A.36 Al­ter­nate Dirac equa­tions

If you look in ad­vanced books on quan­tum me­chan­ics, you will likely find the Dirac equa­tion writ­ten in a dif­fer­ent form than given in chap­ter 12.12.

The Hamil­ton­ian eigen­value prob­lem as given in that sec­tion was

$$\left(\alpha_0 m c^2 + \sum_{i} \alpha_i {\widehat p}_i c\right)\vec\psi = E \vec\psi$$

where $\vec\psi$ was a vec­tor with four com­po­nents.

Now as­sume for a mo­ment that $\psi$ is a state of def­i­nite mo­men­tum. Then the above equa­tion can be rewrit­ten in the form

$$\left(\gamma_0\frac{E}{c} - \sum_{i}\gamma_i p_i\right)\vec\psi = mc\vec\psi$$

The mo­ti­va­tion for do­ing so is that the co­ef­fi­cients of the $\gamma$ ma­tri­ces are the com­po­nents of the rel­a­tivis­tic mo­men­tum four-vec­tor, chap­ter 1.3.1.

It is easy to check that the only dif­fer­ence be­tween the $\alpha$ and $\gamma$ ma­tri­ces is that $\gamma_1$ through $\gamma_3$ get a mi­nus sign in front of their bot­tom el­e­ment. (Just mul­ti­ply the orig­i­nal equa­tion by $\alpha_0^{-1}$$\raisebox{.5pt}{$/$}$​$c$ and re­arrange.)

The par­en­thet­i­cal ex­pres­sion above is es­sen­tially a four-vec­tor dot prod­uct be­tween the gamma ma­tri­ces and the mo­men­tum four-vec­tor. Es­pe­cially if you give the dot prod­uct the wrong sign, as many physi­cists do. In par­tic­u­lar, in the in­dex no­ta­tion of chap­ter 1.2.5, the par­en­thet­i­cal ex­pres­sion is then $\gamma^{\mu}p_\mu$. Feyn­man hit upon the bright idea to in­di­cate dot prod­ucts with $\gamma$ ma­tri­ces by a slash through the name. So you are likely to find the above equa­tion as

$$p\hspace{-5pt}{/}\hspace{.5pt}\vec\psi = mc\vec\psi$$

Isn’t it beau­ti­fully con­cise? Isn’t it com­pletely in­com­pre­hen­si­ble?

Also con­sider the case that $\vec\psi$ is not an en­ergy and mo­men­tum eigen­func­tion. In that case, the equa­tion of in­ter­est is found from the usual quan­tum sub­sti­tu­tions that $E$ be­comes ${\rm i}\hbar\partial$$\raisebox{.5pt}{$/$}$​$\partial{t}$ and ${\skew 4\widehat{\skew{-.5}\vec p}}$ be­comes $\hbar\partial$$\raisebox{.5pt}{$/$}$​${\rm i}\partial{t}$. So the rewrit­ten Dirac equa­tion is then:

$${\rm i}\hbar \left(\gamma_0\frac1c\frac{\partial}{\partial ... ..._i \frac{\partial}{\partial x_i} \right)\vec\psi = mc\vec\psi$$

In in­dex no­ta­tion, the par­en­thet­i­cal ex­pres­sion reads $\gamma^\mu\partial_\mu$. So fol­low­ing Feyn­man

$${\rm i}\hbar \partial\hspace{-6.5pt}{/}\hspace{1.5pt}\vec\psi = mc\vec\psi$$

Now all that the typ­i­cal physics book wants to add to that is a suit­able non-SI sys­tem of units. If you use the elec­tron mass $m$ as your unit of mass in­stead of the kg, $c$ as unit of ve­loc­ity in­stead of m/s, and $\hbar$ as your unit of an­gu­lar mo­men­tum in­stead of kg m$\POW9,{2}$/s, you get

$${\rm i}\partial\hspace{-6.5pt}{/}\hspace{1.5pt}\vec\psi = \vec\psi$$

No out­sider will ever guess what that stands for!