Symmetry breaking and Superfluid - Mott Insulator transition

Question

I know my question is similar to what mentioned in this post: Symmetry breaking in Bose-Hubbard model. Yet, I don't find it clear. I've in mind a 1D Bose-Hubbard Hamiltonian. Moving from the Mott Insulator phase to the Superfluid Phase, a spontaneuos symmetry breaking occurs. What does it mean? Can you provide a clear explanation of it?

Answer 1

$\mathrm{U}(1)$ symmetry is not broken, it is spontaneously broken, meaning that although the Hamiltonian/Lagrangian might enjoy the symmetry, the ground state does not. For example, the ferromagnetic-paramagnetic Hamiltonian has a $\mathrm{SO}(3)$ symmetry that becomes spontaneously broken when you go into the ferromagnetic phase (the magnetic dipoles all point in the same direction), though we can still rotate the ground state with $\mathrm{SO}(3)$ and get another ground state.

So what is physically breaking the $\mathrm{U}(1)$ symmetry in the superfluid phase? It is the quantum-mechanical phase of the bosons; the ground state of the superfluid aligns the quantum-mechanical phases of the bosons in a particular direction. In the insulating phase, the phases do not align in the ground state.