I was reading this post from PSE and it reminded me an old question of mine, in which the use of creation and annihilation operators were discussed. Both questions got answers which agreed on the fact that in QFT one is not always assuming a Fock space as the underlying Hilbert space of the theory. That made me think about the role of Fock spaces to justify path integrals in some contexts. Let me elaborate.

To keep the ideas simple and keep the notations from my previous linked post, I will deal with fermionic models. Suppose our Hilbert space $\mathscr{H}$ has dimension $n < +\infty$. In many-body quantum theory, one considers its associated Fock space $\mathcal{F}^{-}(\mathscr{H})$ which has dimension $2^{n}$. There is a vast literature about rigorous treatments of many-body systems. This Fock space identifies, in a natural way, with a Grassmann algebra with finitely many generators $\psi_{1},...,\psi_{2^{n}}$ by simply identifying each creation operators on the Fock space with a generator of the Grassmann algebra and the vacuum vector $\Omega$ with $1$. In this setting, one can give mathematical meaning to the path integral representation by using:

(i) The fact that the trace of an operator $A$ on $\mathcal{F}^{-}(\mathscr{H})$ has a natural "translation" in terms of Grassmann integrals and

(ii) The Lie-Trotter formula for bounded operators.

At least from my perspective, every reference I know that discusses path integrals for fermions does it in the above setting: finite-dimensional Hilbert spaces of one-particle system, finite dimensional Fock spaces and so on. However, as stressed before, in a more general setting, we couldn't assume a Fock space as the underlying Hilbert space. The fact that (at least to my knowledge) no reference discusses path integral formulations in more abstract setting makes me think this approach either only works for many-body systems, in which Fock spaces can be used, or it is just easier/more intuitive to do this way.

My point is: I don't see why Fock spaces are so much necessary apart from being more intuitive. So, can't we justify the path integral method in a more abstract way, under some additional hypothesis? Suppose $\mathcal{F}^{-}(\mathscr{H})$ is replaced by an arbitrary finite-dimensional complex Hilbert space $\mathscr{H}$, with creation and annihilation operators $\varphi^{\dagger}(x)$ and $\varphi(x)$ and some Hamiltonian $H$ on $\mathscr{H}$. If we assume:

(1) These creation and annihilation operators satisfy CAR relations

(2) The existence of a unique vacuum vector $\Omega$, which is the ground state for $H$

(3) $\mathscr{H}$ is spanned by successive applications of creation operators to $\Omega$


Q1: Doesn't the above hypothesis fully define a fermionic system?

Q2: Aren't (1) to (3) enough to establish a isomorphism between $\mathscr{H}$ and a finitely generated Grassmann algebra?

Q3: Doesn't the path integral formulation, using the Lie-Trotter formula hold in the very same way as before?

In summary: Are the hypothesis (1) to (3) enough to replace the Fock space by an arbitrary Hilbert space and still get the same analysis?

The hypothesis (1) to (3) seems exactly the kinds of hypothesis physicists are assuming when studying their fermionic models, but the lack of discussions about path integrals on these more abstract settings (i.e. outside Fock spaces and many-body quantum mechanics) makes me question whether there is something wrong with the above.

  • $\begingroup$ "At least from my perspective, every reference I know that discusses path integrals for fermions does it in the above setting: finite-dimensional Hilbert spaces of one-particle system, finite dimensional Fock spaces and so on." I have actually never seen this approach to developing fermion path integrals, even in fairly mathematically oriented treatments. I think the reality is that it simply removes too much of what makes realistically useful path integrals nontrivial (and hard to define). $\endgroup$
    – Buzz
    May 27, 2022 at 21:08

1 Answer 1


As soon as you assume the structure of a CAR algebra, then you are automatically dealing with a Fock space. To define a CAR algebra structure, it must be generated by something, and that something is the single-particle Hilbert space $\mathscr{H}$.

Let $\mathscr{H}$ be a complex Hilbert space, and let $\mathscr{A}$ be the CAR algebra it generates. There is a unique state $\Omega$ that is a $0$-eigenstate of all annihilation operators. Then the GNS representation based on $\Omega$ is unitarily equivalent to the Fock space $\mathcal{F}^-(\mathscr{H})$. Now, this is all quite independent of the details of the Hamiltonian $H$. A free field Hamiltonian is compatible with and preserves this structure. We could, if we wanted, consider a more general Hamiltonian, and it would generate an automorphism of the algebra $\mathscr{A}$, but it would not preserve the CAR structure, in the sense that it would not preserve $\mathscr{H}$ or the mapping from $\mathscr{H}$ to creation and annihilation operators.

In infinite-dimensional systems, the situation is even worse. Nothing prevents us from considering the CAR algebra of the field on a particular time-slice. But the interacting Hamiltonians we would want to write down are symmetric but not self-adjoint on the Fock space, so they do not generate a unitary time evolution. No one knows (or at least there is no general agreement) how to write down relations, such as the canonical anticommutation relations, that would define the algebra of interacting fields. So we are forced to approximate with free fields and do perturbation theory.

  • $\begingroup$ Idrinehart, very interesting answer. Does the situation change if the perturbation is quartic in the fields? I ask because I've seen some literature where one construct the Fock space and in this Fock space the Hamiltonian is of the form quadractic term + quartic term which is considered as a perturbation. $\endgroup$
    – IamWill
    May 30, 2022 at 13:28
  • $\begingroup$ @IamWill That sounds like probably an example of what I allude to my last sentence. The Fock space would be that associated with the quadratic part of the Hamiltonian, and the quartic part would be a perturbation to this. $\endgroup$
    – ldrinehart
    May 30, 2022 at 16:07

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