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With the Bose-Hubbard Hamiltonian (BHH) being invariant under a U(1)[tex]\equiv[/tex]O(2) symmetry transformation, it is said that the hopping-term in the BHH tends to break the U(1) symmetry as the system leaves the insulating phase. This is not clear to me.

However within the mean-field description of the BHH, the above statement makes sense because the MF-BHH is no longer invariant under the U(1) transformation due to the superfluid order-parameter term (which comes from the hopping-term) not being gauge invariant. But what is not clear is how this same conclusion is reached even for the original hamiltonian.

Thanks in advance,

Mavi