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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. $(1)$?

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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 $\gamma^{i}=\beta\alpha^{i}$.

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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.

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