# Higgs critical temperature?

Higgs mechanism is defined to operate (give spontaneous symmetry breaking) below some critical temperature. As mentioned, for ex. in Wikipedia:

Below some extremely high temperature, the field causes spontaneous symmetry breaking during interactions. The breaking of symmetry triggers the Higgs mechanism, causing the bosons it interacts with to have mass...

So what is meant by critical temperature quantum mechanically? Is this referring to a 'real' temperature as in a real gas, or some QM (virtual) temperature? Please explain.

• I believe (hence comment, not answer) that the temperature in this case is equivalent to the background energy of the universe. The same way as $k_B T = \langle E \rangle$ for a gas. The early universe was much more energetic than the one we live in right now due to inflation. At certain energies, the symmetry breaking stops playing any role in the scattering amplitudes. – Darkseid Jul 20 '19 at 20:52

In statistical mechanics, temperature is an indirect measure of energy: the higher the temperature, the higher the energy. Here's one way to express the relationship: when we say that the system (the universe, in this case) is in equilibrium with temperature $$T$$, we mean that as far as any local observations are concerned, we might as well use a weighted average over all possible states, weighted by $$e^{-E/kT}$$ where $$E$$ is the energy of the state and $$k$$ is Boltzmann's constnat. The higher the value of $$T$$, the less rapidly these weights fall off with increasing $$E$$, so higher-energy states contribute more and more as $$T$$ increases, especially because the number of high-$$E$$ states is much greater than the number of low-$$E$$ states. This way of thinking about temperature is just as valid in quantum field theory as it is for an ideal gas. In both cases, this $$T$$ is a real temperature: with minor qualifications, it's basically what temperature means.
The Higgs mechanism has a characteristic energy scale, which in the standard model is $$\sim 200$$ GeV. If we consider only states that have low energies compared to that scale, then the consequences of the broken symmetry (experts would say "non-linearly realized" symmetry) are prominent. But when we're averaging over states most of which have energy $$E$$ far larger than the $$W$$ or $$Z$$ boson mass (which are both within a factor of $$\sim 3$$ of the scale $$\sim 200$$ GeV), the lower-energy states where those masses are important are overwhelmed by the far-more-numerous higher-energy states. As a result, the familiar low-energy consequences of electroweak symmetry breaking are less prominent at such high temperatures.