quantum mechanics - QFT Superposition and the Higgs Ground State

In a quantum theory with zero spatial dimensions with Hamiltonian given by: $$H = \pi^\dagger \pi + \mu^2 \phi^\dagger \phi - \lambda (\phi^\dagger \phi)^2,$$ where the non hermitian field, $\phi$, is related to hermitian ones by $$\phi = \frac{1}{\sqrt{2}}\left[\begin{array}{c}\phi_0 + i \phi_1 \\ \phi_2 + i \phi_3 \end{array}\right],$$ and the equal time commutation relations are: $$\begin{align}[\phi_i, \phi_j] & = 0, \\ [\pi_i, \pi_j] & = 0, \ \mathrm{and}\\ [\phi_i, \pi_j] & = i\hbar \delta_{ij}. \end{align}$$ The potential has a classic "Mexican hat" shape, and I would expect the ground state to be a superposition that is invariant under $\operatorname{SU}(2)$ rotations of the complex coordinate, similar to how the ground state of the ammonia molecule is the symmetric superposition of the nitrogen atom being on either side of the hydrogen plane.

The above Hamiltonian is what you would expect for a Higgs field in zero spatial dimensions. When spatial dimensions are added, the Hamiltonian density is:$$\mathcal{H} = \pi^\dagger \pi + \left(\nabla \phi^\dagger\right) \cdot \left(\nabla \phi^{\vphantom{\dagger}}\right) + \mu^2 \phi^\dagger \phi - \lambda (\phi^\dagger \phi)^2.$$

As long as the state is translation invariant, the gradient terms in the Hamiltonian won't affect the above considerations, so why does the ground state have a non-zero vacuum expectation value (vev)? In the symmetric superposition the vev is $0$. Even if the ground state isn't unique, because the $\operatorname{SU}(2)$ degrees of freedom of the field are non-dynamical (i.e. they don't contribute to the kinetic term), why would the state assume a configuration that has a particular vev? Most treatments of the Higgs mechanism calculate the vev as $|\mu|\left/\sqrt{\lambda}\right.$, the vev for a maximally asymmetric state.

The analogy usually given for this process is ferromagnetism, where the direction that a particular magnetic domain points is not fixed by the Hamiltonian, but the part of the Hamiltonian that describes the interactions between the atoms does favor all of them in a domain pointing in the same direction. The difference in this scenario is that the magnetic domain is macroscopic, so its inability to assume a superposition of directions is a matter settled by the measurement problem. Does this mean that decoherence or some other solution of the measurement problem is what causes ground state to have a non-zero vev? If so, a description of the process, or link to one, would be appreciated.