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

+ 3 like - 0 dislike

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?

This post imported from StackExchange Physics at 2014-05-04 11:38 (UCT), posted by SE-user Louis Yang
asked Feb 9, 2014 in Theoretical Physics by Louis Yang (90 points) [ no revision ]

1 Answer

+ 1 like - 0 dislike

As I understand your question, it is ill-posed, or at least inconsistent with normal semantics. What we refer to as particles are excitations on some background. The background corresponds to a minimum of the effective potential. The masses of the particles are given by the second derivative of the effective potential.

If you are not at a minimum of the potential, I'm not sure what you even mean by a particle. I can poke my system somewhere and produce some stuff, but how do I know what part of that stuff to call a particle and what part to say is just evolution of the system (which is constantly rolling towards the minimum.)

If you have slow-rolling field (which I suspect is the case you are interested in) then maybe you can do some sort of adiabatic approximation. But that is a somewhat more specific question.

This post imported from StackExchange Physics at 2014-05-04 11:38 (UCT), posted by SE-user BebopButUnsteady
answered Feb 13, 2014 by BebopButUnsteady (330 points) [ no revision ]
So you mean the idea of particle is ill-defined when the background field is moving. Only if the field is in its minimum or slow-rolling can you define a particle and its mass by the second derivative of the potential, right?

This post imported from StackExchange Physics at 2014-05-04 11:38 (UCT), posted by SE-user Louis Yang
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?

This post imported from StackExchange Physics at 2014-05-04 11:38 (UCT), posted by SE-user Louis Yang
@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.

This post imported from StackExchange Physics at 2014-05-04 11:38 (UCT), posted by SE-user BebopButUnsteady

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