classical_irreproducibility.md

Classical irreproducibility of the DLA-parity asymmetry

The DLA-parity hardware campaign on ibm_kingston measured a non-zero asymmetry between the "even" and "odd" initial-state sectors: peak +17.48 % at depth 6, mean +9.25 % across the eight depths surveyed, with Fisher combined $\chi^{2} = 123.4$ across sixteen degrees of freedom. The observation is real in the statistical sense. This page states precisely which reading of "classical simulators cannot reproduce this" is supported by the evidence and which reading is not, closing audit item D1 with a narrow claim and an explicit, honest limitation.

Two readings of "quantum result beyond classical"

Reading	Claim	This campaign
Narrow — signature irreproducibility	No classical simulator of the idealised Kuramoto-XY Hamiltonian $H = \sum_{i} \omega_{i} Z_{i} + \sum_{\langle i,j \rangle} K_{ij}(X_{i}X_{j} + Y_{i}Y_{j})$ can reproduce the observed asymmetry, because the Hamiltonian conserves total parity exactly.	Proved below.
Broad — quantum computational advantage	No classical algorithm, given polynomial resources, can simulate the hardware evolution at the system size used in the campaign.	Not proved. Phase 1 used $n = 4$ qubits; classical simulation of this system is trivial (we perform it routinely in this repository).

These are different statements. Running them together — "the observed asymmetry is beyond classical" without specifying which reading is meant — is overclaim. The narrow reading is the honest interpretation of the Phase 1 data; the broad reading requires scaling to $n \gtrsim 30$ and closing different open problems tracked as Gap 1 and Gap 2 in GAP_CLOSURE_STATUS.md.

Narrow claim: proof by total-parity conservation

Statement. Let $H$ be the campaign Hamiltonian and let $P = \prod_{i} Z_{i}$ be the total-parity observable. Then $[H, P] = 0$, and consequently any unitary $U(t) = e^{-i H t}$ (including every Lie-Trotter approximation of it built from $H_{Z}$ and $H_{XY}$ separately) commutes with $P$. If the initial state $|\psi_{0}\rangle$ is an eigenstate of $P$ with eigenvalue $\pm 1$, then every subsequent state remains an eigenstate with the same eigenvalue, and the probability of measuring any opposite-parity computational basis string is identically zero.

Proof. Every summand of $H$ commutes with $P$ individually:

• $[Z_{i}, P] = 0$ because $P$ is a product of $Z$ operators and $[Z_{i}, Z_{j}] = 0$.
• $[X_{i}X_{j}, P]$: $X_{i}$ anticommutes with $Z_{i}$ and commutes with every other $Z_{k}$; similarly for $X_{j}$. Conjugating $X_{i}X_{j}$ by $P$ therefore picks up two sign flips, i.e. $(-1)^{2} = +1$. So $P X_{i} X_{j} P^{-1} = X_{i} X_{j}$, hence $[X_{i}X_{j}, P] = 0$.
• $[Y_{i}Y_{j}, P] = 0$ by the same argument — $Y$ also anticommutes with $Z$ on the same site.

Since each summand commutes with $P$, the sum does, so $[H, P] = 0$ and consequently $[e^{-iHt}, P] = 0$ for all real $t$. Because $H_{Z}$ and $H_{XY}$ each commute with $P$ individually, every Lie-Trotter factor $e^{-i H_{Z} \tau}$ and $e^{-i H_{XY} \tau}$ also commutes with $P$, and so does every composed step and every power of the step.

Initial parity eigenvalues. The even sector uses $|\psi_{0}\rangle = |0011\rangle$ with popcount 2 ($P|\psi_{0}\rangle = +|\psi_{0}\rangle$); the odd sector uses $|0001\rangle$ with popcount 1 ($P|\psi_{0}\rangle = -|\psi_{0}\rangle$). Applying the commuting unitary preserves each eigenvalue exactly, so

$$\langle \psi_{0}| U(t)^{\dagger} \Pi_{\text{opp}} U(t) |\psi_{0}\rangle = 0$$

where $\Pi_{\text{opp}}$ is the projector onto computational basis states of opposite parity. Measuring in the computational basis on the evolved state therefore returns zero leakage for every depth and every $t_{\text{step}}$.

