# Charge Conjugation of Dirac equation

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?

• The problem is that by QED equation for $A^\mu$ the potential will accelerate or should I just subtract vacuum state from the $\bar{\psi}\gamma^\mu\psi$? Commented Mar 25, 2022 at 17:07
• @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. Commented Mar 26, 2022 at 13:55