A.35 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:

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

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!