Spin-Statistics Connection for Relativistic Quantum Mechanics

Abstract

The spin-statistics connection has been proved for nonrelativistic quantum mechanics (Jabs in Found Phys 40:776–792, 2010). The proof is extended here to the relativistic regime using the parametrized Dirac equation. A causality condition is not required.

Appendix: Linear Algebra

Given $w_+=\mathsf{w}_+ {\mathbf a}_+=\Lambda _+ w_+$, we seek $w_-$ such that $w_-=\Lambda _- w_-$ and $S(s)w=+(1/2)w$ where $w=w_++w_-$. The required $w_-$ must satisfy

$$\begin{aligned} Q(s)\Lambda _-w_-=-Q(s)w_+, \end{aligned}$$

(19)

which has a solution if and only if $\overline{z}Q(s)w_+=0$ for all $z$ such that $\overline{z}Q(s)\Lambda _-=0$. It suffices to consider a frame in which $s^\mu =(0,0,0,1)$, and so $Q(s)=\mathrm { diag}(0,1,1,0)$. Assuming $p\cdot s= p_3 \ne 0$, it follows that $z_2=z_3=0$ while $z_1$ and $z_4$ are arbitrary. Hence $\overline{z}Q(s)=0$. The solution of (19) for $w_-$ is undetermined up to the addition of $\Lambda _+b$ for any $b$, but $w_-=\Lambda _- w_-$ is uniquely determined.

Consider two free particles with timelike energy-momenta $p^\mu$ and $q^\mu$, where $p \cdot s =p_3 \ne 0$ but $q \cdot s = q_3 =0$. There is a boost $\Omega$ to a new frame where $p'=\Omega p$  and $q'=\Omega q$, with $p_3' \ne 0$ and $q_3' \ne 0$. The common spin axis (0,0,0,1) in the old frame may be replaced with (0,0,0,1) in the new frame. The procedure may be performed any finite number of times for any finite number of particles, with $p_3', q_3',\dots$ remaining bounded away from zero.