Understanding degenerate vacua in a Quantum field theory with sponatneously broken symmetry

$\bullet$ Consider the following Hamiltonian (density)

$$\mathscr{H}=\frac{1}{2}(\partial^\mu\phi)(\partial_\mu\phi)+\mathcal{V}(\phi) \hspace{0.4cm}\text{where}\hspace{0.4cm}\mathcal{V}(\phi)=\frac{\lambda}{4!}(\phi^2-v^2)^2\tag{1}$$ is a double-well potential. Expanding the potential $\mathcal{V}(\phi)=\frac{\lambda}{4!}(\phi^2-v^2)^2$ about one of the minima (say, $\phi=+v$) and retaining up to quadratic term in the potential, one obtains a perturbative vacuum $|0+\rangle$. This is not the exact vacuum of the full Hamiltonian because higher order terms in the potential have been neglected. Similarly, expanding $V(\phi)$ about $\phi=-v$, and again retaining up to quadratic term one obtains another perturbative vacuum $|0-\rangle$. This too, by the same reason, is not the exact vacuum of the full Hamiltonian.

$\bullet$ Similar procedure in a quantum mechanical problem with the Hamiltonian

$$H=\frac{p^2}{2m}+V(x)=\frac{p^2}{2m}+k(x^2-a^2)^2,$$ will lead to two perturbative ground states $|0+\rangle$ and $|0-\rangle$. They are not the exact ground states of $H$. The exact ground state is either given by $|S\rangle$ or $|A\rangle$ which are the symmetric and antisymmetric linear combinations of the perturbative ground states $|0+\rangle$ and $|0-\rangle$.

In this case, one cannot talk about the energy of the states $|0+\rangle$ or $|0-\rangle$ because they are not the eigenstates of H. One can only talk about energies of $|S\rangle$ and $|A\rangle$.

Question 1: If this is true, how is it that in QFT, we talk about the states $|0+\rangle$ and $|0-\rangle$ being degenerate (in energy)? They are just the perturbative vacua and not the exact vacua. Hence, not the eigenstates of $\mathscr{H}$. For a reference, see A Modern Introduction to Quantum Field Theory by Michele Maggiore, page 254-255.

Question 2: What is/are the exact ground states of $\mathscr{H}$ and what is their relation to $|0+\rangle$ and $|0-\rangle$?