Are the $\alpha$ and $\beta$ matrices of the Dirac equation unique? Assume we are dealing wth three spatial dimensions $d=3$ which requires 3 $\alpha$ matrices. Furthermore assume that we are looking for them in the space of 4-dimensional matrices, not in higher dimensions.
I assume there are still different choices, which all fulfill the necessary commutator relations
$$\{\alpha_i,\alpha_j\}=2\delta_{ij}\quad \{\alpha_i,\beta\}=0\quad \beta^2=I.$$
In my lecture notes on assumes
$$\beta=\textrm{diag}(1,1,-1,-1).$$
What are the implications of chosing a different"fixing"? Does the physics change at all?
 A: The $\alpha$ and $\beta$ Dirac matrices are not uniquely defined by their
anticommutation relations
$$\{\alpha_i,\alpha_j\}=2\delta_{ij}, \quad \{\alpha_i,\beta\}=0, \quad \beta^2=1. \tag{1}$$
Dirac basis
The solution from your lecture notes
$$\alpha_i=\begin{pmatrix}0 & \sigma_i \\ \sigma_i & 0\end{pmatrix},
\quad \beta=\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}.$$
(with $0$, $1$, $\sigma_i$ meaning the zero, unity and Pauli $2\times 2$-matrices)
is just one possible choice, known as the Dirac basis.
This choice has the special feature that in the bispinor
$$\psi=\begin{pmatrix} \psi_+ \\ \psi_- \end{pmatrix}$$
the upper two components ($\psi_+$) can be interpreted
as the particle component, and the lower two components
($\psi_-$) as the anti-particle component.
Weyl (or chiral) basis
Another widely used choice is the so-called Weyl or chiral basis
$$\alpha_i=\begin{pmatrix}\sigma_i & 0 \\ 0 & -\sigma_i \end{pmatrix},
\quad \beta=\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}.$$
This choice has the special feature that in the bispinor
$$\psi=\begin{pmatrix} \psi_L \\ \psi_R \end{pmatrix}$$
the upper two components ($\psi_L$) can be interpreted
as the left-handed component, and the lower two components
($\psi_R$) as the right-handed component.
Other representations
You can construct more representations by applying a transformation
(with any constant unitary $4\times 4$-matrix $U$)
$$\begin{align}
\alpha_i&\to\alpha'_i=U\alpha_i U^{-1} \\
\beta&\to\beta'=U\beta U^{-1} \\
\psi&\to\psi'=U\psi
\end{align}$$
Then it can easily be verified that the transformed
$\alpha'_i$ and $\beta'$ again satisfy the anticommutation relations (1),
and that $\alpha'_i$, $\beta'$ and $\psi'$ again satisfy the Dirac equation.
Hence the physics is always the same, independent of the chosen representation.
