12.12 The Rel­a­tivis­tic Dirac Equa­tion

Rel­a­tiv­ity threw up some road blocks when quan­tum me­chan­ics was first for­mu­lated, es­pe­cially for the par­ti­cles physi­cist wanted to look at most, elec­trons. This sec­tion ex­plains some of the ideas.

You will need a good un­der­stand­ing of lin­ear al­ge­bra to re­ally fol­low the rea­son­ing. A sum­mary of the Dirac equa­tion that is less heavy on the lin­ear al­ge­bra can be found in {A.44}.

For zero spin par­ti­cles, in­clud­ing rel­a­tiv­ity ap­pears to be sim­ple. The clas­si­cal ki­netic en­ergy Hamil­ton­ian for a par­ti­cle in free space,

H = \frac 1{2m}\sum_{i=1}^3 {\widehat p}_i^2
{\widehat p}_i = \frac{\hbar}{{\rm i}}\frac{\partial}{\partial r_i}

can be re­placed by Ein­stein's rel­a­tivis­tic ex­pres­sion

H = \sqrt{\left(m c^2\right)^2 + \sum_{i=1}^3 \left({\widehat p}_ic\right)^2}

where $m$ is the rest mass of the par­ti­cle and $mc^2$ is the en­ergy this mass is equiv­a­lent to. You can again write $H\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E\psi$, or squar­ing the op­er­a­tors in both sides to get rid of the square root:

\left(m c^2\right)^2 + \sum_{i=1}^3 \left({\widehat p}_i c\right)^2

This is the “Klein-Gor­don” rel­a­tivis­tic ver­sion of the Hamil­ton­ian eigen­value prob­lem. With a bit of knowl­edge of par­tial dif­fer­en­tial equa­tions, you can check that the un­steady ver­sion, chap­ter 7.1, obeys the speed of light as the max­i­mum prop­a­ga­tion speed, as you would ex­pect, chap­ter 8.6.

Un­for­tu­nately, throw­ing a dash of spin into this recipe sim­ply does not seem to work in a con­vinc­ing way. Ap­par­ently, that very prob­lem led Schrö­din­ger to limit him­self to the non­rel­a­tivis­tic case. It is hard to for­mu­late sim­ple equa­tions with an ugly square root in your way, and surely, you will agree, the rel­a­tivis­tic equa­tion for some­thing so very fun­da­men­tal as an elec­tron in free space should be sim­ple and beau­ti­ful like other fun­da­men­tal equa­tions in physics. (Can you be more con­cise than $\vec{F}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m\vec{a}$ or $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ $mc^2$?).

So P.A.M. Dirac boldly pro­posed that for a par­ti­cle like an elec­tron, (and other spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ el­e­men­tary par­ti­cles like quarks, it turned out,) the square root pro­duces a sim­ple lin­ear com­bi­na­tion of the in­di­vid­ual square root terms:

\sqrt{\left(m c^2\right)^2 + \sum_{i=1}^3 \left({\widehat p...
= \alpha_0 mc^2 + \sum_{i=1}^3 \alpha_i {\widehat p}_i c %
\end{displaymath} (12.18)

for suit­able co­ef­fi­cients $\alpha_0$, $\alpha_1$, $\alpha_2$ and $\alpha_3$. Now, if you know a lit­tle bit of al­ge­bra, you will quickly rec­og­nize that there is ab­solutely no way this can be true. The teacher will have told you that, say, a func­tion like $\sqrt{x^2+y^2}$ is def­i­nitely not the same as the func­tion $\sqrt{x^2}+\sqrt{y^2}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $x+y$, oth­er­wise the Pythagorean the­o­rem would look a lot dif­fer­ent, and adding co­ef­fi­cients as in $\alpha_1x+\alpha_2y$ does not do any good at all.

But here is the key: while this does not work for plain num­bers, Dirac showed it is pos­si­ble if you are deal­ing with ma­tri­ces, ta­bles of num­bers. In par­tic­u­lar, it works if the co­ef­fi­cients are given by

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

This looks like 2 $\times$ 2 size ma­tri­ces, but ac­tu­ally they are 4 $\times$ 4 ma­tri­ces since all el­e­ments are 2 $\times$ 2 ma­tri­ces them­selves: the ones stand for 2 $\times$ 2 unit ma­tri­ces, the ze­ros for 2 $\times$ 2 zero ma­tri­ces, and the $\sigma_x$, $\sigma_y$ and $\sigma_z$ are the so-called 2 $\times$ 2 Pauli spin ma­tri­ces that also pop up in the the­ory of spin an­gu­lar mo­men­tum, sec­tion 12.10. The square root can­not be elim­i­nated with ma­tri­ces smaller than 4 $\times$ 4 in ac­tual size. (A de­riva­tion is in {D.70}. See also {A.36} for al­ter­nate forms of the equa­tion.)

