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

[ingenue]

Consider the simplest \phi^4 with Z_2 breaking. Before the shift, \langle\phi\rangle=0 by symmetry. After the shift, the vev of the shifted field is zero, which means \langle\phi\rangle\neq0, 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?

 

[daschaich]

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

 

[ingenue]

"The shift" isn't an ad hoc procedure. It happened to pick out a true ground state. With \langle\phi\rangle=0, 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?

 

[daschaich]

Why did it happen that the ground state has \langle\phi\rangle\neq 0? That comes from the lagrangian.

 

[ingenue]

No, I meant why did it happen to pick out the right vacuum?

Why did it happen that the ground state has \langle\phi\rangle\neq 0? That comes from the lagrangian.