QTD squared can exceed QJS2

Abstract

A full-rank commuting qubit pair has quantum triangular discrimination squared greater than base-2 quantum Jensen–Shannon divergence.

Yupan Liu, Quantum state testing beyond the polarizing regime and quantum triangular discrimination, arXiv:2303.01952v5; Computational Complexity 34 (2025), article 11. §1.3, Improved inequalities on the quantum triangular discrimination, printed p. 7, verbatim: “We observe that Theorem 1.4(ii) is not tight. Numerical simulations indicate that the tight bound is QTD²(ρ₀, ρ₁) ≤ QJS₂(ρ₀, ρ₁) ≤ QTD(ρ₀, ρ₁) for any states ρ₀ and ρ₁.” After Theorem 3.4, verbatim: “Numerical simulations suggest that the tight bound is QTD²(ρ₀, ρ₁) ≤ QJS₂(ρ₀, ρ₁) ≤ QTD(ρ₀, ρ₁), while we only managed to prove a slightly weaker bound ½QTD²(ρ₀, ρ₁) ≤ QJS₂(ρ₀, ρ₁) ≤ QTD(ρ₀, ρ₁) in Theorem 3.4.”

Definition 1.1 (Quantum Jensen–Shannon divergence).

$$\forall n:\mathbb{N},\forall rho:\mathrm{Matrix}\left(\mathrm{Fin}\left(n\right),\mathrm{Fin}\left(n\right),\mathbb{C}\right),\forall sigma:\mathrm{Matrix}\left(\mathrm{Fin}\left(n\right),\mathrm{Fin}\left(n\right),\mathbb{C}\right),QJS2\left(rho,sigma\right)=vonNeumann2\left((\frac{1}{2})\cdot (rho+sigma)\right)-\frac{vonNeumann2\left(rho\right)+vonNeumann2\left(sigma\right)}{2}$$

Formalization. D5/S3/Divergence/QuantumTriangularJensenShannonRefutation.QJS2 (✓ std3).

Citation. Yupan Liu (2025). Quantum state testing beyond the polarizing regime and quantum triangular discrimination. DOI: 10.48550/arXiv.2303.01952. URL: https://arxiv.org/abs/2303.01952v5.

Commentary.

Definition 2.10 with base-2 logarithms is the entropy of the average state minus the average entropy.

Definition 1.2 (Quantum triangular discrimination).

$$\forall n:\mathbb{N},\forall rho:\mathrm{Matrix}\left(\mathrm{Fin}\left(n\right),\mathrm{Fin}\left(n\right),\mathbb{C}\right),\forall sigma:\mathrm{Matrix}\left(\mathrm{Fin}\left(n\right),\mathrm{Fin}\left(n\right),\mathbb{C}\right),\mathrm{QTD}\left(rho,sigma\right)=\frac{\mathrm{re}\left(\mathrm{trace}\left(\left(rho-sigma\right)\cdot \mathrm{cfc}\left((x\mapsto x^{\frac{-1}{2}}),rho+sigma\right)\cdot \left(rho-sigma\right)\cdot \mathrm{cfc}\left((x\mapsto x^{\frac{-1}{2}}),rho+sigma\right)\right)\right)}{2}$$

Formalization. D5/S3/Divergence/QuantumTriangularJensenShannonRefutation.QTD (✓ std3).

Citation. Yupan Liu (2025). Quantum state testing beyond the polarizing regime and quantum triangular discrimination. DOI: 10.48550/arXiv.2303.01952. URL: https://arxiv.org/abs/2303.01952v5.

Commentary.

Definition 3.1 uses the inverse square root of the sum between two copies of the difference. The function x↦x^(−1/2) has value zero at zero in the real-power convention, so the inverse acts on the support of the sum.

