Eight-copy universal CCZ concentration beyond the two-thirds candidate optimum

An explicit catalyst-free stabilizer protocol turns eight copies of any unknown pure qubit into an exact CCZ state with probability m₃, not the candidate optimum (2/3)m₃.

Yuxuan Zhang

Manuscript, version 1.0, 28 September 2026 (PDF, 7 pages) · Manuscript source, exact checker, mutation tests and review records (ZIP)

Archival source: Manuscript PDF on Zenodo, manuscript 19 in Agentic Proofs for QIQC: Collected Manuscripts, version 1.2 (29 September 2026). QIQC report #121 documents the manuscript; submission does not change the catalog status.

The candidate optimum fails at every non-stabilizer input. A fixed protocol built only from Pauli measurements, classical feedforward, Clifford unitaries and discarding turns \(\psi^{\otimes8}\) into an exact \(\lvert CCZ\rangle\) with probability exactly \(m_3(\psi)=(1-x^6-y^6-z^6)/2\), for every pure qubit \(\psi\) with Bloch vector \((x,y,z)\). The proposed optimum was \(\tfrac23m_3\). With the upper bound of Rizzo and Leone, \(m_3\le p_8^\star\le\tfrac76m_3\).

This is a complete negative answer to the catalog’s eight-copy CCZ question. “Solved (negative)” is the label in this research log. An isolated internal AI review of the complete proof, an independent simulation, 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

In universal magic-state concentration [1] a fixed stabilizer protocol, chosen without knowing the input, must turn copies of an unknown pure qubit into an exact target magic state; it may fail, and it must fail on stabilizer inputs. For eight copies and the target \(\lvert CCZ\rangle=2^{-3/2}\sum_{x}(-1)^{x_1x_2x_3}\lvert x\rangle\), Rizzo and Leone give a protocol with success probability \(\tfrac23m_3(\psi)\), prove that no protocol exceeds \(\tfrac76m_3(\psi)\), and leave eight-copy optimality open. The question [2] asks whether \(\tfrac23m_3\) is the optimum for every pure \(\psi\).

The protocol

  1. Measure \(X^{\otimes8}\) and \(Z^{\otimes8}\). The outcome \((+1,+1)\) is rejected; the other three outcome pairs select a sector \(P\in\{Z,X,Y\}\).
  2. In sector \(P\), measure \(P\) on qubits \(\{0,1,2,3\}\). On \(-1\), measure \(P_0P_1\) and accept either result. On \(+1\), measure \(P\) on qubits \(\{0,3,4,5\}\) and accept only \(-1\).
  3. On each of the nine accepted branches, an explicit Clifford circuit of at most 21 gates, preceded in sectors \(X\) and \(Y\) by one layer of single-qubit Cliffords, maps the state exactly to \(\lvert CCZ\rangle\) on three qubits, with the other five in \(\lvert0\rangle\). They are discarded.

The protocol uses no ancillas and no catalyst, and it does not depend on \(\psi\).

Why it works

On the symmetric subspace, every accepted branch projects onto a single line, so its output does not depend on \(\psi\). In each sector the only part of \(\psi^{\otimes8}\) that can reach an accepted branch is its component along one vector \(E_P\). The three vectors span the three-dimensional part of the symmetric subspace that is orthogonal to \(\tau^{\otimes8}\) for all six single-qubit stabilizer states \(\tau\), and the three squared components add up to \(\tfrac76m_3(\psi)\). Rizzo and Leone’s filter keeps \(\tfrac47\) of each sector direction. The new branch recycles part of the branch they reject and keeps another \(\tfrac27\), so the protocol keeps \(\tfrac67\cdot\tfrac76m_3=m_3\).

The last \(\tfrac17\) of each direction reaches only the rejected branch. On the symmetric subspace that branch’s image is spanned by two stabilizer states, so no continuation of this tree can turn it into \(\lvert CCZ\rangle\). Whether any protocol reaches \(\tfrac76m_3\) remains open. For the \(T\)-type state, the protocol succeeds with probability \(4/9\), against \(8/27\) for the candidate optimum and the upper bound \(14/27\).

Checks

The standalone checker uses only the Python standard library, with exact integer and rational arithmetic. It checks:

  • that every measurement path commutes and that the 13 branches are complete;
  • that all nine accepted branches have rank one on the symmetric subspace, each capturing \(\tfrac27\);
  • the three decoders, gate by gate, and the transfer to sectors \(X\) and \(Y\);
  • the exact \(9\times9\) identity that gives \(p=m_3\), with a second route at exact Gaussian-rational inputs;
  • positive and negative controls.

It prints 21 lines in under a second. Its output is byte-identical on a laptop and in a container without network access. All 25 single-point mutations of the checker are detected. The checker confirms only that the program computes what it prints; the argument itself is in the manuscript.

What this establishes, and what it does not

For every non-stabilizer pure qubit, the optimal eight-copy success probability is at least \(m_3\). This refutes the candidate \(\tfrac23m_3\) and narrows the open window from a factor \(\tfrac74\) to \(\tfrac76\). The exact value of \(p_8^\star\) is not determined, and nothing is claimed for other copy numbers, other targets, catalytic protocols or protocols optimized for a known input.

The protocol was found in round 46 of the AI-assisted campaign. 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. J. Rizzo and L. Leone, Universal magic state concentration, arXiv:2608.13376.
  2. Quantum Information and Quantum Computation Open Problem Zoo, Optimal universal eight-copy concentration to a CCZ state.
  3. S. Bravyi and A. Kitaev, Universal quantum computation with ideal Clifford gates and noisy ancillas, Phys. Rev. A 71, 022316 (2005), doi:10.1103/PhysRevA.71.022316.
  4. 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.