# The Fock space vs the Hilbert space in the context of quantum field theory

Mathematically the definitions are as follows : if $$H_n$$ is a $$n$$-dimensional complex Hilbert space then its two different corresponding Fock space"(s) are often denoted as $$F_{1}$$ and $$F_{-1}$$ defined as, $$F_1 = \oplus_{k=0}^{\infty} \mathrm{Sym}^k(H_n)$$ and $$F_{-1}= \oplus_{k=0}^{\infty} \Lambda^k(H_n)$$.

Physically for a quantum field theory one "defines" its so-called Hilbert space as the dual of an implicit vector space over $$\mathbb{C}$$ whose basis is in bijective correspondence to the set of all possible values for all the classical fields that occur in the underlying Lagrangian.

Now my question is twofold,

• Does this physical notion of a "Hilbert space of a QFT" correspond to the $$H_n$$ or some total Fock space" that can be defined from the first mathematical definition as, $$\otimes_{i \in \text{Fields}} F^i_{p_i}$$ where $$p_i=$$1 if the $$i^{\text{th}}$$ field is bosonic or $$-1$$ if it is fermionic? (..I guess this tensoring is needed because the QFT can have both fermionic as well as bosonic fields..)

• If we agree as above that the states of a QFT live in such a "total Fock space" and not in the Hilbert space defined in the second paragraph then shouldn't the "quantum field" be mapping into a space of Hermitian operators on this total Fock space and not just the Hilbert space?

• Not all Fock space states are necessarily physical: in Yang-Mills theories, the Fadeev-Popov quantization gives ghost fields that generate negative norm states. The BRST prescription then ensures that those cancel against negative norm states from the unphysical polarization states of the gauge bosons, and the physical Hilbert space is actually defined by the cohomology of the BRST operator. – gmvh Feb 26 '20 at 14:00
• Essentially, it is only for free fields that Fock spaces provide a description of the physical Hilbert space. – Abdelmalek Abdesselam Jul 23 '20 at 15:04

By saying "the Hilbert space of a QFT" physicists mean the "total Fock space" for all fields (not the $$H_n$$): $$\cal{F} = \bigotimes_{k} \cal{F}_k$$, where $$\cal{F}_i$$ are Fock spaces for separate quantum fields $$\Psi_i$$. Hence, the corresponding field operators $$\Psi_i$$ will effectively act only in their corresponding Fock sub-spaces $$\cal{F}_i$$ (and trivially on the other parts): $$\Psi_i = \mathbb{1} \otimes \dotsb \otimes \mathbb{1} \otimes \psi_i \otimes \mathbb{1} \otimes \dotsb \otimes \mathbb{1}$$.