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?
m2eff=∂2Veff(ϕ)∂ϕ2
For example, consider a scalar field which has an one-loop effective potential like
Veff(ϕ)=14!λeff(ϕ)ϕ4.
If the field is sitting at some scale
ϕ≠0, 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?
This post imported from StackExchange Physics at 2014-05-04 11:38 (UCT), posted by SE-user Louis Yang