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?

Edit to answer based on question:

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 $\nu$. So, the Feynman rules are different. Does this not mean that different minima give different physical predictions?

  • $\begingroup$ So, what amplitude, experimental measurement, do you propose to detect this difference with? You already noted that you can redefine your fermions (electrons, muons) to eliminate that - sign.... $\endgroup$ Dec 7, 2016 at 20:51
  • $\begingroup$ What do you mean by redefining fermions to eliminate sign? $\endgroup$ Dec 7, 2016 at 21:01
  • $\begingroup$ Ah, I see! My bad! Got it now! $\endgroup$ Dec 7, 2016 at 21:22

1 Answer 1


No, expanding around an equally valid solution that minimises the potential will result in the same physics. For example, in the Lagrangian for the Higgs doublet,

$$\mathcal L \sim (D_\mu H)^\dagger (D_\mu H) - \frac{\lambda}{4}\left(H^\dagger H - \frac{v^2}{2} \right)^2$$

the potential is minimised for any constant $H$ satisfying $H^\dagger H = \frac12 v^2$. The most common choice is to expand around $(0, v/\sqrt{2})^T$, that is,

$$H = \begin{pmatrix} h^+\\ \frac{v}{\sqrt 2} + h^0 \end{pmatrix}.$$

We could however have chosen any other minimum. In terms of insignificance, it is tantamount to choosing a different $\xi$ value in the gauge-fixing term, $\frac{1}{2\xi} (\partial_\mu A^\mu)^2$ in QED. You will get a different value in the propagator for different $\xi$ choices, but in the end anything you compute, from cross sections to decay rates, will be the same.

So, Feynman rules will look different but the physics remains the same. Some choices may also make things computationally just more inconvenient, such as if both components are non-zero, expressions can get messy when expanding the full Lagrangian of the SM.


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