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

  1. Sep 10, 2010 #1
    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: Sep 11, 2010
  2. jcsd
  3. Sep 11, 2010 #2
    "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 gonna work.
     
  4. Sep 11, 2010 #3
    that's exactly why i'm asking. why did it happen to do this?
     
  5. Sep 11, 2010 #4
    Why did it happen that the ground state has [tex]\langle\phi\rangle\neq 0[/tex]? That comes from the lagrangian.
     
  6. Nov 10, 2010 #5
    No, I meant why did it happen to pick out the right vacuum?
     
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