The Fourier LDOI LOCC lower bound is not tight

Johnston and Russo, “Distinguishability of locally diagonal orthogonally invariant quantum states,” Example 17, arXiv:2604.12808v1, ask:

Whether the lower bound in Equation (61) is tight—that is, whether opt_LOCC(E) = 1/2 − (n − 2)/(2n²)—remains an open question.

The uniform ensemble contains $n^2$ states, with prior $1/n^2$ each. Its diagonal vectors are $\phi_{ii}=\sum_k U_{ik}|kk\rangle$, where $U$ is the normalized Fourier matrix. For $i<j$, its two ordered labels are $\phi_{ij}=(|ij\rangle+|ji\rangle)/\sqrt2$ and $\phi_{ji}=(|ij\rangle-|ji\rangle)/\sqrt2$. The question is the equality for every $n\ge3$. Issue #14672 fixes the source conventions and the literal equality claim.

Equation (61) gives a LOCC lower bound $1/2-(n-2)/(2n^2)$ and a PPT upper bound $1/2$. The source constructs a PPT measurement attaining the upper bound. An adaptive finite-round LOCC protocol reaches that same value, closing the optimization interval for this ensemble.

The source still asks the question in Example 17 and its Conclusion. The bounded literature search recorded in #14672 found no settlement among the cited source checks and eleven subsequent papers by the authors; this does not establish exhaustive absence. The exact question is not supplied by a frozen repository owner or the pinned Mathlib searches. The literature locator is Johnston–Russo.

The Lean carrier is finite-round LOCC, $\mathrm{LOCC}_{\mathbb N}$: a finite tree with complete local Kraus instruments and a guessed ordered label at each leaf. Classical communication enters through history-dependent child trees. Success is the sum of correct-leaf squared norms with the uniform prior.

The supremum agrees with the source's full LOCC supremum: Section 2.2 of Chitambar et al. gives $\mathrm{LOCC}{\mathbb N}\subseteq\mathrm{LOCC}\subseteq \overline{\mathrm{LOCC}{\mathbb N}}$, and success is a continuous linear functional of the measurement. Square Kraus operators preserve the statistics after pulling output spaces back by polar decomposition. These correspondences are arguments on paper, not statements in the Lean module.

For $n=3$, Alice selects a pair using its projector divided by $\sqrt2$; Bob tests membership in that pair. Inside, a shared fair choice between X and Y bases resolves the symmetric and antisymmetric labels. Outside, computational outcomes identify the unordered pair, followed by a fair sign guess. The X/Y measurements have a third outcome for the complementary level and reset measured factors with $|0\rangle\langle x|$. An impossible branch receives the fixed guess $(0,1)$.

Completeness and structural induction bound every tree's success by one. The explicit tree has success $1/2$, exceeding $\mathrm{lower}(3)=4/9$; hence the supremum cannot equal the proposed lower bound.

The refutation would fail if any local instrument were incomplete, the source basis or prior differed from the encoding, or the explicit tree's success did not exceed $4/9$. The completeness and success identities for dimension three are checked in the settling module; the simulation separately checks completeness and success for dimensions $3$ through $7$.

D5/S3/Quantum/Measurement/FourierLDOILOCCRefutation.result : ¬ claim is the kernel-checked settlement. Its private success_T establishes success $1/2$ at $n=3$; score_le_mass supplies the induction bound used to justify the supremum. The sole public theorem has no hypotheses. Its axiom closure is contained in ${\mathrm{propext},\mathrm{Classical.choice},\mathrm{Quot.sound}}$.

The Scribe result carries OpenProblemResolutionClaim with resolution Refuted. Admission basis is open-problem-resolution (#14672; Refuted); there is no atom or coverage edge. Computational utility is certified-instance, with refutes targeting this module's claim and result witnessing its negation. Registration is paused under CLAUDE.md §3.9.

What the settlement shows

1. The value — proved on paper using the kernel-checked dimension-three protocol and the source's PPT bound. At $n=3$, the protocol gives $\operatorname{opt}{\mathrm{LOCC}}\ge1/2$; Example 17 gives $\operatorname{opt}{\mathrm{PPT}}=1/2$ and $\mathrm{LOCC}\subseteq\mathrm{PPT}$. Thus $\operatorname{opt}{\mathrm{LOCC}}=\operatorname{opt}{\mathrm{PPT}}=1/2$. The explicit protocol attains the supremum, so it is a maximum. The same conclusion for all $n\ge3$ follows from the paper argument in item 2 and the source's PPT bound. Exact optima and PPT containment are not Lean declarations of this module.

