Dirac Matrices
We can write the Dirac Matrices explicitly as $$\gamma^0 = \begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix},\quad \gamma^i=\begin{pmatrix}0 & \sigma_i\\ -\sigma_i & 0\end{pmatrix}$$ However this is only one representation known as Weyl representation
Just like representations of the Lorentz group are not unique, this representation, this representation of the Dirac matrices is not unique They satisfy the Clifford algebra
Definition: Clifford Algebra
The Dirac algebra is a Clifford algebra is defined by $${\gamma^\mu,\gamma^\nu}=2g^{\mu\nu}$$ It is essentially a Clifford algebra in four dimensions, the Clifford algebra is more general and can be applied to arbirary dimensions.
Using the Dirac matrices we can write the Lorentz transformations of the Dirac spinors $\psi = \begin{pmatrix}\psi_L \\ \psi_R \end{pmatrix}$ as $$\Lambda_s = \exp(i\Theta_{\mu\nu}S^{\mu\nu}),$$ with $$S^{\mu\nu}=\frac{i}{4}[\gamma^\mu,\gamma^\nu]=:\frac{1}{2}\sigma^{\mu\nu},$$ the generators and $\Theta_{\mu\nu}$ containing six real angles (the $\mu=0$ and $\nu=0$ and $\mu=\nu$ elements vanish anyways). So these are the generators for the reducible $(\frac{1}{2},0)\oplus (0,\frac{1}{2})$ representation. However since $$S^{ij\dagger}=S^{ij},\quad S^{0i\dagger}=-S^{0i},$$ we know that $\Lambda_s$ is not unitary!
There are different Dirac representations (representations for $S^{\mu\nu}$) Weyl representation: $$S^{ij}=\frac{1}{2}\epsilon_{ijk}\begin{pmatrix}\sigma_k & 0\\ 0 & \sigma_k\end{pmatrix},\quad S^{0i}=-\frac{i}{2}\begin{pmatrix}\sigma_i & 0\\ 0 & -\sigma_i\end{pmatrix}$$ Majorana representation: $$\gamma^0=\begin{pmatrix}0 & \sigma^2\\ \sigma^2 & 0 \end{pmatrix}, \gamma^1=\begin{pmatrix}i\sigma^3 & 0\\ 0 & i\sigma^3\end{pmatrix},\gamma^2=\begin{pmatrix}0 & -\sigma^2\\ \sigma^2 & 0\end{pmatrix},\gamma^3=\begin{pmatrix}-i\sigma^1 & 0\\ 0 & -i\sigma^1\end{pmatrix}$$
Since the $\Lambda_s$ representation is not unitary, $\psi^\dagger\psi$ is not Lorentz invariant. However one finds that $\bar{\psi}\psi=\psi^\dagger\gamma^0\psi$ is Lorentz invariant. Similarly $\bar{\psi}\gamma^\mu\psi$ transforms like a Lorentz vector. This is because $$\Lambda^{-1}_s\gamma^\nu\Lambda_s=(\Lambda_V)^{\mu\nu}\gamma^\nu,$$ where $\Lambda_V$ is ther Lorentz transformation in vector representation.
This is the reason why $$\bar{\psi}=\psi^\dagger\gamma^0$$ appears in the Dirac Lagrangian.