*As Qmechanic already pointed out: In order to obtain the kinetic momentum you have to take the derivatives (which give you the canonical momentum) and then subtract the interaction with* $A^\mu$.

So everything is ok and the Dirac equation exactly reproduces the classical result. You can gain a deeper understanding of this if you write the Lorentz force in a more advanced way by using the electromagnetic field tensor.

$\frac{\partial j^\mu}{\partial \tau} ~~=~~ \frac{q}{mc}\,F^{\mu}_{~\nu}\,j^\nu~$

Which couples the E field with the boost generators K and the B field with the rotation generators J

$ F^{\mu}_{~\nu} ~~=~~ \Big(\,\mathsf{E}^i\,\hat{K}^i + \mathsf{B}^i\,\hat{J}^i\,\Big) \ =\ \left(
\begin{array}{rrrr}
~\ 0\ \ & ~~\mathsf{E}_x & ~~\mathsf{E}_y & ~~\mathsf{E}_z \ \\
~ \mathsf{E}_x & \ 0\ \ & ~~\mathsf{B}_z & - \mathsf{B}_y \ \\
~ \mathsf{E}_y & - \mathsf{B}_z & \ 0\ \ & ~~\mathsf{B}_x \ \\
~ \mathsf{E}_z & ~~\mathsf{B}_y & - \mathsf{B}_x & \ 0\ \ \
\end{array}
\right)$

For spinors the equivalent interaction generator of time evolution is:

${\cal F}^\mu_{~\nu}\,\varphi ~=~ \left(\,\vec{E}\cdot\hat{\mathbb{K}} + \vec{B}\cdot\hat{\mathbb{J}}\,\right)\varphi$

$\mathbb{K}^i ~=~ -\tfrac12\,\gamma^i\gamma^o, ~~~~~~~~~~ \mathbb{J}^i ~=~ \tfrac{i}{2}\,\gamma^5\gamma^i\gamma^o$

Again the electric field boosts while the magnetic field rotates.

The classical time evolution due to the classical electromagnetic field tensor $F$ operating on the current is exactly the same as when the Spinor field tensor ${\cal F}$ operates on the spinor.

$\exp(F^{\mu}_{~\nu}\,t)\,\bar{\varphi}\,\gamma^\nu\varphi ~~=~~ \overline{\Big(\exp({\cal F}^\mu_{~\nu}\,t)\varphi\Big)}\,\gamma^\mu \, \Big(\exp({\cal F}^\mu_{~\nu}\,t)\varphi\Big)$

If you work out the series expansion of the exponential functions you can find all kind of beauties like.

$\begin{aligned}
&\dot{\bar{\varphi}}\gamma^\mu\dot{\varphi} &=~~~ &\tfrac12\,T^\mu_{~\nu}~\bar{\varphi}\,\gamma^\nu\varphi \\
&\dot{\bar{\varphi}}\gamma^5\gamma^\mu\dot{\varphi} &=~~~ &\tfrac12\,T^\mu_{~\nu}~\bar{\varphi}\,\gamma^5\gamma^\nu\varphi \\
&\dot{\bar{\varphi}}~\mathbb{K}^\mu\,\dot{\varphi} &=~~~ &\tfrac12\,T^\mu_{~\nu}~\bar{\varphi}\,\mathbb{K}^\nu\,\varphi \\
&\dot{\bar{\varphi}}~\mathbb{J}^\mu\,\dot{\varphi} &=~~~ &\tfrac12\,T^\mu_{~\nu}~\bar{\varphi}\,\mathbb{J}^\nu\,\varphi \\
\end{aligned}$

Where T is the symmetric stress energy tensor of the electromagnetic field.

The term ${\cal F}^\mu_{~\nu}\,\varphi$ is just the extra term which occures if you square the Dirac equation with interaction. (although its role there is generally poorly interpreted) The squared Dirac equation contains the second order derivative in time so it should include a term which accounts for the spinor boosts and spinor rotates due to the electromagnetic field. The Klein Gordon equation does not need such a term because it describes a scalar field and scalars are per definition Lorentz invariant.

Regards, Hans

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