Classical Correlation versus Entanglement

Classical correlation and entanglement can give identical statistics for one chosen measurement. They are nevertheless different properties of a quantum state. The difference is not the strength of one correlation, but whether the joint density operator contains nonseparable quantum coherence across a specified subsystem split.

The standard comparison is between the classically correlated separable state

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

and the Bell state

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

Both predict perfectly matching outcomes when both qubits are measured in the computational basis. Only the Bell state is entangled.

Same Correlation in One Basis

For the separable state $\rho_{\mathrm{cc}}$, computational-basis measurement gives

$$P(0,0)=\frac12, \qquad P(1,1)=\frac12, \qquad P(0,1)=P(1,0)=0.$$

For the Bell state $\lvert\Phi^+\rangle$, the same measurement gives the same probabilities:

$$P(0,0)=\frac12, \qquad P(1,1)=\frac12, \qquad P(0,1)=P(1,0)=0.$$

Therefore perfect correlation in one basis is not enough to identify entanglement. It may be caused by a shared classical random variable.

The reduced states also agree: both have $\rho_A=\rho_B=I/2$. This is why the distinction cannot be resolved by one-subsystem measurements alone; it requires joint statistics or a state-level entanglement criterion.

In $\rho_{\mathrm{cc}}$, the preparation can be described as:

$$\text{prepare }00\text{ with probability }1/2, \qquad \text{prepare }11\text{ with probability }1/2.$$

That is a separable preparation. There is correlation, but no entanglement.

Coherence Terms

The Bell-state density operator is

$$\rho_{\Phi^+} = \lvert\Phi^+\rangle\langle\Phi^+\rvert.$$

Expanding,

$$\rho_{\Phi^+} = \frac12 \Bigl( \lvert00\rangle\langle00\rvert + \lvert00\rangle\langle11\rvert + \lvert11\rangle\langle00\rvert + \lvert11\rangle\langle11\rvert \Bigr).$$

The separable state is only the diagonal part:

$$\rho_{\mathrm{cc}} = \frac12 \Bigl( \lvert00\rangle\langle00\rvert + \lvert11\rangle\langle11\rvert \Bigr).$$

The off-diagonal terms

$$\lvert00\rangle\langle11\rvert, \qquad \lvert11\rangle\langle00\rvert$$

are coherence terms between the two product alternatives. They are the terms that let the Bell state show correlations in complementary measurement bases.

One way to say this compactly is:

$$\rho_{\mathrm{cc}} = \mathcal D_Z(\rho_{\Phi^+}),$$

where $\mathcal D_Z$ denotes dephasing in the computational product basis. Dephasing removes the coherence and leaves only the classical record of which matched outcome occurred.

Different Behavior Under Basis Changes

Let

$$\lvert+\rangle = \frac{1}{\sqrt2} \bigl( \lvert0\rangle+\lvert1\rangle \bigr), \qquad \lvert-\rangle = \frac{1}{\sqrt2} \bigl( \lvert0\rangle-\lvert1\rangle \bigr).$$

The Bell state can be rewritten as

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

Thus if both qubits are measured in the $X$ basis, the Bell state again gives perfect agreement:

$$P(+,+)=\frac12, \qquad P(-,-)=\frac12.$$

The classically correlated state $\rho_{\mathrm{cc}}$ does not. For either product state $\lvert00\rangle$ or $\lvert11\rangle$, measuring both qubits in the $X$ basis gives four equally likely outcomes. Hence

$$P(+,+)=P(+,-)=P(-,+)=P(-,-)=\frac14$$

for $\rho_{\mathrm{cc}}$.

Equivalently,

$$\langle Z\otimes Z\rangle_{\rho_{\mathrm{cc}}} = \langle Z\otimes Z\rangle_{\Phi^+} = 1,$$

but

$$\langle X\otimes X\rangle_{\rho_{\mathrm{cc}}} = 0, \qquad \langle X\otimes X\rangle_{\Phi^+} = 1.$$

The Bell state has phase coherence that survives a change of basis. The classical mixture does not.

Reduced Density Matrices

Both states have the same one-qubit reduced states:

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

This is another important warning. Local reduced states alone do not determine whether the joint state is entangled. They describe local statistics, not the full joint coherence.

For $\rho_{\mathrm{cc}}$, the mixed reduced state reflects ordinary ignorance about the shared classical label. For $\rho_{\Phi^+}$, the mixed reduced state arises because the joint pure state is entangled. The same local density matrix can come from physically different joint states.

Entanglement Witness Preview

An entanglement witness is an observable whose expectation value is nonnegative on all separable states but negative on at least one entangled state.

For detecting $\lvert\Phi^+\rangle$, a standard witness is

$$W = \frac12 I - \lvert\Phi^+\rangle\langle\Phi^+\rvert.$$

For any pure product state $\lvert a\rangle\lvert b\rangle$, the overlap with $\lvert\Phi^+\rangle$ is at most $1/\sqrt2$, so

$$\langle W\rangle_{\text{product}} \ge0.$$

By convexity, the same nonnegative bound holds for separable mixed states. For the Bell state,

$$\operatorname{Tr}(W\rho_{\Phi^+}) = -\frac12.$$

For the classically correlated state,

$$\operatorname{Tr}(W\rho_{\mathrm{cc}}) =0.$$

This witness separates the Bell state from the separable boundary, while allowing $\rho_{\mathrm{cc}}$ to sit exactly on the boundary for this test.

