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

+ 6 like - 0 dislike
1401 views

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.

+ 4 like - 0 dislike
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 (7,590 points)
@JiaYiyang: It doesn't have minimum energy really, as you aren't taking the subleading-scaling energy in the strings/domain-walls or whatever gradients into account. Only the energy density is the same as a real edge vacuum.
@RonMaimon, emm, ok. Now I see the logic is consistent if I take your words, I'll read on some statistical field theory material to learn and verify what you said, e.g. domain wall, scaling etc. (Any suggestion of reference?)

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.
@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.
@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?

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.

+ 3 like - 0 dislike

Ron Maimon's first two paragraphs of the initial answer was close to the truth (namely an emergence of a wall), but superselection is a red-herring.

(Disclaimer: it's not my own idea, but I've just learned this from the recent article arXiv:1606.05069.)

It's best to start from the Ising, a stat-mech. Let $F(h)$ be the free energy, and $C(M) = F(h) + hM$. $C$ is the free energy when $M$ is the variable. When $h = \pm 0$, $M = \pm M_{sp}$ and $C(+M_{sp}) = F(+0)$.

The key observation is that $C(M)$ is not convex. Its explicit expression is known for 2d n.n. ferromagnet (see the ref.) but we don't need it. Anyway $C(M) > C(M_{sp})$ when $|M| < M_{sp}$. More precisely $C(M) - C(M_{sp}) \propto L \to \infty$ in thermodynamic limit (TL). (Let the volume $V$ be $L^2$.) As Ron Maimon said, it's the cost of a domain wall.

Notice the last eq has to diverge. If you switch back the variable to $h$, $P(M) = \exp(-\beta C(M)) / Z(h)$, where $P(M)$ is the probability that the magnetization is $M$. In TL, it vanishes. (If you know the rudimentary of Monte Carlo simulation, it should be familiar.)

Yet the order of non-convexity of $C(M)$ inside the region $|M| < M_{sp}$ is $o(V)$. It's because $\partial C/\partial M = h$, er, it's lengthy and see the ref. What's important is that the inside region is $h = 0$, but we control the system by $M$. When there is first-order transition, an extensive variable is a good parameter. (When boiling water, throughout the entire "inside region" p remains 1atm and T = 100°C. Then the volume or energy specifies the state.)

The inside region is not a weird probabilistic mixture of $\pm M_{sp}$ equilibria. (Equilibirium in stat mech = vacuum in QFT.) Otherwise, SSB couldn't happen.

Are wall-states are equilibria? Yes, if you control $M$. No, if you control $h$, and $M$ is not constrained. In the outside region, both are valid, but in the inside region, the only available parameter is $M$. (For a real magnet, $M$ does not conserve, but you should already be familiar with such imaginary objects, like external source coupling to the field.)

The situation is the same for QFT. Intermediate states have a higher energy, but the true vacuum energy is not a "mexican hat." It has only the bump of $o(V)$. The "hat" states are metastable false vacua. (First order phase transition can have metastable states.)

Intermediate states are not a superposition of SSB vacua. $|+M_{sp}>$ and $|-M_{sp}>$ are residents of different theories, given in different Lagrangians $\mathcal{L}[\pm 0]$. They cannot be superposed. When the system is finite, any states can be added. But in TL, no.

The ref claims to prove translational invariance breaks for QFT. I don't know if states with different wall positions can be superposed when the boundary is periodic. Anyway when the boundary is free, translational invariance does not exist, and you see phonons in crystals, so I reckon it's not so much trouble even if translational invariance breaks. (Domain walls can only appear when the symmetry is discrete. For $O(N)$ models, do vortices appear?)

* Coleman-Weinberg model is an example where quantum correction breaks the classical symmetry, so your doubt in the original question has some point. (Good intro is in Peskin & Schroeder, last pages of the part 2.)
* You have to be careful distinguishing results and definitions, including the meaing of "near the classical vacuum"; when there's SSB, there seem to be many coupling constants, but they're related, and independent variables are fewer. Their relations are given by power expansion, and "near the classical vacuum" is the case where a weak-coupling theory is useful, I think.
* In QFT fields naturally have a dimension in power of length, but I'm not sure if their absolute value is meaningful; in classical electromagnetism, you have to introduce a new dimension to describe phenomena, unless you identify the charge element.
* I don't know well how to define coupling constants. For QED, static limit, i.e. Coulomb force can me measured. (The modern, precise way must be QHE or Josephson junction.) I've heard for weak force, it's defined by the tree-level interaction. (It's non-renormalizable.)

answered Jun 21, 2016 by (100 points)

How can C(M) not be convex? The proof I used in the original post looks completely rigorous.

@JiaYiyang: I guess the confusion comes from using the same term "effective potential" for different objects. In statistical physics (and upon a brief check I believe this is also the case of arXiv:1606.05069) one uses the constraint effective potential $U(\varphi)$, defined as $e^{-V U(\varphi)}=\int D\phi\,\delta(\bar\phi-\varphi)e^{-S[\phi]}$, where $\bar\phi$ is the spatially averaged value of the field variable $\phi$. (The term "constraint" refers to integration over field configurations with fixed spatial average.) This object can be non-convex in finite volume, but it flattens out in the thermodynamic limit. In field theory, the effective potential is, however, usually defined as a Legendre transform of the generating functional of the connected Green's functions. This object is by construction convex. The two potentials coincide in the thermodynamic limit. There is a nice discussion of this in an older paper by O'Raifeartaigh et al. (inspirehep.net/record/17920).

The paper mentioned in the preceding comment,  ''The constraint effective potential'' by O'Raifeartaigh et al., Nuclear Physics B 271 (1986),653-680,  is freely available at http://www.tpi.uni-jena.de/qfphysics/homepage/wipf/publications/papers/constreffpot.pdf

@Jia Yiyang: It's because $C(M) = F(h) + hM$ "knows more than" $F(h)$. The convexity of $F(h)$ is rigorous. But it is only defined in the region which corresponds to $|M| > M_{sp}$. So you can't say anything on $C(M)$ for $|M| < M_{sp}$, which is the region of $h = 0$. It's bad to consider $C/V$, which is convex, or you'll be in the old pitfall, superposition of two vacs, superselection etc.

It does not seem to me that Tomáš Brauner has read my answer nor the original arXiv paper by Asanuma. It shows that the assumption that $C$ and the effective action is convex is wrong. The paper by O'Raifeartaigh et al is now obsoleted. The latter defines a false effective action by taking sup, fabricating convexity.

1. If $C(M)$ were really flat, then there would be no SBS. 2. For real substances, $M$, or its counterpart, relative density etc, is conserved. In those cases the inside region is tilted, not horizontal. But it doesn't matter, and anyway the convex free energy with a linear interval is wrong. There's no doubt the criticality of these systems are correctly described by field theory. But if you challenge the case of magnetization, uh, sorry, I need another cheat sheet. :P
 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverfl$\varnothing$wThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.