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RESEARCH / Focused target

A commuting obstruction to the nonlinear measurement monotone

Literature status: not recheckedRepository record: refuted

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Problem

The question

A. Rico and K. Życzkowski, Discrete dynamics in the set of quantum measurements, J. Phys. A 57 (2024) 435302, arXiv:2308.05835v2, Appendix C.1, Conjecture 1, asks whether blockwise bistochastic dynamics preserve the centered-prefix Hilbert–Schmidt norm in the following sense:

Let $P\in\Delta_{n,d}$ and $Q\in\Delta_{n,d}$ be blockwise probability vectors. If there exists a blockwise bistochastsic matrix $B\in B_{n,d}$ such that $Q=B\ast P$, then for any ordering of ${Q_i}$ there exists an ordering of ${P_i}$ such that $\left|\sum_{i=1}^{k}(P_i-1/n)\right|2\geq \left|\sum{i=1}^{k}(Q_i-1/n)\right|_2$ for all $1\leq k\leq n$, where $|A|_2=\sqrt{\operatorname{tr}(A^\dagger A)}$ is the 2-norm.

Definitions 1, 3 (Eq. (19)) and 5 give positive semidefinite effects summing to $I_d$, positive semidefinite blocks with $d\geq2$ and both block row and column sums $I_d$, and the principal-square-root product $Q_i=\sum_j\sqrt{P_j}B_{ij}\sqrt{P_j}$. Here $1/n$ in the centered prefix denotes $(1/n)I_d$. Issue #14670 preregisters this tier-1 question.

Motivation

Our foothold

The frozen declaration D5/S3/Quantum/Measurements/RicoZyczkowskiMonotoneRefutation.result proves ¬ claim. A three-outcome, two-dimensional diagonal measurement is sufficient to refute the universally quantified assertion.

Gap

Missing bridges

Issue #14670 records the source, MathDB p/359456, arXiv searches and later-paper readings. No settlement was found in that searched scope. This bounded literature reading does not establish exhaustive priority. Appendix C.2 settles the two-outcome restriction rather than the full conjecture.

Route

Proposed approach

Using zero-based indices, the formal witness is

$$P_0=\operatorname{diag}(5/12,1/6),\qquad P_1=\operatorname{diag}(1/6,5/12),\qquad P_2=\operatorname{diag}(5/12,5/12).$$

On coordinate zero let $\rho_0$ exchange outcomes zero and one; on coordinate one let $\rho_1$ be the identity. Set

$$B_{ij}=\operatorname{diag}\left( \frac9{10}\mathbf1_{j=\rho_0(i)}+\frac1{30}, \frac9{10}\mathbf1_{j=\rho_1(i)}+\frac1{30}\right).$$

The diagonal positive semidefinite roots give

$$Q_0=\operatorname{diag}(11/60,11/60),\qquad Q_1=Q_2=\operatorname{diag}(49/120,49/120).$$

Every input effect satisfies $|P_j-I_2/3|_2^2\leq5/144$, while $|Q_0-I_2/3|_2^2=9/200>5/144$. Thus the output ordering beginning with $Q_0$ violates the inequality at $k=1$ for every input permutation.

Falsifier

What would falsify this route

The witness satisfies both POVM conditions, both bistochastic identity resolutions and the literal principal-square-root product. All blocks commute. The failure is independent of the input ordering; it does not replace the existential input permutation with a fixed permutation.

Evidence

Evidence to collect

The canonical module is D5/S3/Quantum/Measurements/RicoZyczkowskiMonotoneRefutation.lean. Its public definitions are BlockBistoch, IsBlockProduct, hs and claim; its only public theorem is result : ¬ claim. It reuses the frozen IsPOVM predicate. Private facts P_prob, Q_prob, B_bistoch, block_product, P_hs_sq, Q_hs_sq and prefix_one supply the formal witness and comparison used by result. The axiom closure of every public declaration is contained in {propext, Classical.choice, Quot.sound}.

Triage

Scope assessment

Resolution: Refuted. proof_shape: result: bind-only; escape_witness: none; admission_basis: open-problem-resolution (#14670; Refuted). The fixed numerical certificates have utility kind=certified-instance under refutes, with the literal types claim : Prop and result : ¬ claim.

What the settlement shows

  • Proved in this module: the $n=3,d=2$ witness satisfies the source's assumptions and violates $k=1$ for every input ordering. Evidence: the private witness facts and the kernel-checked theorem result.
  • Computed, with a paper proof of the all-size family: for every $n\geq3,d\geq2,0<a<1/n$ and $\sqrt{5/8}<t<1$, the construction below violates $k=1$. The exact checker covers only the 110 cases $3\leq n\leq12$, $2\leq d\leq12$, with $a=1/(2n)$ and $t=9/10$. Uniform Lean verification of this family is open.
  • Proved by the paper argument below (mechanism): diagonal dynamics is one scalar bistochastic matrix per coordinate. Classical majorization holds separately on each coordinate. Different coordinate permutations align the negative deviations in a single output effect, defeating every common input ordering. The finite instance is also kernel checked through block_product and result.
  • Proved in the source or classical theory (boundary cases): Appendix C.2 proves $n=2$ for the source's matrix dynamics. At $d=1$ the dynamics becomes scalar bistochastic dynamics, and classical majorization holds with conventional decreasing orderings. This classical sorted-prefix conclusion does not by itself establish an arbitrary-output-order absolute-prefix conclusion. The scalar absolute-prefix extension is not claimed here. These boundary results are literature evidence, not new Lean theorems; $d=1$ is outside BlockBistoch's literal $d\geq2$ domain.
  • Open (surviving restrictions): whether the inequality holds for coordinate-independent dynamics $B_{ij}=b_{ij}I_d$, or for an adjusted conclusion allowing orderings chosen separately per coordinate. The present coordinate-dependent, common-ordering obstruction does not settle either restriction.
  • Literature boundary for other source results: Theorems 2 and 3 address linear monotones for sortable measurements. This witness refutes the unrestricted nonlinear Conjecture 1 and gives no refutation of those sortable-measurement theorems. Any conclusion requiring Conjecture 1 as an unconditional hypothesis lacks that hypothesis; no such conclusion is exported by this module.

