Consider a spinor field in $3$ dimensions, coupled to a $\mathrm U(1)$ gauge field. The spinor has charge $q\in\mathbb N$ spin $1/2$.

According to arXiv:1712.00020 (cf. the discussion below eq.2.48), the field $\psi$, when in presence of a monopole, has spin $0$ or $1$, as if the monopole were to change the statistics of the field. In this particular paper, the author is discussing the case $q=1$ and a single monopole, but one may consider the more general situation where both $q$ and the monopole number are arbitrary. I expect the statistics of $\psi$ to depend on $q$ because of the so-called spin/charge relation1.

This notion that a monopole may change the spin of a field seems to be ubiquitous in the condensed matter literature, but I haven't been able to find a clear and self-contained explanation. My question is:

What is the spin of a fermion of charge $q$ in the presence of a monopole background of monopole number $n$, and why?

1: "states of odd electric charge have half-integral spin and states of even electric charge have integral spin", from arXiv:1602.04251.


1 Answer 1


See chapter 5 of Erick Weinberg's book "Classical Solutions in QFT". It seems that, if we turn off orbital angular momentum, the possible total angular momenta for a spin $s$ particle with charge $q$ in the presence of a magnetic monopole with charge $g$ are $$ |qg - s|, \, |qg - s| + 1, \,\ldots , \, qg + s \,,$$ where $q$ and $g$ are normalised such that the Dirac quantization condition is $q_\mathrm{min} g_\mathrm{min} = 1/2$. For a spin $s=1/2$ particle with minimal charge in the presence of a minimal monopole, the allowed states are $j = 0$ and $j=1$ as in your paper.

As far as why the fermion gains additional angular momentum in the presence of the monopole, this can be seen classically from the simple fact that the conventional angular momentum $\vec{r}\times\vec{p}$ of a charged particle in a spherically symmetric magnetic field is not conserved. To retain the maxim that "rotational symmetry implies angular momentum conservation", we need to redefine our angular momentum to include an extra piece proportional to the monopole charge.

Quantum mechanically this is a little bit troubling, on account that we appear to be able to generate a fermion by binding together a bosonic charge with a bosonic monopole. Erick Weinberg covers this matter in some detail, although it's somewhat obscure to me still. I don't know if condensed matter physicists lose much sleep over this sort of thing.

  • $\begingroup$ Hi gj255, I'm just checking out Weinberg's book and, although quite nice, I'm having trouble finding the equation $qg - s, \, qg - s + 1, \,\ldots , \, qg + s$ in there. Can you please tell me the specific equation number, or at least the page? Thank you! $\endgroup$ Commented Jul 19, 2018 at 2:48
  • $\begingroup$ @AccidentalFourierTransform You're welcome. That equation is my guess at the correct answer, given the discussion on page 88. I think there is a typo on this page, so be a bit careful. $\endgroup$
    – gj255
    Commented Jul 19, 2018 at 6:54
  • $\begingroup$ Oh, snap! Well, it's a neat formula; I think I know how to derive it myself, but it would be awesome to have an official reference just to be sure. If you happen to know one, please let me know. I'm going to try to find one myself anyway. Cheers! $\endgroup$ Commented Jul 19, 2018 at 16:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.