I understand how the mechanism of spontaneous symmetry breaking works, and why it produces Goldstone bosons (for global symmetries) and massive gauge bosons (for local ones).

However, I'm confused as to why we identify Goldstone bosons as we do. Take the linear sigma model for example. Why do we say that the fields $\phi^n$ are unphysical, whereas the fields $\phi^n - v^n$ are?

I get the feeling that it's a sort of definition that a physical field must have zero vacuum expectation value. At least in order to be interpreted as a particle, that is. But why is this the case? It's not a Wightman axiom, as far as I know!

There's a related question here but the answers aren't really rigorous enough for my liking. In particular they both rely on "hand-waving" claims that "particles are small oscillations around a vacuum".

I'd rather know that zero VEV is a physical principle, or see a proof that somehow zero VEV follows from a physical requirement. The other question seems to deal with the converse.

Many thanks in advance for your help!


2 Answers 2


I agree that the definition "particles are a small oscillation around a vacuum" is not correct, it's some attempt to apply semi-classical reasoning via the path integral to QFT.

However, it doesn't mean anything for a field to be "physical," fields are not physical. They are a computational tool that allow us to ensure that amplitudes we write down are local and causal. Therefore, zero-VEV cannot be a physical principle, nor can it be derived from one. The physical question is: Does the vacuum of the theory respect the symmetry in question. If the answer is no, the physical consequence is that there exist massless excitations.

To see an example of this, consider the S-Matrix field redefinition theorem. LSZ reduction basically tells you that it doesn't matter which field operator you use to do calculations, as long as the two fields have overlap on the 1- particle states and you compensate your choice with an appropriate normalization factor and put external lines on-shell (this is explained properly in Weinberg vol 1).

The definition of the S-Matrix involves a specification of asymptotic states. Specifically, you assume that the asymptotic states are 1 particle states of some free hamiltonian (usually assumed to be the Hamiltonian you are perturbing about, although this assumption fails for theories like QCD in the IR). In particular, it assumes a particular vacuum state from which particle excitations may be built using creation operators associated with the free Hamiltonian.

So while it does not matter which field you use to calculate correlators, in order for LSZ reduction to work you have to put the external states on shell. Since you understand how SSB works, you know that you can show in many different ways that there must be a massless particle in the spectrum.

So the point is this: it is not that the field with no VEV is any more "physical" than the regular field, this is a meaningless statement. But if you are intending to read off the masses of external states from the tree level Lagrangian, you will be doing the wrong calculation unless some of the fields in the lagrangian are massless. If you have assumed that SSB has occurred and the vacuum is no longer a singlet, then you need to be using m=0 when you compute amplitudes for the goldstone mode scattering. And you need the other Feynman rules to be consistent with this Feynman rule. So you make the expansion you refer to so that you may derive Feynman rules that consistently treat the external state as massless and consistently describe the other interactions. If you do not do this, you are doing the wrong calculation, using asymptotic states that do not actually exist in your theory.

Hope this helps!

  • $\begingroup$ I don't agree with all you said. Fields are not physical, but their expectation values are. In the sense that they are measurable. You have a HEP point of view, which is no wrong it itself, of course, but that renders everything more convoluted (LSZ formula, etc.) to make sense out of them. A stat-phys approach makes everything easier. It's simpler to grasp what is the mean value of a spin (the magnetisation), than that of the Higgs field. Same with the Goldstone mode (that are never there in HEP anyway). $\endgroup$
    – Adam
    Commented Oct 23, 2013 at 2:47
  • $\begingroup$ The author referenced the Wightman axioms, and wanted to know why zero VEV was part of the definition of a particle, so I assumed he wanted the answer for relativistic QFT. In RQFT neither fields nor their expectation values are physical, and the definition of a particle has been settled since Wigner. This much is made clear by the on-shell formulations for scattering amplitudes. I think statmech is less clear. You might disagree, but I find it hard to call an order parameter a particle, and goldstones in stat-mech are really just long-wavelength modes in a classical field, not a particle. $\endgroup$
    – Dan
    Commented Oct 23, 2013 at 5:01
  • $\begingroup$ @Dan - thanks. I still don't really understand why you have to identify the Goldstone bosons as the on-shell particles though. Why can't the original fields $\phi^n$ give rise to these on-shell species? Assuming that the Goldstone bosons represent the particle content of the theory, I can prove in several ways that they must be massless. But I have no idea how to justify this assumption! $\endgroup$ Commented Oct 23, 2013 at 9:16
  • $\begingroup$ Ah - I've reread and have a different interpretation. Is your answer essentially that zero VEV is a mathematical requirement due to the LSZ theorem? If so that makes me very happy (since I'm really a mathematician at heart, so I like mathematical reasoning)! $\endgroup$ Commented Oct 23, 2013 at 9:19
  • $\begingroup$ @Dan: as I said, it's mainly a matter of taste. But I don't think you need the S-matrix and LSZ theorem to describe what is a Goldstone mode, or even what is a particle. You just need to look at the Lehmann representation (spectral function) and see if there is a peak at a given energy/mass. $\endgroup$
    – Adam
    Commented Oct 23, 2013 at 13:30

