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Superconducting pairing correlations #259

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Description

@egranet

Circuit ID

superconducting_pairing_4steps

Circuit file

superconducting_pairing_4steps.py
superconducting_pairing_3steps.py
superconducting_pairing_2steps.py

Qubits

90

Gates

8964

Problem category

Active candidates

Problem description

We consider the Hubbard model on a $L_x\times L_y=6\times 6$ square lattice

$H=-t\sum_{\langle i,j\rangle}\sum_{\sigma =\uparrow\downarrow} c^\dagger_{i\sigma}c_{j\sigma}+c^\dagger_{j\sigma}c_{i\sigma}+U\sum_{j} n_{j\uparrow}n_{j\downarrow}$

at half-filling with $U/t=8$. We use a local fermionic encoding that requires $18$ ancillas, using in total $90$ qubits. We impose "mixed" boundary conditions, that is the average over the four different boundary conditions obtained by imposing periodic/antiperiodic boundary conditions in the vertical/horizontal directions. The circuits involve three different stages:

  1. the preparation of a low-energy state of $H$ at half-filling, with an energy corresponding to a thermal state with temperature $T/t\approx 0.3$
  2. a light-pulse for a time $0.75/t$, where the system is evolved with $H$ in a uniform time-dependent electric field in the diagonal direction, using $2$ Trotter steps
  3. a relaxation for a time at most $0.75/t$, where the system is evolved with $H$ without electromagnetic field, using $2$ Trotter steps.

In total, there are thus at most $4$ Trotter steps of size $0.375/t$ on top of the low-energy state. Then one measures the superconducting pairing correlations $\langle \Delta_i^\dagger \Delta_j \rangle$, where

$\Delta_j= c_{j\downarrow}c_{j\uparrow}$

The output quantity is the staggered average

$P_\eta^{\rm stag}=\frac{1}{L_xL_y}\sum_j \sum_{(x,y)\neq (0,0)} (-1)^{x+y} \langle \Delta_{j}^\dagger \Delta_{j+(x,y)} \rangle$

The circuits are written in Guppy. Three instances are proposed, that differ only by the number of Trotter steps:

  • 2 steps: state preparation + light pulse
  • 3 steps: state preparation + light pulse + 1 step relaxation
  • 4 steps: state preparation + light pulse + 2 steps relaxation

The most challenging circuit to simulate is the 4 steps. The 4-steps circuit is the circuit that we propose for the challenge. The 2-step and 3-step circuits are given for comparison purpose as easier circuits. The values measured for $P_\eta^{\rm stag}$ on the quantum computer are

  • 2 steps: 0.266 +/- 0.025
  • 3 steps: 0.115 +/- 0.039
  • 4 steps: 0.132 +/- 0.043

Name

Superconducting pairing correlations (4 steps)

Observable value

0.132

Error bound (low)

0.089

Error bound (high)

0.175

Method

Quantum computer

Method proof

Preprint
arxiv:2511.02125

Quantum runtime (seconds)

No response

Classical runtime (seconds)

No response

Compute resources (quantum)

No response

Compute resources (classical)

No response

Notes

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Authors

No response

Institutions

Quantinuum

Activity

  1. AIOvernight commented on Oct 21, 2026

    @AIOvernight

    We report a classical estimate for the two-step subcase of issue #259 (6×6 Hubbard, U/t=8, half filling: the specified low-energy-state preparation followed by the two-Trotter-step light pulse; 400-shot Helios experiment). We do not report results for the three- or four-step circuits here; we are working on them.

    Observable convention. The positive number reported for the experiment (0.266) is the statistic

    A = (2/N)(<D> - <eta† eta>), N = 36, D = sum_j n_(j,up) n_(j,down), eta = sum_j (-1)^(x_j+y_j) c_(j,down) c_(j,up).

