# Doubts about spontaneous symmetry breaking

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

$$\mathcal{L}=|D_{\mu}\phi|^2-\frac{1}{4}(F_{\mu\nu})^2-V(\phi^*\phi)$$

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

$$V=-\mu^2\phi^*\phi+\frac{\lambda}{2}(\phi^*\phi)^2$$

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

$$\phi_0=\left(\frac{\mu^2}{\lambda}\right)^{1/2}$$

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

$$\phi_0=\left(\frac{\mu^2}{\lambda}\right)^{1/2}e^{i\alpha(x)}$$

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

$$\phi=v+\psi$$

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?

• 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) Nov 25, 2015 at 19:23
• @Scardinelli regarding your first question: what do you think is the field configuration which minimises the energy of the system? Nov 25, 2015 at 19:29
• @MarkMitchison which definition of energy are you using? Nov 25, 2015 at 19:31
• @ACuriousMind I would appreciate it a lot Nov 25, 2015 at 19:31
• 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. Nov 25, 2015 at 20:01

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.