Metadata

slug: kay-2010-transfer-rate-refutation
bibkey: kay2010perfectstatetransferreview
doi: 10.1142/S0219749910006514
url: https://arxiv.org/abs/0903.4274v3
triage: theorem
motivation_gids:
  - D5/S3/Quantum/Dynamics/KayTransferRateRefutation.result

Kay’s transfer-rate conjecture for perfect state transfer chains

Problem

A. Kay, A Review of Perfect State Transfer and its Application as a Constructive Tool, arXiv:0903.4274v3 (Int. J. Quantum Inf. 8, 641 (2010)), subsection “Transfer Rate”. For the ordered eigenvalues $λ_{1} < \dots < λ_{N}$ of a mirror-symmetric chain the perfect state transfer condition is

$λ_{n} - λ_{n - 1} = (2 m_{n} + 1) π / t_{0}$ where $t_{0}$ is the state transfer time, and $m_{n}$ is a positive integer (which can vary with $n$ ).

and $B^{'} (λ_{n}) = \prod_{m = 1 \neq = n}^{N} (λ_{n} - λ_{m})$ . The lemma of that subsection (the sixth lemma of the source) reads:

If a set of eigenvalues is chosen to fulfill the perfect state transfer condition of Eqn.~(\ref{eqn:st_cond}), then a necessary and sufficient condition to perfectly achieve the rate $M /2 t_{0}$ for integer $M$ is that all the $R_{k}$ for $k = 0 \dots M - 1$ should be equal, where $R_{k} = \sum_{n = 1}^{N} \frac{( - 1 ) ^{n}}{B ^{'} ( λ _{n} )}$ , and the sum is restricted to those terms satisfying the condition $\frac{t _{0}}{π} (λ_{n} - λ_{1}) mod M = k .$

It is followed by the conjecture:

We conjecture that it is impossible to fulfill the condition of Lemma \ref{lemma:rate} for any $M > 2$ , although we only have a proof for $M > N /2$ .

Issue #14475 fixes the reading. A spectrum is a function lam : Fin N → ℝ; the condition is 0 < t₀ together with positive integers m such that every gap between consecutive levels is (2 m + 1) π / t₀; the level n lies in the class k modulo M when the real number (t₀/π)(λ_n − λ_1) is an integer congruent to k; the sign (−1)^n uses the source’s index, which starts at one. The conjecture is the statement that for every N ≥ 1, every such spectrum and transfer time, and every integer M > 2, the sums R_0, …, R_{M−1} are not all equal. The case N = 0 is excluded because all sums of the empty spectrum are empty and therefore equal.

Motivation

The lemma turns the question of how fast a perfect state transfer chain can be reused as a quantum channel into a condition on its spectrum: the standard protocol has M = 2, and M > 2 would mean that a new state can be placed on the first site before the previous one has arrived. The conjecture says this never happens. The declaration D5/S3/Quantum/Dynamics/KayTransferRateRefutation.result proves that the conjecture is false.

Gap

Issue #14475 preregisters the verbatim source, the reading above, the expected settlement and the literature check before any formalization. Its recorded readings: Kay, arXiv:1102.2338, derives the bound $2 l + D \leq M \leq N$ and concludes

Ideally, we want the maximum transfer distance, which would be $N - 1$ (a chain), imposing that $l = 0$ , as conjectured in \cite{akreview}.

so the later paper treats the conjecture as established. Bailey, Derevyagin et al., arXiv:2411.06047, construct perfect state transfer Jacobi matrices whose first-site amplitude vanishes before the transfer time (early state exclusion): a $4 \times 4$ example with eigenvalues $\pm 3/2, \pm 5/2$ and a zero at $arccos (2/3)$ , and, for every even order at least 4 and every $m$ , a matrix with $m$ such zeros; arXiv:2507.18767 and arXiv:2606.04353 continue that study, the latter noting that the phenomenon “can happen multiple times”. Their $4 \times 4$ example already violates the bound of arXiv:1102.2338 ( $N = 4$ , $D = 3$ , $l = 1$ ). For a perfect state transfer spectrum the condition of the lemma with $M = 4$ is equivalent to a zero of the first-site amplitude at exactly $t_{0} /2$ (see Triage); none of the 24 perfect state transfer spectra and three parametrised families extracted from these three papers fulfils the condition of the lemma for any $3 \leq M \leq N$ (second script in Evidence). arXiv:2209.08160 makes no statement on the conjecture; a title and abstract scan of the 285 works that INSPIRE lists as citing the review finds no refutation. These readings are not-found-in-searched-scope; they do not establish an exhaustive literature search or priority.

