# 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?

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).

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.

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.)

• Could the first term of the Dirac equation then be interpreted as a velocity operator acting on the momentum and energy operators? – Matrix001 Nov 20 '19 at 0:57
• Ummm... very, very loosely, the first term is supposed to evoke something like $\vec v \cdot \vec p$ and the second just the rest mass, the leading term in the energy... These are all evocative guides for the math, but, frankly, in the Dirac equation you follow the math and use these shared metaphors for guidance, mostly. – Cosmas Zachos Nov 20 '19 at 1:56