Given a 4-potential $A^\mu=(\phi,\vec{A})$, the Dirac equation with minimal coupling reads $$[i\gamma^\mu(\partial_\mu+ieA_\mu)-m]\psi=0\tag{1}\label{1}$$ By means of a non relativistic reduction, one can prove that in the non relativistic limit we are left with $$\left\{\frac{(\vec{p}-e\vec{A})^2}{2m}-\underbrace{\frac{e}{m}\frac{\vec{\sigma}}{2}}_{\ g=2}\cdot\vec{B}+e\phi\right\}\psi=0\tag{2}\label{2}$$ which is Pauli-Schrödinger equation with the correct gyromagnetic factor.


$$D_\mu:=\partial_\mu+ieA_\mu\qquad\text{and}\qquad \sigma^{{\mu\nu}}:=[\gamma^\mu,\gamma^\nu]$$

As Ref. 1 observes, the non relativistic reduction procedure can also be performed on the squared Dirac equation, the equation we get applying $[i\gamma^\mu(\partial_\mu+ieA_\mu)+m]$ to \eqref{1}, that is

$$[-D_\mu D^\mu-m^2\color{red}{-\frac{e}{2}\sigma^{\mu\nu}F_{\mu\nu}}]\psi=0\tag{3}\label{3}$$ That means that each component of the Dirac spinor satisfies the KG equation with the additional term in red. As pointed out in Ref. 2, such term only appears because we imposed the minimal coupling in \eqref{1} and then we squared. It is straightforward to prove that using the Dirac representation for the $\gamma$ matrices $$\frac{1}{2}\sigma^{\mu\nu}F_{\mu\nu}=(\color{blue}{i\vec{\alpha}\cdot\vec{E}}+\color{green}{\vec{\Sigma}\cdot\vec{B})}\qquad \alpha^i=\gamma^0\gamma^i=\begin{pmatrix}0 &\sigma^i \\ \sigma_i & 0\end{pmatrix}, \Sigma^i=\begin{pmatrix}\sigma^i &0 \\0 & \sigma_i \end{pmatrix}.\tag{4}\label{4}$$ The calculations are quite straightforward, now the problem comes. Then, Ref. 1 and Ref.3 (which is using the chiral representation though) observe that this is a magnetic interaction term, containing the correct (without considering non minimal prescriptions) gyromagnetic factor of Dirac theory $g=2$. I can see why the green piece is a magnetic interaction term and in fact this is the $4\times 4$ version of the term in \eqref{2}. Why is the piece containing the electric - the one in blue - considered as a part of the magnetic dipole term? Also, if I perform the non relativistic reduction in the usual way, like I would do with the KG equation, the electric term will still be there. On the other hand, there is no such term in \eqref{2}, so what does this electric piece represent and why does it only appear if we follow this path?

Addendum: this answer does not have the electric term. I'm not sure this is right, unless they are also assuming the electric field is zero.


  1. Quantum Field Theory, 1987. Itsykson and Zuber. Section $2.2.3$, page 66 eqn $(2.73)$.
  2. QED, Landau&Lifshitz. Section 32, eqn $(32.7a)$.
  3. Quantum Field Theory and the Standard Model, 2013. Matthew D. Schwartz Section 10.4, eqn $(10.4.109)$

1 Answer 1


The coupling to the electric field is there because of Lorentz invariance. A magnetic dipole aquires an electric dipole component when seen from a moving frame. Note that the expectations $\psi^\dagger \alpha_i \psi$ of the $\alpha$ matrices are proportional to the current, and so are effectively zero when the Dirac particle is stationary.

And yes, my old answer had no electric field. My excuse is the "something like".

  • $\begingroup$ For that final part, I'm relieved to have an answer by the author himself. What about the non relativistic limit though? Non relativistic reduction does not kill the other term as I'd expect and I don't remember any particular assumption on the EM field in the case of Schrödinger-Pauli equation. $\endgroup$ Mar 5, 2023 at 16:29
  • 1
    $\begingroup$ Dirac is usualy derived in units where we set $c=1$. I have not done the algebra, but I expect that the $E$ term has a factor of $1/c$ when one uses SI units. If so, it will not be visible in the non-relativistic limit. I always think of the $\alpha$ matrices as being the quantum opertors corresponding to $v/c$ $\endgroup$
    – mike stone
    Mar 5, 2023 at 17:58
  • $\begingroup$ I'm considering the usual trick to make the nR reduction even using natural units. $\Psi(x,t)=e^{-imt}\tilde{\Psi}(x,t)$ and then neglect the less relevant terms... You gave me some great insight about the $\alpha$ matrices with that last sentence but I can't see how it disappears with the non-relativistic limit procedure I am using. It may be one of the neglected terms because it "swaps" the spinors. $\endgroup$ Mar 5, 2023 at 18:08
  • $\begingroup$ Came back to this after some time (Sorry, it slipped my mind). Indeed, one can check esplicitly e.g. in the standard rep - which is more suited for the non relativistic reduction - that $j^\mu=\overline{\Psi}\gamma^\mu\Psi=(\Psi^\dagger\Psi, \Psi^\dagger\alpha^i\Psi)\to(\Psi^\dagger\Psi,0).$ and as you suggested this velocity term is suppressed because only the upper/lower Weyl spinor survives. Since I understood it reading your answer, I'll accept it. $\endgroup$ Jul 13, 2023 at 7:06

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