Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,862 answers , 20,637 comments
1,470 users with positive rep
502 active unimported users
More ...

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

+ 3 like - 0 dislike
74 views

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 10, 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 (325 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

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOverfl$\varnothing$w
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...