Symmetry breaking and Superfluid - Mott Insulator transition

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?

• I invite you to check the following posts, where I gave answer for the case of normal metal / superconducting phase transition : physics.stackexchange.com/a/134410/16689 , physics.stackexchange.com/a/69490/16689 , physics.stackexchange.com/q/306515/16689. Please correct your question such that it becomes clear : you discuss spontaneous symmetry breaking, not symmetry breaking, and the transformation you write down explicitly dos not change the Hamiltonian at all. Hence it's trivial to say its solutions are unchanged ... Commented Feb 13, 2017 at 9:20
• The problem is not about changing the phase locally or globally, it's about the possible conservation law following symmetries of the problem. For superfluidity the symmetry is U(1), i.e. you can change the global phase of the system without changing the Hamiltonian. It results (via Noether) that the number of particles is conserved. The spontaneous symmetry breaking related to a symmetry is associated to Goldstone phenomena. What I misunderstood, is that you want to apply a local phase rotation, which is usually known as a U(1) gauge transform (...) Commented Feb 13, 2017 at 9:26
• If a Hamiltonian is invariant with respect to a gauge transform, one says it presents a gauge redundancy, not a symmetry, though both naming exists because of history reasons. A clear way to distinguish them is in the naming, the Goldstone phenomenon associated to symmetry breaking becomes a Higgs mechanism in the case of gauge redundancy. Anyways, the gauge redundancy is broken in superconductors (or charged superfluids if you prefer), in superfluids only the U(1) symmetry is broken. So your question is not clearly defined to many aspects unfortunately. Commented Feb 13, 2017 at 9:29
• I apologize for the imprecisions. I'm quite new to quantum phase transitions. I try to modify my question in order to make it more clear. Commented Feb 13, 2017 at 9:46
• Thank for editing your question. Yes, the picture of aligning the phase along one direction is a good picture in order to understand the $U\left(1\right)\rightarrow\mathbb{Z}_{2}$ spontaneous symmetry breaking. Commented Feb 13, 2017 at 15:23

1 Answer

$$\mathrm{U}(1)$$ 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.

• Thanks for the answer. But actually you've not answered my question. I know what a spontaneous symmetry breaking is in general and Also concerning ferromagnetic transition. What i am asking is a detailed description of what "symmetry breaking" means in the SF/MI transition. Commented Feb 11, 2017 at 23:21
• Are you asking why the ground state breaks the symmetry in the superfluid phase? Commented Feb 12, 2017 at 0:55
• Yes. For example: in magnetic (Ising) system, when the temperature is decreased below a certain crtitical temperature $T_c$ a non zero spontaneous magnetization shows up. Provided that no external magnetic field is applied, the direction of this spontaneous magnetization can either up or down, with 50% probability. The fact that the spins (partially) align in a specific direction is indeed the symmetry breaking, because in the hamiltonian there is no a priviledged direction. Can you please explain in a similar way the symmetry breaking relevant to SF/MI phase transition? Commented Feb 12, 2017 at 0:59
• I have updated the answer Commented Feb 12, 2017 at 5:57
• Yes. The point is that in the ground state, the relative phase difference between sites is always $0$, and they will evolve in time coherently. At $t=0$, the phases will all lock-on to a particular value (which breaks the symmetry), and from that point onward everything evolves together. Commented Feb 12, 2017 at 18:19