# Dirac equation: covariant form and original form

In my lecture book, the Dirac equation is derived and given as the equation: $$i \hbar \gamma^{\mu} \partial_{\mu} \psi-m c \psi=0 \tag{1}$$ Where: $$\gamma^{0}=\left(\begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array}\right), \quad \gamma^{i}=\left(\begin{array}{cc} 0 & \sigma^{i} \\ -\sigma^{i} & 0 \end{array}\right) \tag{2}$$ However, I have often seen it on the form: $$\left(\beta m c^{2}+c \sum_{n=1}^{3} \alpha_{n} p_{n}\right) \psi(x, t)=i \hbar \frac{\partial \psi(x, t)}{\partial t}, \tag{3}$$ Which, according to Wikipedia, is the original form derived by Dirac. Now, I have seen how eq. $$(1)$$ is derived, but how do I get to eq. $$(3)$$ from eq. $$(1)$$?

Just multiply on the left by $$\beta=\gamma^{0}$$ and divide by $$c$$. Also, recall that $$\beta^{2}=1$$ and $$\vec{p}=-i\hbar\vec{\nabla}$$. The $$\gamma^{i}$$'s for $$i=1,2,3$$ are defined as $$\gamma^{i}=\beta\alpha^{i}$$.
I set $$\hbar=c=1$$. Rewrite (1)
$$\big (i\gamma^0 {\partial \over \partial t} - i \mathbf \gamma \cdot\nabla -m \big) \psi =0$$
Premultiply by $$\gamma^0$$ and use $$\gamma^0\gamma^0 =1$$ (from the defining commutation relation for $$\gamma$$ matrices)
$$i {\partial \psi \over \partial t} =\big ( i \gamma^0\mathbf \gamma \cdot\nabla +m \gamma^0\big) \psi.$$ Require that $$\gamma^0\gamma^a$$ and $$\gamma^0$$ are Hermitian, and define $$\alpha$$ and $$\beta$$ accordinly.