Linearization of the Klein-Gordon equation and decoupling of ''spinners''!
We know that the K-G equation is deduced from the <PERSON> relation:
$E^{2}=m^{2} +\vec{p}^{2} \;\;\;\;$ (with c=1)
It is known that :$E^{2}=\frac{m^{2}}{1-\beta^{2}}=\left(\frac{m}{1-\beta}\right) \left(\frac{m}{1+\beta}\right) \;\;\;\;,\beta=\frac{v}{c} $
we assume that $$ E^{2}=E_{1}E_{2}=\left(\frac{m}{1-\beta}\right)\left(\frac{m}{1+\beta}\right)$$ and$$\vec{p}^{\;2}=\vec{p}{1}.\vec{p}{2}=\left(\frac{m\vec{v}}{1-\beta}\right). \left(\frac{m\vec{v}}{1+\beta}\right)$$ the <PERSON> equation is written $$E_{1}E_{2}-(\vec{\sigma}.\vec{p}{1})\,(\vec{\sigma}.\vec{p}{2})=m^{2}$$ we can deduce the previous equation from : $$ \left(E_{1}-\vec{\sigma}. \vec{p}{1}\right) \left(E{2}+ \vec{\sigma}.\vec{p}{2}\right)=m^{2}$$ because: $\;\;\vec{\sigma}.\left( E{1} p_{2}-E_{2} p_{1} \right)\vec{n}=0$ and $\vec{\sigma}^{\,2}=1$
by the correspondence equations $E=i\hbar\frac{\partial }{\partial t}\;,\;\vec{p}=-i\hbar\vec{\nabla}$, we have ; $$ (i\hbar)^{2}\left(\partial_{t_{1}}+\vec{\sigma}.\vec{\nabla}{1}\right)\left(\partial{t_{2}}-\vec{\sigma}.\vec{\nabla}{2}\right)\psi{1}\psi_{2}=m^{2}\psi_{1}\psi_{2}\;\;\;\;\;\;\;(1)$$ $$\psi(\vec{r},t)=\psi_{1}(\vec{r}{1},t{1})\psi_{2}(\vec{r}{2},t{2})$$ and $$\partial t^{2}=\frac{\partial\tau ^{2}}{1-\beta^{2}}=\left(\frac{\partial\tau }{1-\beta}\right)\left(\frac{\partial\tau }{1+\beta}\right)=\partial t_{1}\partial t_{2}$$ and $$\vec{\nabla}{1}.\vec{\nabla}{2}=\vec{\nabla}^{\;2}$$ for a massless particle, we have $$\left(\partial_{t_{1}}+\vec{\sigma}.\vec{\nabla}{1}\right)\left(\partial{t_{2}}-\vec{\sigma}.\vec{\nabla}{2}\right)\psi{1}\psi_{2}=0$$ i.e:$$\left(\partial_{t_{1}}+\vec{\sigma}.\vec{\nabla}{1}\right)\psi{1}=0$$ or $$\left(\partial_{t_{2}}-\vec{\sigma}.\vec{\nabla}{2}\right)\psi{2}=0$$ the two equations are similar to the the Weyl equations but the two spinners are different.
equation (1) can be written :
$$\left(\partial_{t_{1}}+\vec{\sigma}.\vec{\nabla}{1}\right)\left(\partial{t_{2}}-\vec{\sigma}. 37