There are two definitions of the parity transformation acting on the Dirac spinors: $\Psi_P = \eta \gamma^0 \Psi$ with $\eta = i$ ($P^2=-1$ as in Srednicki) and $\eta=1$ ($P^2=+1$ as in Peskin & Schroeder).

Both definitions result in the same parities of the sesquilinear forms such as $\bar\Psi \Phi$.

However, the bilinear form $\overline{\Psi_C} \Phi$ (à la pseudoscalar diquark) is scalar under $P^2=-1$ inversion and pseudoscalar under the one with $P^2=+1$, since $\Psi_{CP} = -\eta^* \gamma^0 \Psi_C$; $C$ is charge conjugation.

Does it mean that the two definitions are physically inequivalent (and $P^2=-1$ is incorrect)?


Do I miss something that makes $\overline{\Psi_C} \Phi$ a pseudoscalar even for $\eta = i$?

  • $\begingroup$ Is $C$ the same for both P&S and Sred.? $\endgroup$ – AccidentalFourierTransform Apr 11 '16 at 7:23
  • $\begingroup$ Yes, it is. @AccidentalFourierTransform $\endgroup$ – user114001 Apr 11 '16 at 7:41
  • $\begingroup$ Out of curiosity, what is the point of choosing $\eta$ to be imaginary instead of real? $\endgroup$ – my2cts Feb 3 '19 at 16:30

For non-hermitian products of Dirac field operators the parity is not well defined and depends on the phase $\eta=\pm1,\,\pm i$ of the parity transformation $\eta \gamma^0$. For example, $I_P = -\eta^2 I$, where $I = \overline{\Psi_C}\Phi$.

In the $S$-matrix elements, however, all phases go away eventually, because creation and annihilation operators come in pairs there and $|\eta|^2=1$. Thus, for many practical calculations, it appears reasonable to omit the phase $\eta$. Then, we can consider $I$ as a pseudoscalar even for $\eta=\pm i$.


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