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

1. Sep 10, 2010

### 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?

Last edited: Sep 11, 2010
2. Sep 11, 2010

### 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 gonna work.

3. Sep 11, 2010

### ingenue

that's exactly why i'm asking. why did it happen to do this?

4. Sep 11, 2010

### daschaich

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

5. Nov 10, 2010

### ingenue

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