Prove that the $\alpha_i$ and $\beta$ are hermitian, traceless matrices of even dimensionality, with eigenvalues \pm 1 .
The lowest dimensionality matrices satisfying all these requirements are $4 \times 4$. The choice of the four matrices $(\alpha, \beta)$ is not unique. The Dirac-Pauli representation is most frequently used:
$$
\alpha=\left(\begin{array}{ll}
0 & \boldsymbol{\sigma} \\
\boldsymbol{\sigma} & 0
\end{array}\right), \quad \beta=\left(\begin{array}{rr}
I & 0 \\
0 & -I
\end{array}\right)
$$
where $I$ denotes the unit $2 \times 2$ matrix (which is frequently written as 1) and where $\boldsymbol{\sigma}$ are the Pauli matrices:
$$
\sigma_1=\left(\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right), \quad \sigma_2=\left(\begin{array}{rr}
0 & -i \\
i & 0
\end{array}\right), \quad \sigma_3=\left(\begin{array}{rr}
1 & 0 \\
0 & -1
\end{array}\right) .
$$
Another possible representation, the Weyl representation, is
$$
\alpha=\left(\begin{array}{rr}
-\sigma & 0 \\
0 & \sigma
\end{array}\right), \quad \beta=\left(\begin{array}{ll}
0 & I \\
I & 0
\end{array}\right)
$$
Most of the results are independent of the choice of representation. Certainly, all the physics depends only on the properties listed in (5.3). In fact, not until we exhibit explicit solutions of the Dirac equation in Section 5.3 will we use a particular representation. Unless stated otherwise, we shall always choose the Dirac-Pauli representation, (5.4).
A four-component column vector $\psi$ which satisfies the Dirac equation (5.1) is called a Dirac spinor. We might have anticipated two independent solutions (particles and antiparticles), but instead we have four!
Maybe the surprise should not have been total. We know at least one other example where a field with more components appears when linearizing the equation. The covariant Maxwell equations $\square^2 A_\mu=0$ are second-order but can be written in a linear form $\partial_\mu F^{\mu \nu}=0$ by introducing the field strength $F_{\mu \nu}$, which has more components than $A_{\mu^*}$