Recent content by schwarzg

Thank you for your explanation. However, the ground state of the Hubbard model is non-degenerated. Thus, it is not the case you mentioned. Thus I still guess that ##|g\rangle## must be an eigenstate of ##\vec{S}^2## still. Is there are another error in my logic?

Please see this page and give me an advice. https://physics.stackexchange.com/questions/499269/simultanious-eigenstate-of-hubbard-hamiltonian-and-spin-operator-in-two-site-mod Known fact 1. If two operators ##A## and ##B## commute, ##[A,B]=0##, they have simultaneous eigenstates. That means...