# In spontaneous symmetry breaking, why can we be sure the VEV is near the classical one, given the convexity of effective potential?

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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.

edited Sep 18, 2015
I liked it the other way better.
@RonMaimon, ok, reverted.
"as a nonperturbative result, the exact effective potential must be convex, this means our Mexican hat becomes flat-bottomed after the full quantum corrections." Why? What's the reason?
As another comment: the classical VEV is a free parameter of the theory, only determined experimentally through muon decay and muon mass (given by Zeeman transitions in muonion). I guess that in principle the classical VEV might be zero, but muons would be stable, there wouldn't be Higgs mechanism and so on. As for the real (fully quantum) VEV, should it be close to the classical one? Why? What are the implications if they were very distinct and in particular if the real one is zero? Would that make the perturbative approach not justified?
@drake, thanks for the reply. Somehow I didn't get any notification for some of the comments in this post. @dimension10, @polarkernel, has this happened before?

@JiaYiyang This is a known bug, see here

@dimension10, ok, thanks for the explanation.

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The analogous non perturbative thermodynamic theorem is the convexity of the free energy, and the analogous phenomenon is the Maxwell construction. When you have a first order phase transition, the region of coexistence is the flat part of the free-energy function, where the unperturbed pure-phase free energy has the wrong convexity property. The behavior there is completely different from the pure-phase region. The two phases are segregated in bulk, with a phase boundary. The interpolating construction fixes up the nonperturbative free energy to be convex, and there is an additional free energy cost in the middle part from the phase-boundary, it just scales as the area and not the volume, and vanishes in the thermodynamic limit.

In field theory, for scalar $\phi^4$, it's not an analogy, it's the exact same thing. The intermediate region in the phase diagram as a function of the field parameter $\hat\phi$ in the effective potential consists of the two phases separated by domain wall. The domain wall has energy, just with subleading scaling as the area, so it doesn't contribute, so that you have a Maxwell style convex free energy graph which is only qualitatively similar to the perturbation theory free energy at the far ends, and it has a totally flat bottom.

In the Mexican hat case of broken continuous symmetry, the regions of intermediate $\hat{\phi}$ are also described by various superpositions/probabilistic-mixtures of non-uniform fields (depending on whether you are quantum or Euclidean), and this restores convexity, although now, there are field gradients all over. The parameter in the effective potential $\hat{\phi}$ is a constraint telling you what proportion of different vacua are included. This constraint is not physical, and at an actual value, at the minimum value of the effective potential, you are close to the edge. This is where you have the empty vacua we are always interested in perturbing around. The flattened out middle part doesn't describe the same thing.
answered Feb 5, 2015 by (7,730 points)

Wait, after some thought, I'm not quite convinced that

It doesn't have minimum energy really... Only the energy density is the same as a real edge vacuum.

For simplicity let's consider scalar phi 4,suppose we've found the vacuum states $|VAC\pm\rangle$ corresponding to, say, $\langle \phi \rangle=\pm 1$, which are the two degenerate edge vacua. Then all linear combinations $\sin \theta |VAC+\rangle+\cos\theta |VAC-\rangle$ would have the same energy(not energy density), but now the VEV becomes $\sin ^2 \theta - \cos^2\theta$(cross terms vanish by superselection?) Now VEV can take any value in $[-1, 1]$, without any change in energy(again, not energy density), and the state surely is still translation-invariant since $|VAC\pm\rangle$ individually are. What's wrong with this simple deduction?

@JiaYiyang: Yeah, I think you might be right, the intermediate vacua in the middle are probabilistic superpositions of the different edge vacua, not Maxwell construction things with domain-walls and intermediate gradients--- there's no global obstruction to making the superposition global. But it's still nonphysical stuff in the middle region, but I'll update the answer after thinking about it.
@RonMaimon, ok, now I'm almost convinced. Indeed $\{VAC+\rangle, |VAC-\rangle\}$ is a superselcted pair while $\{VAC+\rangle+|VAC-\rangle ,|VAC+\rangle-|VAC-\rangle\}$ is not, I was wrong in my last comment. Now the only thing that still puzzles me a bit is, in general how can I see the superselected vacua must be represented by the edges of the bottom of effective potential? Or equivalently, any vacuum represented by the intermediate region must not be part of a superselected basis?
@JiaYiyang: The effective action is not just a function of the global real parameter $\hat\phi$, it is a functional of the classical parameter field $\hat\phi(x)$, it's the functional Legendre transform of the free energy as a functional of $J(x)$. The global parameter $\hat\phi$ is not the thing that you vary to get local correlation functions, its just the global value. You vary $\hat\phi(x)$ locally. Under local variations, the effective potential is infinitely differentiable, and the one-particle-irreducible (1PI) correlation functions which obey cluster decomposition are those associated with one or another edge vacuum. There are two conditions--- first the global minimum energy condition $\delta \Gamma = 0$, saying the the variation of the effective potential is minimum at the global value of $\hat\phi$, and second the local condition $g(x) * g(x')* ... \delta_x\delta_y... \Gamma$, where the g's are test function and * is convolution, smeared local variations must converge to the appropriate smeared 1PI correlation functions obeying cluster decomposition. The cluster decomposition property of the correlation functions is what ensures you are looking at the S-matrix of one specific vacuum not the unphysical superposition of many vacua. I don't know how to write this formally very well, though, I'll think about it.

I'm still kinda fixated on this question. 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.

@drake, @RonMaimon, then I need elaboration. On the surface the situation seems quite "symmetric", I mean, I can start with $\{VAC+\rangle, |VAC-\rangle\}$ as a set of superselected states, but I can also start with, say, $\{VAC+\rangle+|VAC-\rangle ,|VAC+\rangle-|VAC-\rangle\}$ as a set of superselected states, I don't see why I should prefer one over the other. Weinberg in Vol2 of his QFT argues in presence of external perturbation such as $\phi^3(x)$, the first choice is prefered. However, this reasoning is rather dubious to me, after all the inability to transit between superselected states has been the key in discussions regarding SSB, but physically turning on perturbations such as $\phi^3(x)$ trespasses superselections(Haag theorem), so Weinberg's argument looks like a double standard to me. Though I don't know if some argument along the same line can be made while circumventing Haag theorem.
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