Now if the Hamil­ton­ian is a 4 $\times$ 4 ma­trix, the wave func­tion at any point must have four com­po­nents. As you might guess from the ap­pear­ance of the spin ma­tri­ces, half of the ex­pla­na­tion of the wave func­tion split­ting into four is the two spin states of the elec­tron. How about the other half? It turns out that the Dirac equa­tion brings with it states of neg­a­tive to­tal en­ergy, in par­tic­u­lar neg­a­tive rest mass en­ergy.

That was of course a cu­ri­ous thing. Con­sider an elec­tron in what oth­er­wise is an empty vac­uum. What pre­vents the elec­tron from spon­ta­neously tran­si­tion­ing to the neg­a­tive rest mass state, re­leas­ing twice its rest mass in en­ergy? Dirac con­cluded that what is called empty vac­uum should in the math­e­mat­ics of quan­tum me­chan­ics be taken to be a state in which all neg­a­tive en­ergy states are al­ready filled with elec­trons. Clearly, that re­quires the Pauli ex­clu­sion prin­ci­ple to be valid for elec­trons, oth­er­wise the elec­tron could still tran­si­tion into such a state. Ac­cord­ing to this idea, na­ture re­ally does not have a free choice in whether to ap­ply the ex­clu­sion prin­ci­ple to elec­trons if it wants to cre­ate a uni­verse as we know it.

But now con­sider the vac­uum with­out the elec­tron. What pre­vents you from adding a big chunk of en­ergy and lift­ing an elec­tron out of a neg­a­tive rest-mass state into a pos­i­tive one? Noth­ing, re­ally. It will pro­duce a nor­mal elec­tron and a place in the vac­uum where an elec­tron is miss­ing, a hole. And here fi­nally Dirac's bold­ness ap­pears to have de­serted him; he shrank from propos­ing that this hole would phys­i­cally show up as the ex­act an­tithe­sis of the elec­tron, its anti-par­ti­cle, the pos­i­tively charged positron. In­stead Dirac weakly pointed the fin­ger at the pro­ton as a pos­si­bil­ity. Pure cow­ardice, he called it later. The positron that his the­ory re­ally pre­dicted was sub­se­quently dis­cov­ered any­way. (It had al­ready been ob­served ear­lier, but was not rec­og­nized.)

The re­verse of the pro­duc­tion of an elec­tron/positron pair is pair an­ni­hi­la­tion, in which a positron and an elec­tron elim­i­nate each other, cre­at­ing two gamma-ray pho­tons. There must be two, be­cause viewed from the com­bined cen­ter of mass, the net mo­men­tum of the pair is zero, and mo­men­tum con­ser­va­tion says it must still be zero af­ter the col­li­sion. A sin­gle pho­ton would have nonzero mo­men­tum, you need two pho­tons com­ing out in op­po­site di­rec­tions. How­ever, pairs can be cre­ated from a sin­gle pho­ton with enough en­ergy if it hap­pens in the vicin­ity of, say, a heavy nu­cleus: a heavy nu­cleus can ab­sorb the mo­men­tum of the pho­ton with­out pick­ing up much ve­loc­ity, so with­out ab­sorb­ing too much of the pho­ton's en­ergy.

The Dirac equa­tion also gives a very ac­cu­rate pre­dic­tion of the mag­netic mo­ment of the elec­tron, sec­tion 13.4, though the quan­tum elec­tro­mag­netic field af­fects the elec­tron and in­tro­duces a cor­rec­tion of about a tenth of a per­cent. But the im­por­tance of the Dirac equa­tion was much more than that: it was the clue to our un­der­stand­ing how quan­tum me­chan­ics can be rec­on­ciled with rel­a­tiv­ity, where par­ti­cles are no longer ab­solute, but can be cre­ated out of noth­ing or de­stroyed ac­cord­ing to the mass-en­ergy re­la­tion $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ $mc^2$, chap­ter 1.1.2.

Dirac was a the­o­ret­i­cal physi­cist at Cam­bridge Uni­ver­sity, but he moved to Florida in his later life to be closer to his el­der daugh­ter, and was a pro­fes­sor of physics at the Florida State Uni­ver­sity when I got there. So it gives me some plea­sure to in­clude the Dirac equa­tion in my text as the cor­ner stone of rel­a­tivis­tic quan­tum me­chan­ics.