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At some point Dirac writes:

(OpA)(OpB)Y = 0

where OpA and OpB are those two brackets that differ only in the sign of m, then he deduces:

(OpA)Y = 0 OR (OpB)Y = 0

(or is that AND).

I don't get this. If he'd started with:

(OpA)Y . (OpB)Y = 0

then fair enough, but that's not the same as (OpA)(OpB)Y = 0

(OpA)(OpB) is just Klein-Gordon, right, so that can't be what he means.

Perhaps he means that he had the choice of writing:

(OpB)(OpA)Y = 0

and that the whole thing wouldn't be zero unless the result of applying the right hand operator alone was zero, but why should that be?

Why can't the right hand operator applied to Y output something non-zero that just so happens to go to zero under the left hand operator, and vice versa?

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    $\begingroup$ can you LaTeX your equations? e.g. $$\hat A_i Y = 0$$ gives $$\hat A_i Y = 0$$ $\endgroup$ – innisfree Jul 21 '15 at 8:07
  • $\begingroup$ Which reference by Dirac? $\endgroup$ – Qmechanic Jul 21 '15 at 8:11
  • $\begingroup$ @innisfree I assume he means $$\require{cancel}(i\cancel{\partial}-m)$$ and $$(i\cancel{\partial}+m)$$ (As the operators) $\endgroup$ – Omry Jul 21 '15 at 8:41
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    $\begingroup$ Hi Adrian May and welcome to Physics.SE! Please see this help post to learn how to write your equations in a way nicer way i.e. in $\LaTeX$, in order to improve legibility. Thanks! $\endgroup$ – Gonenc Jul 21 '15 at 9:06
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Sorry I didn't get around to learning the latex.

I read this paper (the only one accessible to dummies like me): http://www2.warwick.ac.uk/fac/sci/physics/current/teach/module_home/px435/dirac.pdf

It was a fairly stupid question. The answer is that there are solutions where OpB outputs non-zero and then OpA takes it to zero, and those are just solutions of the KG equation. If OpB outputs zero, we have other solutions which KG didn't find because they didn't know what OpB was.

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