A counterexample to the negative-q commutator bound

Abstract

A traceless complex matrix of size three and a rank-one projector violate the proposed q-deformed commutator inequality at q = −1.

Definition 1.1 (The dimension-dependent coefficient).

$$\forall n:\mathbb{N},g\left(n\right)=\frac{(n:\mathbb{R}{)}^2-3\cdot (n:\mathbb{R})+3}{(n:\mathbb{R})\cdot ((n:\mathbb{R})-1)}$$

Formalization. D5/S3/QuantumBounds/QDeformedCommutatorNegativeQRefutation.g (✓ std3).

Citation. D. Chruściński, G. Kimura, H. Ohno, T. Singal (2022). Bounding the Frobenius norm of a q-deformed commutator. DOI: 10.48550/arXiv.2202.11520. URL: https://arxiv.org/abs/2202.11520v2.

Commentary.

Equation (24) defines the real coefficient g(n) = (n² − 3n + 3)/(n(n − 1)). The natural number n is embedded in the real numbers in every arithmetic operation. For n at least two, the denominator is positive. As a total real-valued function, g uses the real division convention also at n = 0 and n = 1; the inequality below only concerns n at least two.

Definition 1.2 (The inequality in Conjecture 2).

$$claim⟺\forall n:\mathbb{N},2\le n\to \forall q:\mathbb{R},q\le 0\to \forall AB:Matrix(Finn)(Finn)\mathbb{C},(\mathrm{trace}\left(A\right)=0\vee \mathrm{trace}\left(B\right)=0)\to \mathrm{mass}\left(A\cdot B-(q:\mathbb{C})\cdot (B\cdot A)\right)\le \max \left(g\left(n\right)\cdot (1-q{)}^2,1+q^2\right)\cdot \mathrm{mass}\left(A\right)\cdot \mathrm{mass}\left(B\right)$$

Formalization. D5/S3/QuantumBounds/QDeformedCommutatorNegativeQRefutation.claim (✓ std3).

Citation. D. Chruściński, G. Kimura, H. Ohno, T. Singal (2022). Bounding the Frobenius norm of a q-deformed commutator. DOI: 10.48550/arXiv.2202.11520. URL: https://arxiv.org/abs/2202.11520v2.

Commentary.

Chruściński, Kimura, Ohno and Singal define [A,B]q = AB − qBA. Conjecture 2 in Section 3 states that, for any q ≤ 0, if A or B is traceless, the sharp bound has coefficient max[g(n)(1 − q)², 1 + q²]. Here mass M = ∑ᵢ ∑ⱼ |M i j|² is the squared Frobenius norm. Fin n indexes rows and columns by 0,…,n−1. The real scalar q is embedded in the complex numbers for scalar multiplication of matrices. The displayed claim is the inequality clause for all dimensions n at least two, all nonpositive real q and all complex matrix pairs satisfying the disjunction of trace conditions. A violation of the inequality also refutes the assertion that it holds and is sharp.

Theorem 1.3 (A three-dimensional counterexample).

$$\neg claim$$

Proof. Machine-checked in Lean as D5/S3/QuantumBounds/QDeformedCommutatorNegativeQRefutation.result (✓ std3). ∎

Resolves. Problems/chruscinski-2022-qdeformed-commutator-negative-q-refutation (refuted) by D5/S3/QuantumBounds/QDeformedCommutatorNegativeQRefutation.result.

Source. Repository-derived.

Commentary.

Take n = 3, q = −1, A = diag(2,−1,−1) and B = diag(1,0,0). The trace of A is 2 − 1 − 1 = 0, while the trace of B is 1. Their squared Frobenius norms are mass A = 6 and mass B = 1. The products satisfy AB = BA = diag(2,0,0), so AB − qBA = diag(4,0,0) and its squared Frobenius norm is 16. Equation (24) gives g(3) = 1/2, hence max[g(3)(1 − (−1))², 1 + (−1)²] = 2. The proposed bound becomes 16 ≤ 2 · 6 · 1 = 12, a contradiction. The disjunction of trace conditions admits this pair because A is traceless; the pair does not satisfy the stronger condition that both matrices are traceless.

References

• Truth anchor: D5/S3/QuantumBounds/QDeformedCommutatorNegativeQRefutation.claim
• Truth anchor: D5/S3/QuantumBounds/QDeformedCommutatorNegativeQRefutation.g
• Truth anchor: D5/S3/QuantumBounds/QDeformedCommutatorNegativeQRefutation.result
• Dependency: D5/S3/Quantum/Entanglement/MoreauYosidaFormationSelectiveLoccRefutation