In SSB, why shifting a field suffices to pick a corresponding vaccum?

  • Level: Graduate 
  • Thread starter Thread starter ingenue
  • Start date Start date
  • Tags Tags
    Field
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
Sign up Log in
4 replies · 2K views
ingenue
Messages
9
Reaction score
0
Consider the simplest [tex]\phi^4[/tex] with [tex]Z_2[/tex] breaking. Before the shift, [tex]\langle\phi\rangle=0[/tex] by symmetry. After the shift, the vev of the shifted field is zero, which means [tex]\langle\phi\rangle\neq0[/tex], which in turn means we have picked the corresponding vacuum out of two possibilities. However, through the calculation of path integral, shifting a field by a constant only has done nothing as to fixing the boundary condition. Then why did this happen?
 
Last edited:
Physics news on Phys.org

Discussion

"The shift" isn't an ad hoc procedure. It happened to pick out a true ground state. With [tex]\langle\phi\rangle=0[/tex], you're trying to perturbatively expand around a false vacuum, and it ain't going to work.
 
daschaich said:
"The shift" isn't an ad hoc procedure. It happened to pick out a true ground state. With [tex]\langle\phi\rangle=0[/tex], you're trying to perturbatively expand around a false vacuum, and it ain't going to work.

that's exactly why I'm asking. why did it happen to do this?
 
Why did it happen that the ground state has [tex]\langle\phi\rangle\neq 0[/tex]? That comes from the lagrangian.
 
No, I meant why did it happen to pick out the right vacuum?
daschaich said:
Why did it happen that the ground state has [tex]\langle\phi\rangle\neq 0[/tex]? That comes from the lagrangian.