Physical interpretation of gamma matrices Just out of plain curiosity, I want to ask: What are/is the physical interpretation(s) of the gamma matrices? If there is none, is it right to assume that it is just a mathematical fudge-factor?
 A: They are not just mathematical definitions; there is some physics in them. Actually, they are intrinsically connected with the spin structure of the fields. You can see this from the fact that the spin 1/2 representations of the Lorentz group, namely (1/2,0) and (0,1/2) in $SU(2)\times SU(2)$ classification, naturally introduce the definition of the gamma matrices. In particular, the Lorentz transformations of a bispinor (which transforms as $(1/2,0)\oplus(0,1/2)$) are generated by the commutator of gamma matrices.
A: Here are interpretations for at least two gamma matrices:


*

*$\gamma_0$ is the spinor metric. It's role is analogous to the role of the Minkowski metric for four-vectors. We need the Minkowski metric to write down the scalar product of two four-vectors. Analogously, we need $\gamma_0$ to write down the scalar product of Dirac spinors.

*$\gamma_5 \equiv i \gamma_0\gamma_1\gamma_2\gamma_3$ is the chirality operator. If we act with $\gamma_5$ on a spinor, it tells us its chirality (whether it's left-chiral, right-chiral or a superposition). 

A: In natural units, the Hamiltonian your text tells you was introduced by Dirac is 
$$
H=\vec{\alpha}\cdot \nabla /i+ \beta m , \implies  i\partial_t \psi = H \psi ,\\
\beta =\gamma_0 , \qquad \vec {\alpha}= \gamma_0 \vec{\gamma} .
$$
You then have a bona-fide continuity equation
$$
\partial_t \rho + \nabla \cdot \vec j =0, \\
\rho =\psi^\dagger \psi > 0, \qquad \vec j=  \psi^\dagger \vec \alpha \psi, 
$$
so the above $\vec \alpha$ is the velocity operator  for the (positive) probability fluid flow, so the flow velocity when sandwiched between two $\psi$s, and you  might think of it that way.  
$\beta$ is just a conversion of spinors, useful in relativistic covariance considerations, so you might think of it as a mathematical fudge-factor, but math rules here -- Dirac was a math undergraduate, after all. 
In any case, from the above gradient expressions,  you work out that
$$
i[H,\vec x]= \vec \alpha ,
$$
a velocity entity! (At low momenta, recall $H= m+ \vec p\cdot \vec p /2m+ ...$, alright.)
