One might think that the Hubbard model on the square lattice with infinite U would be relatively boring. For example, a "simple" theory like Brinkman-Rice [or equivalently slave bosons] would predict that the ground state is a metal, except at half filling where it is a Mott insulator.

There is an interesting preprint Phases of the infinite U Hubbard model by Liu, Yao, Berg, and Kivelson. Here are a just a couple of the results concerning the ground state on a ladder, that I found interesting.

For 3/8 filling [3 electrons per 4 sites] they find the ground state is an insulator with a charge gap (0.24t) and plaquette bond order. The spin degrees of freedom are equivalent to those of a spin-3/2 antiferromagnetic Heisenberg model. I think this can be "understood" this by starting from the limit of weakly coupled placquettes with 3 electrons per plaquette.

For 1/4 filling the ground state is an insulator with a charge gap (~0.1t) and a small spin gap and "dimerisation".

The authors argue these phases will also be stable for the actual isotropic square lattice.

It is fascinating to me that one can produce such rich and diverse broken symmetry states starting from such a relatively simple model which has no competing interactions in the Hamiltonian.

**competing phases are a generic feature of strongly interacting systems.**

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