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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]

Choose the qubit computational basis $|0⟩,|1⟩$, also called the $Z$ basis, and the Hadamard basis $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$, also called the $X$ basis. Compare a coherent preparation of $|+⟩$ with choosing $|0⟩$ or $|1⟩$ at random with equal probabilities and discarding that label:These are original ideal preparation models. Their $Z$-basis probabilities are both $(1/2,1/2)$, because their quantum state populations agree. Their $X$-basis probabilities are respectively $(1,0)$ and $(1/2,1/2)$. Indeed $⟨+|{\rho }_{+}|+⟩=1$, whereas $⟨+|I/2|+⟩=1/2$. The missing off-diagonal quantum coherence therefore changes a prediction that the common populations cannot determine.

For a relative quantum phase $\varphi$, the coherent family instead hasThe last formula follows by adding or subtracting the two probability amplitudes before taking the squared modulus. At $\varphi =\pi$, the state is $|-⟩$ and the $X$ outcomes switch to $(0,1)$, while the $Z$ outcomes remain $(1/2,1/2)$. In contrast, $-|+⟩$ changes both components by the same global phase and retains the $X$ outcomes $(1,0)$. In photon polarization, the identification $|0⟩=|H⟩$, $|1⟩=|V⟩$ makes $|+⟩$ and $|-⟩$ the two orthogonal diagonal linear polarizations; these labels describe measurement bases, not spatial axes for photon spin.

Even perfect statistics in one basis cannot certify a preparation. The above $Z$ data fit both density matrices. Conversely, distinct preparation histories can yield the same quantum state:The off-diagonal terms cancel in the right-hand ensemble. Every measurement on the qubit alone then has identical probabilities for those two histories. A retained classical label would allow conditioning on different members of the quantum state ensemble; it is extra information beyond the unlabelled density operator.

A dephasing channel suppresses off-diagonal entries of a density matrix in a preferred basis while preserving quantum state populations. Complete dephasing is identical to a projective measurement in that basis with its outcome discarded.

For the Bell correlations versus a classical mixture example, completely dephasing Alice's qubit in the computational basis maps ${\rho }_{\mathrm{Bell}}$ to ${\rho }_{\mathrm{mix}}$. It preserves their perfect $Z$ agreement and each reduced density matrix, yet destroys the $X$ correlation. This is why testing only populations can miss damage to a quantum memory: the stored quantum coherence may have disappeared while every preferred-basis measurement still looks correct. The joint Von Neumann entropy rises from zero to one bit, although each local entropy stays fixed.

A projective measurement has self-adjoint orthogonal projections $P_i$ satisfying $P_i{P}_j={\delta }_{ij}{P}_i$ and ${\sum }_i{P}_i=I$. Measuring an observable with discrete spectral decomposition $A={\sum }_i{a}_i{P}_i$ reports eigenvalue $a_i$ with Born rule probability $p_i=\mathrm{Tr}(P_i\rho )$. A projector can have rank greater than one: the outcome then identifies an eigenspace, rather than a particular eigenvector inside it. Projectors specify the outcome statistics; the usual ideal conditional update is the additional Lüders rule.

In the qubit example from measurement in quantum mechanics, $Z=|0⟩⟨0|-|1⟩⟨1|$. At $\theta =\pi /3$, the zero and one probabilities are $3/4$ and $1/4$, regardless of the relative quantum phase $\varphi$. Under the Lüders rule, obtaining zero prepares $|0⟩$; obtaining one prepares $|1⟩$, up to global phase. Repeating the same measurement returns the same label with certainty, provided no intervening evolution or operation changes the measured sector. Repeatability concerns the conditional output, and does not mean the first outcome was predetermined.

Change to $X$ and its Pauli X eigenstates $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$. Adding the two probability amplitudes before squaring givesAt $\theta =\pi /3$ and $\varphi =0$, $p_{+}=(2+\sqrt{3})/4\simeq 0.9330$ and $p_{-}\simeq 0.0670$. At $\varphi =\pi /2$, both are $1/2$; at $\varphi =\pi$, the first pair is exchanged. The $Z$ statistics remain $3/4,1/4$ throughout. This is the same addition-and-cancellation principle as wave interference: relative quantum phase changes how amplitudes reinforce or cancel in another basis.

The beauty of this basis dependence is that information invisible to one measurement becomes accessible to another. A $Z$ measurement of an ensemble estimates its quantum state populations, while $X$ probes one component of its quantum coherence. These are experiments on separate identically prepared copies. Sequentially measuring $Z$ and then $X$ on one copy instead probes the disturbed post-measurement state. Quantum state tomography combines suitable ensemble statistics to reconstruct the original state.