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?