_i r_i\sigma_i\otimes I +\sum_j s_jI\otimes\sigma_j +\sum_{i,j}T_{ij}\sigma_i\otimes\sigma_j\right].$$

The formula follows from orthogonality of the Pauli operator basis under the trace inner product. The marginals supply six real coefficients, and the nine entries of $T$ supply the remaining information. Arbitrary coefficient choices need not produce a positive density operator.

For states with $\mathbf r=\mathbf s=0$, three examples are:

Joint state	Correlation matrix $T$
Uncorrelated $I_4/4$	$0$
Correlated mixture $\rho_{\mathrm{cc}}$	$\operatorname{diag}(0,0,1)$
Bell projector $\lvert\Phi^+\rangle\langle\Phi^+\rvert$	$\operatorname{diag}(1,-1,1)$

All three have identical marginals. Their correlation matrices differ, but no single entry is an entanglement measure. The two-correlation bound shows how a combination of entries can certify entanglement.

Mutual Information Preview

In finite dimensions, the quantum mutual information is

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

where

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

is the von Neumann entropy in bits. In infinite dimensions, individual entropies may diverge; use the relative-entropy definition of mutual information rather than subtracting infinities.

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 a useful starting point. Connected correlation functions explains the ordering and clustering assumptions; continuum field correlations also require their distributional or regularization prescriptions.

Common pitfalls

Perfectly matching fixed outcomes need not be correlated. If both outcomes are already certain separately, their connected correlation is zero.

One zero covariance does not establish a product state. The complete product criterion tests a spanning set of local operators on both subsystems.

Correlations do not imply a causal influence. A joint preparation can correlate outcomes while every unconditioned local marginal obeys no-signalling.

References

• Alexander Altland and Ben Simons, Condensed Matter Field Theory, 2nd ed., Cambridge University Press, 2010, correlation functions.
• Michael A. Nielsen and Isaac L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010, chapters 2 and 11.
• John Watrous, The Theory of Quantum Information, Cambridge University Press, 2018, chapters 2 and 5. Author-hosted text.

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. For the biased mixture $\rho_p$ above with $p=0.9$, compute the raw and connected $Z$ correlations.

Solution

The raw correlation is $1$, and both local means are $2p-1=0.8$. Therefore the connected value is $1-0.8^2=0.36$. The decrease from the unbiased value $1$ reflects that much of the agreement is already explained by the strong local bias.

3. Reconstruct the state with $\mathbf r=\mathbf s=0$ and $T=\operatorname{diag}(0,0,c)$. Determine the physical range of $c$ and whether the state is separable.

Solution

$$\rho=\frac14(I\otimes I+cZ\otimes Z) =\frac14\operatorname{diag}(1+c,1-c,1-c,1+c).$$

Positivity requires $-1\le c\le1$. In this range the diagonal entries are probabilities of four computational-basis product states, so every state in the family is separable. A nonzero connected $ZZ$ correlation alone cannot certify entanglement.

4. Show that a pure marginal $\rho_A=|a\rangle\langle a|$ forces $\rho_{AB}=|a\rangle\langle a|\otimes\rho_B$.

Solution

Set $Q=(I_A-|a\rangle\langle a|)\otimes I_B$. Then $\operatorname{Tr}(Q\rho_{AB})=0$. Positivity implies $Q\rho_{AB}^{1/2}=0$, since its squared Hilbert–Schmidt norm equals that trace. Hence the support of $\rho_{AB}$ lies inside $\operatorname{span}\{|a\rangle\}\otimes\mathcal H_B$. The state must therefore be $|a\rangle\langle a|\otimes\tau_B$; tracing $A$ identifies $\tau_B=\rho_B$.

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.