-1
$\begingroup$

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?

Improve this question
$\endgroup$
1

1 Answer 1

Reset to default
0
$\begingroup$

The difference between the two is in the potential term (Ryder divides by 4! whereas Zee divides by 4). The physics doesn't change, provided the potential term has a nonzero minima. Really, that's where spontaneous symmetry breaking happens for a scalar field.

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.

Improve this answer
$\endgroup$
4
  • $\begingroup$ How will I write in factorial form? I mean !. $\endgroup$
    – user12906
    Commented Apr 20, 2013 at 17:04
  • $\begingroup$ @UnlimitedDreamer beg pardon? I don't quite understand your question... $\endgroup$
    – Alex Nelson
    Commented Apr 20, 2013 at 23:10
  • $\begingroup$ I meant why two different authors choice two different potential? Arbitrary choosing wouldn't be a problem? $\endgroup$
    – user12906
    Commented Apr 21, 2013 at 3:15
  • $\begingroup$ @UnlimitedDreamer Oh, well, the difference is purely mathematical and doesn't impact the physics. And in either case, there's no symmetry breaking until you add another term to the potential. $\endgroup$
    – Alex Nelson
    Commented Apr 21, 2013 at 19:53