Metadata

slug: rexiti-memarzadeh-mancini-2022-positive-kraus-side-entanglement
bibkey: rexiti2022dephasing
doi: 10.1088/1751-8121/ac6d2c
url: https://arxiv.org/abs/2201.00388v2
triage: theorem
motivation_gids:
  - D5/S3/QuantumChannels/PositiveKrausSideEntanglementRefutation.result

Positive Kraus operators do not exclude a side-entanglement advantage

Problem

M. Rexiti, L. Memarzadeh and S. Mancini, Discrimination of dephasing channels, J. Phys. A 55, 245301 (2022), arXiv:2201.00388v2, Section 7, conjecture:

In contrast, if all Kraus operators have simply the identity as unitary part in the polar decomposition, there is no benefit from side entanglement.

The complete source passage is in the Library note. Issue #16535 preregisters the negative clause, its representation-dependent reading and the exact finite-dimensional conventions.

Statement echo

Source clause or convention	Lean counterpart and scope
Kraus operators have identity polar unitary	`claim` assumes every Kraus operator is strictly positive definite, which makes that unitary uniquely the identity; this stronger hypothesis suffices to refute the source statement.
Channels and side entanglement	Both channels act on complex qubit matrices; the ancilla is a qubit and the first tensor factor. These are instances of the source’s unrestricted systems.
Trace-preserving Kraus representations	Both right Gram sums equal the identity, with the same finite number of operators.
No benefit	For every assisted density and binary effect, some unassisted density and binary effect achieve at least the same success probability, with equal priors (`successAs` and `successUn`).
Density and effect	Frozen `GHZMeasureBiseparableBound.IsDensity` is the positive-semidefinite, trace-one predicate. The preregistered effect condition and the two Kraus sums are inlined literally; proofs use the frozen Kraus and Bell owners.
Negative clause fails	`result : ¬ claim`, with four operators per channel.

Polar factors describe a Kraus representation. The success probabilities depend on the channels represented. A positive-definite representation therefore needs a separate argument before it can rule out an entanglement advantage.

Gap

The source’s phase-damping theorem does not prove the general negative clause. The literature screen recorded by the orchestrator in #16535 covers the source v2 and all ten citing works returned by Semantic Scholar: 2607.03757, 2506.01851, 2411.11313, 2407.14935, 2402.18941, 2402.05793, 2401.15634, 2212.10139, 2211.15108 and 2205.05736. That screen reports no treatment of this polar-factor criterion. The two same-group follow-ups do not settle it. The positive-Kraus and assisted-discrimination ingredients are published separately; their combination refutes this particular clause. Literature status: not-found-in-searched-scope, with the scope and attribution above.

Route

For the tetrahedron $T = {(1, 1, 1), (1, - 1, - 1), (- 1, 1, - 1), (- 1, - 1, 1)}$ use

$K_{s} (a) = \frac{a I + s _{x} X + s _{y} Y + s _{z} Z}{2 a ^{2} + 3}, a \in {2, 3} .$

The consumed private helpers prove positive definiteness, completeness, and

$Φ_{a} (A) = λ_{a} A + \frac{1 - λ _{a}}{2} Tr (A) I, λ_{a} = \frac{a ^{2} - 1}{a ^{2} + 3} .$

Two nonnegative trace pairings bound every unassisted strategy by $47/84$ . For the Bell density $P$ , the assisted input and effect are both $P$ , and the success probability is $33/56$ . Since $33/56 - 47/84 = 5/168 > 0$ , the asserted universal comparison fails.

Falsifier

The refutation requires the literal right Gram normalization, positive definiteness of all eight operators, a valid Bell density and effect, the universal unassisted upper bound, and the assisted success equality. Changing the source interpretation to require positivity in every representation is a different conjecture. No such representation-invariant statement is settled here.

Evidence

Canonical source: PositiveKrausSideEntanglementRefutation.lean, with public definitions successUn, successAs, claim, and the single public theorem result. The helpers below are private compiled declarations in that module. The axiom closure of every public declaration is contained in {propext, Classical.choice, Quot.sound}.

The frozen Kraus owner FiniteKrausChannel.PhyslibLeaf.MatrixMap.of_kraus supplies the linear-map expression used in the proofs. Frozen BroadcastingChannelExtremaRefutation.density_le_one and RHLinalg.trace_mul_nonneg_of_posSemidef supply the density-complement and trace-pairing bounds. Frozen BellPureStateMixedMarginal.bell_pure_state_has_maximally_mixed_marginal supplies Bell idempotence, and CHSHWitness.bell_density_is_state supplies positivity and trace one. The density-complement bound is reused from its published owner.

