sqd-spin-referee

es sense. I'm still not sure it's correct and I didn't miss any detail, but it's not stupid-AI-slop.

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.

thout 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