D.70 Emer­gence of spin from rel­a­tiv­ity

This note will give a (rel­a­tively) sim­ple de­riva­tion of the Dirac equa­tion to show how rel­a­tiv­ity nat­u­rally gives rise to spin. The equa­tion will be de­rived with­out ever men­tion­ing the word spin while do­ing it, just to prove it can be done. Only Dirac’s as­sump­tion that Ein­stein's square root dis­ap­pears,

\begin{displaymath}
\sqrt{\left(m c^2\right)^2 + \sum_{i=1}^3 \left({\widehat p...
...^2}
= \alpha_0 mc^2 + \sum_{i=1}^3 \alpha_i {\widehat p}_i c,
\end{displaymath}

will be used and a few other as­sump­tions that have noth­ing to do with spin.

The con­di­tions on the co­ef­fi­cient ma­tri­ces $\alpha_i$ for the lin­ear com­bi­na­tion to equal the square root can be found by squar­ing both sides in the equa­tion above and then com­par­ing sides. They turn out to be:

\begin{displaymath}
\alpha_i^2 = 1 \mbox{ for every $i$}
\qquad
\alpha_i\alpha_j+\alpha_j\alpha_i = 0
\mbox{ for $i\ne j$} %
\end{displaymath} (D.45)

Now as­sume that the ma­tri­ces $\alpha_i$ are Her­mit­ian, as ap­pro­pri­ate for mea­sur­able en­er­gies, and choose to de­scribe the wave func­tion vec­tor in terms of the eigen­vec­tors of ma­trix $\alpha_0$. Un­der those con­di­tions $\alpha_0$ will be a di­ag­o­nal ma­trix, and its di­ag­o­nal el­e­ments must be $\pm1$ for its square to be the unit ma­trix. So, choos­ing the or­der of the eigen­vec­tors suit­ably,

\begin{displaymath}
\alpha_0=\left(\begin{array}{cc} 1 & 0 \ 0 & -1 \end{array}\right)
\end{displaymath}

where the sizes of the pos­i­tive and neg­a­tive unit ma­tri­ces in $\alpha_0$ are still un­de­cided; one of the two could in prin­ci­ple be of zero size.

How­ever, since $\alpha_0\alpha_i+\alpha_i\alpha_0$ must be zero for the three other Her­mit­ian $\alpha_i$ ma­tri­ces, it is seen from mul­ti­ply­ing that out that they must be of the form

\begin{displaymath}
\alpha_1 =
\left(\begin{array}{cc}0&\sigma^\dagger_1\\ \si...
...n{array}{cc}0&\sigma^\dagger_3\\ \sigma_3&0\end{array}\right).
\end{displaymath}

The $\sigma_i$ ma­tri­ces, what­ever they are, must be square in size or the $\alpha_i$ ma­tri­ces would be sin­gu­lar and could not square to one. This then im­plies that the pos­i­tive and neg­a­tive unit ma­tri­ces in $\alpha_0$ must be the same size.

Now try to sat­isfy the re­main­ing con­di­tions on $\alpha_1$, $\alpha_2$, and $\alpha_3$ us­ing just com­plex num­bers, rather than ma­tri­ces, for the $\sigma_i$. By mul­ti­ply­ing out the con­di­tions (D.45), you see that

\begin{displaymath}
\alpha_i \alpha_i = 1
\Longrightarrow \sigma_i^\dagger\sigma_i = \sigma_i\sigma_i^\dagger = 1
\end{displaymath}


\begin{displaymath}
\alpha_i \alpha_j + \alpha_j \alpha_i = 0
\Longrightarrow
...
..._i =
\sigma_i\sigma_j^\dagger + \sigma_j\sigma_i^\dagger = 0.
\end{displaymath}

The first con­di­tion above would re­quire each $\sigma_i$ to be a num­ber of mag­ni­tude one, in other words, a num­ber that can be writ­ten as $e^{{\rm i}\phi_i}$ for some real an­gle $\phi_i$. The sec­ond con­di­tion is then ac­cord­ing to the Euler for­mula (2.5) equiv­a­lent to the re­quire­ment that

\begin{displaymath}
\cos\left(\phi_i - \phi_j\right) = 0 \mbox{ for $i\ne j$};
\end{displaymath}

this im­plies that all three an­gles would have to be 90 de­grees apart. That is im­pos­si­ble: if $\phi_2$ and $\phi_3$ are each 90 de­grees apart from $\phi_1$, then $\phi_2$ and $\phi_3$ are ei­ther the same or apart by 180 de­grees; not by 90 de­grees.

It fol­lows that the com­po­nents $\sigma_i$ can­not be num­bers, and must be ma­tri­ces too. As­sume, rea­son­ably, that they cor­re­spond to some mea­sur­able quan­tity and are Her­mit­ian. In that case the con­di­tions above on the $\sigma_i$ are the same as those on the $\alpha_i$, with one crit­i­cal dif­fer­ence: there are only three $\sigma_i$ ma­tri­ces, not four. And so the analy­sis re­peats.

