language. Direct computation yields the correspondences $$\gamma_t \gamma_a 
  = \pmatrix{-\sigma_a& 0\cr \noalign{\smallskip} 0& \sigma_a\cr}
  \longleftrightarrow
  \sigma_a$$ and $$\gamma_t \longleftrightarrow \sigma_k$$ and of course also $$\gamma_t\gamma_t \longleftrightarrow \sigma_t$$ since the left-hand-side is the $4\times4$ identity matrix and the right-hand-side is the $2\times2$ identity matrix. Direct translation of (\[DiracI\]) now leads to the quaternionic Dirac equation $$\label{DiracIV}
({{\bf p}}- m\sigma_k) (\eta + \sigma_k\theta) = 0$$ Working backwards, we can separate this into an equation not involving $k$, which is precisely (\[DiracIIIb\]), and an equation involving $k$, which is $$\label{DiracV}
{{\bf p}}\, \sigma_k \theta - m \sigma_k \eta = 0$$ Multiplying this equation on the left by $\sigma_k$, and using the remarkable identity $$\label{Remarkable}
\sigma_k \, {{\bf p}}\, \sigma_k = - {\widetilde}{{\bf p}}$$ reduces (\[DiracV\]) to (\[DiracIIIa\]), as expected.

So far, we have done nothing more or less than rewrite the usual momentum-space Dirac equation in 2-component quaternionic language. However, the appearance of the term $m\sigma_k$ suggests a way to put in the mass term on the same footing as the other terms, which we now exploit. Multiplying (\[DiracIV\]) on the left by $-\sigma_k$ and using (\[Remarkable\]) brings the Dirac equation to the form $$\label{DiracVI}
({\widetilde}{{\bf p}}+m\sigma_k) \, \psi = 0$$ where we 