Route

Take N = 8, t₀ = π, M = 4 and the levels

λ = (0, 31, 46, 65, 88, 107, 122, 153).

The gaps are 31, 15, 19, 23, 19, 15, 31, that is 2m + 1 with m = 15, 7, 9, 11, 9, 7, 15, all positive.
B'(λ_n) is −16291106900640, 760363989840, −273136152240, 203460697440, −203460697440, 273136152240, −760363989840, 16291106900640.
Since t₀ = π, the number (t₀/π)(λ_n − λ_1) is λ_n, with residues 0, 3, 2, 1, 0, 3, 2, 1 modulo 4. The classes are {0, 88}, {65, 153}, {46, 122}, {31, 107} for k = 0, 1, 2, 3.
Every term (−1)^n / B'(λ_n) is positive, and 1/16291106900640 + 1/203460697440 = 1/760363989840 + 1/273136152240 = 194/38984495395755, so R_0 = R_1 = R_2 = R_3 = 194/38984495395755.

Here M = 4 = N/2, outside the range M > N/2 for which the source has a proof.

Falsifier

The refutation would be undone by a reading of the lemma’s condition under which the four sums of this spectrum differ. The sums are nonzero, so every class is nonempty; a sign convention with (−1)^{n−1} in place of (−1)^n changes all four sums by the same sign; admitting m_n = 0 enlarges the set of spectra and leaves this one admissible. The source’s proof for M > N/2 is not contradicted, because M = N/2.

Evidence

The canonical source is D5/S3/Quantum/Dynamics/KayTransferRateRefutation.lean. Its declarations are SpectrumCondition, derivativeAt, InClass, rateSum, RateCondition, claim and result : ¬ claim. result instantiates claim at the spectrum above, proves the perfect state transfer condition with the multipliers 15, 7, 9, 11, 9, 7, 15, evaluates the eight values of derivativeAt, decides the class of each level for each k < 4, and evaluates the four sums by exact rational arithmetic. It uses the axioms propext, Classical.choice and Quot.sound; no sorry, native_decide or new axiom.

The following script recomputes the values quoted in this dossier with exact rational arithmetic (python3 kay.py, Python 3.9):

from fractions import Fraction as F
from itertools import product

def levels(gaps):
    return [sum(gaps[:i]) for i in range(len(gaps) + 1)]

def derivative(lam, n):
    b = 1
    for j, y in enumerate(lam):
        if j != n: b *= lam[n] - y
    return b

def sums(gaps, M):
    lam = levels(gaps)
    r = [F(0)] * M
    for n, x in enumerate(lam):
        r[x % M] += F((-1) ** (n + 1), derivative(lam, n))
    return r

def hits(N, G, sym=False):
    odd = range(1, G + 1, 2)
    tuples = (t + t[-2::-1] for t in product(odd, repeat=N // 2)) if sym else product(odd, repeat=N - 1)
    return [(M, g) for g in tuples for M in range(3, N + 1) if len(set(sums(g, M))) == 1]

