Restoration of spontaneously broken symmetry at high energy

It is common to find books saying that above a certain energy, a certain symmetry in particle physics is restored, e.g. the $SU(2)\times U(1)$ electroweak symmetry was unbroken between $10^{-36}$ to $10^{-32}$ seconds after the Big Bang, around when the universe was in an extremely hot and highly energetic state. Going even further back in time, the symmetry of a GUT which encompasses the strong and electroweak interactions would be restored, and going even further back (the Planck epoch), the symmetry of a quantum theory which unifies all four fundamental forces including gravity would be restored.

From what I read, it seems that at energies above those of the masses of (e.g.) the Weak gauge bosons, the electroweak symmetry is restored.

My question is, how does one understand this from a technical, quantum field theoretic point of view? In other words, in a general quantum field theory, how does a spontaneously broken symmetry get restored at higher energies, via quantum effects? Does it have something to do with the renormalization group flow of the parameter which enters the mass term of the Higgs field?

Preface: high temperatures instead of high energies

It seems that you confuse the words "energy" and "temperature" when reading about the "high scale" in the context of your question. As ACuriousMind have noted in the comment section, if you fix the temperature to be zero, then, of course, the EW symmetry group is broken independently on typical energies of processes. EW theory continue to be in the broken phase, since the symmetry breaking value $v$, which denotes Higgs doublet VEV, is nonzero independently on energies.

However, in the question you write about the early Universe, and from this follows that you talk not about the energy scale, but about the temperature scale. Precisely, we have the time-temperature relation from the Friedmann equation, and instead of conception of time for the age of the Universe we can use the conception of the temperature. You may think that you live in the Universe with zero temperature; but the early Universe is high temperature one. So now your question is about high temperature restoration of electroweak symmetry, i.e., about temperature dependence of Higgs VEV $v(T)$.

Nonzero temperatures: the free energy instead of the usual energy

When you deal with nonzero temperatures, thermal effects take the part. Precisely, thermodynamical system is described by the minimum not the energy $E$, but the large thermodynamical potential $G$. Heuristically it is explained by changing the structure of the ground state and excitations for nonzero temperature. (added) At zero temperature the vacuum is the state with zero "real" particles; it is filled, however, by quantum fields and their "virtual" excitations. At nonzero temperatures, however, these excitations become real, and instead of zero particles state we have the state with "thermal bath". This, of course, makes the contribution in observed quantities; for example, it definitely changes VEVs, since they are depended on the definition of vacuum. Such difference makes impossible to interpret QFT at finite temperature $T$ as QFT with typical energies $E \sim T$ of processes, as we may do with classical phycics due to equipartition theorem. (added)

In realistic case chemical potentials above the scale of $1 \text{ GeV}$ are small, so that $G \approx F$, where $$F \equiv F[\varphi ,T]$$ denotes the free energy, and $\varphi$ denotes the set of fields of theory.

When the space is homogeneous, then $$F[\varphi , T] = \Omega V_{eff}[T, \varphi ]$$ The Higgs VEV $$v(T) \equiv\langle H \rangle,$$ which breaks $SU_{L}(2)\times U_{Y}(1)$ down to $U_{EM}(1)$, is therefore the minimum of $V_{eff}[\varphi , T]$ instead of $V_{eff}[\varphi , 0]$. While the latter is given by one-loop effective action $\Gamma = \Omega V_{eff}[\varphi , 0]$ at zero temperature (and the usual zero temperature QFT is used), the former has to be computed by using nonzero temperature QFT methods.

Thus it is possible that instead of $v(0) = v$ at zero temperature, we will obtain that at some nonzero temperature $T_{0}$ (and at larger ones) $$v(T_{0}) = 0,$$ which means the exact restoration of symmetry (i.e., gauge bosons are massless exactly at temperatures above $T_{0}$).

