Transform the Gordon-Klein equation to a system of first order PDEs but not using Dirac's approach

Question

This is the Gordon-Klein equation: $\frac{\partial^2 \psi }{\partial t^2} - c^2 (\frac{\partial^2 \psi }{\partial x^2} +\frac{\partial^2 \psi }{\partial y^2} + \frac{\partial^2 \psi }{\partial z^2} ) +m^2c^4 \psi =0$ , where the reduced Planck constant is left out for convenience. The Dirac equation is the result of factorising the Gordon-Klein equation into a system of first order PDEs. But I feel that something has been left out of the original Gordon-Klein equation by this factorisation. I wonder if there is an alternative way of expressing the Gordon-Klein equation into another system of first order PDEs which corresponds exactly to the original Gordon-Klein equation.

Answer 1

Set $c=1$. I'll do the real-$\psi$ case; complex $\psi$ is similar.

The Klein-Gordon equation follows from the Lagrangian density$$\mathcal{L}=(\dot{\psi}^2-(\nabla\psi)^2-m^2\psi^2)/2.$$Physicists usually note this has momentum $\pi=\dot{\psi}$, so the Hamiltonian density is$$\mathcal{H}=\dot{\psi}\pi-\mathcal{L}=(\pi^2+(\nabla\psi)^2+m^2\psi^2)/2.$$Hamilton's equations are then$$\dot{\psi}=\pi,\,\dot{\pi}=\nabla^2\psi-m^2\psi.$$We can remove $\nabla^2$ by defining an alternative Hamiltonian density in terms of polymomenta$$\pi^\mu:=\frac{\partial\mathcal{L}}{\partial\partial_\mu\psi}=\partial^\mu\psi$$viz.$$\mathcal{H}=\partial_\mu\psi\cdot\pi^\mu-\mathcal{L}=\frac12\left(\pi_\mu\pi^\mu+m^2\psi^2\right)$$(in $+---$), giving first-order equations$$\partial^\mu\psi=\pi^\mu,\,\partial_\mu\pi^\mu=-m^2\psi.$$