Choose to de­scribe the wave func­tion in terms of the eigen­vec­tors of the $\sigma_3$ ma­trix; this does not con­flict with the ear­lier choice since all half wave func­tion vec­tors are eigen­vec­tors of the pos­i­tive and neg­a­tive unit ma­tri­ces in $\alpha_0$. So you have

\begin{displaymath}
\sigma_3=\left(\begin{array}{cc} 1 & 0 \ 0 & -1 \end{array}\right)
\end{displaymath}

and the other two ma­tri­ces must then be of the form

\begin{displaymath}
\sigma_1=\left(\begin{array}{cc}0&\tau^\dagger_1\\ \tau_1&0...
...\begin{array}{cc}0&\tau^\dagger_2\\ \tau_2&0\end{array}\right)
\end{displaymath}

But now the com­po­nents $\tau_1$ and $\tau_2$ can in­deed be just com­plex num­bers, since there are only two, and two an­gles can be apart by 90 de­grees. You can take $\tau_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $e^{{\rm i}\phi_1}$ and then $\tau_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $e^{{\rm i}(\phi_1+\pi/2)}$ or $e^{{\rm i}(\phi_1-\pi/2)}$. The ex­is­tence of two pos­si­bil­i­ties for $\tau_2$ im­plies that on the wave func­tion level, na­ture is not mir­ror sym­met­ric; mo­men­tum in the pos­i­tive $y$-​di­rec­tion in­ter­acts dif­fer­ently with the $x$ and $z$ mo­menta than in the op­po­site di­rec­tion. Since the ob­serv­able ef­fects are mir­ror sym­met­ric, do not worry about it and just take the first pos­si­bil­ity.

So, the goal of find­ing a for­mu­la­tion in which Ein­stein's square root falls apart has been achieved. How­ever, you can clean up some more, by re­defin­ing the value of $\tau_1$ away. If the four-di­men­sion­al wave func­tion vec­tor takes the form $(a_1,a_2,a_3,a_4)$, de­fine $\bar{a}_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $e^{{\rm i}\phi_1/2}{a}_1$, $\bar{a}_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $e^{-{\rm i}\phi_1/2}a_2$ and sim­i­lar for $\bar{a}_3$ and $\bar{a}_4$.

In that case, the fi­nal cleaned-up $\sigma$ ma­tri­ces are

\begin{displaymath}
\sigma_3 =
\left(\begin{array}{rr} 1 & 0\ 0 & -1\end{arra...
...(\begin{array}{rr} 0 & -{\rm i}\ {\rm i}& 0\end{array}\right)
\end{displaymath} (D.46)

The s word has not been men­tioned even once in this de­riva­tion. So, now please ex­press au­di­ble sur­prise that the $\sigma_i$ ma­tri­ces turn out to be the Pauli (it can now be said) spin ma­tri­ces of chap­ter 12.10.

But there is more. Sup­pose you de­fine a new co­or­di­nate sys­tem ro­tated 90 de­grees around the $z$-​axis. This turns the old $y$-​axis into a new $x$-​axis. Since $\tau_2$ has an ad­di­tional fac­tor $e^{{\rm i}\pi/2}$, to get the nor­mal­ized co­ef­fi­cients, you must in­clude an ad­di­tional fac­tor $e^{{\rm i}\pi/4}$ in $\bar{a}_1$, which by the fun­da­men­tal de­f­i­n­i­tion of an­gu­lar mo­men­tum dis­cussed in ad­den­dum {A.19} means that it de­scribes a state with an­gu­lar mo­men­tum $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\hbar$. Sim­i­larly $a_3$ cor­re­sponds to a state with an­gu­lar mo­men­tum $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\hbar$ and $a_2$ and $a_4$ to ones with $-\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\hbar$.

For nonzero mo­men­tum, the rel­a­tivis­tic evo­lu­tion of spin and mo­men­tum be­comes cou­pled. But still, if you look at the eigen­states of pos­i­tive en­ergy, they take the form:

\begin{displaymath}
\left(
\begin{array}{c}
\vec a\\
\varepsilon ({\skew0\vec p}\cdot\vec\sigma) \vec a
\end{array} \right)
\end{displaymath}

where $\varepsilon$ is a small num­ber in the non­rel­a­tivis­tic limit and $\vec{a}$ is the two-com­po­nent vec­tor $(a_1,a_2)$. The op­er­a­tor cor­re­spond­ing to ro­ta­tion of the co­or­di­nate sys­tem around the mo­men­tum vec­tor com­mutes with ${\skew0\vec p}\cdot\vec\sigma$, hence the en­tire four-di­men­sion­al vec­tor trans­forms as a com­bi­na­tion of a spin $\frac12\hbar$ state and a spin $-\frac12\hbar$ state for ro­ta­tion around the mo­men­tum vec­tor.