- #1
tommy01
- 40
- 0
Hi together ...
I wonder how one can deduce the existence of antiparticles from the Klein-Gordon equation.
Starting from [tex](\frac{\partial^2}{\partial t^2} - \nabla^2 + m^2) \Psi(t,\vec{x})=0[/tex]
one gets solutions [tex]\Psi(t,\vec{x})=\exp(\pm i (- E t + \vec{p} \cdot \vec{x}))[/tex] leading to [tex]E^2=p^2 + m^2 [/tex]. You need the plus/minus in the exponential to get a complete system of solution functions but why you can't just ignore the negative root of the enegry-momentum relation and why you interpret the exp(- ...) as negative momentum and negative energy? why you can't just say the energy is always positive but the functional dependence is exp(+ ...) respectively exp(- ...)?
thanks and merry christmas.
I wonder how one can deduce the existence of antiparticles from the Klein-Gordon equation.
Starting from [tex](\frac{\partial^2}{\partial t^2} - \nabla^2 + m^2) \Psi(t,\vec{x})=0[/tex]
one gets solutions [tex]\Psi(t,\vec{x})=\exp(\pm i (- E t + \vec{p} \cdot \vec{x}))[/tex] leading to [tex]E^2=p^2 + m^2 [/tex]. You need the plus/minus in the exponential to get a complete system of solution functions but why you can't just ignore the negative root of the enegry-momentum relation and why you interpret the exp(- ...) as negative momentum and negative energy? why you can't just say the energy is always positive but the functional dependence is exp(+ ...) respectively exp(- ...)?
thanks and merry christmas.