Consider the potential $$V(\phi)=\frac{1}{2}\mu^2\phi^2+\lambda\phi^4$$ where $\phi=\phi(t,\textbf{x})$ is a real scalar field. Let, $\mu^2<0$ and $\lambda>0$ then the potential is bounded from below with two global minima at $$\phi=\pm\sqrt{\frac{-\mu^2}{\lambda}}.$$ Can quantum corrections change this picture? Apparently, if the "running coupling constant" $\lambda(\Lambda)$ becomes negative at some higher energy scale $\Lambda$ then these two minima become local minima and the vacuum becomes metastable.

  1. Is this argument correct? I mean, is it correct to just substitute $\lambda(\Lambda)$ in the potential $V(\phi)$ and conclude the potential is metastable?

  2. Or should I calculate the effective potential $V_{eff}(\phi)$ for this theory and investigate $V_{eff}(\phi)$ to conclude about the stability of the potential?

  3. If both approaches are equivalent, what is the connection between these calculations?

  4. With $\mu^2<0$, the vacuum is always metastable and cannot become unstable. Is this statement correct?

  • $\begingroup$ If your coupling gets negative, you have more serious problems than symmetry breaking. $\endgroup$ – ACuriousMind Mar 31 '16 at 11:59
  • $\begingroup$ Probably you're referring to the fact that the potential becomes unbounded from below and there is no stable minimum. But I've technical questions related to this issue. $\endgroup$ – SRS Mar 31 '16 at 13:00

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