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Can someone explain to me the physical meaning of $\bar{\psi}=\psi^\dagger\gamma^0$ in the Dirac equation? I understand it is obtained as one of the solutions of Dirac equation and it is used to build the scalar $\bar{\psi}\psi$ but I am not sure I understand its physical meaning. I see that when we calculate propagation amplitudes we use $<0|\psi(x)\bar{\psi(y)}|0>$ and not $<0|\psi(x)\psi(y)|0>$ (as we would do in Klein-Gordon equation), so I guess from this that $\bar{\psi(x)}$ applied on the vacuum state creates a particle at position x, but $\psi(x)$ does exactly the same thing (and I mean both create a particle, not antiparticle) so what is the point of that, then?

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The Dirac adjoint is defined in order to construct a Lorentz scalar, as $\psi^\dagger \psi$ does not transform as a scalar unless the representation were unitary, which it is not.

When we quantise the theory and expand the fields in terms of operators, we have schematically,

$$\psi \sim b + c^\dagger, \quad \psi^\dagger \sim b^\dagger + c$$

ignoring the integration and other factors. This means that $\psi$ containing $c^\dagger$ creates particles so to speak with the spinor $v(p)$ corresponding to negative frequency solutions and $b^\dagger$ in $\psi^\dagger$ contributes positive frequency solutions $u(p)$ of the Dirac equation.

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  • $\begingroup$ Thank you for your answer. I kinda got this part. The main thing I don't understand is why do they define the propagation of a particle from y to x in that way, and not as they do in KG case? $\endgroup$
    – Silviu
    Jul 7 '17 at 14:33

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