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What are the consequences of this?

Relevant literature...

• Can massless particles be charged? http://journals.aps.org/pr/abstract/10.1103/PhysRev.125.1055

There's a compelling connection to electromagnetic waves here.

My thought on expressing $F$ as a bilinear covariant may not be the best route; more useful may be identifying, as Hestenes suggested in his to-be-published paper, $A = \psi \gamma_1 \widetilde \psi$.

I am no longer sure that $\dot \nabla \psi p_0 \gamma_1 \dot{\widetilde{\psi}} = 0$. I mistakenly believed $\gamma^\mu \partial_\mu \psi = \gamma^\mu m_\mu \psi \implies \partial_\mu \psi = m_\mu \psi$.

Equation (32) must be incorrect. It cannot be true that the only conserved bilinear covariant is $\rho p$. The (spacelike) current $J = \nabla F = \nabla \cdot F$ is curl free, where $F = \psi p_0 \gamma_1 \widetilde \psi$, so $\nabla \cdot J = 0$ by construction.

Off by a factor of 2

If I want to use $\psi_\pm$ rather than $\langle \psi \rangle_\pm$, then I need to consider conjugations for Equations 17 and 19, instead of Equation 16. C and T conjugations carry an extra change in helicity ($p_0 \mapsto \bar{p_0}$), so it's incorrect to say that Equations 17 and 19 have charge and time conjugations. They simply do not exist.

On the other hand, what ultimately matters is how observables change. While $p_0$ maps to $\bar p_0$, $J = \psi p_0 \widetilde \psi$ simply maps to $-J$, the helicity of which is unchanged. This would suggest that this works out fine.

Not iff, only <=.