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. Entanglement means that no mixture of product states describes the joint density operator across the chosen subsystem split. Off-diagonal coherence in one basis, by itself, is not an entanglement criterion.

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.

Helpful background. Separable mixed states explains mixtures of products; Bell states introduces the two-qubit state compared here.

Identical correlations in one basis

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

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

Squaring the two Bell-state amplitudes gives exactly the same probabilities.

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 two states therefore respond differently to the new measurement basis. This is not a claim that the classical mixture has no off-diagonal entries in every basis: coherence is basis dependent, whereas entanglement is invariant under local basis changes.

Two correlations give an entanglement test

Knowing that two candidate states differ is weaker than excluding every separable state. For two qubits there is a useful bound:

$$|\langle X\otimes X\rangle|+|\langle Z\otimes Z\rangle|\le1 \qquad\text{for every separable state}.$$

For a product state, let the local Bloch components be $a_x,a_z$ and $b_x,b_z$. Cauchy–Schwarz gives

$$|a_xb_x|+|a_zb_z| \le\sqrt{a_x^2+a_z^2}\sqrt{b_x^2+b_z^2}\le1.$$

A convex mixture obeys the same bound by the triangle inequality. The Bell state gives $2$, while $\rho_{\mathrm{cc}}$ gives $1$. Thus sufficiently strong correlations in these two complementary settings certify entanglement without reconstructing the entire density matrix. Values at or below $1$ are inconclusive.

The axes abbreviate $XX=X\otimes X$ and $ZZ=Z\otimes Z$. Every separable state lies in the gray diamond; points outside certify entanglement, while points inside are inconclusive. Along $\rho_v=v|\Phi^+\rangle\langle\Phi^+|+(1-v)I_4/4$, the correlations are $(v,v)$ and this test detects entanglement only for $v>1/2$.

This test assumes that the measured observables are the calibrated qubit Pauli operators. A Bell test makes a different claim under a different set of assumptions.

A fidelity witness

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 normalized product factors, Cauchy–Schwarz gives $|\langle\Phi^+|a,b\rangle|^2=|a_0b_0+a_1b_1|^2/2\le1/2$. Therefore

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

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,

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

The threshold is fidelity $\langle\Phi^+|\rho|\Phi^+\rangle>1/2$. An expectation of zero or greater does not establish separability. See entanglement witnesses for how the observable is chosen.

Comparison with Bell inequalities

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 even admit local hidden-variable models for all projective measurements. Bell violation is therefore a stronger requirement than entanglement in that measurement scenario; it is not the definition. The Bell theorem states the locality and measurement assumptions.

Common pitfalls

An $X$ correlation alone is not enough. The product state $|++\rangle$ has $\langle X\otimes X\rangle=1$. The separable bound above combines information from incompatible local observables.

Local mixedness is inconclusive for a mixed joint state. The Bell state, the correlated mixture, and $I_4/4$ all have marginals $I/2$.

Separable is broader than fully classical. A separable preparation may use nonorthogonal local states and have nonzero quantum discord. The nonorthogonal preparation example illustrates why absence of entanglement does not remove every quantum feature.

References

• Ryszard Horodecki, Paweł Horodecki, Michał Horodecki, and Karol Horodecki, “Quantum entanglement,” Reviews of Modern Physics 81, 865–942 (2009). DOI.
• Michael A. Nielsen and Isaac L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010, section 2.5.
• Reinhard F. Werner, “Quantum states with Einstein–Podolsky–Rosen correlations admitting a hidden-variable model,” Physical Review A 40, 4277–4281 (1989). DOI.

Exercises

1. Compute $\langle X\otimes X\rangle$ and $\langle Z\otimes Z\rangle$ in the product state $|++\rangle$. Does it saturate the separable bound?

Solution

The local expectations are $\langle X\rangle=1$ and $\langle Z\rangle=0$, so the joint expectations are $1$ and $0$. Their absolute values sum to $1$, exactly the separable bound. Strong correlation in the $X$ setting alone does not certify entanglement.

2. Consider partial dephasing with a possible phase shift,

$$\rho_c=\frac12\bigl(|00\rangle\langle00|+|11\rangle\langle11| +c|00\rangle\langle11|+c^*|11\rangle\langle00|\bigr), \quad c\in\mathbb C,\quad |c|\le1.$$

Check positivity and compute the $XX$, $XY$, and $ZZ$ correlations. When can the fixed $XX,ZZ$ test miss entanglement? Show how replacing Bob’s $X$ axis by $B_\phi=\cos\phi\,X+\sin\phi\,Y$ detects every entangled member of this family.

Solution

The nonzero block has eigenvalues $(1+|c|)/2$ and $(1-|c|)/2$, both nonnegative. Directly taking the Pauli traces gives

$$\langle ZZ\rangle=1,\qquad \langle XX\rangle=\operatorname{Re}c,\qquad \langle XY\rangle=-\operatorname{Im}c.$$

The fixed test violates its separable bound when $\operatorname{Re}c\ne0$, but misses a purely imaginary nonzero coherence. Rotating Bob’s calibrated measurement axis recovers it:

$$\langle X\otimes B_\phi\rangle =\operatorname{Re}(c e^{i\phi}).$$

Because $B_\phi$ and $Z$ are perpendicular Bloch axes, the same Cauchy–Schwarz argument gives $|\langle X\otimes B_\phi\rangle|+|\langle Z\otimes Z\rangle|\le1$ for every separable state. For $c\ne0$, choose $\phi=-\arg c$; the score is $1+|c|>1$. At $c=0$, the state is explicitly the separable mixture $\rho_{\mathrm{cc}}$.

Thus every nonzero coherence in this particular family implies entanglement, even when its phase hides it from the original settings. This does not make an arbitrary off-diagonal entry an entanglement criterion, nor does failure of a fixed witness establish separability.

3. Show that for any two-qubit density matrix in the ordered computational basis,

$$\langle X\otimes X\rangle=2\operatorname{Re}(\rho_{00,11}+\rho_{01,10}).$$

Solution

$X\otimes X$ exchanges $00\leftrightarrow11$ and $01\leftrightarrow10$. Taking the trace selects the four corresponding entries of $\rho$. Hermiticity pairs each entry with its complex conjugate, giving the stated real parts. This also shows why $\langle XX\rangle$ is not a measurement of just one coherence for an arbitrary state.

4. Let $\rho_v=v|\Phi^+\rangle\langle\Phi^+|+(1-v)I_4/4$, with $0\le v\le1$. Compare the noise thresholds of the two-correlation bound and the fidelity witness.

Solution

Both $XX$ and $ZZ$ expectations equal $v$, so their sum exceeds $1$ when $v>1/2$. The Bell fidelity is $v+(1-v)/4=(1+3v)/4$, hence

$$\operatorname{Tr}(W\rho_v)=\frac{1-3v}{4}<0 \quad\Longleftrightarrow\quad v>1/3.$$

The fidelity witness detects more of this noisy family. The weaker test misses entanglement for $1/3<v\le1/2$, illustrating that a failed witness test is inconclusive.