When applying the Dirac quantization rule for electric and magnetic charge, I assume one is considering unit electric charges such as electrons. How does the Dirac quantization rule apply for the fractional electric charges of quarks?


The Dirac quantization rule comes from integrating the angular momentum of the superposed electromagnetic field of a charge and a monopole. This angular momentum turns out to be finite and independent of the distance $h$ between the charge and the monopole. The argument then goes that if it's possible to isolate a single fundamental charge $e$ and a single fundamental monopole $g$ in some region of space, then the total angular momentum in that region has to be a multiple of $\hbar$. Here "isolation" means that the distance to any other particle is $\gg h$.

But note that the isolation of the particles is critical. If you put a monopole near a hydrogen atom, the total angular momentum of the electromagnetic field vanishes, because the angular momentum density is of the form $\textbf{E}\times\textbf{B}$, which is bilinear in the fields.

Since quarks are confined, the argument can never be applied to a quark.

  • $\begingroup$ This is a good point. Nonetheless, the argument does apply to QFTs where the number of flavors is high enough that confinement doesn't happen. $\endgroup$ – user1504 Jun 12 '13 at 14:55
  • $\begingroup$ What if you raised the temperature above the Hagedorn temperature? Then quarks would deconfine, and you could transport a single quark around the Dirac string. $\endgroup$ – tparker Jul 14 '16 at 18:59
  • $\begingroup$ Your argument doesn't apply in the case where you move a quark around a monopole in a circle smaller than the confinement radius, or above the Hagedorn radius. See physics.stackexchange.com/questions/268709/… for the complete answer. $\endgroup$ – tparker Jul 26 '16 at 0:16

The logic is the same: If $q$ is an electric charge and $g$ is a magnetic charge, one must have $gq \in 2\pi\mathbb{Z}$ (in lazy theorist units). So if there is a largest magnetic charge, then there is necessarily a smallest electric charge. You don't have to assume that you are dealing with unit charges to make this argument.


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