# Is the second derivative of the effective potential always the mass square of the particle?

In quantum field theory, we can calculate the effective potential of a scalar field. My question is whether the second derivative of the effective potential always represents the mass square of the particle? Why or why not? $$m_{eff}^{2}=\frac{\partial^{2}V_{eff}(\phi)}{\partial\phi^{2}}$$ For example, consider a scalar field which has an one-loop effective potential like $$V_{eff}(\phi)=\frac{1}{4!}\lambda_{eff}(\phi)\phi^{4}.$$ If the field is sitting at some scale $\phi\neq0$, would the second derivative still be the physical mass square of the particle? If this is true, then the particle will have different masses as it rolls down the potential to its minimum. Would this affect the decay channel of the particle?

• But if the field $\phi$ can decay into other particles, does that mean the field $\phi$ won't decay while it is sliding down the potential in a non-slow-rolling way? – Louis Yang Feb 14 '14 at 7:53
• @LouisYang: The field $\phi$ will still decay if it's coupled to other fields but it just doesn't seeem to make sense to talk about that process as particles on top of some moving background. As I said how do you distinguish between the 'particles' and the 'background'? Your field is just some complicated, non-linearly evolving function. – BebopButUnsteady Feb 14 '14 at 16:25