# How is the ground state chosen in a spontaneous symmetry breaking process?

In relativistic QFT, this cannot be a process in time. The unstable initial state does not exist at all: An unstable ground state is impossible in relativistic QFT at temperature T=0 (i.e., the textbook theory in which scattering calculations are done) since it would be a tachyonic state with imaginary mass, while the Kallen-Lehmann formulas require $m^2\ge 0$. (In particular, supercooling the symmetric phase down until $T=0$ is impossible.)

The unstable state can be defined only with a cutoff, but diverges into meaninglessness as the cutoff is removed. After renormalization, only the state of broken symmetry makes sense. All these are equivalent, so describe exactly the same physics. The ground state is equally described by any of these - they all produce equivalent physics. (Whenever a gauge symmetry is broken, the symmetry is reflected in the broken phase by the impossibility to distinguish between the different broken symmetry ground states. These form a single orbit under the gauge group, and hence any choice of gauge is equally valid.)

In the bending of a bar under a sufficiently strong longitudinal force - which bends in a random direction determined by the unmodeled details in the bar, the surrounding is fixed, hence the direction makes a physical difference. But the standard model doesn't float in a fixed ether - so there is nothing at all which could describe the difference between the various ground states. This means that the person doing explicit computations can choose the symmetry-breaking direction arbitrarily, for convenience, which is what is indeed done in the textbooks.

If the QFT is described by causal perturbation theory (the conceptually most - though not fully - rigorous construction), one never sees the unbroken symmetry.

At sufficiently high temperature $T$, the symmetry may sometimes be restored - but the particle interpretation is then quite different as we have instead only quasiparticles. In this case, lowering the temperature below the phase transition temperature $T_c$ breaks the symmetry, and essentially the same happens as in the case of a superconductor - the nature of the quasiparticles changes at the phase transition.

Dynamically, what happens when the temperature drops slightly below $T_c$ the physical state moves away from the previously stable and now unstable symmetric ground state to move towards the just barely emerged ring of (equivalent and indistinguishable) new ground states with broken symmetry. While it is moving it is out of [local] equilibrium and hence not easily described - in the equilibrium formulation this is visible as a discontinuity in the response functions. Once [local] equilibrium is recovered at a new broken symmetry ground state, the changes when lowering the temperature further are smooth again, as the ring of global minimizers changes smoothly. In the limit of zero temperature, it will have converged to the textbook theory with broken symmetry.