The Dirac Equation

It's time to put everything together. Insert (9) and (10) of §12 into (11) of §12. Take advantage of the block structure by writing the Dirac (4-component) spinor $\Psi$ in terms of two Penrose/Weyl (2-component) spinors $\theta$ and $\eta$ as \begin{equation} \Psi = \begin{pmatrix}\theta\cr \noalign{\smallskip} \eta\end{pmatrix} \end{equation} This yields a particularly nice form of the Dirac equation, namely \begin{equation} \begin{pmatrix} p^0\II-p^a\SIGMA_a& -m\II\cr \noalign{\smallskip} -m\II& p^0\II+p^a\SIGMA_a \end{pmatrix} \begin{pmatrix}\theta\cr \noalign{\smallskip} \eta\end{pmatrix} = 0 \label{DiracII} \end{equation} We can do even better. The $2\times2$ matrix corresponding to the momentum vector $p^\mu$ is just \begin{equation} \PPP = p^\mu \SIGMA_\mu \end{equation} and we introduce the notation \begin{equation} \tilde\PPP = \PPP - \tr(\PPP) I \end{equation} for trace reversal, which reverses the sign of $p^0$. 1) This finally reduces the Dirac equation to the two equations \begin{align} -\tilde\PPP\theta - m\eta &= 0 \label{DiracIIIa}\\ -m\theta + \PPP\eta &= 0 \label{DiracIIIb} \end{align} which no longer contain any gamma matrices! Furthermore, inserting (\ref{DiracIIIb}) into (\ref{DiracIIIa}) yields the constraint \begin{equation} -\tilde\PPP \PPP = m^2 \II \end{equation} or equivalently \begin{equation} p_\mu p^\mu = m^2 \end{equation} which is just the condition that the norm of the momentum vector be the mass.

While all of this works over any of the 4 division algebras, over the complex numbers we recover the usual formalism in 4 spacetime dimensions. In this case, $\theta$ and $\eta$ have 2 complex components. But since \begin{equation} \HH^2 = \CC^2 \oplus \CC^2 \end{equation} we can replace a pair of complex 2-component spinors by a single quaternionic 2-component spinor. We choose the identification \begin{equation} \label{IdentifyC} \begin{pmatrix}A\cr B\cr C\cr D\cr\end{pmatrix} \longleftrightarrow \begin{pmatrix}C - k B\cr \noalign{\smallskip} D + k A\cr\end{pmatrix} \end{equation} with $A,B,C,D\in\CC$. Equivalently, we can write this identification in terms of the Penrose/Weyl spinors $\theta$ and $\eta$ as \begin{equation} \label{IdentifyK} \Psi = \begin{pmatrix}\theta\cr \noalign{\smallskip} \eta\cr\end{pmatrix} \longleftrightarrow \psi = \eta + \SIGMA_k \theta \end{equation} where the generalized Pauli matrix $\SIGMA_k$ is given by \begin{equation} \SIGMA_k = \begin{pmatrix}0& -k\cr \noalign{\smallskip} k& 0\cr\end{pmatrix} \end{equation}

Since (\ref{IdentifyK}) is clearly a vector space isomorphism, there is also an isomorphism relating the linear maps on these spaces. We can use the induced isomorphism to rewrite the (4-dimensional, complex) Dirac equation (11) of §12 in 2-component quaternionic language. Direct computation yields the correspondences \begin{equation} \gamma_0 \gamma_a \longleftrightarrow \SIGMA_a \label{corr1} \end{equation} and \begin{equation} \gamma_0 \longleftrightarrow \SIGMA_k \label{corr2} \end{equation} One way to see that these equivalences make sense is to notice that \begin{equation} \Psi = \begin{pmatrix}\theta\cr \noalign{\smallskip} 0\cr\end{pmatrix} + \gamma_0 \begin{pmatrix}\eta\cr \noalign{\smallskip} 0\cr\end{pmatrix} \end{equation}

Direct translation of (11) of §12 now leads to the quaternionic Dirac equation \begin{equation} \label{DiracIV} (\PPP - m\SIGMA_k) (\eta + \SIGMA_k\theta) = 0 \end{equation} Working backwards, we can separate this into an equation not involving $k$, which is precisely (\ref{DiracIIIb}), and an equation involving $k$, which is \begin{equation} \label{DiracV} \PPP \SIGMA_k \theta - m \SIGMA_k \eta = 0 \end{equation} Multiplying this equation on the left by $\SIGMA_k$, and using the remarkable identity \begin{equation} \label{Remarkable} \SIGMA_k \PPP \SIGMA_k = - \tilde\PPP \end{equation} reduces (\ref{DiracV}) to (\ref{DiracIIIa}), as expected.

So far, all we have done is rewrite the usual (4-dimensional, complex) 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 (\ref{DiracIV}) on the left by $-\SIGMA_k$ and using (\ref{Remarkable}) brings the (4-dimensional) Dirac equation to the form \begin{equation} \label{DiracVI} (\tilde\PPP+m\SIGMA_k) \, (\theta + \SIGMA_k\eta) = 0 \end{equation} The spinor in parentheses is just $\SIGMA_k\psi$, which we will henceforth relabel as simply $\psi$, that is, from now on we write 2) \begin{equation} \psi = \SIGMA_k (\eta + \SIGMA_k\theta) = \theta + \SIGMA_k\eta \label{newpsi} \end{equation} When written out in full, (\ref{DiracVI}) takes the form \begin{equation} \begin{pmatrix} -p^t+p^z& p^x-\ell p^y-km\cr \noalign{\smallskip} p^x+\ell p^y+km& -p^t-p^z \end{pmatrix} \psi = 0 \label{DiracWeyl} \end{equation} This clearly suggests viewing the mass as an additional spacelike component of a higher-dimensional vector. Furthermore, since the matrix multiplying $\psi$ has determinant zero, this higher-dimensional vector is null. We thus appear to have reduced the massive Dirac equation in 4 dimensions to the massless Dirac, or Weyl, equation in higher dimensions, thus putting the massive and massless cases on an equal footing. This expectation is indeed correct, as we will show in subsequent chapters in the more general octonionic setting.

1) We can therefore think of $-\tilde\PPP$ as the 1-form (covariant vector) dual to $\PPP$. This interpretation is strengthened by noting that \begin{align} -\tilde\PPP \PPP = \det(\PPP) I = p_\mu p^\mu I \nonumber \end{align}
2) This also inserts a minus sign into the correspondence (\ref{corr1}), but leaves (\ref{corr2}) intact. We could of course have simply started with (\ref{newpsi}) in (\ref{IdentifyC}), but this makes it slightly more difficulty to obtain (\ref{DiracIV}).

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