Your question - computation of higgs self-interaction potential

The problem now is to calculate nonzero effective potential of higgs field $V_{eff}(\varphi , T)$. Suppose you are interested only in the higgs VEV self-interaction potential, $V_{eff}[v(T), T]$. Then you have to intergate out the other SM degrees of freedom, and simultaneously taking into the account nonzero temperature effects. One of the way to do this is to introduce Matsubara formalism, for which effective potential $V_{eff}[v(T), T]$ (in so-called thermal one-loop approximation, when we just neglect the interaction between higgs doublet and gauge fields, but leave gauge fields mass term) is given from the thermal path integral as $$\tag 1 e^{-\beta V_{eff}[v(T) , T]} = \int D[\varphi]e^{-S^{\beta}[\varphi]},$$ where $S_{\beta}[\varphi]$ is thermal action, $$S_{\beta}[\varphi] \equiv \int \limits_{0}^{\beta}d\tau \int d^{3}\mathbf x \left( \frac{1}{4}F_{\mu \nu}^{a}F_{\mu \nu}^{a} + \frac{m^{2}(v(T))}{2}A_{\mu}^{a}A_{\mu}^{a}\right)$$ $\tau$ is euclidean time and $\beta$ is inversed temperature.

Here I write only SM boson degrees of freedom since in infrared zone, which determines $V_{eff}[v(T), T]$, only they are relevant. This is true since Matsubara frequencies, which are the analog of the usual fourier frequencies (and thus energies) for nonzero temperature, for bosons start from zero, while for the fermions the start not from zero, and thus the latter are irrelevant. You then need to integrate $A_{\mu}$ out.

The fact that below such scale $v(T)$ may be zero gives rise to the statement that there is such phase transition in SM (first-second kind, or just the crossover).

Unfortunately, perturbative treatment for calculating $V_{eff}[v(T), T]$ fails. Here I use the "perturbative" in a sense that we naively expect that the expansion constant for the effective potential is $\frac{v(T)}{T}$, which is definitely small near the transition. If it is true, than we may use one-loop calculation of effective action, which is simple. However, so-called infrared problem arise: the true expansion constant is $\frac{T}{v(T)}$. This can be shown by using the Matsubara formalism $(1)$. Thus the calculation of $V_{eff}[T, v(T)]$ is the big challenge. We need to use nonperturbative methods for exaluating it. One of powefrul methods is lattice quantum field theory.

There is, however, heuristical argument of what we expect from the EW phase transition. The value of $v(T)$ itself isn't gauge invariant. It can be given, however, as $$v^{2}(T) \equiv \langle \text{vac}| H^{\dagger}H|\text{vac}\rangle$$ The above quantity, however, can't be used as order parameter, since it is invariant under all explicit symmetry transformations in SM. We expect then that phases with broken and unbroken symmetries are indistinguishable, and so there is no phase transition of the second kind). The lattice simulations show that this is true (i.e., the phase transition is, really, the first kind for small Higgs bosons masses and crossover for realistic Higgs boson mass $m \approx 125\text{ GeV}$).

• This is really nice---a perspective I've never realized is relevant. Thanks for the insightful answer. It is clear from your answer that energy and temperature are not strictly related notions: Indeed, I understand that "the usual" methods of QFT are only valid for zero temperature. However, as high energy physicists are used to always exchanging (by virtue of using simple unit systems) notions like $E$ and $T$ without second thought, I would like to ask you in which situations this is actually allowed? Surely, for e.g. an ideal gas the equipartition theorem relates the two. Care to comment? – Danu Mar 6 '16 at 19:06
• Furthermore, I assume that questions such as calculating the Higgs potential at different temperatures have actually been addressed in e.g. the lattice QFT literature. Are you aware of the general outline of the main results in this direction? – Danu Mar 6 '16 at 19:09
• @Danu : in classical physics, people identify the temperature with the averaged energy of particle because of equipartition theorem (however, Boltzmann-like weight whatever exist). The number $N$ of particles, however, doesn't depend on the temperature: at nonzero temperature it is the same as at zero temperature. But when we deal with QFT, this is false. The zero-temperature vacuum is filled by quantum fields and their "virtual" excitations, but it doesn't contain real particles. But at nonzero temperatures, the vacuum is filled by the real particles. This modifies all of computations. – Name YYY Mar 7 '16 at 10:56
• @Danu : modification of vacuum state surely changes VEVs, so that finite temperature effects are important for discussing SSB in QFT. – Name YYY Mar 7 '16 at 10:58
• @Danu : I've modified an answer. – Name YYY Mar 7 '16 at 11:20