# Goldstone's theorem. What's the catch?

A theory of scalar field with SO(3) symmetry and Higg's potential is presented by Lagrangian

$$L=\frac{1}{2}\partial_{\mu}\phi^{a}\partial^{\mu}\phi^{a}-\frac{\lambda}{4}(\phi^{a}\phi^{a}-\mu^{2})^2$$ where $a={1,2,3; \, \lambda>0; \, \mu^{2}>0}$ and is real number.

Let's consider a small excitation of vacuum

$$\phi^{1}=\theta^{1}$$ $$\phi^{2}=\theta^{2}$$ $$\phi^{3}=\mu+\chi.$$

After simple calculations one can obtain that fields: $\theta^{1}, \, \theta^{2}$ - massless and field $\chi$ has mass $m^{2}=2\lambda\mu^{2}$.

So the initial symmetry SO(3) violates to SO(2) and this residual symmetry tends chosen vacuum into itself. This symmetry has one generator ($\hat{\Gamma}_{3}$ - a generator of rotation around axis 3 in isotopic space {$\phi^{a}$} ), that is, $\hat{\Gamma}_{3}\vec{\phi^{0}}=0$, $\\\$ $\hat{\Gamma}_{1}\vec{\phi^{0}}=\vec{\phi^{1}}$, $\\\$ $\hat{\Gamma}_{2}\vec{\phi^{0}}=\vec{\phi^{2}}$.

Assuming that $$\vec{\phi}=\tilde{\theta}^{1}\vec{\phi^{1}}+\tilde{\theta}^{2}\vec{\phi^{2}}+\vec{\phi^{0}},$$ where $\tilde{\theta}^{1}=\tilde{\theta}^{1}(x^{\mu})$ and $\tilde{\theta}^{2}=\tilde{\theta}^{2}(x^{\mu})$. Then one can rewrite $\vec{\phi}$ as $$\vec{\phi}=\vec{\phi^{0}}+\tilde{\theta}^{1}\hat{\Gamma}_{1}\vec{\phi^{0}}+\tilde{\theta}^{2}\hat{\Gamma}_{2}\vec{\phi^{0}}=(1+\tilde{\theta}^{1}\hat{\Gamma}_{1}+\tilde{\theta}^{2}\hat{\Gamma}_{2})\vec{\phi^{0}}$$ In the first order of $\epsilon$ this can be shown as $$\exp(\tilde{\theta}^{1}\hat{\Gamma}_{1}+\tilde{\theta}^{2}\hat{\Gamma}_{2})\vec{\phi^{0}}=\tilde{\phi}.$$ But then $V(\vec{\phi})=V(\vec{\phi^{0}})=V(\tilde{\phi})=0$.

So, the questions are:

1. Can it be shown that fields $\tilde{\theta^{1}}$ and $\tilde{\theta^{2}}$ are massless?
2. How one can introduce massive fields in discussion and would be there any degeneracy of masses?
• They are massless since due to exponential representation they appear only in derivative terms of the Lagrangian. – Name YYY Dec 29 '15 at 13:10
• If you break SO(3) -> SO(2), a symmetry with 3 generators is broken to one generators. In a Lorentz-invariant QFT this leads to two massless bosons exactly by Goldstone theorem (Weinberg Vol II gives a good account of that). There is also one massive scalar $\chi$, as you correctly say. So i'm confused what your question is? – Lorenz Mayer May 11 at 5:31