# What are some good resources to learn about Hilbert spaces corresponding to QFT?

I want to deepen my understanding about Hilbert spaces corresponding to QFT. Could you please introduce some good resources?

• Does this help? physicsforums.com/threads/… Commented May 3, 2021 at 21:03
• spaFor non-interacting fields they are Fock space. For interacting fields, they have been constructed for 1+1 D theories, but not explicitly. Commented May 4, 2021 at 16:57

## 1 Answer

For say a real scalar Bosonic QFT in $$d$$ dimensional flat spacetime, I believe that regardless of interacting or not, the Hilbert space should be $$\mathcal{H}=L^2(\mathscr{S}'(\mathbb{R}^{d-1}),d\nu)$$. Here $$\mathscr{S}'(\mathbb{R}^{d-1})$$ is the space of real-valued Schwartz temperate distributions. The measure $$\nu$$ is a Borel probability measure on $$\mathscr{S}'(\mathbb{R}^{d-1})$$.

In case the Euclidean path integral can be realized as integration with respect to an honest probability measure $$\mu$$ on $$\mathscr{S}'(\mathbb{R}^{d})$$ and if one has the so-called Markov property, then one should be able to get $$\nu$$ as follows. One must show that for $$\mu$$-almost all $$\phi\in\mathscr{S}'(\mathbb{R}^{d})$$ one can define the restriction of $$\phi(\tau,x_1,\ldots,x_{d-1})$$ to the $$\tau=0$$ hyperplane. Here $$\tau$$ is the Euclidean or imaginary time, while $$x_1,\ldots,x_{d-1}$$ are the spatial coordinates. Then this restriction operation gives a Borel-measurable map $$R: \mathscr{S}'(\mathbb{R}^{d})\rightarrow\mathscr{S}'(\mathbb{R}^{d-1})$$ and $$\nu$$ is just the push-forward measure $$R_{\ast}\mu$$, namely the law of the time zero restriction.

This can be done rather explicitly for the free field while recovering the isomorphism of $$\mathcal{H}$$ with Fock space. See:

1. J. Glimm and A. Jaffe, Quantum Physics, 2nd ed., Springer-Verlag, New York, 1987
2. B. Simon, The $$P(\phi)_2$$ Euclidean (Quantum) Field Theory, Princeton University Press, Princeton, N.J., 1974. Princeton Series in Physics.
3. A. Klein, "The semigroup characterization of Osterwalder-Schrader path spaces and the construction of Euclidean fields", J. Funct. Analysis, vol. 27, no. 3, pp. 277-291, 1978.
• If we consider lattice quantum field theory, the Hilbert space corresponding to each site is L^2(R). Now, for the whole system we have a tensor product of L^2(R). what happens when we take the continuum limit? Commented May 17, 2021 at 8:36
• On an infinite lattice like $\mathbb{Z}^d$, the Hilbert space would be $L^2(s'(\mathbb{Z}^{d-1}),d\nu)$ where $s'(\mathbb{Z}^{d-1})$ is the space of lattice fields with at most polynomial growth at infinity. I don't think it is a good idea to try to realize this as an infinite tensor product of $L^2(\mathbb{R})$'s. You can map these discrete spaces into distributions on the continuum. When you do a continuum limit, you have weak convergence of a sequence of probability measures... Commented May 18, 2021 at 14:10
• ...You can learn about that at mathoverflow.net/questions/362040/… and mathoverflow.net/questions/364470/… on the example of the free field, in order to get the conceptual picture right. Commented May 18, 2021 at 14:13