Dimensionality of Gamma Matrices

Question

If I express the Dirac equation in the form of

$$i\hbar \frac{\partial}{\partial t} \psi_a(x) = \left(-i\hbar c(\alpha^j)_{ab}\partial _j + mc^2(\beta)_{ab}\right)\psi_b(x),$$

with the constraints

$$\{\alpha^j,\alpha^k\}_{ab} = 2\delta^{jk}\delta_{ab}, \qquad \{\alpha^j,\beta\}_{ab}=0, \qquad (\beta^2)_{ab} = \delta_{ab},$$

since the Gamma matrices are traceless and their eigenvalues are either +1 or -1, we can show that the Gamma matrices are of even dimension.

My question is: how can I show $2 \times 2$ matrices are not enough?

Answer 1

You need four linearly independent matrices. Write down the representation:

a + ib   c + id
e + if   g + ih

The traceless condition gives $a = -g$, $h=-b$. Now we have

a + ib   c + id
e + if  -a - ib

This has six unknowns (six possible independent matrices).

Then the eigenvalue conditions for the four matrices give four equations in six unknowns, which means that there is no linearly independent solution with $2\times2$ matrices, since that would require four degrees of freedom, but we only have two left.

Incidentally this question is often an exercise in books on quantum field theory or relativistic quantum mechanics.