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The pure-state HTMB bound

Abstract

Pure preparations satisfying both trit parity equivalences have HTMB success probability at most two thirds in every finite complex dimension.

Definition 1.1 (The average trit-guessing score).

Formalization. D5/S3/Quantum/Contextuality/PureStateHTMBBound.htmbValue (✓ std3).

Citation. Armin Tavakoli; Emmanuel Zambrini Cruzeiro; Roope Uola; Alastair A. Abbott (2021). Bounding and simulating contextual correlations in quantum theory. DOI: 10.1103/PRXQuantum.2.020334. URL: https://arxiv.org/abs/2010.04751v3.

Commentary.

The preparation label x is a pair of residues modulo three. The two matrix families m and n represent the measurements used to guess its first and second coordinates, respectively. For a vector v, projector v is the complex matrix with entries vᵤ times the conjugate of vᵥ. Each summand takes the real trace of this matrix multiplied by the effect for the correct outcome. The factor 1/18 averages over nine preparation labels and two measurement choices. This is the score in equation (33), lines 492–499 of Manuscript_infocontext_reresubmission.tex in arXiv:2010.04751v3. The function is defined for arbitrary vectors and matrices; unit length and measurement positivity and completeness are hypotheses of the bound.

Definition 1.2 (The pure-state conjecture).

Formalization. D5/S3/Quantum/Contextuality/PureStateHTMBBound.claim (✓ std3).

Citation. Armin Tavakoli; Emmanuel Zambrini Cruzeiro; Roope Uola; Alastair A. Abbott (2021). Bounding and simulating contextual correlations in quantum theory. DOI: 10.1103/PRXQuantum.2.020334. URL: https://arxiv.org/abs/2010.04751v3.

Commentary.

For every natural dimension d, all nine preparation vectors have norm one and each of the two three-outcome measurements consists of positive semidefinite matrices summing to the identity. For each coefficient a equal to one or two, the sums of preparation projectors on the three fibres of x₁ + a x₂ are equal. Equality with the zero fibre expresses all these equalities; division by the common fibre size three gives equality of the uniform mixtures. Under both equivalences, the score is at most 2/3. This is the finite-dimensional unit-vector and POVM formulation of the conjecture stated at line 506, following the numerical pure-state bound at lines 502–505 of Manuscript_infocontext_reresubmission.tex in arXiv:2010.04751v3. There is no dimension lower bound or assumption that either measurement is projective.

Theorem 1.3 (Pure states obey the HTMB inequality).

Proof. Machine-checked in Lean as D5/S3/Quantum/Contextuality/PureStateHTMBBound.result (✓ std3). ∎

Resolves. Problems/tavakoli-zambrini-cruzeiro-uola-abbott-2021-pure-state-htmb (proved) by D5/S3/Quantum/Contextuality/PureStateHTMBBound.result.

Source. Repository-derived.

Acknowledgement. Armin Tavakoli; Emmanuel Zambrini Cruzeiro; Roope Uola; Alastair A. Abbott (2021). Bounding and simulating contextual correlations in quantum theory. DOI: 10.1103/PRXQuantum.2.020334. URL: https://arxiv.org/abs/2010.04751v3.

Commentary.

Fix d, ψ, m and n satisfying the hypotheses, and write Pᵢⱼ for the projector of ψ(i,j). First the two parity equivalences force Pᵢⱼ = Pᵢ₀ + P₀ⱼ − P₀₀. To see this entry by entry, let Lₐₖ be the sum of Pᵢⱼ over i + a j = k, and put D₁ = L₁₁ − L₁₀, D₂ = L₁₂ − L₁₀, D₃ = L₂₁ − L₂₀ and D₄ = L₂₂ − L₂₀. All four differences vanish. For Rᵢⱼ = Pᵢⱼ − Pᵢ₀ − P₀ⱼ + P₀₀, expanding the fibre sums gives 3R₁₁ = −2D₁ + D₂ − D₃ − D₄, 3R₁₂ = −D₁ − D₂ − 2D₃ + D₄, 3R₂₁ = −D₁ − D₂ + D₃ − 2D₄ and 3R₂₂ = D₁ − 2D₂ − D₃ − D₄. Thus these four remainders are zero; those with a zero coordinate vanish directly.

Define Dcol = ∑ᵢ Re tr(Pᵢ₀ mᵢ) and Drow = ∑ⱼ Re tr(P₀ⱼ nⱼ). Summing the additive formula over j gives ∑ⱼ Pᵢⱼ = 3Pᵢ₀ + T, where T = ∑ⱼ P₀ⱼ − 3P₀₀ has trace zero because every unit-vector projector has trace one. Since ∑ᵢ mᵢ is the identity, the total contribution of T is Re tr(T) = 0. The first-coordinate part of the score numerator is therefore 3Dcol. Summing in the other order and using ∑ⱼ nⱼ equal to the identity gives 3Drow for the second part. Consequently the score is (Dcol + Drow)/6.

Positivity makes each unit-vector expectation against an effect nonnegative. The sum of the three expectations for a fixed unit vector is one, by measurement completeness, so each individual expectation is at most one. In particular Dcol and Drow are each at most three. If all Pᵢ₀ equal P₀₀, completeness of m gives Dcol = Re tr(P₀₀) = 1, and hence Dcol + Drow is at most four. If all P₀ⱼ equal P₀₀, completeness of n instead gives Drow = 1 and the same bound.

Otherwise choose i and j with Pᵢ₀ different from P₀₀ and P₀ⱼ different from P₀₀. The key geometric fact is that if unit-vector projectors satisfy P(a) − P(b) = P(c) − P(e) ≠ 0, then a and b lie in the span of c and e. Indeed, a and b are linearly independent: the unit-ray projector identity makes their projectors equal whenever the unit vectors are scalar multiples. For w orthogonal to c and e, applying the matrix equality to w gives ⟨a,w⟩a = ⟨b,w⟩b. Independence forces both coefficients to vanish. Taking the orthogonal complement twice in finite dimension puts a and b in the stated span.

Take s to be the complex span of ψ(0,j) and ψ(0,0), whose dimension is at most two. For every k the additive identity gives Pₖⱼ − Pₖ₀ = P₀ⱼ − P₀₀ ≠ 0, so the geometric fact puts ψ(k,0) in s. Likewise Pᵢₖ − P₀ₖ = Pᵢ₀ − P₀₀ ≠ 0 puts ψ(0,k) in the span of ψ(i,0) and ψ(0,0). Both generators already lie in s. Thus all vectors on the two coordinate axes lie in the same s.

For a unit vector v in s and an arbitrary w, orthogonal projection onto s followed by Cauchy–Schwarz bounds |⟨v,w⟩|² by the squared norm of that projection. Parseval expresses this squared norm as ∑ₖ |⟨eₖ,w⟩|² for an orthonormal basis of s. A positive semidefinite matrix is a finite sum of matrices z z*. Applying the overlap bound to each z and summing gives Re tr(P(v) A) ≤ ∑ₖ Re tr(P(eₖ) A) for every positive semidefinite A. Apply this to each of the three effects and its corresponding axis vector. Interchange the sums and use completeness: each basis vector contributes Re tr(P(eₖ)) = 1. The resulting discrimination sum is at most dim(s), hence at most two. Using m for the first axis and n for the second proves Dcol ≤ 2 and Drow ≤ 2. In every case Dcol + Drow ≤ 4, so the score is at most 4/6 = 2/3.

The bound concerns all finite complex dimensions with both parity equivalences. It does not assert an infinite-dimensional extension or a bound for mixed preparations.

References