I was looking at Dirac's factorization of the Klein Gordon Equation and became inspired to see if there was an alternative way to yield some of its results without resorting to 4x4 matrices. I came up with something interesting and so I wanted to ask whether my interpretation of it is correct.

Derivation of Klein Gordon, You can Skip if Familiar

We can start off with the Schrodinger Equation in momentum space (free potential) as:

$$ i\hbar \frac{\partial \phi(p,t)}{\partial t} = \frac{p^2}{2m} \phi(p,t) $$

Now we define our energy operator as $ \phi \rightarrow i\hbar \frac{\partial \phi}{\partial t} $ and our momentum operator as $\phi \rightarrow p\phi$ .

Recall the relativistic energy formula which is

$$ (KE)^2 = (mc^2)^2 + (pc)^2 $$

So we can assemble Klein Gordon from this as:

$$ -\hbar^2 \frac{\partial^2 \phi}{\partial t^2}= ((mc^2)^2 + (pc)^2)\phi $$

We can re-arrange terms to yield:

$$ \left( (mc^2)^2 + (pc)^2 + \hbar^2 \frac{\partial^2}{\partial t^2} \right) \phi = 0$$

My Idea:

So here is where I want to take a different route than Dirac. Instead of the taking the square root of that operator, I am just as free, to factor this as:

$$ \left( A + B \frac{\partial}{\partial t} \right) \left( C + D \frac{\partial}{\partial t} \right) \phi = 0$$

And the interpretation then is that a relativistic wave function of a single particle satisfies either:

$$ i \hbar \frac{\partial \phi}{\partial t} = - i \hbar \frac{A}{B} \phi $$


$$ i \hbar \frac{\partial \phi}{\partial t} = - i \hbar \frac{C}{D} \phi $$

We then have that:

$$ AC = (mc^2)^2 + (pc)^2, AD + BC = 0, BD = \hbar^2$$

This has solutions:

$$ A = \pm \frac{i \hbar}{D} \sqrt{(mc^2)^2 + (pc)^2} \\ B = \frac{\hbar^2}{D} \\ C = \mp \frac{i D}{\hbar}\sqrt{(mc^2)^2 + (pc)^2} \\ D = D $$

Fixing $D = i\hbar$ we yield a nice equation of the form

$$ i \hbar \frac{\partial \phi}{\partial t} = \mp \sqrt{(mc^2)^2 + (pc)^2} \phi $$


$$ i \hbar \frac{\partial \phi}{\partial t} = \pm \sqrt{(mc^2)^2 + (pc)^2} \phi $$

I feel that these are valid wave equations for a particle, (although the negative energy solutions are a bit odd), but they can be interpreted classically (ex: to find probability of a particle's momentum in some space S, assuming particle's momentum satisfies one of the above, just integrate the absolute value squared of a phi over S).


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