Bell Inequality Preview

Bell-inequality violations are another way to distinguish some entangled states from classical explanations of correlations. The singlet state and the Bell state $\lvert\Phi^+\rangle$ can violate a CHSH inequality with suitable measurement settings.

The state $\rho_{\mathrm{cc}}$ cannot do so. It can be modeled by a shared classical bit that tells both parties whether the preparation was $00$ or $11$ in the computational basis.

The logical relations are subtle:

$$\text{Bell violation} \quad\Longrightarrow\quad \text{entanglement},$$

but the converse is not true for every mixed state and every measurement scenario. Some entangled mixed states do not violate a given Bell inequality. Bell violation is therefore a strong nonclassicality test, not the definition of entanglement.

Common Mistakes

• Treating perfect correlation in one basis as proof of entanglement.
• Looking only at reduced density matrices and ignoring the joint state.
• Forgetting the off-diagonal coherence terms in a Bell projector.
• Thinking that all correlations in quantum mechanics are entanglement.
• Thinking that every entangled mixed state must violate a simple Bell inequality.
• Confusing a dephased Bell state with the original Bell state.

Cross-Links

• Separable Mixed States
• Entangled States
• Bell States
• Entanglement in Foundations
• Local Unitary Equivalence
• Entanglement Depends on a Decomposition
• Local Measurement Statistics
• Marginals and Correlations
• Subsystem Entropy
• Mutual Information
• Reduced Density Operators
• Partial Trace
• Schmidt Decomposition
• Concurrence for Two Qubits
• Negativity and PPT Criterion
• Entanglement Witnesses
• Density Operators
• Born Rule
• Glossary: Entanglement

References

• J. S. Bell, “On the Einstein Podolsky Rosen Paradox,” Physics 1, 195-200, 1964.
• J. F. Clauser, M. A. Horne, A. Shimony, and R. A. Holt, “Proposed Experiment to Test Local Hidden-Variable Theories,” Physical Review Letters 23, 880-884, 1969.
• R. F. Werner, “Quantum States with Einstein-Podolsky-Rosen Correlations Admitting a Hidden-Variable Model,” Physical Review A 40, 4277-4281, 1989.
• 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.
• R. Horodecki, P. Horodecki, M. Horodecki, and K. Horodecki, “Quantum Entanglement,” Reviews of Modern Physics 81, 865-942, 2009.

Exercises

1. Show that $\rho_{\mathrm{cc}}$ and $\rho_{\Phi^+}$ give the same computational-basis probabilities.

Solution

For

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

the nonzero probabilities are $P(0,0)=1/2$ and $P(1,1)=1/2$.

For

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

the Born-rule probabilities in the computational basis are also $1/2$ for $00$, $1/2$ for $11$, and zero for the other outcomes.

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

Solution

For $\rho_{\mathrm{cc}}$,

$$\operatorname{Tr} \bigl[ \rho_{\mathrm{cc}}(X\otimes X) \bigr] = \frac12\langle00\vert X\otimes X\vert00\rangle + \frac12\langle11\vert X\otimes X\vert11\rangle =0,$$

because $X\otimes X$ maps $\lvert00\rangle$ to $\lvert11\rangle$ and $\lvert11\rangle$ to $\lvert00\rangle$.

For $\lvert\Phi^+\rangle$,

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

so

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

3. Identify the coherence terms in $\rho_{\Phi^+}$ that are absent from $\rho_{\mathrm{cc}}$.

Solution

The Bell projector expands as

$$\rho_{\Phi^+} = \frac12 \Bigl( \lvert00\rangle\langle00\rvert + \lvert00\rangle\langle11\rvert + \lvert11\rangle\langle00\rvert + \lvert11\rangle\langle11\rvert \Bigr).$$

The terms absent from $\rho_{\mathrm{cc}}$ are

$$\frac12\lvert00\rangle\langle11\rvert \qquad \text{and} \qquad \frac12\lvert11\rangle\langle00\rvert.$$

They encode coherence between the two product alternatives.

4. For $W=\frac12 I-\lvert\Phi^+\rangle\langle\Phi^+\rvert$, compute $\operatorname{Tr}(W\rho_{\Phi^+})$ and $\operatorname{Tr}(W\rho_{\mathrm{cc}})$.

Solution

For $\rho_{\Phi^+}=\lvert\Phi^+\rangle\langle\Phi^+\rvert$,

$$\operatorname{Tr}(W\rho_{\Phi^+}) = \frac12-1 = -\frac12.$$

For $\rho_{\mathrm{cc}}$,

$$\langle\Phi^+\vert\rho_{\mathrm{cc}}\vert\Phi^+\rangle = \frac12,$$

so

$$\operatorname{Tr}(W\rho_{\mathrm{cc}}) = \frac12-\frac12 =0.$$