# Dirac equation derivation

I am working through a set of lecture notes containing a derivation of the Dirac equation following the historical route of Dirac. It states that Dirac postulated a hermitian first-order differential equation for a spinor field $$\psi(x) \in \mathbb{C}^{n}$$,

$$\begin{equation} i \partial^{0} \psi(x)=\left(\alpha^{i} i \partial^{i}+\beta m\right) \psi(x),\tag{1} \end{equation}$$

where the sum is over spacial indices only, and the hermitian property of $$H_D:=\left(\alpha^{i} i \partial^{i}+\beta m\right)$$ means that the coefficient matrices $$\alpha^{i}, \beta \in \mathbb{C}^{m \times m}$$ are hermitian. Next, the notes go on to derive that

\begin{align}\label{EQadawdwww} \left\{\alpha^{i}, \alpha^{j}\right\}=2 \delta^{i j} I, \quad\quad\left\{\alpha^{i}, \beta\right\}=0,\quad\text{ and} \quad \beta^{2}=I. \end{align}

It then goes on to state that, as expected for a Hamiltonian formulation of a theory, the ansatz above does not treat space and time on equal footing. This problem is claimed to be fixed by multiplying through by $$\beta$$ and rearranging: \begin{align} 0=\left(i\left(\beta \partial^{0}-\beta \alpha^{i} \partial^{i}\right)-m\right) \psi(x).\tag{2} \end{align}

First off, I don't see how (1) does not treat space and time on equal footing. Unlike the Schrodinger equation, (1) has space and time derivatives that are of the same order, so I don't see the problem. I'm also curious about the claim that Hamiltonian formulations generically have similar issues. I can't think of a convincing argument.

The next claim is equally puzzling. What have we actually changed in going from (1) to (2) that remedies the proposed unequal treatment of space and time?

## 1 Answer

The problem is that the matrix $$\beta$$ will mix the components of $$\psi$$ in the $$m\beta\psi$$ term, but not in the terms with $$\partial ^i \psi$$ in equation 1. That will prevent you from putting the components of the 4-derivative operator $$\partial^\mu = (\partial^0, \partial^i)$$ into a Lorentz-invariant product that operates like a scalar (roughly speaking) on $$\psi$$.

Multiplying through by $$\beta$$ eliminates the problem. Since $$\beta^2 = I$$, the matrix is "taken off" of the term proportional to the scalar $$m$$, and it now pre-multiplies all of the components of the derivative operator (space and time) in the same way. The relative sign between the time and space derivatives point to an inner-product-type construction that you could make more explicit if you like, and $$m$$, as noted, is already a scalar. Now you have a "scalar-like" object multiplying your spinor. Again, you could formalize this.