gaps = (31, 15, 19, 23, 19, 15, 31)
lam = levels(gaps)
print(lam, [derivative(lam, n) for n in range(8)], sums(gaps, 4))
w = [F(1, abs(derivative(lam, n))) for n in range(8)]
w = [x / sum(w) for x in w]
print(w[:4], [sum(w[n] for n in range(8) if lam[n] % 4 == k) for k in range(4)])
p0, p1, norm0, diag, coup = [F(0)] * 8, [F(1)] * 8, None, [], []
for k in range(8):
    norm = sum(w[i] * p1[i] ** 2 for i in range(8))
    a = sum(w[i] * lam[i] * p1[i] ** 2 for i in range(8)) / norm
    b = norm / norm0 if k else F(0)
    diag.append(a); coup.append(b)
    p0, p1, norm0 = p1, [(lam[i] - a) * p1[i] - b * p0[i] for i in range(8)], norm
print(diag, coup[1:])
print([hits(N, 31) for N in range(2, 7)], hits(7, 15), hits(8, 9), hits(8, 41, sym=True))

Its output: the levels, the eight values of B' and the four sums as in the Route; the weights 2527/819456, 2461/37248, 6851/37248, 202337/819456 of the first four levels and the class weights 1/4, 1/4, 1/4, 1/4; the diagonal 153/2 (eight times) and the squared couplings 834785/1164, 1197437/1455, 34047/20, 2500, 34047/20, 1197437/1455, 834785/1164; and the search result [[], [], [], [], []] [] [] [(4, (31, 15, 19, 23, 19, 15, 31))].

The second script evaluates the condition of the lemma, for every $3 \leq M \leq N$ , on perfect state transfer spectra taken from the examples and theorems of arXiv:2411.06047, arXiv:2507.18767 and arXiv:2606.04353 (the symmetric spectra are listed by their non-negative levels) and on three of their parametrised families with parameter below 80. An output [] means that no such $M$ fulfils the condition, None that the listed levels are not a perfect state transfer spectrum; the last two lines are the spectrum of this dossier, with the single hit $M = 4$ , and the families, with no hit (python3 lit_spectra.py, Python 3.9):

from fractions import Fraction as F
from math import gcd
from functools import reduce
def check(lam):
    lam = sorted(lam); N = len(lam)
    gaps = [lam[i+1] - lam[i] for i in range(N-1)]
    g = reduce(gcd, gaps)
    q = [(x - lam[0]) // g for x in lam]          # (t0/pi)(lam_n - lam_1) at the earliest candidate time pi/g
    if any(((b - a) % 2) == 0 for a, b in zip(q, q[1:])):
        return None                               # not a PST spectrum
    B = []
    for n in range(N):
        p = 1
        for j in range(N):
            if j != n: p *= q[n] - q[j]
        B.append(p)
    out = []
    for M in range(3, N + 1):
        R = [F(0)] * M
        for n in range(N):
            R[q[n] % M] += F((-1) ** (n + 1), B[n])
        if len(set(R)) == 1: out.append(M)
    return out
def sym(pos, zero=True):
    return sorted(([0] if zero else []) + list(pos) + [-x for x in pos])
named = {
 "0,1,2,3": sym([1,2,3]), "0,3,4,5": sym([3,4,5]), "0,1,4,7": sym([1,4,7]), "0,3,4,9": sym([3,4,9]),
 "0,3,6,11": sym([3,6,11]), "0,1,4": sym([1,4]), "0,3,12,13": sym([3,12,13]), "0,3,12": sym([3,12]),
 "0,3,8,9,12": sym([3,8,9,12]), "0,1,12..15": sym([1,12,13,14,15]), "0,1,12,13,14": sym([1,12,13,14]),
 "0,1,14..17": sym([1,14,15,16,17]), "0,1,14,15,16,19": sym([1,14,15,16,19]), "0,1,14,15,16": sym([1,14,15,16]),
 "0,1,2,13,14,15": sym([1,2,13,14,15]), "0,1,2,3,20,25": sym([1,2,3,20,25]), "0,1,2,3,4": sym([1,2,3,4]),
 "0,1,2,3,82..86": sym([1,2,3,82,83,84,85,86]), "0,1,2,81..86": sym([1,2,81,82,83,84,85,86]),
 "0,3,6,9": sym([3,6,9]), "0,3,8,9": sym([3,8,9]), "0,3,4": sym([3,4]), "0,1,2,81..85": sym([1,2,81,82,83,84,85]),
 "0,1,2,4,13,14,15": sym([1,2,4,13,14,15]), "0,1,2,3,13,14,15": sym([1,2,3,13,14,15]),
 "4x4 Bailey +-3/2,+-5/2 (x2)": [-5,-3,3,5],
 "delivery": [0, 31, 46, 65, 88, 107, 122, 153],
}
for k, v in named.items():
    print(k, check(v))
hits = []
for m in range(1, 80):
    for name, lam in [("0,1,2m,2m+1", sym([1, 2*m, 2*m+1])), ("0,2m+1,2m+2,2m+3", sym([2*m+1, 2*m+2, 2*m+3]))]:
        r = check(lam)
        if r: hits.append((name, m, r))
    for n in range(2, 9):                          # Bailey et al. family, doubled to integers
        up = [2*m + 2*k + 1 for k in range(n)]
        r = check(sym(up, zero=False))
        if r: hits.append(("bailey", n, m, r))
print("family hits", hits)

