In all explainations of spontaneous symmetry breaking that I've seen, the scalar field doing the breaking is redefined around its vacuum expectation value (VEV), and similarly for the higgs mechanism - for example with a complex scalar field and U(1) symmetry:

$\phi(x) = (v + \rho(x))e^{i\chi(x)/v}$ where v is the VEV

Similarly for the higgs mechanism: the field is written as above, and then expanded out. This makes it very clear where the gauge field's mass term comes from, or in the case of just SSB, where the mass term for the goldstone boson goes. But my question is as follows: is this redefinition necessary?

Firstly, is there some fundamental reason why we need to expand around the VEV? I've seen the phrase "we perturb around the ground state" thrown around a few times, but my understanding was that when we do pertubation theory, we expand in the coupling constant, not some specific value of the fields - in fact in the path integral formalism, don't we even integrate over all of the field configs, even in pertubation theory? I feel I may have some fundamental misunderstanding here relating to the nature of pertubation theory.

Additionally, and this is my main question, if we don't make this field redefinition, the physical content of the theory seems to me like it shouldn't change. Specifically for the Higgs mechanism, my concern is as follows: when we don't make the field redefinition, there isn't an explicit mass term for the gauge boson, only an interaction term with the scalar. In this form of the lagrangian, does the mass of this vector boson (and the non-existence of the goldstone boson) manifest itself in some other way, and if so, whats the mechanism? If not, why not? Does it show up from renormalisation or loop corrections? Is there some non-perturbative effect which accounts for it? It puzzles me that the mass term can be seemingly eliminated by a field redefinition.

Also, even if there is no VEV, is it allowed to make the redefinition above anyway, with v a random parameter? This gives the gauge boson a mass term from the $\gamma \gamma \phi \phi$ term in the lagrangian, so what is fundamentally different about this situation? I sense that it is to do with the fact that the above redefinition is actually singular around $\phi$ = 0, and since that is the average value in the vacuum, we run into problems. But if we use something like $\phi = v + a + ib$ instead, the same idea seems to apply. What am I missing here?

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    $\begingroup$ I guess the accepted answer to this question might help you: physics.stackexchange.com/questions/605486 $\endgroup$ Oct 20, 2023 at 11:33
  • $\begingroup$ @AlmostClueless Ah, so its in some sense so that we can have the 'a' operators anihilate the vacuum? That clears up why the redefinition is fundamentally different for zero vs non-zero VEV, thanks. Still wondering how it works out if the redefinition is not done, hopefully someone can answer that, but thanks for the response. $\endgroup$ Oct 20, 2023 at 11:40
  • $\begingroup$ Well, maybe think about it in the following way. The theory might be parameterized in different ways, which all should agree upon when compared. What parameterizations are the useful ones? I would argue that the theory should be parameterized in a way making direct contact to the objects we are studying in experiment. Hence, particle states characterized by electric charge, mass and spin. Thus, we should try to formulate the theory in terms of fields whose quante are eigenstates of said quantities. This is done using the usual formulation you find in most textbooks. $\endgroup$ Oct 20, 2023 at 12:08
  • $\begingroup$ Ah so it will work no matter which way you formulate it but this one is just most convenient, as opposed to there being some reason why the theory shouldn't work unless you define the shifted field? In hindsight this seems obvious so I am not sure why I was confused :). I guess I just couldn't think of the mechanism which delivers the same results regardless - though I can see now how the propagator and other similar objects for the unshifted field will be altered by the VEV since usual canonical quantisation doesn't work out the same way - so that part seems plausible now. Thanks! $\endgroup$ Oct 20, 2023 at 12:16

1 Answer 1


In the framework of quantum field theory, a field can be expanded in terms of creation and annihilation operators. In the expansion the annihilation operator appears to the right of creation operator. So the action of the field on vaccum state is zero, ie, the VEV will be zero. This means that a field with non zero VEV cannot be a quantum field. That's why we expand the field around the VEV.

If $\phi(x)$ is a field with VEV $v$, we define a quantum field $h(x)$ such that \begin{equation} \phi(x)=v+h(x) \end{equation} Note that the VEV of the field $h(x)$ is zero.

Now if the VEV of $\phi$ is already zero and you are trying to redefine the field as, \begin{equation} \phi(x)=v+h(x) \end{equation}The VEV of the new field $h(x)$ won't be quantum field.


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