Barrier Penetration in Spontaneous Symmetry Breaking In spontaneous symmetry breaking discussion in Weinberg Chapter 19 section 19.1, he says that the off-diagonal elements between two vacua $$|VAC, +> \pm |VAC->$$ is suppressed by the factor of $$\exp({-C\mathcal{V}}).$$ For a scalar field theory ($-\frac{1}{2}(\partial \phi)^2 - V(\phi)$) by analogy to wave mechanical problem of barrier penetration we have $$C = \int_{-\bar{\phi}}^{\bar{\phi}} \sqrt{2V(\phi)} d\phi.$$  I am trying to understand how to get this result. Please let me know if you know how to obtain this mathematically.
 A: Following Weinberg, we'll consider a QFT with a symmetry $\phi \rightarrow -\phi$.
We'll use the usual phi4 theory so common in describing such systems, described by the action,
\begin{equation}
\mathcal{S} = \int dt \, d^d x \left\{ \frac{1}{2}\left( \partial_t \phi \right)^2 - \frac{1}{2}\left( \nabla \phi \right)^2 - \frac{m^2}{2} \phi^2 - \frac{\lambda}{4!} \phi^4 \right\},
\end{equation}
and the path integral
\begin{equation}
\mathcal{Z} = \int \mathcal{D}\phi \, e^{i \mathcal{S}}.
\end{equation}
We'll be interested in $m^2 \ll 0$ where the symmetry-broken ground states are favored.
Unlike usual, however, we'll take this to be in a finite spatial volume, $V = L^d$, with periodic boundary conditions.
In this case, the Fourier transform is a function of discrete momenta $k_i=2 \pi n_i/L$.
To focus on the low-energy physics, it will be useful to concentrate on the zero-momentum mode,
\begin{equation}
\phi(x,t) = L^{-d/2} \varphi(t) + \sum_{k \neq 0} \tilde{\phi}(k,t),
\end{equation}
where the $k=0$ term in the mode expansion is equivalent to a spatial average over the field
\begin{equation}
\varphi(t) = L^{-d/2} \int d^d x \, \phi(x,t).
\end{equation}
The ground state(s) of this theory will have zero momentum, so we can study these by integrating out the finite-momentum modes:
\begin{equation}
\mathcal{Z} = \int \mathcal{D}\varphi \, e^{i \mathcal{S}_{eff}[\varphi]}, \qquad \mathcal{S}_{eff}[\varphi] = - i \log \int \mathcal{D}\tilde{\phi} \, e^{i \mathcal{S}}.
\end{equation}
If you follow this procedure, you'll find that the zero-mode action is
\begin{equation}
\mathcal{S}_{eff}[\varphi] = \int dt  \left\{ \frac{1}{2}\left( \partial_t \varphi \right)^2 - \frac{m'^2}{2} \varphi^2 - \frac{\lambda'}{ L^d 4!} \varphi^4 \right\} + \cdots.
\end{equation}
Integrating out the $\tilde{\phi}$ will cause a shift in the coupling constants (which we denote by $m'^2$ and $\lambda'$ now), and generate an infinite number of extra terms, $\varphi^6$, $\varphi^8$, and higher time derivatives. We will ignore these terms, which don't change the physics (especially when we can appeal to perturbation theory).
The point of this procedure is that the resulting theory $\mathcal{Z}$ is given precisely by the Feynman path integral for a simple and familiar model in quantum mechanics. This path integral is that obtained from the Hamiltonian
\begin{equation}
H = \frac{p^2}{2} + \frac{m'^2}{2} x^2 + \frac{\lambda'}{L^d 4!} x^{4},
\end{equation}
with $[x,p] = i$. In the limit we are interested in, $m'^{2} \ll 0$, this is precisely the Hamiltonian of a non-relativistic particle in a one-dimensional double-well potential, and you can read off all of your knowledge from that problem (see previous posts here and here for example).
The exact eigenstates of the system are symmetric under $x \rightarrow -x$, with the ground state being the symmetric combination of the particle in both wells, and the first excited state is the antisymmetric combination. As Weinberg says in his Section 19.1, the energy difference between the states is (half of) the matrix element between these states.
If you prepare a particle in a state where it is initially in one of the two wells, with a position centered at $\pm x_0 = \pm\sqrt{-6L^{d}m'^2/\lambda'}$, this will be a linear combination of the ground state and the first excited state, so it will oscillate between the two wells with a period given by the inverse energy difference. There's an asymptotic expression for the energy difference given on Wikipedia. This period can alternatively be estimated by the WKB expression for tunneling between the wells:
\begin{equation}
T \sim \exp\left[ 2 \int_{-x_0}^{x_0} \sqrt{2 [V(x) - V(x_0)]} dx \right].
\end{equation}
Either way, we find that the two lowest ground states have a splitting proportional to $\exp(-CL^d)$, as stated by Weinberg.
You can go through this procedure for a more general field potential and see how it's not particular to phi4 theory.
