In the picture below of the potential of the Higgs field, we can see that the field has its highest energy if the field is zero, i.e. if no Higgs particles are present.

The symmetry is restored at an extremely high temperature, leaving us with a Higgs field that is zero everywhere, i.e. no Higgs particles are present at that high energy.

I can't imagine how the Higgs field "knows" that the temperature is so high that the Higgs field evaporates. Of course, heat is transferred to the Higgs field but how can the field "sense" this high temperature. Are the energies involved in the collisions of particle fields with the Higgs field high enough to let the broken symmetry evolve into the non-broken symmetry?

For almost all phase transitions (like the one to a superconducting state, to a Bose-Einstein condensate, the ones involving solid-liquid-gas or the ones here, except the ones involving SSB) I can see how they relate to temperature. But not for this one.

So, what's going on in the transition from the presence of the Higgs field to the non-presence (evaporation)when the temperature is high enough? And what stuff is the "carrier" of this temperature? The Higgs field itself?

Higgs potential

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    $\begingroup$ Hi. I don't hold a definite answer to your question but may I ask this: why specifically the Higgs field? How would you describe other phase transition in other contexts- how do the fields there "know" what's happening and the transition happens? Just a question, to understand what difference you see. I ' m not specifying a certain phase transition example, for you to choose whatever you want. Thanks. $\endgroup$ – Constantine Black Jun 14 at 7:38
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    $\begingroup$ At a high enough temperature, there will be Higgs particles everywhere, because the average collision in the general plasma of particles will be energetic enough to produce them... In some sense, the answer to the question is, that the Higgs field interacts with the other fields. However I admit I can't answer the question to my own satisfaction, because I don't understand how to interpolate between thermal field theory (formalism for finite temperature QFT) and ordinary QFT (formalism for zero temperature). The vev is zero in thermal field theory, and nonzero in ordinary QFT... $\endgroup$ – Mitchell Porter Jun 14 at 8:47
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    $\begingroup$ ... but there must be some interpolating formalism. First four references in section 2.2 of dark.ft.uam.es/mununiverse/images/tesis_andrew.pdf are classic works where this was first studied. $\endgroup$ – Mitchell Porter Jun 14 at 8:48
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    $\begingroup$ @descheleschilder Really? I am assuming you can understand I am referring to such situations even if in general not all transitions come from SSB. $\endgroup$ – Constantine Black Jun 14 at 12:41
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    $\begingroup$ I am not asking you that. And ice to water is not a simple example, is one we are used to. Take Bose-Einstein condensation, take a feromagnet, take hard disks in 2D. And if you want, yes, take water. The fields are the molecules as you call them; what are the fields in the Ising model? The spins of the atoms. You can do field theory without doing high-energy particle theory. $\endgroup$ – Constantine Black Jun 14 at 13:10

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