Solution to the Klein Gordon Equation

benk99nenm312
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Hey guys, I was reading up on the Klein Gordon equation and I came across an article that gave a general solution as: \psi(r,t)= e^i(kr-\omegat), under the constraint that -k^2 + \omega^2/c^2 = m^2c^2/\hbar^2, forgive my lack of latex hah.

Through Euler's law this does give a solution tantamount to cos(kr-\omegat)+isin(kr-\omegat).

My question is simply.. is this valid? I ask because if you were to integrate the square over an interval you should get a probability, however the imaginary term will carry through from the de Moivre formula. I'm terribly confused.

Thanks guys!
 
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hey benk99nenm312! :smile:
benk99nenm312 said:
… if you were to integrate the square over an interval you should get a probability, however the imaginary term will carry through from the de Moivre formula. I'm terribly confused.

no, the probability is ψ*ψ, not ψ2 :wink:
 
tiny-tim said:
hey benk99nenm312! :smile:


no, the probability is ψ*ψ, not ψ2 :wink:

Omg wowww, lol. Thank you hah.
 
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