The spectral gap of random Pauli rotations on four qubits is 1/15

Yuxuan Zhang

Manuscript, version 1.0, 28 September 2026 (PDF, 8 pages) · Manuscript source, exact checker, mutation tests and review records (ZIP)

Archival source: Manuscript PDF on Zenodo, manuscript 20 in Agentic Proofs for QIQC: Collected Manuscripts, version 1.2 (29 September 2026). QIQC report #122 documents the manuscript; submission does not change the catalog status.

The conjectured gap fails at four qubits. Consider the random walk on \(\mathrm{SU}(16)\) that applies \(e^{i\theta P}\) with a uniformly random non-identity Pauli operator \(P\) and a uniform angle \(\theta\). Its spectral gap is exactly \(1/15\), below the conjectured \(26/255\). The slowest mode lives in the representation \(\Lambda^8\mathbb C^{16}\), not in the fourth moment.

This is a complete negative answer to the catalog’s random-Pauli-rotation question. “Solved (negative)” is the label in this research log. An isolated internal AI review of the complete proof, an independent computation over all 255 Pauli operators, a requirements audit against the catalog record and a blind reconstruction from the bare statements found no defect. External specialist review, historical priority and QIQCOP acceptance remain unconfirmed.

The question

Baer and Haah [1] proved

for the gap \(\Delta_n\) of this walk on \(L^2_0(\mathrm{SU}(2^n))\). Their Conjecture 3.48 states that the upper bound, the value of the fourth-moment representation, is exact for \(n\ge3\). The question [2] asks whether it holds for every \(n\ge4\). At \(n=4\) the bounds read \(1/15\le\Delta_4\le26/255\).

The slow mode

Label the basis of \(\mathbb C^{16}\) by \(\mathbb F_2^4\). Each of the 30 affine hyperplanes \(A\) of \(\mathbb F_2^4\) has eight points. Let \(e_A\) be the wedge product of their basis vectors in increasing binary order, and let

Then \(Mw=\tfrac{14}{15}w\) for \(M=\mathbb E_{P,\theta}\Lambda^8(e^{i\theta P})\), and \(\|M\|=\tfrac{14}{15}\) on \(\Lambda^8\mathbb C^{16}\). This irreducible representation occurs in \(L^2(\mathrm{SU}(16))\), so \(\Delta_4\le1/15<26/255\), which already answers the question. The Baer–Haah lower bound equals \(1/15\) at \(n=4\), so \(\Delta_4=1/15\) exactly.

The mode can be written as a function: with \(B_0=\{0,\dots,7\}\), \(f(U)=\sum_A\det U[A,B_0]\), a sum of \(8\times8\) minors, has mean zero and satisfies \(K_4f=\tfrac{14}{15}f\).

Why it works

For a \(Z\)-type Pauli operator, 28 of the 30 hyperplane wedges have charge zero, so \(P\) keeps \(\tfrac{28}{30}=\tfrac{14}{15}\) of \(w\) in its fixed space. The Clifford group fixes \(w\) and acts transitively on the non-identity Pauli operators, so every \(P\) behaves in the same way. The Hadamard step uses the self-duality of the Reed–Muller code \(\mathrm{RM}(1,3)\).

Two independent arguments give the matching upper bound \(\|M\|\le\tfrac{14}{15}\): a quadratic Casimir identity, and a selection rule based on Walsh-charge divisibility and the Bose–Burton theorem. The selection rule also shows that no irreducible representation \(\lambda\) with \(8\nmid|\lambda|\) can violate the conjectured value, for any \(n\ge3\).

Checks

The standalone checker uses only the Python standard library. It prints 22 tagged checks: 21 are exact and one is a floating-point check of the eigenfunction. They include:

• the charge counts;
• the order lemma and the wedge signs under all affine maps of \(\mathbb F_2^4\);
• Clifford invariance for 20 generators by exact exterior-algebra expansion;
• an independent Lie-algebra computation of \(Mw\) on the full 12,870-dimensional space;
• the Casimir identity and the selection rule;
• positive controls reproducing \(\Delta_3=5/63\) and the fourth-moment value \(229/255\);
• negative controls.

It runs in about ten seconds, and its output is byte-identical on a laptop and in a container without network access. All 32 single-point mutations of the checker are detected.

What this establishes, and what it does not

At \(n=4\) the conjectured value fails and the gap is \(1/15\), so the answer to the question is no. The slow mode is an unbalanced representation: the centre of \(\mathrm{SU}(16)\) acts on it nontrivially. It therefore says nothing about convergence of moments or unitary designs, which is governed by balanced representations. That question at \(n=4\) and the exact gap for \(n\ge5\) remain open.

The witness was found in round 49 of the AI-assisted campaign, after earlier rounds settled the fourth-moment sector. The manuscript discloses AI assistance, the internal review and the independent checks. K-Dense’s Scientific Agent Skills guided the evidence record and writing.

References

1. T. Baer and J. Haah, Random unitary circuits with constant spectral gap, arXiv:2607.20919.
2. Quantum Information and Quantum Computation Open Problem Zoo, Exact spectral gap of random Pauli rotations.
3. J. Haah, Y. Liu and X. Tan, Efficient approximate unitary designs from random Pauli rotations, Commun. Math. Phys. 406, 309 (2025), arXiv:2402.05239.
4. R. C. Bose and R. C. Burton, A characterization of flat spaces in a finite geometry and the uniqueness of the Hamming and the MacDonald codes, J. Combin. Theory 1, 96–104 (1966).
5. T. Kassis, V. Agarwal, Y. He, D. Patel and A. M. Brueckner, Scientific Agent Skills: A Library of Procedural Knowledge for Research Agents, arXiv:2609.00065v2.