I don't think that what you're saying is true. VEVs are physical, and in principle (i.e. if you do the calculation non-perturbatively), there is no need to expand the field around the VEV. Usually, the expansion is useful in a mean-field picture, where one assume that the fluctuations around the VEV are small, and therefore the physics is well described by keeping only the quadratic terms in the fluctuation.

For instance, let's write the potential as $V(\phi)=\lambda(\phi^2-v_0^2)$ (I'll just discuss the O(2) model, with $\phi^2=\phi_1^2+\phi_2^2$). This potential has a minimum at a mean-field level at say $\phi_1=v_0$. However, due to fluctuations, the real value of the VEV, that is $\langle \phi\rangle$ has a different value, call it $v$ (which can be zero !).

If the fluctuation are small, we can expand the potential around its minimum, that is $\phi=\langle \phi\rangle+\delta\phi$. Note that there is a priori no need to do that, if you're smart enough to compute everything non-perturbatively. At a mean-field level, $\langle \phi\rangle=v=v_0$ and the expansion to order two in $\delta\phi$ will give rise to the usual goldstone mode, plus an amplitude mode (along the direction 1), which is well defined and has a mass of order $\lambda v_0$. If you want to compute now the effects of the fluctuation beyond mean-field, you still want to expand the potential around its true minimum, order by order in perturbation theory, that is $\langle \phi\rangle=v$ order by order, which means $\langle \delta\phi\rangle=0$ order by order. The VEV of the real field is not zero, it is $v$. Its the VEV of the "fluctuation field" which is zero, by definition.

Finally, let's note that there is no reason a priori that the amplitude mode stays well defined (in particular in strong coupling). That is, it is well possible that the coupling between the amplitude and the Goldstone mode makes changes completely its properties, such that we can't say that there is a well defined particle. In fact, that's exactly what happens, as one can show that in fact $ \langle \delta\phi_1(p)\delta\phi_1(-p)\rangle^{-1}\to p^{4-d}$ as $p\to 0$ !

  • $\begingroup$ Okay - so a Goldstone boson is really just an approximation in a particular choice of variables then? I've genuinely never seen that written before, but it makes a whole lot of sense. So we only see Goldstone bosons because we're operating perturbatively close to the vacuum in our experiments, right? And similarly with the masses of the $W$ and $Z$ bosons? Could this then be evidence that at high energies our current theory of electroweak interactions will be insufficient to explain physics? $\endgroup$ Commented Oct 22, 2013 at 23:35
  • $\begingroup$ (Thanks for your answer btw! I'll try to have another go at answering your Wick rotation question tomorrow!) $\endgroup$ Commented Oct 22, 2013 at 23:36
  • $\begingroup$ No, the Goldstone exists if there is a spontaneous (continuous) symmetry breaking. Always. I'm not talking about HEP, because the Higgs mechanism changes a lot of things. But Goldstone mode are really important in cond-mat/stat-phys, where they play a dominant role in the low energy physics. Their direction is "perpendicular" to the order parameter (the direction of the VEV), by definition. And the direction of the VEV is arbitrary (but any direction works), but has to be chosen "spontaneously". You should look at some statistical field theory textbook, it's usually clearer than in HEP books. $\endgroup$
    – Adam
    Commented Oct 23, 2013 at 0:57

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