# Geometrical interpretation of the Dirac equation

Is there an intuitive geometrical picture behind the Dirac equation, and the gamma matrices that it uses? I know the geometric algebra is a Clifford algebra. Can the properties of geometric algebra, be used to paint a picture of the Dirac equation? I searched the web for this, but I couldn't find a credible source, that explains everything properly.

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It depends what you mean by "geometric". Spinors in the Dirac equation are geometric objects in the sense that they have a well-defined behavior under geometric operations, rotations and Lorentz transformations. But they're not vectors or their tensor products, they're more elementary than that, so any picture is bound to be either misleading or distorting the linearity of the spinor space. One may visualize spinors via their tensor squares which are vectors and other p-forms but that obscures the fact that spinors may be naturally added before they're squared. The best picture is an equation. – Luboš Motl Dec 26 '12 at 10:40
Search for papers by David Hestenes (Arizona State) - he has spent a lot of time on exactly this kind of question. – DarenW Apr 8 '14 at 8:22

$$\nabla \psi \gamma_{21} = m \psi \gamma_0$$
Where $\nabla = \gamma^\mu \partial_\mu$, of course, and $I = \gamma_{0123}$. I would just interpret this as a differential equation on a Minkowski spacetime, no different from how $\nabla F = - \mu_0 J$ captures electromagnetism in a 3+1 flat spacetime. The authoritative resource for further information on doing quantum mechanics with geometric algebra is probably Doran and Lasenby's book. Both Dirac and Pauli problems can be handled with geometric algebra, obviating the need for a true complex imaginary--objects that square to -1 naturally arise as a result of the geometry.