Spinor representations
Spinor representations are the representations with half spin. There are two representations of the Lorentz Group which generate a single spin $\frac{1}{2}$ particle, $(\frac{1}{2},0)$ and $(0,\frac{1}{2})$. They act on a vector space with $2\frac{1}{2}+1=2$ degrees of freedom, so we need $2\times 2$ matrices that satisfy the Lorentz algebra $$[J^+_i,J^+_j]=i\epsilon_{ijk}J^+_k$$ $$[J^-_i,J^-_j]=i\epsilon_{ijk}J^-_k$$ $$[J^+_i,J^-_j]=0$$ We might therefore take: For $(\frac{1}{2},0)$: $\vec{J}^-= \vec{\sigma}/2, \vec{J}^+=0$ For $(0,\frac{1}{2})$: $\vec{J}^-= 0, \vec{J}^+=\frac{1}{2}$ If we then obtain the usual Lorentz generators $\vec{J}$ and $\vec{K}$ from $\vec{J}=\vec{J}^-+\vec{J}^+$ and $\vec{K}=i(\vec{J}^–\vec{J}^+)$ we would get the following representations:
$(\frac{1}{2},0)$: Representation: $$S_L(\Theta,\beta)=e^{\frac{1}{2}(i\Theta_j\sigma_j - \beta_j\sigma_j)}$$ Representation space: $2$ dimensional vectors, called left-handed Weyl spinors $\psi_L$, which transform under Lorentz transformation as: $$\psi_L\to S_L(\Theta,\beta)\psi_L$$
$(0,\frac{1}{2})$: Representation: $$S_R(\Theta,\beta)=e^{\frac{1}{2}(i\Theta_j\sigma_j + \beta_j\sigma_j)}$$ Representation space: $2$ dimensional vectors, called right-handed Weyl spinors $\psi_R$, which transform under Lorentz transformation as: $$\psi_R\to S_R(\Theta,\beta)\psi_r$$
The problem with this representation is that it is not unitary. This is because the generators are not hermitian. We would need a hermitian generator so that $(e^{iK})^\dagger = e^{-iK}$ so that $\braket{\psi|S^\dagger S|\psi}=\braket{\psi|\psi}$ leaving matrix elements invariant.
One can show that:
Unitary representations of the Lorentz Group
There are no finite-dimensional unitary representations of the Lorent group.
To construct an infinite dimensional representation we need the transformations to also act on the degrees of freedom of momentum (which are of course infinite dimensional), which means the basis needs to depend on $p$.
This means we get an infinite dimensional representation by just promoting the two component spinors to fields:
$$\psi_R(x)=\begin{pmatrix}\psi_1(x)\\psi_2(x)\end{pmatrix}$$ $$\psi_L(x)=\begin{pmatrix}\psi_1(x)\\psi_2(x)\end{pmatrix}$$