Does the Dirac equation put charge and spin on the same footing? I read somewhere that, when represented as 4 complex numbers, the wavefunction in the Dirac equation can be thought of as the respective probabilities of (1) spin up electron, (2) spin down electron, (3) spin up positron, and (4) spin down positron.  But then again, I'm also under the impression that you can change your Dirac matrix basis and thus end up with different complex components in your wavefunction.  So it kind of sounds like, by changing your representation, you can effectively "rotate" charge and spin into each other.  As though charge and spin are merely two components of a higher-order invariant quantity/concept.
Does that notion have any validity, however slight, either in the Dirac theory or QFT?
 A: One of the misconceptions that exists on the Dirac-equation is that the 2 components of the 4 in total describe an electron and the other 2 describe a positron. This cannot be simply because if one couples the Dirac equation to the electromagnetic field, all components couple with the same coupling constant including its sign.
$$ (i\gamma^\mu p_{\mu} -m)\psi =0 \quad\rightarrow  (i\gamma^\mu (p_{\mu}-eA_\mu) -m)\psi =0$$
where $p_\mu$ are the components of the 4-momentum and $A_\mu$ are the components of the 4-vector electromagnetic potential and $e$ is the elementary electric charge $e$ which is the same including sign for all 4 components.
For the following we define
positive frequency solutions of the Dirac-equation that are associated with electrons, and negative frequency solutions that are "associated" with positrons (but "association" does not mean that they are identical to positrons). What this association really means it explained in the following.
Actually, positron solutions are only obtained when the negative-frequency solutions of the Dirac-equation are charge-conjugated with the charge conjugation operator $C$:
$$\psi_{positron}^{positive frequency} = C\bar{\psi}_{electron}^{negative frequency}$$
According to the interpretation of Feynman diagrams negative frequency solutions look like electrons moving backwards in time. These can then be interpreted as positrons moving forward in time.
But these solutions still correspond to electrons which only by interpretation can be considered as positrons.
However, to do so Feynman rules have to be strictly followed, so that negative frequency solutions are in-going from the future to the past (instead for a positive frequency solutions from the past to the future) into the interaction point. They can by the virtue of interpretation considered as out-going positrons running into the future.
The formal expression of a scattering of a particle at a interaction potential  $V$ according to non-relativistic QM is
$$\psi^\ast_{outgoing} V \psi_{ingoing}$$
and this is true for positive and negative charged particles.
This does not principally change in relativistic QM. For instance the formal expression of a vertex of a scattered u-quark with negative charge $-1/3$ would look like:
$$\bar{\psi}_u \gamma^\mu \psi_u$$
whereas the the formal expression of a vertex of a scattered d-quark with positive charge $2/3$ would look like the same:
$$\bar{\psi}_d \gamma^\mu \psi_d$$
although the particles are of different charge. The solution of the out-going particle stands on the left side and the solution the in-going particle stands on the right side of the expression (the gamma-matrix). Note that $\psi_{u/d} \sim \exp(-ipx)$, so these are in-going particles.
However a  vertex of a scattered anti-electron(=positron) has to be written according to the Feynman rules like
$$\bar{v}\gamma^{\mu} v$$
Remarkable is here that the negative frequency solution $v$  interpreted as "out-going positron"  $v$  (it is out-going because $v\sim \exp(ipx)$) stands on the right side of the formal expression of the vertex and the so interpreted "in-going positron " $\bar{v} \sim\exp(-ipx) $  on the left side of the gamma-matrix in contrast to what was shown before.
This shows that negative frequency solutions of the Dirac equations are not really positrons, the true positron solutions would follow the rule which was demonstrated for the quarks. Actually one can use the charge-conjugated negative-frequency solutions of the Dirac-equation for a formal expression of a vertex. These  are true positron solutions and follow the same rule as that demonstrated for the quarks.
The bottom line is that a base change applied on a solution of the Dirac-equation does not mix spin with charge. The charge is the same for all 4 components and is not changed by such a transformation.
