Invertibility of Dirac matrices

Question

Suppose we have Dirac equation in the following form :

$i\partial_t\psi = (-i\vec{\alpha}\cdot \nabla +m\beta)\psi$

and assume that the Klein-Gordon equation is satisfied, i.e.

$\partial_t^2 \psi =(\nabla^2-m^2)\psi$.

Suppose that we treat the RHS of Dirac equation as a Hamiltonian in Schrödinger equation, i.e. in particular it is hermitian.

Under these assumptions we can obtain many properties of Dirac matrices $\gamma^0=\beta, \gamma^k=\gamma^0\alpha_k$, for example anticommutation relations, hermiticity of $\gamma^0$ etc.

My question : Can obtain fact that these matrices are invertible $\textbf{only}$ using above assumptions ? Of course, I know other formalisms which are use to introduce this equation, but I'm interested in what we can formally derive from such simple assumptions of our equation.

Answer 1

The Dirac matrices are invertible because of the defining Dirac anticommutators $$ \gamma^\mu\gamma^\nu+\gamma^\nu \gamma^\mu = 2g^{\mu\nu}\cdot 1 $$ For $\mu=\nu$, one simply obtains $(\gamma^\mu)^2 = g^{\mu\mu}\cdot 1$ (no summation) which means that the inverse of $\gamma^\mu$ is $\gamma^\mu g^{\mu\mu}$ i.e. basically $\gamma_\mu$ if $g_{\mu\nu}$ is diagonal. Up to the sign, the inverse is the same matrix.

If you were worried that I am assuming too much, the Dirac algebra actually follows from the assumptions you have made, up to an overall normalization. The first, Dirac equation already fully determines the evolution of $\psi$. If the second equation doesn't make the system overdetermined, it must follow from the first one. So the operator "box" must be a multiplication of the Dirac operator by another one (from the left). The only solution is when the "other one" is the same Dirac operator and the Dirac algebra holds (perhaps up to an overall scaling).