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Follow the closed question When does the spinor need to be in a grassmann variable?

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Does the spinor in the spinor representation of the space-time symmetry

  1. Lorentz space-time symmetry, like $so(1,d)$ or $spin(1,d)$

  2. Euclidean space-time symmetry, like $so(1+d)$ or $spin(1+d)$

necessarily need to be in a Grassmann variable? So $\psi \eta = -\eta \psi$ for Grassmann variable $\psi$ and $\eta$?

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Марина Marina S
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    The 2-spinors used in spinor representation of spacetime symmetry are grassmann even. The spinors used in qft of fermionic fields uses grassmann odd numbers because of spin-statistics / pauli-exclusion principle – KP99 May 17 '23 at 23:13

1 Answers1

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  1. When considering an equation that is linear in $\psi$ (such as, e.g. the transformation rule for a spinor representation), it becomes agnostic to whether we treat $\psi$ as Grassmann-odd or Grassmann-even.

  2. A physical spinor field $\psi$ is Grassmann-odd in an interacting relativistic theory, cf. the spin-statistics theorem.

  3. For more information, see e.g. this related Phys.SE post.

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