I am wondering if the form of the Dirac equation given in the case of minimal coupling can be "squared" to give back the corresponding Klein-Gordon equation as in normally done in the field-free case. Specifically, I am trying to reproduce the equation, $$(\hat{H} - q\phi)^2 = c^2 (\hat{p}-q\vec{A})^2 + m^2c^4.$$

I am starting from the Dirac equation including minimal coupling motivated by the relativistic Hamiltonian written out as,

$$\hat{H} - q\phi = \vec{\alpha} \cdot c(\hat{p} - q\vec{A} ) + \beta mc^2 $$

where $\vec{\alpha}$ is a vector of matrices with the same size as $\beta$ and $\phi$, $\vec{A}$ are the scalar and vector potentials, respectively. Then, once I square the equation in order to give the conditions on the various $\alpha_j$ and $\beta$ by consistency with the Klein-Gordon equation, I run into a problem where terms such as,

$$\alpha_x \alpha_y (p-qA)_x (p-qA)_y + \alpha_y \alpha_x (p-qA)_y (p-qA)_x$$

do not completely vanish even after enforcing the anticommutator relationships $\{ \alpha_j, \alpha_k\} = 2\delta_{jk}$ of the various $\alpha_j$ (which I motivated by showing that the cross terms in just the momentum must vanish). Based on the terms I have left it seems there may be something that looks sort of like $\nabla \times \vec{A} \equiv \vec{B}$, but this would still not reproduce the Klein-Gordon equation. So my main question is, can we justify the form of the Dirac equation in the case of minimal coupling by reference to the Klein-Gordon equation, or will there simply be new terms that show up? If this is the case, how can we actually justify this form of the Dirac equation? If I have made an error, I am also very open to correction on that. It would be nice to fully motivate this form of the Dirac equation, since it is very easy to show that this equation obeys the same $U(1)$ gauge symmetry as the Schrodinger equation.

EDIT: I found a source (https://www.hep.phy.cam.ac.uk/theory/webber/GFT/gft_handout2_06.pdf) that says that my concerns were valid; squaring the minimally coupled Dirac equation does not reproduce the Klein-Gordon equation with minimal coupling. It seems that the "extra" terms have something to do with the interaction of the fermion spin with the magnetic field, where I have read elsewhere that the Klein-Gordon equation is meant to describe spineless particles.

  • $\begingroup$ If you expand out the terms (using the anticommutator relations for the matrices) you can't seem to get rid of the pieces like $i \hbar q \alpha_x \alpha_x (\partial_x A_y - \partial_y A_x)$. At least, I have no idea how one could presume that to be 0. In other words, the fact that the momentum terms no longer commute under minimal coupling seems to mean that the Klein-Gordon equation is no longer consistent with the Dirac equation. $\endgroup$ Jan 22, 2021 at 16:36
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    $\begingroup$ For an excellent treatment of the squared Dirac equation see Itzykson and Zuber. $\endgroup$
    – my2cts
    Jan 22, 2021 at 20:10
  • $\begingroup$ Thanks for the recommendation! I am worried I may not be able to follow much of that book. I am only a lowly physical chemist and am thus trying to avoid getting into quantum field theory (for now). $\endgroup$ Jan 22, 2021 at 20:33

1 Answer 1


The square of the static Dirac Hamiltonian is something like $$ H^2= -\nabla^2 +m^2+e {\boldsymbol \sigma}\cdot {\bf B} $$ where $\nabla$ is the gauge covariant derivative.

  • $\begingroup$ I have just been reading up on this after realizing that I was on to something with the $\nabla \times \vec{A}$ looking terms! Would it be right to say that the "extra" terms we find in the squared Dirac equation are a result of the Dirac equation describing spin 1/2 fermions and the Klein-Gordon equation describing spineless particles? $\endgroup$ Jan 22, 2021 at 20:26
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    $\begingroup$ Yes. They encode the fact that a Dirac electron has magnetic moment with a $g$-factor of 2, and so has its energy altered in a magnetic field. $\endgroup$
    – mike stone
    Jan 22, 2021 at 20:42
  • $\begingroup$ Wonderful, thank you. I will mark your answer as accepted with that clarification. If I may also ask, is there any (simple) way to justify this, or is it just a mathematical consequence that "falls out" of the way we attempt to derive the Dirac equation with minimal coupling? $\endgroup$ Jan 22, 2021 at 21:47

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