Why does the Higgs field fall into the same ground state at all points across space?

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?

• It is not true that the Higgs field necessarily fall in the same vacuum. The magnitude of its vacuum expectation value (vev) is of course invariant but Higgs itself can fall in different vacua for different points in space. This may give rise to topological solutions such as kinks, vortices and monopole. – Diracology Jan 28 '17 at 16:58

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.

• Perhaps I'm being naïve here, but I don't see why it's so obvious. Generally, when you have spontaneous symmetry breaking of a spatially extended field, it does not break in the same direction everywhere, and you end up with domain walls or other topological defects where regions with different phase meet. Sure, such an inhomogeneous state must be metastable at best, since eliminating the domain walls would reduce the total energy, but the relaxation timescale can be very long. – Ilmari Karonen Oct 13 '16 at 13:42
• Anyway, while looking to see what, if anything, had been written about this topic already, I stumbled across this question, whose answers seem to claim that the real answer is that there really isn't any actual symmetry breaking going on with the Higgs field. – Ilmari Karonen Oct 13 '16 at 13:43

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?

The shape of the potential that gives rise to the degenerate minima, forms this spherical shell as a collection of continuously connected points that all give the same VEV. So, to say this differently, the value of the VEV would not have been different if another point on the shell has been chosen. So then why this particular point?

Actually, one should perhaps not assume that the point on the shell is the same throughout space. We already know that Goldstone bosons represent excitations of the motion along the valley, in other words, the motion of the point as it remains on the shell. So, in principle the point could be moving around on the shell as we move from point to point through space. However, as long as it remains on the shell, the VEV would be the same everywhere.

I like to answer this question after reading the comment of @Diracology.
Before the unification between the weak and e.m. force broke, very early in the history of the Universe (when the temperature was about $$10^{15}(K)$$), the field connected to the Mexican Hat potential (present at every point in space) was zero and corresponding to a maximum, constant energy throughout space.
When the temperature dropped the unification between the two forces broke (after which the weak and e.m. force became two distinguishable forces), because the zero field connected to the MH potential came to lie on a random point on the circle at the rim of the MH. Each of these points corresponded (corresponds) to the same vev. As is said in another answer:

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.

Now look at this picture:

Left one obviously sees the MH potential, while on the left one can see the cross-section of a cosmic string. The arrows represent the omnipresent Higgs field. The diameter is about the same as that of a proton ($$1(fm)$$). Inside the string, the same conditions are present as the conditions in the early Universe, just before the electroweak symmetry breaking so the temperature inside the string is about $$10^{15}(K)$$. The configuration of the Higgs field has, in this case, a zero gradient, which means the total energy is minimized.

This one example shows that the Higgs field falls does not have to fall in the same ground state at all points in space. Though such a string (of which it is thought that there is one in each Hubble volume) has (probably) never been observed, there is a theoretical possibility that it exists.