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Consider the fermionic field operators $\psi_a(x), \psi^{\dagger}_b(y)$ with the canonical anti-commutation relations $$\{\psi_a(x),\psi_b(y)\} = 0 $$ and $$\{\psi^{\dagger}_b(t,\vec{x}),\psi_a(t,\vec{y})\} = \delta_{ab} \delta(\vec{x}-\vec{y}).$$

What can we say about their eigenvalues? Are they real or Grassmann-numbers?

I'm a bit confused about this at first I would guess they are Grassmann-numbers since $$\psi_a(x)\psi_a(x) = -\psi_a(x)\psi_a(x) = 0$$ but I'm not sure if this conclusion is true.

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  • What's the eigenvalue of a field operator? Do you consider that $\psi\psi=0$ says something about eigenvalues? – Demosthene Feb 23 '17 at 09:09
  • If there is an eigenvactor ${a}$ with $\psi{a} = \lambda_a {a}$ then $\psi^2 {a} = \lambda_a ^2 {a} = 0$. But I'm not sure if there is an eigenvector at all. – StrangeField Feb 23 '17 at 09:12
  • Comment to the post (v2): Note that Grassmann-numbers can be real. – Qmechanic Feb 23 '17 at 23:03

1 Answers1

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  1. An eigenvalue $\lambda$ of an operator $\hat{A}$ (with definite Grassmann-parity $|\hat{A}|$) is a complex supernumber of the same Grassmann-parity. See also this Phys.SE post and links therein.

  2. Note however that an annihilation operator $\hat{a}$ and a creation operator $\hat{a}^{\dagger}$ of definite Grassmann-parity $|\hat{a}|$ do not supercommute $$ [\hat{a}, \hat{a}^{\dagger}]~:=~\hat{a}\hat{a}^{\dagger}-(-1)^{|\hat{a}|}\hat{a}^{\dagger} \hat{a}~=~ \hbar~\hat{\bf 1}~\neq~0, $$ and are therefore not supernormal, and hence not diagonalizable, cf. e.g. this Phys.SE post. OP's fermionic fields are field-theoretic versions of Grassmann-odd annihilation & creation operators.

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