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Give Alice the first qubit $A$ and Bob the second $B$. Compare the Bell state $|{\Phi }^{+}⟩$ with a preparation that tosses a fair classical bit and gives both parties its value. Their density operators, written in the computational basis ordered $00,01,10,11$, areThey have identical diagonal quantum state populations. The off-diagonal entries of the Bell state retain quantum coherence between $00$ and $11$; the classical mixture has none in this basis.

If both parties perform a Pauli measurement of $Z$, the Born rule gives $P(00)=P(11)=1/2$ and $P(01)=P(10)=0$ for both preparations. Perfect agreement here therefore cannot establish entanglement: an ordinary shared random bit already explains it.

Now both measure $X$, whose normalized eigenstates are $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$. Changing basis givesThe Bell state again has matching outcomes, each with probability $1/2$. For either product state $|00⟩$ or $|11⟩$, each of the four $X$ outcome pairs has probability $1/4$, so their classical mixture does too. Equivalently, the Pauli correlator $⟨X\otimes X⟩$ is one for the Bell state and zero for this mixture, while $⟨Z\otimes Z⟩$ is one for both.

Taking a partial trace gives ${\rho }_A={\rho }_B=I_2/2$ for both preparations. Thus every measurement confined to Alice's system has the same statistics in the two cases, and so does every measurement confined to Bob's. They reveal the difference by comparing records through classical communication. Conditional on Alice's result, Bob can predict a matching $X$ result for the Bell state; without that result, his outcomes remain evenly split. An uncommunicated local quantum channel, including a measurement with its outcome discarded, leaves his reduced density matrix unchanged. This is quantum no-signalling, even when conditional states change. ^[ref]

The Von Neumann entropy makes the joint-versus-local distinction numerical. With base-two logarithms, the Bell state has eigenvalues $1,0,0,0$, giving $S(AB)=0$, whereas the mixture has $1/2,1/2,0,0$, giving $S(AB)=1$. Both have $S(A)=S(B)=1$. Their quantum mutual information $I(A:B)=S(A)+S(B)-S(AB)$ is consequently two and one bits, respectively. The value two is not two readable classical bits in either local qubit: the matching $Z$ records have ordinary mutual information one bit in both cases. Random records can come from a globally pure state. ^[ref]

These two matching-axis measurements distinguish the specified preparations; they alone are not a test of a Bell inequality. A local hidden-variable theory could assign shared random answers separately to these two axes. Bell theorem concerns incompatible correlations across a suitable larger collection of measurement settings. Bell's 1964 paper established the obstruction for local hidden-variable descriptions of the predicted correlations. ^[ref]

By the Schmidt decomposition theorem, suitable local unitary operations put the normalized pure state into the formIt is entangled precisely when both Schmidt coefficients are nonzero. Put $s=2{\lambda }_0{\lambda }_1$, so $0<s\le 1$ for the case of interest. The Pauli operatorssatisfy $⟨Z\otimes Z⟩=1$ and $⟨X\otimes X⟩=s$ in this pure state: $Z\otimes Z$ fixes both basis terms, and $X\otimes X$ exchanges them.

Choose Alice's two observables as $A_0=Z$, $A_1=X$, and Bob's asThese are valid dichotomic projective measurements. Indeed, $XZ+ZX=0$ and $X^2=Z^2=I$, so $B_0^2=B_1^2=I$; each observable has outcomes $\pm 1$. For the CHSH inequality convention $|⟨A_0\otimes (B_0+B_1)+A_1\otimes (B_0-B_1)⟩|\le 2$, the quantum value isFor the original pure state, conjugate these four observables by the corresponding local unitary operators. Their outcomes and Pauli correlators are unchanged, and each party still measures only their own qubit. Thus every entangled pure two-qubit state violates CHSH for suitable local measurements, proving Gisin's theorem directly. The CHSH axes for an entangled pure two-qubit state approach a nonviolating configuration continuously as one Schmidt coefficient tends to zero; the strict violation therefore need not be large for weak entanglement.

The Schmidt decomposition of an entangled pure two-qubit state has two nonzero terms. Absorb both complex phases into the local basis vectors and interchange the labels of the second qubit to obtainIn this state the only nonzero same-axis two-qubit Pauli correlators areAll mixed-axis correlators vanish. Expanding $(\mathbf{a}\cdot \mathbf{\sigma })\otimes (\mathbf{b}\cdot \mathbf{\sigma })$ therefore gives

For the four stated vectors this becomesIt follows immediately thatEvery separable quantum state obeys the corresponding CHSH inequality with upper bound two. Set $s=2c_0{c}_1>0$, choose ${\beta }^{\prime }=\pi -\beta$, and take $\tan \beta =s$ with $0<\beta <\pi /2$. The displayed expression is thenThus every entangled pure two-qubit state has local measurement correlations that no separable state can reproduce, which is Gisin's theorem. In an ideal Bose--Marletto--Vedral experiment, optimized local measurements can therefore witness any nonzero pure-state entanglement generated during the gravitational interaction. If gravity is the only interaction between the masses, such a violation shows that the mediator cannot be described by a purely classical local variable under the assumptions of the proposal; experimentally, control of decoherence and nongravitational forces is essential to that inference.

For the pure state ${\lambda }_0|00⟩+{\lambda }_1|11⟩$ with real normalized Schmidt coefficients, the two-body Pauli correlators form $T=\mathrm{diag}(s,-s,1)$, $s=2{\lambda }_0{\lambda }_1$. Thus arbitrary local axes give $⟨(\mathbf{a}\cdot \sigma )\otimes (\mathbf{b}\cdot \sigma )⟩=s(a_x{b}_x-a_y{b}_y)+a_z{b}_z$. The negative $yy$ component follows from the phases in the Pauli operators.