Is Dirac theory just a real Clifford algebra?

Question

The gamma matrices $\gamma^\mu$ appearing in the Dirac equation span the Clifford algebra ${\cal Cl}_{1,3}$ over real numbers. They are generators of Clifford algebra in that sense that their products: $\gamma^\mu\gamma^\nu,\; \gamma^\mu\gamma^\nu\gamma^\sigma$ and $\gamma^0\gamma^1\gamma^2\gamma^3$ make basis of the 16-dimensional ${\cal Cl}_{1,3}$. I have two questions:

1. what is the physical interpretation of these products ?

2. The $\gamma$'s products acquire the physical interpretation only after applaying the complex numbers, for example: $\gamma^5=i\gamma^0\gamma^1\gamma^2\gamma^3$ is the pseudoscalar operator, $\sigma_{\mu\nu} = \frac{i}{2}[\gamma^\mu,\gamma^\nu]$ is the tensor operator. It means that Dirac theory, in fact, is based on a wider structure. What is the structure ?

Answer 1

There is an approach to Dirac theory that gives the gamma matrices an explicitly geometric interpretation - namely Geometric Algebra or Spacetime Algebra.

The approach builds the Real Clifford Algebra from geometric entities - vectors representing oriented 1D lengths, bivectors representing oriented 2D areas, trivectors representing oriented 3D volumes, and so on. They also represent geometric transformations: reflections, rotations, dilations, and so on. The geometric product combines the geometrically-defined dot product and wedge product into one.

From this perspective, the gamma matrices are just an arbitrarily-chosen matrix representation designed to map matrix multiplication onto the Geometric Algebra's geometric product. The gamma matrices represent the unit vectors of a coordinate basis of spacetime, the products of gamma matrices you mentioned represent bivectors for the coordinate planes, and so on.

The entire Dirac theory can be presented without requiring any use of scalar imaginary numbers at all - where they appear in the equations, they can all be implemented using geometric entities that naturally have the property of squaring to give $-1$. Or to put it another way, the complex numbers appearing in physics are actually just a particular sub-algebra of the real Clifford algebra and have a real geometric interpretation.

The entire Geometric Algebra approach is an excellent one from the point of view of giving an intuitive geometric insight into the abstract algebraic results, which some physicists prefer. But from an abstract mathematician's point of view, it doesn't add anything new to the already well-established theory of Clifford algebras, and requires learning a whole new set of notations and conventions, so it has rather mixed reviews.

For a more detailed introduction and starting point for further exploration, I'd recommend David Hestenes paper 'Real Dirac Theory' in J. Keller and Z. Oziewicz (Eds.), The Theory of the Electron, UNAM, Facultad de Estudios Superiores, Cuautitlan, Mexico (1996), p. 1-50.

Answer 2

The gamma matrices $\gamma^\mu$ appearing in the Dirac equation span the Clifford algebra ${\cal Cl}_{1,3}$ over real numbers. They are generators of Clifford algebra in that sense that their products: $\gamma^\mu\gamma^\nu,\; \gamma^\mu\gamma^\nu\gamma^\sigma$ and $\gamma^0\gamma^1\gamma^2\gamma^3$ make basis of the 16-dimensional ${\cal Cl}_{1,3}$.

The 16 combinations $\gamma^0$, $i\gamma_1$, $\ldots$, $\gamma^0\gamma^1$, $\gamma^0\gamma^2$, $\gamma^0\gamma^3$, $\ldots$, $i\gamma^0\gamma^2\gamma^3$, $\ldots$, $i\gamma_5$ (or similar combinations, depending on conventions) are sometime called the "big gamma" ($\Gamma$) matrices. These combinations can be used to prove a number of results, for example, that the gamma matrices must be at least 4x4 in size. Further, any 4x4 matrix $X$ can be written as a linear combination of the big gamma matrices: $$ X = \sum_{k=1}^{16}x_k \Gamma_k\;, $$ where $$ x_k = \frac{1}{4}\text{Tr}(X\Gamma_k) $$

I have two questions:

1. what is the physical interpretation of these products ?

The products $\gamma^0\gamma^1$, $\gamma^0\gamma^2$,$\gamma^0\gamma^3$ are sometimes called the "velocity operator" (divided by c). They are identical with the matrix usually called "alpha" $\vec{\alpha}$, where the Dirac equation is: $$ \frac{1}{c}\frac{\partial \psi}{\partial t} + \vec{\alpha}\cdot{\vec \nabla}\psi + \frac{imc}{\hbar}\psi = 0 $$

One physical interpretation, at least for this subset of Gamma matrices, is that they are related to the velocity. Of course, it doesn't help the interpretation that the eigenvalues are $\pm 1$. But nevertheless, we often see c$\vec \alpha$ being called the velocity operator (since, for example, $\frac{\partial H}{\partial p} = c\vec \alpha$.)

...It means that Dirac theory, in fact, is based on a wider structure. What is the structure ?

This part of OP's question is not entirely clear to me.

The Dirac equation is derived by treating $ct$, $x$, $y$, and $z$ symmetrically and demanding that only the first derivative wrt time occur. This can only be accomplished by an equation that involves at least 4x4 matrices. You can generalize to larger matrices, if this is what you are asking about.

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General Reference: "Intermediate Quantum Mechanics" (3rd Edition) by Bethe and Jackiw. Chapter 22 "Dirac Equation, Formal Theory."

Answer 3

The conventional form of the Dirac matrices are generators of the Clifford Algebra Cl(1,3) with metric (1,-1,-1,-1). There exists no real matrix representation of Cl(1,3). However, there exists a 4x4 real matrix representation of the Clifford Algebra Cl(3,1) with metric (-1,1,1,1). This real representation allows to write the Dirac equation without explicit use of the unit imaginary. Furthermore the real representation can be used for the description of classical (Hamiltonian) phase spaces, specifically that of two coupled harmonic oscillators. (see my publications: https://journals.aps.org/prab/abstract/10.1103/PhysRevSTAB.14.114002, https://arxiv.org/abs/2008.13409).