Experiment: pinned experiment directory. Command: python3 check.py in that directory; exit 0, ALL_OK; script SHA-256 d35ec2a9e0f1c617aa69b2a4808b61dfc108dccdc119ebe7967ef3d831cf2a60. Tested scope: exactly $a = 2, 3$ , with symbolic entries of a qubit matrix. It checks Hermiticity, positive eigenvalues, completeness, independence, the channel formula, $λ_{2} = 3/7$ , $λ_{3} = 2/3$ , $δ = 5/21$ , the scalar unassisted value $47/84$ , the Bell-output difference spectrum ${5/28, - 5/84, - 5/84, - 5/84}$ and the Bell success $33/56$ . The universal unassisted inequality is supplied by Lean, rather than by a finite sample of states.

Triage

Tier 1 external named conjecture, resolution Refuted, preregistered in #16535. Module admission_basis: open-problem-resolution (#16535; Refuted). Every theorem has proof_shape: bind-only; the private helpers have live consumers. No escape witness is claimed. The computational certificates have utility kind=certified-instance; basis=refutes the literal claim. Delivery form: deposit-uncovered, no atom and no cover.

The public result is the designated refutation result (basis=refutes) and is exempt from four-slot escape registration (CLAUDE.md §3.9).

What the settlement shows

Target item	Status and evidence
Positive tetrahedral representations and side-entanglement advantage	Kernel-checked: `kraus_pos_def`, `kraus_complete`, `kraus_apply_formula`, `unassisted_bound`, `assisted_success`, and `result` prove the concrete representation and strict advantage for $a = 2, 3$ .
Channel-pair versus representation mechanism; Pauli representation of the same channels	Proved in prose: the Pauli calculation below exhibits the same channel with nonidentity polar factors. It explains representation dependence; the Pauli comparison is not an additional Lean theorem.
Published ingredients	Literature reading: DeBrota–Stacey, arXiv:1907.10999, Proposition 1, treats positive Kraus representations of depolarizing channels; Sacchi, quant-ph/0505174 (also quant-ph/0506072), treats entanglement-assisted channel discrimination. These ingredients are prior art, not new standalone results.
Representation-invariant criterion	Open: neither the universal quantification over representations nor a general criterion from canonical Kraus operators is proved here.
Orthogonal/canonical reading for a fixed support decomposition	Proved in prose: the pinching argument below leaves no channel parameter to discriminate within that fixed decomposition. Comparisons between different decompositions are outside that argument.
Exact numerical check	Computed: the pinned experiment entry gives the command, exit, hash, exact values and tested scope; no extension to all $a > 3$ is attributed to the experiment.
Consequences for the source’s other results	Literature reading: the source’s proved phase-damping no-advantage result and its cited Pauli and amplitude-damping examples are unaffected. The failed negative clause supplies no general criterion for further channels.

Pauli representation (prose derivation). For $λ = λ_{a}$ , put $p_{0} = (1 + 3 λ) /4 = a^{2} / (a^{2} + 3)$ and $p_{x} = p_{y} = p_{z} = (1 - λ) /4 = 1/ (a^{2} + 3)$ . The weighted Pauli Kraus operators $p_{0} I, p_{x} X, p_{y} Y, p_{z} Z$ have unitary polar factors $I, X, Y, Z$ . For any qubit matrix, $X A X + Y A Y + Z A Z = 2 Tr (A) I - A$ , by its four entries. Substitution gives exactly the displayed $Φ_{a}$ . Thus changing these representations changes polar factors while preserving both channel maps and their discrimination probabilities, including the tensor extension.

Unassisted sharpness (prose derivation). Take the input and effect both equal to the projector onto the first computational basis vector. Then $Tr (N σ) - Tr (N) /2 = 1/2$ . The expression in success_un_formula gives $47/84$ . Lean proves the universal upper bound; this paragraph supplies attainability without claiming an additional compiled theorem.

Orthogonal positive Kraus operators (prose derivation). If positive semidefinite $K_{i}$ are Hilbert–Schmidt orthogonal, $Tr (K_{i} K_{j}) = 0$ for $i \neq = j$ . This trace is the squared Hilbert–Schmidt norm of $K_{i}^{1/2} K_{j}^{1/2}$ , so their supports are orthogonal. On the support of $K_{i}$ , the completeness equation $\sum_{j} K_{j}^{2} = I$ forces $K_{i}^{2} = I$ ; positivity makes $K_{i}$ the support projector. The channel is the pinching $A \mapsto \sum_{i} P_{i} A P_{i}$ . Fixing those support projectors fixes the channel, so a pair with that same fixed decomposition is identical and has no discrimination advantage. Positivity in an arbitrary nonorthogonal representation does not impose this restriction, as the concrete refutation shows.

ASSUMED-UNVERIFIED

Literature completeness outside the source and ten citing works recorded in #16535 is unverified. The general $a > 3$ family, a representation-invariant criterion, and comparison of arbitrary distinct canonical pinching decompositions are not settled by the delivered Lean theorem. The Pauli-representation and pinching arguments are prose mathematics, not additional kernel-checked claims.

trureturing — Discovering truth