2
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.