# A Spontaneous symmetry breaking - potential minima

1. Dec 7, 2016

### spaghetti3451

In spontaneous symmetry breaking, you expand the Lagrangian around one of the potential minima and write down the Feynman rules using this new Lagrangian.

Will it make any difference to your Feynman rules if you expand the Lagrangian around different minima of the potential?

2. Dec 7, 2016

### Orodruin

Staff Emeritus
This depends on what you mean, it is just a question of which basis you write down the Feynman rules in. In other words, we generally chose the basis in the higgs representation in such a way that the vev is real and appears in the second position.

If you chose another basis you will just get SU(2) rotated Feynman rules.

3. Dec 7, 2016

### spaghetti3451

Let's say we have the following theory with $\phi \rightarrow -\phi$ symmetry under $\psi_{i}\rightarrow \gamma_{5}\psi_{i}$:

$$\mathcal{L} = \bar{\psi}_{e}(i\gamma^{\nu}{\partial_{\nu}}-y_{\mu}\phi)\psi_{\mu}+\frac{1}{2}(\partial_{\mu}\phi)^{2}-V(\phi)$$

with $$V(\phi) = -\frac{1}{2}|\kappa^{2}|\phi^{2} + \frac{\lambda}{24}\phi^{4}$$

It can be shown that this theory has two true vacua at $\phi = \pm \nu$, and after expanding about $\phi(x) = \nu + h(x)$, we get

$$\underbrace{\bar{\psi}_{e}(i\gamma^{\nu}{\partial_{\nu}}-y_{e}\nu)\psi_{e}}_{\text{Dirac Lagrangian for field \psi_{e} with mass y_{e}\nu}} \qquad \underbrace{-y_{e}\bar{\psi}_{e}h\psi_{e}}_{\text{interaction term coupling field \psi_{e} with field h with Yukawa coupling y_{e}}} +\underbrace{\frac{1}{2}(\partial_{\mu}h)(\partial^{\mu}h)-\frac{1}{2}\left(2|\kappa^{2}|\right)h^{2}}_{\text{Klein-Gordon Lagrangian for field h with mass \displaystyle{\sqrt{2|\kappa^{2}|}}}}\\ \\ \underbrace{-\frac{\lambda}{6}\nu h^{3}}_{\text{cubic self-interaction term for field h with coupling constant \displaystyle{\frac{\lambda}{6}\nu}}}\qquad \underbrace{-\frac{\lambda}{24}h^{4}}_{\text{quartic self-interaction term for field h with coupling constant \displaystyle{\frac{\lambda}{24}}}}$$

I can see clearly that the cubic self-interaction term as well as the masses of the electron and the muon depend on the value of $\nu$. Does this not mean that different minima give different physical predictions?

Last edited: Dec 7, 2016