Obtaining the 16 elements of the Clifford algebra from the $\gamma^\mu$ generators

Question

In my study of the Dirac equation, I have fully understood the "linearization" of the relativistic energy to obtain a matrix-valued equation that reduces to the Klein-Gordon equation if the matrix coefficients satisfy

$$ \left\{ \gamma^\mu,\gamma^\nu \right\}=2\eta^{\mu\nu}\mathbf{1} $$

What happens next is that we use the $\gamma^\mu$ as generators of the 16 elements of the Clifford algebra. Specifically, we obtain a maximal set of linearly independent products of $\gamma$-matrices: there are 16 of them. After we do it, everything makes sense, and it is nice that every operator on a Dirac state can be expressed as a linear combination of the 16 basis matrices. However, I do not see where we get the idea in the first place to generate the 16 basis matrices from the four $\gamma^\mu$. The fact that we can do this is why the $\gamma^\mu$ are called generators, but I do not see where the idea to generate comes from. The purpose of this post is to solicit clarifications on this issue.

An algebra is a vector space $\mathcal{V}$ equipped with a bilinear form. In the present case of the Dirac algebra, the bilinear is the anti-commutator $\{\,,\}:\mathcal{V}\times\mathcal{V}\to \mathcal{V}$, so I understand that we would want to find a basis for $\mathcal{V}$. Still, what motivation do we have to express the 16 basis matrices of the (matrix) vector space associated with the anti-commutation relation for Dirac matrices as products of those matrices. Where does this idea come from? Every one of several textbooks I've looked just form the 16 independent matrices from the $\gamma^\mu$ without saying why we should do that, or even why we would have the idea to do that. Even the Wiki for the Dirac algebra jumps straight into defining a basis generated from the $\gamma^\mu$ without giving any hint as to why we would prefer that basis over the basis of 16 different matrices with one 1 and 15 zeros, for example, or any other random basis of 16 linearly independent matrices.

THANKS!

Answer 1

A partial argument of extending from the vector space of $\gamma^\mu$ to the algebra of $\gamma^\mu$ goes as follows:

1. The (span of) commutators $M^{\mu\nu}\sim [ \gamma^\mu,\gamma^\nu]$ form a representation of the Lorentz algebra $so(3,1)$ (when it acts on spinors).

2. The exponentiation of the Lorentz algebra $so(3,1)$ maps into the the restricted Lorentz group $SO^+(3,1)$.

3. Using exponentiation, the corresponding representation of the restricted Lorentz group $SO^+(3,1)$ therefore sits in the even part ${\rm Cl}(3,1)_{\rm even}$ of the Clifford algebra ${\rm Cl}(3,1)$.