I have been exposed to the usual treatment about spontaneous symmetry breaking in the standard model but it shames me to admit that there are some loose ends I still have to tie up. For simplicity, instead of the standard model let's consider a $U(1)$ gauge theory with a complex scalar $\phi$ given by the Lagrangian


The $V$ part is called the scalar potential and we take it to be


where both $\mu$ and $\lambda$ are positive and whose shape is the logo of this very site. It is straightforward to check that the minimums of the potential occur at the field value


or at any other related to this one by the $U(1)$ symmetry $\phi_0=$


Until here I have no problem. In the next step it is assumed that $\phi_0=\left(\frac{\mu^2}{\lambda}\right)^{1/2}$ is the vacuum expectation value (I will use the letter $v$ henceforth) of the field $\phi$. FIRST QUESTION. How does this follow? why does the minimum of the scalar potential give the vacuum expectation value of the field?

Be that as it may, we have that $\phi$ has a vacuum expectation value. The next step is to expand $\phi$ around its VEV


and by introducing this in the Lagrangian we get a massive gauge boson that eats a degree of freedom from $\phi$. My SECOND QUESTION is, why do we have to expand around the VEV of $\phi$ to get the spectrum of the theory?

  • $\begingroup$ You need to think about what you're doing when you're doing perturbation theory - what are you perturbing around? (It's the VEV, always, but usually it is assumed to be zero. I might write a full answer later) $\endgroup$
    – ACuriousMind
    Nov 25, 2015 at 19:23
  • $\begingroup$ @Scardinelli regarding your first question: what do you think is the field configuration which minimises the energy of the system? $\endgroup$ Nov 25, 2015 at 19:29
  • $\begingroup$ @MarkMitchison which definition of energy are you using? $\endgroup$
    – Yossarian
    Nov 25, 2015 at 19:31
  • $\begingroup$ @ACuriousMind I would appreciate it a lot $\endgroup$
    – Yossarian
    Nov 25, 2015 at 19:31
  • 1
    $\begingroup$ In canonical QFT, the VEV is what you expand around because the VEV of the quantum field needs to be zero to have the usual asymptotic limit towards a free field with creators/annihilators. In the path integral version, you expand around a minimum of the exponential, anyway, to get the usual perturbative diagrams and interpretations. While looking around, I found that this question has already been asked, and also here and here. $\endgroup$
    – ACuriousMind
    Nov 25, 2015 at 20:01

1 Answer 1


Here are two facts -

  1. A vacuum expectation value of a quantum field is equal to the minimum of the effective potential (taken from the 1PI effective action). The effective potential takes the general form $$ V_{\text{eff}}(\phi) = V_{\text{classical}} (\phi) + \text{quantum corrections} $$ In perturbation theory, where quantum corrections are assumed to be small, the minimum of the effective potential is given by the minimum of the classical potential. In other words $$ \langle \phi \rangle = \phi_0 + \text{quantum corrections} $$ where $\phi_0$ is the minimum of the classical potential.

In the case of spontaneous symmetry breaking, we usually have more than one vacuum. All these vacua are related non-trivially by a symmetry transformation. However, the physics in each vacuum is identical and it is therefore irrelevant which one we choose. In the example you showed, there are a whole bunch of vacua given by $\phi_0 e^{i \alpha}$. However, under a $U(1)$ transformation, I can shift $\alpha \to \alpha + \lambda$. I can choose to work in any vacuum I want and would therefore like to choose one that is particularly convenient - which in this case turns out to be the choice $\alpha = 0$.

  1. Next, for us to be able to use the LSZ theorem for fields, two things must be true for all fields that are used in the application of the theorem $$ \langle \phi \rangle = 0, \qquad \langle 0 | \phi(0) | p \rangle = 1 $$ This must be true at the full quantum level (see Srednicki for a derivation of this fact).

When there is spontaneous symmetry breaking, the first condition is no longer true. Thus we need to define a new field $$ {\tilde \phi} = \phi - \phi_0 $$ and we have $$ \langle {\tilde \phi} \rangle = \langle \phi \rangle - \phi_0 = 0 $$ as required.

Thus, we need to expand around the VEV to truly understand the dynamics of the theory.

ASIDE: The second condition is also not generally true for any field. More generally, we have $$ \langle 0 | \phi(0) | p \rangle = Z^{-1} $$ for some number $Z$. To fix this issue, we need to renormalize the fields and define $$ {\tilde \phi}(x) = Z \phi(x) $$ This is the process of field renormalization.


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