Bounds on 2HD potential

  • Thread starter Safinaz
  • Start date
  • #1
219
2

Main Question or Discussion Point

Hi all,

I found that any two Higgs doublets potential should be bounded from below - at ## V \to - \infty ##. I want to know why this bound is assumed or what does it mean ?

Also are there any textbooks to learn how to make this bound on any other general potential and so to constrain the potential's parameters ?

Best.
 

Answers and Replies

  • #2
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,671
6,455
If the potential is not bounded from below, your theory has no ground state.
 
  • #3
ChrisVer
Gold Member
3,331
438
The bounds are obtained just by looking at the potential...
For example [itex]V=a |\phi|^2 [/itex] is bounded for [itex]a>0[/itex] and unbound for [itex]a<0[/itex]. If you draw the potential you will see that.
In case you have more than one fields, I guess one has to look at each direction independently ... So the potential for two Higgses could be described by a sheet [itex]V(\phi_1, \phi_2)[/itex] on the space of [itex]\phi_1, \phi_2[/itex], if it's bounded it means that it can't go to [itex]-\infty[/itex] anywhere.

If there was no bound then even if you had a vacuum at some potential's minimum, this vacuum would not be stable. At some point the field would "escape" the well, and then start rolling down forever.
 

Related Threads on Bounds on 2HD potential

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
901
  • Last Post
Replies
11
Views
2K
  • Last Post
2
Replies
35
Views
17K
Replies
5
Views
1K
  • Last Post
Replies
1
Views
2K
Replies
4
Views
691
  • Last Post
Replies
3
Views
3K
Replies
2
Views
1K
Replies
3
Views
671
Top