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In contituation of this question

In answers of this question people mentioned charged conjugation and formula below

$\bar{\psi}\gamma^\mu\psi=u^2-v^2$

With $u$ for particles and $v$ for antiparticles

If I got everything right

$\bar{\psi}\gamma^\mu\psi=\bar{\psi}_+\gamma^\mu\psi_+-\bar{\psi}_-\gamma^\mu\psi_-$

Where $\psi_+$ are particles and $\psi_-$ are antiparticles

$\psi_-=C\bar{\psi}_+$

So

$\bar{\psi}\gamma^\mu\psi=\bar{\psi}_+\gamma^\mu\psi_+-\bar{C}\psi_+\gamma^\mu C\bar{\psi}_+$

$\bar{C}=C^{T*}\gamma^0$

Since $C=\gamma^2\gamma^0=\begin{bmatrix} 0 & 0 & 0 & i\\ 0 & 0 & -i & 0\\ 0 & -i & 0 & 0\\ -i & 0 & 0 & 0 \end{bmatrix}$

$\bar{C}=\begin{bmatrix} 0 & 0 & 0 & -i\\ 0 & 0 & i & 0\\ 0 & i & 0 & 0\\ -i & 0 & 0 & 0 \end{bmatrix}=\gamma^2$

So

$\bar{\psi}\gamma^\mu\psi=\bar{\psi}_+\gamma^\mu\psi_+-\gamma^2\psi_+\gamma^\mu \gamma^2\gamma^0\bar{\psi}_+$

For electric field $\mu=0$

$\bar{\psi}\gamma^0\psi=\bar{\psi}_+\gamma^0\psi_+-\gamma^2\psi_+\gamma^0 \gamma^2\gamma^0\bar{\psi}_+$

Substituting the Dirac matrices we get

$\bar{\psi}\gamma^0\psi=0$

So no electric potential is created at all.

It seems that I didn't get something the right way. Where's my mistake?

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Your third equation has no basis. And you should decide first if you assume that the components of the Dirac spinor are complex numbers or Grassmaniann numbers, or operators with canonical anticommuting relations. Because if they are complex numbers, your first equation is also wrong.

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  • $\begingroup$ I assume that they're complex numbers. What is the equation in that case? $\endgroup$ Commented Mar 24, 2022 at 18:16
  • $\begingroup$ @JavaGamesJAR : Then you don't have antiparticles, you only have positive- and negative-energy states. The charge density has the same sign for positive- and negative-energy states. $\endgroup$
    – akhmeteli
    Commented Mar 24, 2022 at 18:55
  • $\begingroup$ So I have to use Dirac sea for antiparticles? $\endgroup$ Commented Mar 25, 2022 at 16:58
  • $\begingroup$ The problem is that by QED equation for $A^\mu$ the potential will accelerate or should I just subtract vacuum state from the $\bar{\psi}\gamma^\mu\psi$? $\endgroup$ Commented Mar 25, 2022 at 17:07
  • $\begingroup$ @JavaGamesJAR : I don't know what you're trying to achieve, so I don't know what you should do. If you assume that the components of the Dirac spinor are complex numbers, then the charge density will always be negative (if $e<0$). And, by the way, your second equation is also wrong under this assumption. $\endgroup$
    – akhmeteli
    Commented Mar 26, 2022 at 13:55

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