2
$\begingroup$

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.

Improve this question
$\endgroup$
3
  • $\begingroup$ What's the eigenvalue of a field operator? Do you consider that $\psi\psi=0$ says something about eigenvalues? $\endgroup$
    – Demosthene
    Commented Feb 23, 2017 at 9:09
  • $\begingroup$ 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. $\endgroup$
    – StrangeField
    Commented Feb 23, 2017 at 9:12
  • $\begingroup$ Comment to the post (v2): Note that Grassmann-numbers can be real. $\endgroup$
    – Qmechanic ♦
    Commented Feb 23, 2017 at 23:03

1 Answer 1

Reset to default
2
$\begingroup$

  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.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.