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?

  • 1
    $\begingroup$ 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. $\endgroup$
    – gmvh
    Feb 26, 2020 at 14:00
  • $\begingroup$ Essentially, it is only for free fields that Fock spaces provide a description of the physical Hilbert space. $\endgroup$ Jul 23, 2020 at 15:04

2 Answers 2


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): $\Psi_i = \mathbb{1} \otimes \dotsb \otimes \mathbb{1} \otimes \psi_i \otimes \mathbb{1} \otimes \dotsb \otimes \mathbb{1}$.

This article treats on that question: Back and forth from Fock space to Hilbert space: a guide for commuters, arXiv:1805.04552


Not sure what you mean by the quantum field to be "mapping" somewhere, but it is absolutely correct that in scalar and other simple ("gauge-free") theories that can be interpreted quantum-mechanically, the hamiltonian as well as the observables are hermitian operators on Fock space. These operators are generally written as Weyl algebra elements, in terms of creation and annihilation operators. While the Hilbert space of states is not Lorenz equivariant (it corresponds to a time zero slice), the Weyl algebra can be made Lorenz equivariant, making the Heisenberg picture more convenient.

What you are perhaps talking about is the fact that the path integral formulation is written in terms of the fields themselves, not the Weyl algebra. But here the things that correspond to the quantum hamiltonian and observables are the lagrangian and the path integral observables. These are functionals which depend on both the field and its first time derivative, which is "the same size" as the Weyl algebra.

For non-free theories, I'm only aware of all the correspondence between quantum mechanical and field theoretic notions being explicitly worked out perturbatively.


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