By extending the Hilbert space, using grassmann numbers instead of complex numbers, we can write down eigenstates of the fermionic annihilation operator $a$ without getting into trouble with the anticommutation relations: $$\begin{align} |\eta \rangle = e^{\eta a^{\dagger}} |0\rangle = (1 + \eta a^{\dagger}) |0\rangle = |0\rangle + \eta |1\rangle. \end{align}\tag{1}$$ And $$\begin{align} a|\eta\rangle = \eta |0\rangle = \eta |0\rangle + 0 |1\rangle = \eta |0\rangle + \eta \eta |1\rangle = \eta |\eta\rangle. \end{align}\tag{2}$$

Now I am wondering whether we can also write down eigenstates of the complete field $\Psi(x)$, using those grassmann numbers. I tried to write down an expression similar to what has been done in this answer, but up until now, I wasn't able to get to a result.

So, in general, is it possible to write down a state that satisfies $$\begin{align} \Psi(x) |\psi\rangle = \psi(x) |\psi\rangle \forall x~? \end{align}\tag{3}$$

In the book "Quantum field theory of point particles and strings" by Brian Hatfield, chapter 10.3, he simply assumes such a state and writes it down in the way we want it here. Can one do it in the way it is proposed here? And would it even matter whether one is able to express that state in the usual known fock state basis?

To put it differently: Does representing the fermionic field operator as $$\begin{align} \hat{\Psi}(x) \hat{=} \psi(x) \end{align}\tag{4}$$ with $\psi(x)$ being a (set) of grassmann numbers, and representing the state $f$ as $$\begin{align} | f \rangle = f[\phi] \end{align}\tag{5}$$ being a functional on such functions, imply, that there actually are eigenstates like $\hat{\Psi}(x) |\Psi \rangle = \psi(x) |\Psi \rangle$?

  • $\begingroup$ Good question. I think the answer is no, because the field operator combines the fermion and anti-fermion operators. $\endgroup$ Commented Sep 21, 2022 at 14:25
  • $\begingroup$ But that would mean one couldn't use those states in the way it is done in that book, when working in the functional schroedinger picture. $\endgroup$ Commented Sep 22, 2022 at 6:07
  • $\begingroup$ There may be ways around the problem. $\endgroup$ Commented Sep 22, 2022 at 10:34
  • $\begingroup$ Would it work out if I restricted the question to 1 component of the field? $\endgroup$ Commented Sep 22, 2022 at 10:47
  • $\begingroup$ Not sure I understand what you mean. If one would exclude anti-fermion fields, then the field operators won't just be Fourier transformed fermion ladder operators and their eigenstates would be those related to the ladder operators. $\endgroup$ Commented Sep 22, 2022 at 13:33

1 Answer 1


The Dirac field operator consists of a linear combination of creation and annihilation operators, one of which is associated with the fermion field and the other with the anti-fermion field. If this field operator has eigenstates then the creation operator would simply add another particle to the state. It would need to be cancelled by the part produced by the annihilation operator. This can work if the eigenstate contains a kind of "anti-vacuum" state that produces an extra particle when the annihilation operator operates on it. However, there is another problem that prevents the cancellation. The two terms in the field operator are respectively multiplied by different Dirac spinors. Those for the fermion field are orthogonal to those for the anti-fermion field. As a result, the resulting state produce by one of the two terms in the field operator cannot be cancelled by those produced by the other term.

This argument is not a mathematical proof. So perhaps there is some subtle way in which the problem can be solved to allow eigenstates for the Dirac field operator. However, I doubt that. It would require a very complicated mechanism, which means it is not justified to simply assume that such eigenstates exist.

Nevertheless, there may be ways get along without such eigenstates, depending on what the eigenstates are needed for. Sometimes, the purpose of such eigenstates is to define a Wigner functional formalism for fermions [1]. That can be done in a different way. I recently prepared a manuscript [2] in which a Wigner functional formalism for fermions is based on the identification of suitable operators that can act like quadrature operators for fermion fields. The natural choice would have been the Majorana operators, but they don't work like their bosonic equivalents. Instead, I realized that the Hermitian adjoint needs to be augmented by a spin transformation, which changes the notion of a self-adjoint operator. The set of operators that are self-adjoint under this new fermionic adjoint has eigenstates that form complete orthogonal bases. Using these bases, we find that things work out in much the same way than it does for the bosonic Wigner functionals.

[1] S. Mrowczynski, “Wigner functional of fermionic fields,” Phys. Rev. D 87, 065026 (2013).

[2] F. S. Roux, “Fermionic Wigner functional theory,” arXiv:2209.13223 (2022).

  • $\begingroup$ I'm very excited to read this. Appearently, what I called "Real-Part" and "Imaginary-Part" does seem to be the quadrature operators you talk about. $\endgroup$ Commented Sep 30, 2022 at 7:55
  • $\begingroup$ I have a question regarding eq. (22) and (23): Do I get right that the the problem here is the uncountable infinite degrees of freedom? If one would have countable infinite degrees of freedom, one could proceed as is done above with the spin degrees of freedom. There is a sentence that follows eq. (15) that indicates this as well. $\endgroup$ Commented Sep 30, 2022 at 11:26
  • $\begingroup$ Yes, however, the aim is to develop a functional theory for which continuous degrees of freedom are required. In the end it is a matter of convenience. $\endgroup$ Commented Sep 30, 2022 at 13:07
  • $\begingroup$ Could your approach be extended to include anti-fields? I don't see an immediate showstopper. $\endgroup$ Commented Sep 30, 2022 at 14:05
  • $\begingroup$ The anti-fermion field would be a duplication of the same formalism. So the full Dirac field lives on a product of two such phase spaces. $\endgroup$ Commented Oct 1, 2022 at 3:08

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