Triage

Tier 1, an explicitly stated conjecture of a 2010 review; settlement Refuted. result has proof_shape: bind-only (an exact finite evaluation of the definitions at one explicit spectrum); admission_basis: open-problem-resolution under the preregistration #14475; utility kind=certified-instance; basis=refutes with the typed claim and result. There is no atom and no digestion coverage edge.

What the settlement shows

Proved by D5/S3/Quantum/Dynamics/KayTransferRateRefutation.result: the condition of the lemma can be fulfilled with M > 2. The spectrum 0, 31, 46, 65, 88, 107, 122, 153 with t₀ = π fulfils the perfect state transfer condition and has R_0 = R_1 = R_2 = R_3 for M = 4.
Computed (the script in Evidence), mechanism. The weights of the eigenvectors on the first site are w_n = (1/|B'(λ_n)|) / Σ_m (1/|B'(λ_m)|), the source’s Eqn. eqn:hoch; all eight terms (−1)^n/B'(λ_n) have the same sign, so the sums R_k are the class weights up to a common factor. The four residue classes modulo 4, {0, 88}, {65, 153}, {46, 122}, {31, 107}, each carry weight exactly 1/4. Hence the first-site amplitude γ_1(t) = Σ_n w_n e^{−iλ_n t} satisfies γ_1(jπ/2) = (1/4) Σ_{k<4} (−i)^{jk}, which is 0 for j = 1, 2, 3 and 1 for j = 4: the amplitude vanishes at t₀/2, t₀ and 3t₀/2 and returns at 2t₀.
Computed, why one equation suffices. The spectrum is mirror symmetric, λ_n + λ_{9−n} = 153, so |B'(λ_n)| = |B'(λ_{9−n})|; and 153 ≡ 1 modulo 4, so the mirror exchanges the classes 0 ↔ 1 and 2 ↔ 3. Thus R_0 = R_1 and R_2 = R_3 hold for every mirror-symmetric spectrum with these residues, and the only remaining condition is 1/|B'(λ_1)| + 1/|B'(λ_4)| = 1/|B'(λ_2)| + 1/|B'(λ_3)|, which the four gaps 31, 15, 19, 23 solve. The source’s displayed argument for large M uses a class with a single level; at M = N/2 every class can have two levels and that argument has nothing to act on.
Computed, the chain. The inverse eigenvalue problem for this spectrum gives the mirror-symmetric chain with constant diagonal 153/2 and squared couplings 834785/1164, 1197437/1455, 34047/20, 2500, 34047/20, 1197437/1455, 834785/1164; its first-site weights are the w_n above by construction. The chain is not in Lean.
What survives. The source states a proof for M > N/2; it is not formalised or checked here. If it holds, M = 4 needs N ≥ 8, so the spectrum above has the least possible number of levels for M = 4. Computed (the script in Evidence, last line): for 3 ≤ M ≤ N there is no spectrum with N ≤ 6 and odd gaps at most 31, none with N = 7 and odd gaps at most 15, and none with N = 8 and odd gaps at most 9; among the mirror-symmetric spectra with N = 8 and odd gaps at most 41 the spectrum above is the only one, and only for M = 4. Gaps equal to 1 (m_n = 0) are included in these searches.
Open. Whether an odd M > 2 can occur; whether M = 4 can occur with N = 9, 10, 11 or with a spectrum that is not mirror symmetric; whether some M ≥ 5 can occur; a proof that N = 8 is least for M = 4. Issue #14475 reports an unformalised 2-adic argument and a computation excluding 4 | M for N ≤ 7 and N = 9, 10, 11; neither is reproduced here and neither is a proof in this repository.