    Equivalently, the reported value is −Σ_{(x,y)≠0} (−1)^{x+y} P_η(x,y), with P_η as in the paper's Eq. (6) (which includes the Hermitian conjugate); summing the authors' released per-distance CSV values this way gives ≈0.265 (the CSVs pool pairs per displacement; the shot-level estimator gives 0.2658). The off-site pair sum displayed in the issue has neither the Hermitian conjugate nor the overall minus sign, so in a state of fixed total particle number it equals −A/2 (hardware: −0.133 ± 0.013). Values must be compared in the same convention; below we use A throughout.

    Classical value. Our estimate of the noiseless two-step value, averaged with equal weight over the four boundary-sign sectors of the mixed preparation, is

    A = 0.38364 (finite-model value 0.3836428102).

    It is obtained by propagating the readout operator backwards through the two pulse steps (the final onsite layer cancels; the result has particle rank ≤ 6 on Gaussian-dressed modes of at most four sites) and evaluating it on the prepared state through connected preparation cumulants on supports of up to six sites, summed over all supports before reconstructing moments.

    Uncertainty, and its status. Supports of seven or more sites are omitted. We have no proven bound on that omission. Under an empirical transfer hypothesis calibrated on the exactly solvable 4×2 lattice (worst-sector cutoff-six error 7.4×10⁻⁴, which sets the calibration constant) and tested on 4×4 and 8×2 — omitted tail at most four times the larger of that calibration error and the last cutoff increments — plus a 0.002 numerical allowance, the estimated envelope is ±0.005. This is a conditional estimate, not a confidence interval and not a certified bound. Supporting controls at the specified parameters:

    • exact simulation of the same gate sequence and parameters on 4×4 (four sectors, mean 0.3844142, reproduced by two independently written fixed-particle-number simulators to 2×10⁻¹⁶) and 8×2 lattices, where the same method at cutoff six errs by at most 1.9×10⁻⁴ and 9.2×10⁻⁵ per sector respectively;
    • an exact 4×2 encoded-qubit simulation of the circuit agreeing with the fermionic simulation to 6×10⁻¹⁵;
    • an independent reimplementation of the reduction, which reproduces the finite-model rows (the complete 4×2 hierarchy through cutoff six, all four 4×4 sectors at cutoff four, the complete PP sector of 6×6 at cutoff four, and selected 6×6 contractions at cutoff six) to ~10⁻¹⁴. The full 6×6 cutoff-six production itself was not independently recomputed.

    The method is not uniformly accurate: doubling the weak preparation angles produces a 0.022 mean error (0.031 in the worst sector) on 4×2, accompanied by visibly larger cutoff increments. On 6×6 the last two cutoff increments are both ≈5×10⁻⁴ and positive in every sector (mean change between them −3 %, +13 % in one sector), so they show no clear decrease; this is why we do not claim a convergence-based bound.

    Experimental value. Re-parsing the released Zenodo shots, we reproduce the two-step value A = 0.2658 ± 0.0255 (sample standard error, 400 shots). That estimator rescales the usable system-site pairs of each shot (only 48 of 400 shots have all 36 sites usable), so relating it to the ideal unconditioned expectation requires assumptions about leakage and missingness. We do not attribute the difference between 0.384 and 0.266 to any specific noise mechanism, and we do not claim a statistical incompatibility, because the systematic and selection effects of the estimator are not quantified.

    Request. If the authors have a noiseless reference for the full 6×6 two-step A (or an emulator value), a comparison would directly test this estimate. Code, stored results and reproduction scripts are public: https://github.com/AIOvernight/hubbard-pairing-classical (./reproduce.sh verify checks the stored numbers on CPU in seconds; ./reproduce.sh cheap reruns the exact 4×2/4×4 controls and the 6×6 PP sector at support cutoff four in a few GPU-minutes).

    AI disclosure. This calculation, the code and this text were produced by AI agents (Claude by Anthropic and GPT by OpenAI) coordinated by @AIOvernight. There has been no human or institutional review; see docs/AI_AND_PROVENANCE.md in the repository for which checks are independent in implementation.

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