1

The commutation relation for neutral Klein Gordan field is

$$[\phi(x,t),\pi(x',t)]=i\delta^3(x-x')$$ with all other commutators zero;

The commutation relation for charged Klein Gordan field is $$[\phi(x,t),\pi(x',t)]=[\phi^\dagger(x,t),\pi^\dagger(x',t)]=i\delta^3(x-x')$$ with all other commutation zero;

but the anticommutation relation for the Dirac field is

$$\{\psi_a(x,t),\psi_b^\dagger(x',t)\}=\delta_{ab}\delta^3(x-x').$$

My question is, why is the anticommutation relation for the Dirac field between $\psi_a$ and $\psi^\dagger_b$, not between $\psi_a$ and $\pi_b$ as one would expect looking at the Klein Gordan field?

Qmechanic
  • 201,751
  • Possible duplicates: https://physics.stackexchange.com/q/43502/2451 , https://physics.stackexchange.com/q/186952/2451 , https://physics.stackexchange.com/q/57964/2451 and links therein. – Qmechanic May 07 '22 at 04:29
  • 2
    Actually I just realized I'm being silly. $\psi_b^\dagger$ agrees withs $\pi_b$ up to a constant. – Simplyorange May 07 '22 at 04:31

0 Answers0