Fixed-cost quantum transport metrics

The problem asks whether one state-independent cost operator can make coupling-based quantum transport a metric, or its square root a metric, on all quantum states. The new theorem answers no for dimensions $d \ge 3$, but identifies a positive result for $d=2$.

Known results

• Friedland et al., 2022: the antisymmetric-projector cost gives a semidistance; for single-qubit states, its square root satisfies the triangle inequality.
• For $d=2$, admissibility forces the cost to be a positive multiple of $P_-=(\mathbb{1}-F)/2$, whose square-root transport is a metric.

2026 no-go theorem

For every admissible fixed Hermitian cost and $d \ge 3$, explicit mutually commuting states violate the triangle inequality for both $T_C$ and $W_C=\sqrt{T_C}$. The obstruction is uniform over costs and arises from coherence in optimal couplings. Miller’s earlier diagonal-qutrit counterexamples refuted the corresponding SWAP conjectures. A separate result rules out channel-induced fixed linear costs as semidistances for $d \ge 2$.

Current status (as of July 2026): The fixed-cost coupling question is settled negatively for $d \ge 3$ and positively for $d=2$ within the stated framework; suitable replacements for higher-dimensional quantum transport remain open.