How does spontaneous symmetry breaking (SSB) happen? I've just finished studying for an exam on the Standard Model (so electroweak theory and symmetry breaking) and I can't figure out how this question never crossed my mind. I'm now studying the QCD chiral symmetry breaking, but I think my question applies to any (physical) SSB.
I know what SSB is (symmetry of the Lagrangian but not of the states) and I also know how one implements it in a theory (scalar sector with a mexican-hat potential) and it's clear to me the implications of the two different phases, the broken one and the restored one (vev for the nonbroken scalars, Goldstone bosons eventually "eaten" by the gauge bosons).
What I don't understand is, how does the phase transition work? How did the universe go from one phase to the other?
Does the potential just "switch on"? Is it always on but at high energies the quantum/thermal fluctuations don't "see" its structure?
 A: I'm just sketching the trail map of where your question might wish to go... it is a subject of limitless complexity.
The universe cools down, and thermal QFT dictates mutation of the Higgs potential with temperature, section 3.
This mutation of the effective Higgs potential from the form favoring the symmetric phase to one favoring SSB at lower temperature describes when the phase transition is likely to switch on, at $T_c$.

A: Good question.
Well in any QFT, the origin of SSB can be traced back to the loop Quantum corrections as was pointed out beautifully by Sidney Coleman and Eric Weinberg for the first time.
The analogy with solid state physics is helpful but not precisely and (necessarily) what takes place in QFT.
There are several variants of SSB, Global, Local, Chiral, Dynamical etc. But in all these cases SSB is due to quantum fluctuations.(At least at the level of Peskin Schroeder which is all written at zero temperature)
The effective action formalism(used for the first time by Jona Lasinio in the context of QFT) can shed enormous light on the physics of SSB in QFT as makes the solid state analogy even closer to real while one should keep in mind, SSB in QFT happens at zero temperature(though one can extend it to finite temperature QFT in a more advanced level) so the anology with solid state shouldn't be taken literally, and the role of temperature in QFT is basically played by a controlling parameter like vacuum expectation value of a Quantum Field(say scalar).
The following paper might help:

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*S. Coleman and E. Weinberg, Radiative Corrections as the Origin of Spontaneous Symmetry Breaking, Phys. Rev. D 7, 1888 (1973).

EDIT:
In fact "why the initial state is at the tip of the Mexican hat" can be answered this way in my opinion:
I'd say perhaps the question can be answered from a more diplomatic point of view. One can assume the initial temperature is so high that quantum fluctuations are dominated by thermal fluctuations and one has a prabolic potential, and this preseves the vacuum at the bottom of the well, and as the temperature decreases then at a certain point the quantum fluctuations can dominate the statistical ones, where the old vacuum tunnels into the new one.[In fact it can be even more complicated, the potential might not be necessarily parabolic but that "suffices" to make my point]
The first regime is where $kT>ℏω$ and the second one is where $kT<ℏω$. And all this is to avoid nonsensical negative mass terms needed for thermal phase transition! That's the root cause of my objection to thermal phase transition as the ultimate proposal for SSB in particle physics!
