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

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.

• By something missing, do you mean the cross terms we'd naively get by squaring (in the sense of operators) the Dirac equation? Anticommutation relations on matrices prevent such cross terms, which the Klein-Gordon equation isn't meant to have anyway.
– J.G.
Jan 7, 2020 at 14:41
• I am aware of this. The Gamma matrices take care of the unwanted terms. But I feel that factorisation may leave something out from the original equation. I have seen other ways of transforming a linear second order PDE into a system of linear first order PDEs, e.g., math.stackexchange.com/questions/1391707/… where the original equation can be obtained in the first order system without a further application of the operator which is required with the Dirac equation. Jan 7, 2020 at 15:51

## 1 Answer

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$$\partial^\mu\psi=\pi^\mu,\,\partial_\mu\pi^\mu=-m^2\psi.$$

• But we still have second order differentials in the del square. It is not yet first order. Jan 7, 2020 at 16:56
• @Damon You could get around that with a non-standard technique called polymomenta.
– J.G.
Jan 7, 2020 at 17:04
• Thanks. I did a quick search for polymomenta but there is no obvious articles to look at. Some are rather thick and dense. Any suggestion? Jan 7, 2020 at 18:27
• I looked at the approach on core.ac.uk/download/pdf/82716428.pdf, from page 508 onward, but things become rather complicated. There may be another way. Jan 7, 2020 at 18:29
• @Damon I think I had something different in mind; see edit.
– J.G.
Jan 7, 2020 at 18:36