Dirac vs KG propagation amplitude

[Silviu]

Hello! Can someone explain to me the physical meaning of $\bar{\psi}=\psi^\dagger\gamma^0$ in the Dirac equation? For example when calculating propagation amplitude I see that what we calculate is $<0|\psi(x)\bar{\psi(y)}|0>$ and not $<0|\psi(x)\psi(y)|0>$ (as we do for KG equation) and I am not sure I understand why. Can someone explain? Thank you!

 

[LightHero]

The Dirac equation describes the behavior of a spin-1/2 particle, such as an electron. The Dirac spinor, \(\psi\), is a four-component wavefunction that contains all of the information about the particle's spin and momentum. The conjugate spinor, \(\bar{\psi}\), is defined as \(\bar{\psi}=\psi^\dagger \gamma^0\). This definition ensures that the inner product between two Dirac spinors is Lorentz invariant, which in turn ensures that the Dirac equation is also Lorentz invariant. When calculating the propagation amplitude for a Dirac spinor, you are computing the inner product between two spinors, which is why you are computing $<0|\psi(x)\bar{\psi(y)}|0>$ instead of $<0|\psi(x)\psi(y)|0>$. This inner product is Lorentz invariant, meaning it will be the same in all frames of reference, whereas an inner product between two Dirac spinors without the conjugate spinor would not be Lorentz invariant.