Definition 1.3 (The proposed lower inequality).

$$claim\iff \forall n:\mathbb{N},\forall rho:\mathrm{Matrix}\left(\mathrm{Fin}\left(n\right),\mathrm{Fin}\left(n\right),\mathbb{C}\right),\forall sigma:\mathrm{Matrix}\left(\mathrm{Fin}\left(n\right),\mathrm{Fin}\left(n\right),\mathbb{C}\right),(\mathrm{IsState}\left(rho\right))\Rightarrow ((\mathrm{IsState}\left(sigma\right))\Rightarrow (\mathrm{QTD}{\left(rho,sigma\right)}^2\le QJS2\left(rho,sigma\right)))$$

Formalization. D5/S3/Divergence/QuantumTriangularJensenShannonRefutation.claim (✓ std3).

Citation. Yupan Liu (2025). Quantum state testing beyond the polarizing regime and quantum triangular discrimination. DOI: 10.48550/arXiv.2303.01952. URL: https://arxiv.org/abs/2303.01952v5.

Commentary.

For every finite dimension and every pair of quantum states, the assertion requires QTD²≤QJS₂. This is the lower conjunct of the proposed two-sided bound.

Theorem 1.4 (A commuting qubit counterexample).

$$\neg (\forall n:\mathbb{N},\forall rho:\mathrm{Matrix}\left(\mathrm{Fin}\left(n\right),\mathrm{Fin}\left(n\right),\mathbb{C}\right),\forall sigma:\mathrm{Matrix}\left(\mathrm{Fin}\left(n\right),\mathrm{Fin}\left(n\right),\mathbb{C}\right),(\mathrm{IsState}\left(rho\right))\Rightarrow ((\mathrm{IsState}\left(sigma\right))\Rightarrow (\mathrm{QTD}{\left(rho,sigma\right)}^2\le QJS2\left(rho,sigma\right))))$$

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

Resolves. Problems/liu-2025-qtd-qjs-lower-bound-refutation (refuted) by D5/S3/Divergence/QuantumTriangularJensenShannonRefutation.result.

Source. Repository-derived.

Acknowledgement. Yupan Liu (2025). Quantum state testing beyond the polarizing regime and quantum triangular discrimination. DOI: 10.48550/arXiv.2303.01952. URL: https://arxiv.org/abs/2303.01952v5.

Commentary.

Take ρ=diag(255/256,1/256) and σ=diag(1/256,255/256). Both states have strictly positive diagonal entries and unit trace. Their sum is the identity, so its inverse square root is the identity. The difference has diagonal entries 127/128 and −127/128, giving QTD=(127/128)². The average is the maximally mixed qubit state with entropy one. Both input entropies equal h(1/256), where h(a)=−a log₂ a−(1−a) log₂(1−a). Since log₂(1/256)=−8 and log₂(255/256)<0, h(1/256)≥1/32. Thus QJS₂≤31/32<(127/128)⁴=QTD², and QTD²−QJS₂≥97793/268435456. The matrices commute, so the failure already occurs for classical probability distributions. The general family diag(1−a,a), diag(a,1−a) has Q=(1−2a)² and J=1−h(a). Mathematical reasoning, not additional formal theorems here, gives (1−J)/(1−Q)=h(a)/(4a(1−a))→∞ as a→0⁺. Consequently no fixed p>0 makes Q^p≤J universal, and no function F with F(1)=1 that is locally Lipschitz at one makes F(Q)≤J universal. The logarithmic entropy deficit near orthogonal supports causes this obstruction. The upper inequality QJS₂≤QTD and the metric assertion for √QTD are not addressed.

References

• Truth anchor: D5/S3/Divergence/QuantumTriangularJensenShannonRefutation.QJS2
• Truth anchor: D5/S3/Divergence/QuantumTriangularJensenShannonRefutation.QTD
• Truth anchor: D5/S3/Divergence/QuantumTriangularJensenShannonRefutation.claim
• Truth anchor: D5/S3/Divergence/QuantumTriangularJensenShannonRefutation.result
• Dependency: D5/S3/Divergence/CoherenceL1RelativeEntropyBound