Let us consider the following term

$$\bar\psi(p_1)M\psi(p_2) \bar\psi(p_3)N\psi(p_4)$$

According to Ortin's book (equation (D.65)), from the Fierz identity we would have something like

$$\bar\psi(p_1)M\psi(p_2) \bar\psi(p_3)N\psi(p_4)= -\frac12 \sum_I \bar\psi(p_1)MO_IN\psi(p_4) \ \bar\psi(p_3)O^I \psi(p_2)$$

where $O^I=\{1,\gamma^a,\gamma^{ab},\ldots\}$, and the minus arises from the fact that the spinors anticommute.

However, assuming that the anticommutation relations of, for example, a Dirac field, are (depending on notation, something like)

$$\{\psi_\alpha(p),{\bar\psi}^\beta(q)\}=(\gamma^0)_\alpha{}^\beta \delta(p-q),$$

to recover the above result, I would obtain two more extra terms which are quadratic in fermions and which arise from anticommuting $\psi(p_2)$ ($\propto\delta(p_2-p_3)$) and $\psi(p_4)$ ($\propto\delta(p_4-p_3)$) with $\bar\psi(p_3)$.

Why those terms are never considered in the literature?


1 Answer 1


Fierz identity concerns the product of multiple $\psi_i$ as Grassmann-valued classical fields, NOT as quantum fields. The minus sign arises from the Grassmann nature of the classical fermion fields.

You can interpret it in two ways:

  • In the path integral formulation, what appear in the Langrangian are classcial fields to be path-integrated. Thus quantum anticommutation relation does not apply to the Langrangian.
  • In the canonical quantization formulation, you can regard the Fierz-rearranged Hamiltonian as the "right" Hamiltonian, similar to the "normal ordering" procedure. Thus the delta terms drop out.

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