For concreteness let's nail the context to be Mexican hat potential. To deal with spontaneous symmetry breaking(SSB), one usually starts with assuming the vacuum expectation value(VEV) to be near the classical one, i.e., a minimum of the Mexican hat, and then quantum correct it by perturbation theory. However, as a nonperturbative result, the exact effective potential must be convex, this means our Mexican hat becomes flat-bottomed after the full quantum corrections. Then it means the VEV can be any where in this flat bottom(it can even be 0!), possibly very very far away from the classical one, so why is it reasonable at all to start the calculation from the classical VEV as if it must be approximately correct?
Update: After the long discussion, what's clear to me now is that, the physical vacuum must be chosen from a list of superselected vacua, and superpositions of vacua in this list are not viable choices, so the VEV probably cannot be freely chosen from the flat bottom. However, what's not yet clarified is, how do we know in general, that the superselected vacua must be represented by the edge of the flat bottom?
Update: After talking to people, it kinda surprised me that this is not a widely known result even to field theorists, so I'll just put the standard proof here for reference:
Consider an Euclidean scalar theory, we have the free energy functional that generates connected diagrams $W[\rho]$, defined as
$$\exp W[\rho]= \int \mathcal{D}\phi \exp\{-\int d^n x [L-\rho \phi] \}.$$
We can define a new positive measure for notational convenience
$$\mathcal{D}\mu= \mathcal{D}\phi \exp\{-\int d^n x L\}$$
Now we can write
$$\exp W[\rho]= \int \mathcal{D}\mu \exp\{\int d^n x \rho \phi \}.$$
Now Holder's inequality asserts
$$\int \mathcal{D}\mu F^\alpha G^\beta \leq [\int \mathcal{D}\mu F]^\alpha[\int \mathcal{D}\mu G]^\beta,$$
where $F$ and $G$ are arbitrary positive functionals, $\alpha$ and $\beta$ are positive numbers satisfying $\alpha+\beta=1$.
Now apply Holder's inequality to $F=\exp\{ \int \rho_1\phi \}$ and $G=\exp\{ \int \rho_2\phi \}$ then we have
$$\exp W[\rho_1+\rho_2]\leq \exp \{W[\rho_1]+W[\rho_2]\}$$
And this means $W$ is a convex functional. Since Legendre transform preserves convexity, we conclude $\Gamma[\langle \phi \rangle]$ is a convex functional.
But then I'm not yet sure what the counter part of this theorem is in Minkowski QFT.
And I copy one of my comments under Ron's post since it might be relevant, but still quite possibly very flawed:
I think if somehow we can make the analogy toward Gibbs free energy of phase transitions, we may still sensibly argue VEV should be at the edge: in this analogy, superselected vacua should correspond to pure phases, while maybe linear superpositions of superselected vacuua should correspond to coexisting phases. Then when one tries to goes across the phase transition lines, it would be expected that we first start by hitting the coexisting phase with 0 mixture, then gradually to the ones with finite mixture, then enter the zone of 0 mixture again, for example: water-->water vapour mixture-->vapour; it would be just very weird to suddenly hit a point of finite mixture at the very beginning, then at some point in the middle it becomes pure again.
This is not yet a well thought of argument, for example the immediate flaw I can think of would be that the correspondence to Gibbs free energy can't be exact. To be exact, one must have a path integral over periodic boundary conditions, but the effective potential we are talking about really is derived from the partition function with vacuum as fixed boundary conditions, not integrated over.
Q&A (4869)