Mechanical verification in the repository

The proof is not left as prose. It is enforced by the test suite and the numerical reference.

• tests/test_classical_irreproducibility.py (28 tests) performs three independent algebraic checks at every relevant system size:

  1. [H, P] = 0 as a SparsePauliOp identity — each term pair is reduced symbolically, and every residue coefficient is numerically below $10^{-12}$ (in practice 0.0 exactly). Run at $n = 3, 4, 6$.
  2. [H_{Z}, P] = [H_{XY}, P] = 0 individually, i.e. each Lie-Trotter generator commutes with $P$ by itself. This is the stronger statement that makes the Trotter decomposition exact on $P$, not just first-order accurate.
  3. For a parity-eigenstate initial condition, the opposite-parity mask applied to $U(t) |\psi_{0}\rangle$ has total probability below $10^{-18}$ at depth 1, 4, 10, and 30 and at $t_{\text{step}}$ = 0.1, 0.3, and 1.7.

• src/scpn_quantum_control/dla_parity/baselines.py computes the noiseless leakage reference across the published depth sweep using two independent backends (numpy dense and optional QuTiP). Both report max_abs_leakage < 1e-10 and agree with each other to $10^{-12}$.

• tests/test_cross_validation_qutip_dynamiqs.py separately confirms that the Hamiltonian matrix itself agrees between our internal Qiskit-based builder, QuTiP, and Dynamiqs to $10^{-10}$ relative. The invariant $[H, P] = 0$ is therefore a property of the physics we specified, not an artefact of the specific Hamiltonian-assembly code.

What this does not say

• Nothing about quantum advantage at scale. The classical evolution at $n = 4$ runs in milliseconds on a laptop. The statement here is strictly about reproducibility of the specific observed signature, not about complexity-class separations. The broad claim ("no efficient classical algorithm can simulate this for any $n$") remains open and is tracked under Gap 2 in GAP_CLOSURE_STATUS.md.
• Nothing about "hidden" classical models. The statement covers classical simulators that faithfully implement the idealised Hamiltonian. It does not rule out noise-matched classical models that deliberately inject a biased depolarising channel to recreate the observed asymmetry empirically. Such a model would have to reproduce the exact depth and sector dependence, and would be measurement-driven fitting rather than first-principles simulation.

Why the hardware asymmetry is non-zero anyway

Because the parity-conservation proof depends on $U(t)$ being exactly the idealised unitary, and hardware circuits are not exactly unitary. Any physical channel that allows a single bit to flip — relaxation during an idle, measurement crosstalk, a miscalibrated single-qubit rotation — breaks parity conservation in the measured distribution. The observed asymmetry between even and odd sectors is therefore a signature of the hardware's noise channel interacting with the chosen initial state, not a property of the Hamiltonian. The short paper paper/submissions/submission_002_phase1_dla_parity/phase1_dla_parity.tex presents this as a hardware-calibration observation, not as a quantum-advantage claim, and the distinction made in the table above is the reason.

Relationship to falsification criterion C8

docs/falsification.md registers a Phase 1 reproducibility criterion: the 342-circuit dataset, rerun through the analysis pipeline, must recover $\chi^{2} = 123.4 \pm 1$ and the ${+}17.5 %$ peak asymmetry at depth 6. That criterion guards against regression in our analysis code. The present document guards against a different failure mode: overclaim of scope. Both are load-bearing: the reproduction proves our analysis is stable; the parity-conservation argument proves the hardware signature is the only source of the result; the open Gap 2 proves we are not yet claiming computational advantage.

Audit closure

This document and tests/test_classical_irreproducibility.py together close the narrow reading of audit item D1 from the internal audit index. The broad reading remains open and is tracked in GAP_CLOSURE_STATUS.md as Gap 2, with the concrete next step being scaling the campaign via circuit cutting and/or MPS-falsifying regimes.