For a potential $V$, how do we define the mass of a small fluctuation around its vacuum? For example I have the potential

$$ V_\mathrm{eff}(\phi) = \frac{1}{2} \left(\frac{\rho}{M^2} - \mu^2\right) \phi^2 + \frac{1}{4} \lambda \phi^4. $$

What is the definition of the mass of small fluctuations around the symmetry-breaking vacuum and what is the physical meaning of it?

  • $\begingroup$ Hi milli, and welcome to Physics Stackexchange! Just so you know, we support MathJax for LaTeX-like formatting so you don't have to include equations as images. Most basic stuff is supported, as covered in this extensive guide. $\endgroup$ – user10851 Sep 2 '14 at 9:36
  • $\begingroup$ thank you. I'm not so familiar with LaTeX but next time I'll try. thanks for tutorial link. $\endgroup$ – milli Sep 2 '14 at 10:27

The mass of a small fluctuation is usually defined as $$ \pm m^2= \frac{d^2V}{d\phi^2}\biggr|_\text{VEV}$$

The sign depends on your conventions. This makes sense in analogy with the canonical free field potential $$V_\text{free}=\pm \frac{1}{2}m^2\phi^2$$ for which the above formula is clearly right. More generally, we can expect any (reasonably smooth, [insert other obscure mathematical assumptions here]) potential to be well-approximated by a quadratic potential when it is close to an extremum, so we can define a mass in analogy with the harmonic oscillator - the free field is of course just that!

In your case, it yields, $$\pm m^2=\frac{\rho}{M^2} -\mu^2 +3\lambda \phi_\text{VEV}$$

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  • $\begingroup$ thanks for your answer. but I think that the right form of the first equation must be $$m^2= \pm \frac{d^2V}{d\phi^2}\biggr|_\text{VEV}$$. because I have result that are agree with this $\endgroup$ – milli Sep 2 '14 at 10:07
  • $\begingroup$ Ehh of course! Squares everywhere! Sorry about that $\endgroup$ – Danu Sep 2 '14 at 13:01

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