This is somewhat of a follow-up question to my previous question on the Dirac-Hestenes equation. In that question, I asked whether the equation could be written in a form that omits the dangling indices.
Context
In free space, using natural units, the Dirac-Hestenes equation takes the form $$ \vec{\nabla} \psi I \sigma_3 = m \psi \gamma_0 $$ where $\psi$ is an even multivector field, $I\sigma_3$ is an arbitrarily chosen unit bivector, $m$ is the mass, and $\gamma_0$ is a future-pointing timelike unit vector. It's possible to remove one index by notating the unit bivector term as $\hat{B}$, resulting in the form $$ \vec{\nabla} \psi \hat{B} = m \psi \gamma_0 $$ This leaves one final index, associated with $\gamma_0$.
The realization
At the same time, I had another issue with $\gamma_0$: because the left side of the equation contains a spatial derivative, I expected that the right side should multiply $\psi$ with something that has inverse spacetime units, so I would expect to be written $\gamma^0$.
Then I remembered some basic relativistic physics: $$ E^2 - p^2 = m^2 $$ Perhaps, when deriving the Dirac-Hestenes equation from the Klein-Gordon equation by taking the square root of the operators on both sides, the intended square root on the right side is not simply a mass scalar, but the energy-momentum vector associated with the field. For real particles, this energy-momentum vector is guaranteed to be timelike, and it would have the correct inverse spacetime units if interpreted as a wavevector $\vec{k}$. (However, it wouldn't necessarily be future pointing: perhaps corresponding to antiparticle states?)
I proposed that it was possible to write an index-free Dirac-Hestenes equation of the form $$ \vec{\nabla} \psi \hat{B} = \vec{k} \psi $$ This seemed to be the most compact form I could fit the equation into, and it seemed to suggest an interpretation of $\psi \mapsto \vec{\nabla} \psi \hat{B}$ as the momentum operator, with $\vec{k}$ being the associated eigenvalue.
The problem
I've been told that such an equation could not be a valid equation of motion, as wavevectors arise from $\psi$, and not the other way around. Additionally, although the equation would admit planewave solutions in free space as its eigenstates, it would not admit a linear combination of planewaves generated from different wavevectors as a solution.
Perhaps the equation I derived is the relativistic equivalent to the time-independent Schrödinger equation, with eigenvalues being states with uniform energy-momentum distributions over spacetime. But at the same time, it seems like the Dirac-Hestenes equation could be be interpreted as the action of an operator with the resulting eigenvalue being $m \gamma_0$, and from a dimensional analysis perspective, $\vec{k}$ and $m \gamma_0$ seem to have equivalent units. The only major difference is that $\gamma_0$ can be chosen arbitrarily, whereas $\vec{k}$ depends on $\psi$.
The question
So my question is: how are the wavevector $\vec{k}$ and the $m \gamma_0$ term related? Or equivalently, is the equation I have written meaningfully different from the Dirac-Hestenes equation as first presented?
