Quantum computing's flagship chemistry results don't say which state they found

I've submitted the underlying paper here twice before as a Zenodo link, but it didn't get much discussion. I wrote this new post to be more accessible to non-chemists, plus a couple of things have happened since the last time.

First, IBM merged two PRs (364 and 366) to qiskit-addon-sqd on Sept 9 after I reported the issue. The docs for the symmetrize_spin option now clarify that it "does not guarantee that the state is an eigenvector of the total spin operator S^2". That covers one of the options. IBM hasn't commented on the paper itself, which I've recently updated to v2 on ChemRxiv.

The short version of the post: IBM's released random-input control actually matches their quantum hardware on the smaller iron-sulfur cluster. Also, IBM's flagship paper, the published critique of it, and the RIKEN-IBM Fugaku follow-up all report energies without stating which spin state those energies belong to. If you rebuild the data for the largest [2Fe-2S] size they published, the lowest state has S^2 = 1.3711. They were aiming for a singlet, which should be 0.

You can verify this without relying on my code:

Hardware vs random (takes about 10 seconds in a browser, reading IBM's files directly): https://colab.research.google.com/github/PureStateLabs/sqd-s...

Spin check (takes about 35 seconds, running PySCF on IBM's public integrals): https://github.com/PureStateLabs/sqd-spin-referee/blob/main/...

The last section of the post outlines exactly what would change my mind. If you try to reproduce a number and it doesn't match, let me know here.

Paper: https://doi.org/10.26434/chemrxiv.15006382/v2 Data and code: https://doi.org/10.5281/zenodo.21359922

gus_massa took the time to actually read this on the last thread and asked three things. That thread locked before I answered, which was my fault for being slow, so here they are.

You suggested a short version, two or three pages, that goes straight to the Table 1 results. I wrote one. It's about three pages and it's here: https://github.com/PureStateLabs/sqd-spin-referee/blob/main/...

You asked how many states sit near that energy, and whether there's a graphic. Yes. The four lowest roots of the flagship subspace span 12.5 mHa, with the ground gap at 0.285 mHa, and their spin values are S^2 = 1.371, 2.363, 4.738 and 11.590. Climbing the energy ladder climbs the spin ladder. The figure is fig_flagship_s4.png in the archive: panel A is that ladder, B is the residual against S^2, C is the per-root variance.

You said you couldn't picture how S greater than 1 arises, and wondered if it needed extra freedom in the d orbitals. It doesn't. [2Fe-2S] is two high-spin Fe(III) centres, each d5 with S = 5/2. Coupling 5/2 with 5/2 gives total S = 0,1,2,3,4,5, so S^2 runs 0, 2, 6, 12, 20, 30. That's the whole Heisenberg ladder and it's ordinary Clebsch-Gordan, no d-orbital subtlety needed. It also puts your objection on solid ground: 4.7 isn't an exotic value, it sits between the S=1 and S=2 rungs, which is exactly why it reads as a mixture rather than a contaminated singlet.

You wrote that some contamination is expected but 4.7 was too much in your opinion. I've since measured the second moment on those states, and the variance is the part that settles it. A spin eigenstate has Var(S^2) = 0. The low-lying roots here run 3.5 to 7.4. They aren't slightly perturbed eigenstates, they're broad mixtures, and a scalar mean of 4.7 understates how broad.