You know how there are no antiparticles for the Schrödinger equation, I've been pushing around the equation and have found a solution that seems to indicate there are - I've probably missed something obvious, so please read on and tell me the error of my ways...
Schrödinger's equation from Princeton Guide to Advanced Physics p200, write $\hbar$ = 1, then for free particle
$$i \psi \frac{\partial T}{\partial t} = \frac{1}{2m}\frac{\partial ^2\psi }{\partial x^2}T$$
rearrange
$$i \frac{1}{T} \frac{\partial T}{\partial t} = \frac{i^2}{2m}\frac{1}{\psi }\frac{\partial ^2\psi }{\partial x^2}$$
this is true iff both sides equal $\alpha$
it can be shown there is a general solution (1)
$$\psi (x,t) \text{:=} \psi (x) e^{-i E t}$$
But if I break time into two sets, past -t and future +t and allow energy to have only negative values for -t, and positive values for +t, then the above general solution can be written as (2)
$$\psi (x,t) \text{:=} \psi (x) e^{-i (-E) (-t)}$$
and it can be seen that (2) is the same as (1), diagrammatically
And now if I describe the time as monotonically decreasing for t < 0, it appears as if matter(read antimatter) is moving backwards in time. Its as if matter and antimatter are created at time zero (read the rest frame) which matches an interpretation of the Dirac equation.
This violates Hamilton's principle that energy can never be negative, however, I think I can get round that by suggesting we never see the negative states, only the consequences of antimatter scattering light which moves forward in time to our frame of reference.
In other words the information from the four-vector of the antiparticle is rotated to our frame of reference.
Now I've never seen this before, so I'm guessing I've missed something obvious - many apologies in advance, I'm not trying to prove something just confused.
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