I'm struggling to understand the concept of spontaneous symmetry breaking. I understand that the sign of the coefficient $\mu^2 > 0$ in the Higgs potential:
$$ V(\phi) = \mu^2 \phi^{\dagger} \phi – \frac{\lambda}{4} (\phi^{\dagger}\phi)^2 $$
Leads to the minimum of the classical potential being nonzero and the Higgs developing a non-zero VEV $\langle \phi\rangle = \frac{\nu^2}{2}$
My troubles follow: am I correct in saying that the Higgs potential is 'replicated' at every point in space, and that the gauge symmetry of the SM is spontaneously broken when the Higgs field selects the same ground state across all points in 3-space?
If so, why should the Higgs field at space point $x_1$ collapse into the same ground state as the Higgs field at point $x_2$? I understand that the circle of degenerate minima form a spherical shell $\phi^{\dagger}\phi = \frac{\nu^2}{2}$, so why does the Higgs field choose the same point on this shell across all space?
A similar question could be asked about minimization of free energy in the Ginzburg-Landau theory of superconductivity, in this case there is a $U(1)$ symmetry and the ground state picks a unique phase at every point across the system. But why?
Best Answer
If I understand it correct, then in a nutshell, you are asking why is the VEV independent of spacetime. If the Higgs field had different values at different points in space i.e., if it had a spacetime variation, then the gradient term would give a positive contribution to the Hamiltonian, and hence, the total energy will not be minimized.