2. Every $n\ge3$ — computed for $n=3,4,5,6,7$; proved on paper for all $n\ge3$. Set $P_{ij}=|i\rangle\langle i|+|j\rangle\langle j|$ for $i<j$. Alice's instrument is $K_{ij}=P_{ij}/\sqrt{n-1}$, complete since each basis coordinate belongs to exactly $n-1$ pairs, giving $\sum_{i<j}K_{ij}^{\dagger}K_{ij}=I$. Bob's pair/complement projections are complete. The X/Y instruments include every complementary basis coordinate; the computational instruments and the fair coin are complete. No branch reports a diagonal label, so every diagonal state's correct score is zero.

   For an off-diagonal state on ${a,b}$, selection of that pair has probability $1/(n-1)$ and the inside X/Y correlations determine its sign perfectly. There are $2(n-2)$ other selected pairs containing exactly one of $a,b$, each with probability $1/(2(n-1))$. Bob is outside these pairs and the computational readout followed by a fair sign guess succeeds with probability $1/2$. All remaining pair outcomes have probability zero. Each off-diagonal label therefore has success $1/(n-1)+(n-2)/(2(n-1))=n/(2(n-1))$. Averaging the $n(n-1)$ such states among $n^2$ labels gives $1/2$.

   Computation: python3 /tmp/op-jr/check.py, exit 0; script SHA-256 c0df960df8154c3189b7a35e56c63e5db74088fca90759c73641d2f56f77aad8. It constructs local Kraus effects and checks that their sum is the identity using numpy.allclose. Scope is exactly $n=3,\ldots,7$; the floating-point checks alone prove no uniform extension.

   | $n$ | computed success | proposed lower bound | | --- | --- | --- | | 3 | 0.5 | 0.4444444444444444 | | 4 | 0.5 | 0.4375 | | 5 | 0.5 | 0.44 | | 6 | 0.5 | 0.4444444444444444 | | 7 | 0.5 | 0.4489795918367347 |

3. The extension — proved on paper. Item 2 ignores $U$ because it never reports a diagonal label. It therefore gives success $1/2$ for every balanced-amplitude LDOI basis $A=(1/\sqrt2)1_n$. When $|u_{ki}|^2\le1/2$ for all indices, the source's Theorem 10, Eqs. (27)–(28), admits $c_i=1/2$: $c_i\ge\max_k|u_{ki}|^2$ and $c_ic_j=1/4=|a_{ij}|^2|a_{ji}|^2$. Its upper bound is $[n(n-1)/2+n/2]/n^2=1/2$. Hence $\operatorname{opt}{\mathrm{LOCC}}=\operatorname{opt}{\mathrm{PPT}}=1/2$, attained by this protocol. This includes normalized complex Hadamard bases for $n\ge3$, since their entry squares are $1/n$. This is a paper argument using Theorem 10, not an additional Lean theorem.

4. The Corollary 16 gap bound — computed for $n=3,\ldots,7$; its failure to be attained here is proved on paper for all $n\ge3$. The simulation returns success $1/2$ strictly above the proposed lower bound in every tested dimension. The difference is $(n-2)/(2n^2)>0$. Items 1–2 and the source PPT bound give actual PPT–LOCC gap zero for this basis, so it does not attain that positive gap estimate. Corollary 16 remains a valid upper bound; the protocol improves the LOCC lower bound for this basis. Open: whether any LDOI basis has a positive PPT–LOCC gap.

5. The mechanism — proved in the concrete Lean construction and explained on paper for the general protocol. Pair selection followed by Bob's membership test isolates each Bell-like pair. Inside it, symmetric and antisymmetric states have respectively equal and unequal outcomes in a shared X or Y basis, resolving their signs perfectly. The source's Theorem 10 lower bound uses a product-measurement assignment and does not express this adaptive pair-conditioned discrimination. The formal completeness and score evidence is pair_complete, split_complete, basis_complete, cross_score and success_T, all private and consumed on result's proof path.

