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My question is simple: Does not the standard differential wave equation from Maxwell's relations lead to both positive and negative energy solutions for a photon's E field? If so, then why do we always throw away the negative energy solutions? Is this just custom? I suspect it is. But when considering excitations of the vacuum, it appears to me that the negative energy solutions for the photon would be important. Are they thrown away by QED theorists?

Could it be possible to create a zero energy superposition state with positive energy photons superposed to negative energy photons? And would that state not take zero energy to make, by definition? What would the wave function of such a state look like then?

Maybe QED theorists treat negative energy photons as positive energy photons moving backwards in time.

And when dealing with negative energy photons, is there not the apparent necessity of understanding how the E and B fields transform under energy inversion? So, perhaps the E and B fields of positive and negative photons don't superpose in such a simple manner.

I believe this all relates to an important real-world physics problem, and I believe I know some of the answers I am seeking, but because I am an experimentalist, I am seeking advice from a "higher authority."

Thanks for your help.

RoKo