Computed, the zeros. For the spectrum above the first-site amplitude with the Hochstadt weights vanishes at t₀/2, t₀ and 3t₀/2 and equals 1 at 2t₀ (first script: the four class weights are 1/4).
Proved on paper, not in Lean: M = 4 is a zero at t₀/2. For a perfect state transfer spectrum put z_n = (t₀/π)(λ_n − λ_1); then γ_1(t₀/2) is proportional to Σ_k R_k (−i)^k over the classes modulo 4. The parity of z_n alternates with n, so R_0 + R_2 and R_1 + R_3 are the two parity sums, which are equal because Σ_n 1/B'(λ_n) = 0 (equivalently, the amplitude vanishes at t₀); hence γ_1(t₀/2) = 0 holds exactly when R_0 = R_2 and R_1 = R_3, that is when all four R_k are equal. The zero at 3t₀/2 then follows from |γ_1(t₀+s)| = |γ_1(t₀−s)|.
Open, conditional on the source’s lemma. By the lemma of the source, which is not formalised here, the chain above achieves the rate M/2t₀ = 2/t₀, that is t_r = t₀/2; this would contradict the stronger conjecture stated before the lemma, “that there are no chains with $t_{r} < t_{0}$ ”, and attain at N = 8 the bound 1/t_r ≤ N/4t₀ that the source derives from the lemma.
Computed, the count for arXiv:1102.2338. That paper’s bound 2l + D ≤ M ≤ N, with D the transfer distance, M the number of distinct eigenvalues and l the number of times before t₀ at which the amplitude on the input site vanishes, gives l = 0 for a chain (D = N − 1). The spectrum above has N = 8, D = 7 and a zero at t₀/2, so 2l + D ≥ 9 > 8. The published 4×4 example of arXiv:2411.06047 violates the same bound (N = 4, D = 3, l = 1), so this failure is not new here. Open: which step of the derivation of the bound fails.
Literature reading with a computed check: early state exclusion. arXiv:2411.06047 constructs perfect state transfer Jacobi matrices whose first-site amplitude vanishes before t₀, with any prescribed number of such zeros, and proves that equally spaced spectra have none; arXiv:2507.18767 and arXiv:2606.04353 continue that study. By the equivalence above, the spectrum of this dossier is an example with a zero at exactly t₀/2. Computed (second script): none of the spectra extracted from those three papers fulfils the condition of the lemma for any 3 ≤ M ≤ N.

ASSUMED-UNVERIFIED

The source’s lemma (equal R_k if and only if the rate M/2t₀ is achieved perfectly), its proof for M > N/2 and its Eqn. eqn:hoch are used as stated; none is formalised here. The consequences listed for the rate, for t_r and for arXiv:1102.2338 depend on them.
The journal text was not compared with arXiv v3; the statement formalised is that of the arXiv v3 source review.tex.
The literature readings in Gap: the two quoted sentences of arXiv:1102.2338 and the theorems and the 4×4 example of arXiv:2411.06047 were read in their sources; the spectra used by the second script were extracted from arXiv:2411.06047, arXiv:2507.18767 and arXiv:2606.04353 by a review seat and were not compared example by example with the papers; arXiv:2209.08160 and the INSPIRE scan are as recorded in #14475. No exhaustive novelty or priority is asserted.
The 2-adic argument and the computation for N = 9, 10, 11 reported in #14475 were not reproduced.

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