Theorem 10's bounds, the PPT/separable constructions in Eqs. (62)–(65), and Corollary 16 remain valid. The tightness question for Example 17 is refuted; its exact LOCC value is $1/2$ by the argument above. The source's broader Conclusion question, equality of LOCC and PPT for every uniform orthonormal LDOI ensemble outside the stated balanced-amplitude class, remains open.

Reproducible simulation source

Save the following exact UTF-8 source as /tmp/op-jr/check.py and run python3 /tmp/op-jr/check.py with NumPy installed. SHA-256: c0df960df8154c3189b7a35e56c63e5db74088fca90759c73641d2f56f77aad8.

import numpy as np, itertools
def ensemble(n):
    U=np.array([[np.exp(2j*np.pi*i*k/n) for k in range(n)] for i in range(n)])/np.sqrt(n)
    e=lambda i,j: np.kron(np.eye(n)[i],np.eye(n)[j])
    vecs={}
    for i in range(n): vecs[(i,i)]=sum(U[i,k]*e(k,k) for k in range(n))
    for i in range(n):
        for j in range(i+1,n):
            vecs[(i,j)]=(e(i,j)+e(j,i))/np.sqrt(2); vecs[(j,i)]=(e(i,j)-e(j,i))/np.sqrt(2)
    return vecs
def protocol_success(n):
    vecs=ensemble(n); N=n*n
    # build LOCC protocol effects explicitly from local Kraus operators
    e=np.eye(n); succ=0.0; total=np.zeros((N,N),complex)
    Xp=lambda i,j,s:(e[i]+s*e[j])/np.sqrt(2)          # X basis on pair
    Yp=lambda i,j,s:(e[i]+s*1j*e[j])/np.sqrt(2)       # Y basis on pair
    for i,j in itertools.combinations(range(n),2):
        Ka=(np.outer(e[i],e[i])+np.outer(e[j],e[j]))/np.sqrt(n-1)   # Alice picks pair
        Pin=np.outer(e[i],e[i])+np.outer(e[j],e[j])
        # Bob inside: basis choice X/Y each prob 1/2, both measure; sym if product of signs... compute effects
        for basis in ['X','Y']:
            for sa in (1,-1):
                for sb in (1,-1):
                    va=Xp(i,j,sa) if basis=='X' else Yp(i,j,sa)
                    vb=Xp(i,j,sb) if basis=='X' else Yp(i,j,sb)
                    KA=np.outer(va,va.conj())@Ka; KB=np.outer(vb,vb.conj())@Pin
                    K=np.kron(KA,KB)/np.sqrt(2)
                    E=K.conj().T@K; total+=E
                    # guess: X basis: sym if sa==sb ; Y basis: |ij>+|ji> ... choose label maximizing for consistency
                    guess=None
                    best=-1
                    for lab in [(i,j),(j,i)]:
                        v=vecs[lab]; p=np.real(v.conj()@E@v)
                        if p>best: best=p; guess=lab
                    succ+=np.real(vecs[guess].conj()@E@vecs[guess])
        # Bob outside pair: both read computational labels, guess sign fairly
        Pout=np.eye(n)-Pin
        for a in (i,j):
            for b in range(n):
                if b in (i,j): continue
                KA=np.outer(e[a],e[a])@Ka; KB=np.outer(e[b],e[b])@Pout
                E=np.kron(KA,KB).conj()[email protected](KA,KB); total+=E
                lab=(a,b) if a<b else (b,a)
                lab2=(lab[1],lab[0])
                succ+=0.5*np.real(vecs[lab].conj()@E@vecs[lab])+0.5*np.real(vecs[lab2].conj()@E@vecs[lab2])
    assert np.allclose(total,np.eye(N)), "not complete"
    return succ/N
for n in range(3,8):
    s=protocol_success(n); print(n, round(s,12), "LB", 0.5-(n-2)/(2*n*n))

The bounded literature search does not establish exhaustive novelty or priority. The LOCC carrier correspondence, the PPT bound, the general-dimension counting argument and the balanced-amplitude extension are arguments on paper, not kernel-checked declarations of this delivery. The simulation is floating-point and checks completeness with NumPy's default allclose tolerance; it is not an exact rational certificate.