For the set of all fermionic field operators $\Psi(x) | x \in \mathbb{R}^{3 +1}$, we won't find a $|\phi \rangle$ that is an eigenstate to the complete set of field operators, unless we make use of Grassmann numbers and construct a fermionic coherent state. However, we can do canonical quantization or nonrelativistic many-body quantum mechanics with fermionic field operators just fine, without ever hearing anything about Grassmann numbers.

Now I am wondering: Do I NOT encounter Grassmann numbers because I don't dig deep enough, or is it that a usual Hilbert space over complex numbers is fully fine and sufficient to also describe the set of $\Psi(x) | x \in \mathbb{R}^{3+1}$, with the $\Psi(x)$ all anticommuting?

I tried to specify ladder operators using only complex numbers, and for a finite number of states, this works pretty well (see for example this answer), but I don't know if I run into problems when I try the same with an uncountably infinite number of base states (one for every position, for example).

To make the question more specific: If I try to write down matrix elements, would I encounter Grassmann numbers? Is there another way that Grassmann numbers might appear?

In this question one answer is that the operators should already be operators on a supervectorspace - Is that a contradiction to my guess that we don't need them in general, or is it that due to the very specific question? (The OP of the other question specifically asked wether it's possible to construct creation and annihilation operators that also anticommute with grassmann numbers)?


2 Answers 2


It is possible to define the antisymmetric Fock space, and related fermionic creation and annihilation operators, without any need to introduce Grassmann numbers.

Grassmann numbers are only necessary to define a "classical" fermionic field theory, in which the basic observables must anti-commute.

So, roughly speaking, as soon as one has advantages in using classical paths or other objects (Lagrangians, actions) for the calculations, it is necessary to introduce Grassmann numbers.

If one would be able to study all interesting fermionic QFTs purely on Hilbert spaces, there may be no need of introducing such numbers; however, at present, there is no way to do that, and Grassmann numbers are very commonly used (and useful).

  • $\begingroup$ We can define the annihilation operators action on states from the "usual" hilbert space over C without the use of grassmann numbers. But its eigenvector is from a larger hilbert space with supernumbers as coefficients, and its eigenvalues then contains supernumbers. Doesn't that hint that the annihilation operator itself is a general operator on this larger space from the start? $\endgroup$ Jun 21, 2022 at 13:35
  • $\begingroup$ Fermionic creation and annihilation operators are bounded operators on the Fock space. And many operators in Hilbert spaces do not have eigenvectors (the momentum and position operators in QM, for example). That alone does not imply much, and surely not that Grassmann numbers are necessary to define fermionic fields. $\endgroup$
    – yuggib
    Jun 21, 2022 at 14:50

Supernumbers are needed & used in (among other things):

  1. the path/functional integral formalism of a QFT with fermionic fields, cf. e.g. this Phys.SE post.

  2. fermionic coherent states, cf. e.g. this Phys.SE post.

  3. the fermionic version of the theorem by Stone & von Neumann, cf. e.g. this Phys.SE post.

  4. the super-Poisson algebra for classical fermionic systems, cf. e.g. this Phys.SE post.

OP is asking if supernumbers are also needed in the operator formalism of fermionic fields? Are they e.g. needed in the fermionic Fock space? No, they are not, but they are clearly easy to implement under the hood, which enables relations to the other formalisms.

  • $\begingroup$ I see that they are easy to implement under the hood - but shouldn't the fact that we need them to talk about eigenstates of the annihilation operator already tell us that they are there from the start? $\endgroup$ Jun 21, 2022 at 13:30
  • $\begingroup$ $\uparrow $ No. $\endgroup$
    – Qmechanic
    Jun 21, 2022 at 13:41
  • $\begingroup$ Why? When an operator has eigenvalues from a certain field and its eigenvectors come from a hilbert space over that field, isn't it natural to assume that the operator itself is a linear map on that hilbert space? $\endgroup$ Jun 21, 2022 at 13:46
  • $\begingroup$ The operator formalism works without supernumbers. $\endgroup$
    – Qmechanic
    Jun 21, 2022 at 14:00

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