This question is related to my previous post here.
According to Wienberg's Volume I (9.2.9), we have the result $$ \langle \phi_a | 0 \rangle = \mathcal{N} \mathrm{exp}\left( - \frac{1}{2} \int d^{3}\mathbf{x} \int d^{3}\mathbf{y}\ \mathcal{E}(\mathbf{x}, \mathbf{y}) \phi_{a}(\mathbf{x}) \phi_{a}(\mathbf{y}) \right) $$ where $| 0 \rangle$ is the vacuum state for the free theory of a real scalar field $\phi$, and $| \phi_a \rangle$ denotes an eigenstate of the Schrodinger picture field operator where $\hat{\phi}(0,\mathbf{x}) | \phi_a \rangle = \phi_{a}(\mathbf{x}) |\phi_a \rangle$. The kernel $\mathcal{E}$ is given by: $$ \mathcal{E}(\mathbf{x}, \mathbf{y}) = \int \frac{d^3 \mathbf{p}}{(2\pi)^3} \sqrt{|\mathbf{p}|^2 + m^2 }\ e^{i\mathbf{p} \cdot ( \mathbf{x} - \mathbf{y}) } $$ which you can write in terms of a MacDonald function as $\mathcal{E}(\mathbf{x},\mathbf{y}) = - \frac{m^2}{2\pi^2|\mathbf{x} - \mathbf{y}|^2} K_{2}\big(m|\mathbf{x} - \mathbf{y}|\big)$ if you'd like.
My question is: what is the value of the normalization constant $\mathcal{N}$?
Weinberg says to use the normalization of the vacuum state to determine $\mathcal{N}$ so I write: $$ 1 = \langle 0|0\rangle = \int d\phi_a(\mathbf{x}) \ \langle 0|\phi_a\rangle\langle \phi_a|0\rangle $$ This leads me to an identity $$ 1 = |\mathcal{N}|^2 \int d\phi_a(\mathbf{x}) \ \mathrm{exp}\left( - \int d^{3}\mathbf{x} \int d^{3}\mathbf{y}\ \mathcal{E}(\mathbf{x}, \mathbf{y}) \phi_{a}(\mathbf{x}) \phi_{a}(\mathbf{y}) \right) $$ I am unsure what to do with this. Lists of Gaussian integrals has me guessing that the integral in the RHS is equal to something like $\sqrt{ \frac{(2\pi)^{M}}{\det( \mathcal{E}/2)} } \big|_{M \to \infty}$, but I have no idea what the determinant would mean here since we're dealing with infinite-dimensional spaces. Is there a formal way to write down what $\mathcal{N}$ means?
