# Can the QED action be expressed using geometric algebra?

Hestenes et al. have been able to rewrite the Dirac equation in terms of the "spacetime algebra" (Clifford algebra over Minkowski space), as laid out here. I was wondering if the same can be done for the QED action as written here.

Consider the form written near the end of that section in the second linked article, which if you expand the shorthands becomes:

$$\mathcal{L} = -\frac14 F_{\mu\nu}F^{\mu\nu} + \psi^{\dagger}\gamma^0(i\gamma^{\mu}\partial_{\mu} - m)\psi - e\psi^{\dagger}\gamma^0\gamma^{\mu}\psi A_{\mu}.$$

It is obvious that the Faraday tensor is just a bivector, and the first term is its squared magnitude. But the matter terms are not nearly so clear-cut.

We know from the first linked article above that $$\psi$$ is an even multivector, which can be written as $$R(\rho e^{i\beta})^{\frac12}$$, with $$R$$ a Lorentz transform. I would then naively guess that $$\psi^{\dagger}$$ is the "reverse" of that multivector, $$\gamma^0$$ and $$\gamma^\mu$$ are spacetime basis vectors -- and $$i$$, where it appears both directly in the action and within the gauge covariant derivative, is the unit pseudoscalar.

The problem is, this leads to a Lagrangian that is a multivector rather than a scalar. For example, the term $$\psi^{\dagger}\gamma^0m\psi$$ would be a vector, whereas something like $$\psi^{\dagger}\gamma^0 i \gamma^x\partial_x\psi$$ would be an even multivector. I could try to take the real part, but that would get rid of entire terms, such as $$e\psi^{\dagger}\gamma^0\gamma^x\psi A_x$$, which is a pure bivector. So that approach does not seem right.

Has this translation into spacetime algebra been done before, or does anyone have any suggestions as to how to go about it? Or perhaps is it true that the action is a multivector, and the way it is applied in the path integral must somehow change to compensate?

The problem is, this leads to a Lagrangian that is a multivector rather than a scalar.

According Hestenes, traditional expressions like $$\bar{\psi}\psi = \psi^{\dagger}\gamma^0\psi$$ are translated to $$<\bar{\psi}\psi>_{1, I}$$ where $$<...>_{1, I}$$ means you only retain scalar and pseudoscalar parts.

For example, the term $$\psi^{\dagger}\gamma^0m\psi$$ would be a vector

As per Hestenes, traditional matrices like $$\gamma^\mu$$ are translated into bivectors, therefore you don't have the vector issue.

I could try to take the real part, but that would get rid of entire terms, such as $$e\psi^{\dagger}\gamma^0\gamma^x\psi A_x$$, which is a pure bivector.

Please use the prescription I mentioned above and try again.

• Do you have a reference for this? I've only seen the $\gamma^\mu$ referred to as the orthonormal frame of vectors, not bivectors. For example in this document. Jan 4, 2023 at 21:06
• Please check out eq. 190 in the document you referenced, which indicates that traditional vector operation $\gamma^\mu\psi$ is translated to bivector operation $\gamma^\mu\psi\gamma_0$ Jan 4, 2023 at 23:18