# What is the spin of an electron in the presence of a monopole?

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.

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.
• 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! – AccidentalFourierTransform Jul 19 '18 at 2:48