All-size construction and paper proof

Put $c=1/n$. The first two diagonal coordinates of the first three input effects are $(c+a/2,c-a)$, $(c-a,c+a/2)$ and $(c+a/2,c+a/2)$. Every remaining entry is $c$. Each coordinate sums to one, and every entry is positive because $0<a<c$. Let $\rho_0$ swap outcomes zero and one, and let every other $\rho_r$ be the identity. Define

$$B_{ij}^{(r)}=t\mathbf1_{j=\rho_r(i)}+(1-t)c.$$

Each coordinate matrix is bistochastic and every entry is strictly positive for $0<t<1$. Hence its diagonal blocks are positive definite and both block resolutions equal $I_d$. Commutation gives $\sqrt{P_j}B_{ij}\sqrt{P_j}=P_jB_{ij}$, so $Q_i^{(r)}=tP_{\rho_r(i)}^{(r)}+(1-t)c$. In particular, the centered first output has coordinates $(-ta,-ta,0,\ldots,0)$. The centered squared input norms are $5a^2/4$, $5a^2/4$, $a^2/2$ and zero for all remaining effects. Thus

$$|Q_0-cI_d|_2^2=2t^2a^2>\frac54a^2 =\max_j|P_j-cI_d|_2^2,$$

where the strict inequality is exactly $t>\sqrt{5/8}$. This proves failure at the first prefix for all the stated parameters on paper. It also explains why the commuting setting does not force a common permutation across coordinates.

Exact computation

Command: python3 /tmp/op-rz/check.py; exit code: 0. Script SHA-256: 02876a92308857348a8f3ed9d4b4896cf9acb436efbbe757ba17ab401043060b. The exact-rational result is $Q_0=(11/60,11/60)$, $\max_j|P_j-I_2/3|_2^2=5/144$, $|Q_0-I_2/3|_2^2=9/200$ and gap $37/3600$. All 110 stated finite cases violate the first-prefix inequality. The complete script source follows; save it as check.py and run python3 check.py to reproduce the same scope.

from fractions import Fraction as F
from itertools import permutations
def family(n,d,a,t):
    c=F(1,n)
    P=[[c]*d for _ in range(n)]
    P[0][0]=c+a/2; P[0][1]=c-a
    P[1][0]=c-a;   P[1][1]=c+a/2
    P[2][0]=c+a/2; P[2][1]=c+a/2
    # remaining effects absorb: coordinates sum check
    for k in range(d):
        s=sum(P[i][k] for i in range(n)); assert s==1,(n,d,k,s)
    # B0: coordinate 0 swaps outcomes 0,1; identity elsewhere; B_t = t B0 + (1-t) J/n
    def B(i,j,k):
        perm = (1 if (k==0 and {i,j}=={0,1}) else 0) if (k==0 and (i in (0,1) or j in (0,1))) and i!=j else (1 if i==j and not (k==0 and i in (0,1)) else 0)
        return t*perm + (1-t)*c
    for k in range(d):
        for i in range(n):
            assert sum(B(i,j,k) for j in range(n))==1 and sum(B(j,i,k) for j in range(n))==1
            for j in range(n): assert B(i,j,k)>0
    # diagonal commuting: sqrt(P_j) B_ij sqrt(P_j) = P_j B_ij entrywise
    Q=[[sum(P[j][k]*B(i,j,k) for j in range(n)) for k in range(d)] for i in range(n)]
    dev=lambda X: sum((x-c)**2 for x in X)
    return P,Q,max(dev(p) for p in P),dev(Q[0])
P,Q,mp,q0=family(3,2,F(1,6),F(9,10))
print("n=3 d=2 Q1:",Q[0],"maxP dev",mp,"Q1 dev",q0,"violation",q0>mp, "gap", q0-mp)
bad=0
for n in range(3,13):
    for d in range(2,13):
        a=F(1,2*n); t=F(9,10)
        P,Q,mp,q0=family(n,d,a,t)
        bad+= (q0>mp)
print("violations over 3<=n<=12, 2<=d<=12:", bad, "of", 10*11)

ASSUMED-UNVERIFIED

Unverified assumptions

The bounded literature absence, bibliographic metadata and boundary results are external evidence. Only the concrete refutation is verified by this Lean module; the all-size parameter family is proved on paper and finitely checked, without a uniform Lean theorem. The two surviving restrictions remain open.

Information-escape registration is paused under CLAUDE.md §3.9.

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