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The Dirac equation as derived by Hestenes is $$ \hbar \nabla \psi I \sigma_3 = m \psi \gamma_0 $$ where $I \sigma_3 = \gamma_2 \gamma_1$. The equation is claimed to be Lorentz invariant, because the choice of $\gamma_0$ and $I \sigma_3$ are arbitrary, provided $\gamma_0$ is a future timelike unit vector and $I \sigma_3$ commutes with $\gamma_0$.

My question is whether there is a fundamental limitation that prevents us from rephrasing the Dirac equation in spacetime algebra without reference to indices, especially specifically enumerated indices as shown here. I find it surprising that the STA treatment of the Dirac equation isn't as concise as, say, that of the classical electromagnetic field (Maxwell's equations simplify to $\nabla F = J$).

[edit] removed extraneous factor of $c$ in the equation

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    $\begingroup$ Do note that the standard treatment actually hides the appearance of indices too. $\bar\psi=\psi^\dagger\gamma_0$. What is nice and interesting in this form is that basically all 4 indices appears once. time, x y appears explicitly in the above form, and the spin you see are in the z direction (actually, xy spin evolving in time t in GA lingo). So this is quite good. $\endgroup$
    – naturallyInconsistent
    Commented Jun 1, 2023 at 23:53
  • $\begingroup$ This probably should have been part of the question, but are there any indices that should be included with the spinor components? I have been assuming there are not, but I realize that it's probably good to clarify that explicitly. $\endgroup$
    – brainandforce
    Commented Jun 4, 2023 at 21:48

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The weird format of Hestenes's version of Dirac equation $$ \hbar \nabla \psi I \sigma_3 = m \psi \gamma_0 $$ stems from the restriction of $\psi$ to the even part of the Clifford algebra $Cl(1,3)$. In other words, $\psi$ is composed of scalar $1$, bi-vectors $\gamma_a\gamma_b$, and the pseudoscalar $I=\gamma_0\gamma_1\gamma_2\gamma_3$. The "leftover indices" such as $\sigma_3=\gamma_3\gamma_0$ are unfortunate artifacts when Hestenes was trying too hard to fit spinor $\psi$ into the cramped Clifford-even subspace.

However, one can lift the restriction and look at a transformed spinor $\Psi$ (a left-ideal) derived from the original Clifford-even spinor $\psi$ $$ \Psi = \frac{1}{2}\psi(1+\gamma_0)(1-\sigma_3) $$ which obeys a Dirac equation more like the conventional form $$ \hbar \nabla \Psi = m \Psi I $$ As you can see, the Dirac equation above for $\Psi$ does not have those "leftover indices" such as $\sigma_3$. The price to pay is that $\Psi$ is now a left-ideal that can take values in both Clifford-even (scalar, bi-vector, pseudoscalar) and Clifford-odd (vector and tri-vector) components. See more details here (specifically eq 5 and eq 12 therein).

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    $\begingroup$ But this version of the Dirac equation still references specifically enumerated indices in the definition of the ideal. You can again argue that it's Lorentz invariant because the choice of indices doesn't matter, but it seems to be no improvement on the other form. $\endgroup$
    – benrg
    Commented Jun 2, 2023 at 19:50
  • $\begingroup$ The updated version is a major improvement, since the "index" in the ideal definition is a bonus and has the concrete meaning of differentiating a electron from a neutrino via right-sided idempotent projections (see eq. 23 in referenced link), while in the original version the "index" does not have this physical meaning. $\endgroup$
    – MadMax
    Commented Jun 2, 2023 at 20:20
  • $\begingroup$ Actually, in the updated version, you can get rid of the "index" all together (without the right-sided idempotent projection as in eq. 23). Then the left-chiral part of the index-free spinor $\Psi$ describes the electron-neutrino weak $SU(2)$ doublet. In other words, the electroweak unification pops out of the Clifford algebra $Cl(1,3)$ automatically. Is that neat! $\endgroup$
    – MadMax
    Commented Jun 2, 2023 at 20:29
  • $\begingroup$ Electroweak unification certainly doesn't pop out. There's nothing to connect this to the electroweak theory beyond the existence of an SU(2) or U(2) group action on the spinors. There's no explanation of why the left-chiral symmetry is dynamically broken but the right-chiral symmetry is explicitly broken. You still have to pick a seemingly Lorentz-violating axis for the latter, and the former suggests that the Higgs field breaks Lorentz symmetry also (in an undetectable way). Besides all that, the SM gauge group isn't SU(2) or U(2), but S(U(2)×U(3)). $\endgroup$
    – benrg
    Commented Jun 2, 2023 at 21:06
  • $\begingroup$ Well, there are lot of questions that Clifford algebra $Cl(1,3)$ alone can not answer. As you rightfully pointed out: "There's no explanation of why the left-chiral symmetry is dynamically broken but the right-chiral symmetry is explicitly broken". The last time I checked, the standard model can not answer this question either. And for that matter, there are a lot other questions the Clifford algebra can not answer: for instance, out of so many able and talented American politicians why the 2024 presidential election is a duel between two 80+ old dudes ;) $\endgroup$
    – MadMax
    Commented Jun 2, 2023 at 21:24

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