Whew! We spent a considerable amount of wordage developing the Dirac Equation. Now, it’s time to tie this development back to the supersymmetry material we studied earlier in the non-relativistic context. The result will be a surprising mapping between relativistic and non-relativistic quantum mechanics. Today, we’ll just get the gist of it, and to get started, we’ll begin with the final equation we had before,

$$(i\displaystyle{\not} \partial – m)\psi = 0.$$

Recalling Feynman’s notation of slashed quantities,

$$\displaystyle{\not} a = \gamma^\mu a_\mu,$$

we can unpack this a little to

$$\left(i\gamma^\mu\partial_\mu – m\right) \psi = 0,$$

which we can elaborate to include an electromagnetic field as follows:

$${\left[i\gamma^\mu(\partial_\mu + iA_\mu) - m\right] \psi = 0.$$

The Dirac Hamiltonian $$H_D$$ has a rich SUSY structure, of which we can catch a glimpse even having pared the problem down to its barest essentials. To take the simplest possible case, consider a Dirac particle living in one spatial dimension, on which there also lives a scalar potential $$\phi(x^1)$$. (We could call this a “1+1-dimensional” system, to remind ourselves of the difference between time and space.) The SUSY structure can be seen most clearly when we look at the limit of a massless particle; this eliminates the $$m$$ term we had before.

All that fuss over the matrices $$\gamma^\mu$$ has its practical benefit now, because writing our formulas in terms of them allows us to write a new Dirac Equation embodying the analogous physics in a different number of dimensions. Just as our first $$\alpha$$ matrices were defined by the anticommutator relation

$$\anticomm{\alpha_\mu}{\alpha_\nu} = 2\delta_{\mu\nu}\idmat,$$

we now consider the matrices $$\gamma^\mu$$ defined by their anticommutator,

$$\anticomm{\gamma_\mu}{\gamma_\nu} = 2\eta_{\mu\nu}.$$

The only difference between our original, 3+1-dimensional problem and our current one is that indices like $$\mu$$and $$\nu$$ will only run over 0 and 1 instead of running up to 3. A useful choice satisfying the anticommutator constraint is

$$\gamma^0 = \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array}\right),$$

$$\gamma^1 = \left(\begin{array}{cc}i & 0 \\ 0 & -i \\ \end{array}\right).$$

What was a position four-vector is now a “two-vector,” but we must remember that moving an index from up to down involves changing the sign on the zeroth component. (The oneth component remains the same: $$x_1 = x^1$$.) Writing an unadorned $$x$$ to stand for the entire vector, the Dirac Equation is

$$i\gamma^\mu \partial_\mu \psi(x) – \phi(x^0)\psi(x) = 0.$$

As usual, we separate out the time dependence, leaving a wavefunction which depends only on $$x_1$$:

$$\psi(x) = \exp(-i\omega x^0) \psi(x^1) = \exp(i\omega x_0)\psi(x_1).$$

Substituting in this ansatz reduces the 1+1-dimensional Dirac Equation to

$$\gamma^0\omega\psi(x^1) + i\gamma^1 \partial_1 \psi(x^1) – \phi(x^1)\psi(x^1) = 0.$$

If we write the two-component wavefunction $$\psi(x^1)$$ as

$$\psi(x^1) = \left(\begin{array}{c} \xi(x^1) \\ \chi(x^1) \\ \end{array}\right),$$

then the reduced Dirac Equation becomes the two coupled equations

$$A\xi(x^1) = \omega \chi(x^1)$$
$$A^\dag \chi(x^1) = \omega \xi(x^1),$$

where the operators $$A$$ and $$A^\dag$$ are given by

$$A = \partial_1 + \phi(x^1),\ A^\dag = -\partial_1 + \phi(x^1).$$

Decoupling our equations gives a familiar-looking result:

$$A^\dag A \xi(x^1) = \omega^2\xi(x^1),$$

$$AA^\dag \chi(x^1) = \omega^2 \chi(x^1).$$

Our result identifies readily with the SUSY partner potential concepts developed earlier. $$\phi(x^1)$$ is just the superpotential of our old, Schrödinger formalism. What’s more, $$\xi$$ and $$\chi$$ are the eigenfunctions of the two Hamiltonians $$H_1 = A^\dag A$$ and $$H_2 = AA^\dag$$, which we already know are isospectral, except that one has an extra state at zero energy.

Looking back to our results on shape invariance, we see that every Schrödinger problem with a potential $$V_1(x)$$ which has a shape-invariant partner $$V_2(x)$$ can become the scalar potential $$\phi(x^1)$$ in a Dirac problem!

Of course, physical problems using the Dirac Equation exist in more than one dimension, and the electron is not a massless particle. (Well, electrons trapped within materials like sheets of graphene can behave as if they were massless Dirac fermions. . . .) If we soup up our machinery just a little, however, we can apply it to physical situations, such as figuring out what a relativistic charged particle in a Coulomb field will do. This provides the relativistic corrections to our earlier solution of the hydrogenic atom. Without derivation, the result is that for a nucleus of atomic number $$Z$$ and angular momentum $$J$$,

$$E_n = m\left(1 + \frac{Z^2 e^4}{(s + n)^2}\right)^{-1/2},$$

where $$n$$ is a non-negative integer and

$$s = \sqrt{(J+\frac{1}{2})^2 + Z^2 e^4}.$$

This legitimately relativistic result was deduced from the non-relativistic analogous potentials and SUSY. The details can be found in Cooper, Klein and Sukhatme, section 11.2.

It is also interesting to note that in 4D Euclidean space — where the metric $$\eta_{\mu\nu}$$ is just the identity, and we don’t care whether four-vector indices are up or down — the Dirac Hamiltonian always exhibits SUSY structure.

SUSY QM SERIES