Peres–Mermin Square

The Peres–Mermin square organizes nine two-qubit observables into six compatible contexts. Quantum mechanics fixes the product in each context, but no context-independent assignment of nine signs reproduces all six products. This page turns the Kochen–Specker parity proof into a statistical witness and explains the extra work required to interpret imperfect experiments.

Helpful background. Kochen–Specker Theorem establishes the parity contradiction and dimension assumptions.

Compatible rows and columns

With $X,Y,Z$ the Pauli matrices and juxtaposition denoting a tensor product, use

$$\begin{pmatrix} XI&IX&XX\\ IY&YI&YY\\ XY&YX&ZZ \end{pmatrix}.$$

Each row and each column consists of commuting observables with eigenvalues $\pm1$. The three row products and first two column products are $+I$; the last column product is $-I$. For example, $XX\,YY=(XY)\otimes(XY)=-ZZ$, so multiplying by $ZZ$ gives $-I$. The different notation $XY$ inside a tensor-product label must not be confused with the single-qubit matrix product $X Y=iZ$ used in that calculation.

For a joint measurement of any context, the product of the three outcomes consequently has expectation $+1$ or $-1$ for every density operator. State independence here means no specially chosen input state is needed for these operator identities; it does not remove assumptions about how the observables are implemented.

From an impossible assignment to a bound

Denote the random row products by $R_1,R_2,R_3$ and column products by $C_1,C_2,C_3$. Define

$$W=\langle R_1\rangle+\langle R_2\rangle+\langle R_3\rangle +\langle C_1\rangle+\langle C_2\rangle-\langle C_3\rangle.$$

For a deterministic noncontextual assignment, every cell has one sign reused in its row and column. Multiplying all six context products counts each cell twice, hence gives $+1$. The six signed terms in $W$ have product $-1$, so at least one equals $-1$. Their sum is at most $4$. Convex mixtures of such assignments retain $W\le4$. Quantum theory predicts $W=6$.

This bound belongs to a model of sharp, compatible observables with context-independent values. Generalized operational noncontextuality for unsharp, noisy procedures needs explicit operational-equivalence assumptions and may require a different inequality. Merely measuring six correlations and finding $W>4$ is insufficient unless the correspondence to the hypothesized contexts has been justified.

A transparent noise benchmark

As a diagnostic model, suppose each of the three recorded signs in every context is independently flipped with probability $e$. A recorded triple product then has expectation reduced by $(1-2e)^3$. The ideal witness becomes

$$W(e)=6(1-2e)^3.$$

The inequality is violated for

$$e<\frac{1-(2/3)^{1/3}}2\approx0.0632.$$

This is not a universal experimental error threshold. Correlated errors, context-dependent recalibration, or disturbance among sequential measurements are not independent flips. The calculation is useful because it makes one error model reproducible and shows why small imperfections cannot be treated as a single unspecified visibility.

Compatibility and identity of observables

Each cell occurs in two contexts. Its operational implementation must correspond to the same observable in both, within the assumptions and tolerances of the analysis. Sequential measurements can disturb later outcomes even if their nominal matrix representatives commute. Tests of repeatability, order dependence and marginal statistics provide evidence about the implemented procedures; they cannot be replaced by the matrix commutator alone.

The square uses the four-dimensional Hilbert space of two qubits. It does not contradict the projective single-qubit exception to Kochen–Specker. Nor is every context a set of measurements independently available to spatially separated parties, so this witness should not be relabeled a Bell test without specifying a different operational scenario.

Exercise: one flipped cell cannot repair parity

Start with all nine assigned signs $+1$. Flip the sign of one cell. How many of its row and column products change, and can this produce the quantum pattern?

Solution

Exactly two context products change: the row and column containing that cell. The product of all six stays $+1$. Repeating any number of cell flips preserves this invariant, whereas the target quantum pattern has product $-1$. The obstruction concerns reuse of the same cell value across contexts, not an inability to assign signs within any single context.

References

• Cabello, Adán, “Experimentally Testable State-Independent Quantum Contextuality,” Physical Review Letters 101, 210401 (2008), doi:10.1103/PhysRevLett.101.210401 — inequalities suitable for state-independent witnesses.
• Mermin, N. David, “Simple unified form for the major no-hidden-variables theorems,” Physical Review Letters 65, 3373–3376 (1990), doi:10.1103/PhysRevLett.65.3373 — the finite operator construction.
• Spekkens, Robert W., “Contextuality for preparations, transformations, and unsharp measurements,” Physical Review A 71, 052108 (2005), doi:10.1103/PhysRevA.71.052108 — the generalized operational framework.