Linearization of the Klein-Gordon equation and decoupling of ''spinors''! [closed]

Question

We know that the K-G equation is deduced from the Einstein 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 Einstein 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 spinors are different.

equation (1) can be written:

$$\left(\partial_{t_{1}}+\vec{\sigma}.\vec{\nabla}_{1}\right)\left(\partial_{t_{2}}-\vec{\sigma}.\vec{\nabla}_{2}\right)\psi_{1}\psi_{2}=\frac{m^{2}}{(i\hbar)^{2}}\psi_{1}\psi_{2}$$ we see that it is the product of two equations : $$\begin{cases}\left(\partial_{t_{1}}+\vec{\sigma}.\vec{\nabla}_{1}\right)\psi_{1}=\frac{m}{i\hbar}\psi_{1} \\ \left(\partial_{t_{2}}-\vec{\sigma}.\vec{\nabla}_{2}\right)\psi_{2}=\frac{m}{i\hbar}\psi_{2}\end{cases}$$

$$\begin{cases}\left(\partial_{t_{1}}+\vec{\sigma}.\vec{\nabla}_{1}+i\frac{m}{\hbar}\right)\psi_{1}=0 \\ \left(\partial_{t_{2}}-\vec{\sigma}.\vec{\nabla}_{2}+i\frac{m}{\hbar}\right)\psi_{2}=0\end{cases}$$ we have two decoupled ''spinors'' and their product is the wave function that verifies the K-G equation.

Is there something that I have missed?

Answer 1

Dirac originally arrived at spin-1/2 field equations by taking "square root" of the wave operator: $$\square=\not{\partial}\cdot\not{\partial}$$where $\not{\partial}=A\partial_x+B\partial_y+C\partial_z+iD\partial_t$ (see this Wikipedia article). After taking square and equating with the wave operator, we observe that the coefficients satisfy the same commutation relation as the Pauli matrices.

A Dirac field is a positive frequency solution for the equation $(i\not{\partial}-m)\psi=0$. From this expression it is not clear how a spin-1/2 wave function $\psi$ is related to a scalar field $\phi$ which satisfies free K.G. field equation. OP claims that it is possible to write a scalar field as product of two spinor like quantities, which requires some explanation. This is not possible in general, for (intuitive)reasons that I have mentioned in the comments. However, if $\psi$ satisfies the Dirac's equation, then it is also a solution for K.G. field equation: $$(i\not{\partial}-m)\psi=0\\ (i\not{\partial}+m)(i\not{\partial}-m)\psi=0\\ (\square+m^2)\psi=0$$which is expected since the primary motivation for solving Dirac equation was to arrive at restrictive class of solution for free K.G. field which could allow for positive definite current density. There is a more general result that any free massless spin-n/2 field in flat space-time will also satisfy the massless K.G. equation (see section 6.8 of Spinors and Space-Time Volume-II). It uses the idea of 2-spinor formalism (see Spinors and Space-Time Volume-I; An Introduction to Twistor Theory - S.A.Huggett & K.P.Tod). In flat space-time, if we are working with Minkowski null tetrad, then the Infeld-van der Waerden symbols are the Pauli matrices. It means that the 2$\times$2 operator: $$\sigma_a\nabla^a={\sigma_a}^{AA'}\nabla^a:\iff\nabla^{AA'}$$We can extend this construction for a curved space-time. It is more intuitive to arrive at tensorial equation starting from the 2-spinor algebra (to get the motivation see my answer in an earlier post here). In the language of 2-spinors, a free z.r.m. spin-n/2 field equation is given by $\nabla^{AA'}\phi_{\underbrace{AB\cdots L}_n}=0$. On contracting with $\nabla_{MA'}$ operator we get (after some manipulation): $$(\square+(n+2)R/12)\phi_{AB\cdots L}=2(n-1){\Psi_{(AB}}^{MN}\phi_{C\cdots L)MN}+2ie{\varphi_A}^M\phi_{MB\cdots L}$$You can find this expression in equation 6.8.35 of Spinors and space-time volume II. For flat space-time and if the spinor field is chargeless, the RHS is zero and the Ricci scalar is zero, hence we have $\square\phi_{AB\cdots L}=0$. Note that this expression is exact for $n>0$ only. However, it is possible to extend this definition to scalar field (n=0) by defining the limit where $n\to 0$ . This limit is well defined provided that $\Psi_{ABCD}=0$ (space-time is patch wise conformally flat) and $e\varphi_{AB}=0$ (no electromagnetic interaction), then we essentially end up with the conformally coupled scalar field equation: $$\left(\square+\frac{R}{6}\right)\phi=0$$