Symmetry breaking with Lagrangian

I have been studying the spontaneous symmetry braking from Zee (Quantum Field theory ) and found in the page 224, he wrote the lagrangian as
$$\mathcal{L}= \frac{1}{2}\{ λ (∂φ)^2 + μ^2φ^ 2\} − \frac{\lambda}{4}(φ^4)$$

But according to sclar field theory I got from Ryder $$\mathcal{L}=\frac{1}{2}\{ λ (∂φ)^2 + μ^2φ^ 2\} − \frac{\lambda}{4!}(φ^4)$$ Am I doing wrong?

My another question is, the potential we have inserted in the Lagrangian is look like Mexican hat. What conditions will change the shape of the potential? Symmetry breaking means ,changing the Lagrangian with kinetic energy terms?

• – Alex Nelson Apr 20 '13 at 16:40

You have your Lagrangian $$\mathcal{L} = \frac{1}{2}\left(\partial^{\mu}\varphi\partial_{\mu}\varphi-\mu^{2}\varphi^{2}\right)-V(\varphi).$$ If it has a nonzero vacuum energy, then $V(\varphi)\not=0$ for any $\varphi$. If this happens, it breaks symmetry since the vacuum is nonunique (egads!).
We find this happening if $\varphi_{0}$ satisfies $V'(\varphi_{0})=0$ but $V(\varphi_{0})\not=0$.
Addendum: The potential $V(\varphi)=c_{1}\varphi^{2}+c_{2}\varphi^{4}$ would experience spontaneous symmetry breaking, for nonzero constants $c_{1}$, $c_{2}$...but if $c_{1}=0$ -- as your model has -- there is no symmetry breaking.