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  Fermionic field configuration from magnetic monopole?

+ 1 like - 0 dislike

In the quantum mechanics problem of a charged particle interacting with a magnetic monopole field, the orbital angular momentum ends up being quantized in units of half integer spin. This appears for instance in Dirac's 1931 paper on the topic.

Using the idea that magnetic charge $g$ and electric charge $e$ obey a quantization condition $$eg = 2\pi n \hbar,$$

we can see that this has a semi-classical analogue in that we can define the angular momentum of the total field of the magnetic monopole at the origin and the point charge at position $\vec{r}$ as $$\vec J_{field}=\int \vec r'\times(\vec E\times \vec B)dV' = -\frac{eg}{4\pi}\hat{r}=-\frac{n}{2}\hbar \hat{r}.$$

So for odd values of $n$ this is quantized in half integers.

So I'm wondering what is going on here in terms of spin and statistics? Can a charged particle and magnetic monopole (both bosons, say) together be treated as a fermionic dyon?

I've read a 1968 paper by Finkelstein and Rubinstein showing how topological defects of bosonic fields could be fermionic, is there a relation to this case?

asked Feb 13, 2018 in Theoretical Physics by octonion (145 points) [ revision history ]
edited Feb 14, 2018 by octonion

What field do you insert in your integral $\vec J_{field}=\int \vec r\times(\vec E\times \vec B)dV$? If it is of a resting electron, then this integral is equal to zero since $\vec B=0$.

It is also unclear why your integral is equal to $-\frac{eg}{4\pi}\hat{r}=-\frac{n}{2}\hbar \hat{r}$. Is it a definition?

B is the magnetic Coulomb field of a magnetic monopole, E is an electric Coulomb field. Evaluate the integral as an exercise and use the quantization condition $eg=2\pi n \hbar$ and you will see the result.

So your magnetic monopole is electrically charged too?

There are two particles, "a charged particle interacting with a magnetic monopole field." I edited the sentence about the fields and put a prime on the variables in the integral so hopefully it's more clear if you want to try it yourself.

Thank you for your explanation (there are two particles). But I feel myself uncomfortable about this integral because it uses the instant values of the fields. I wonder whether these two particles will stay at the same distanse when the time is running. Because in absence of other fields the angular momentum $\vec{J}$ may be a conserved quantity.

They won't stay the same distance. This $J_{field}$ is not separately conserved, the total angular momentum which includes the orbital angular momentum of the point charge about the monopole is conserved. You can see that just by taking the time derivative of $\vec{r}\times(m\vec{v})$ and using the Lorentz force law.

That integral is not something I came up with, by the way. It goes back to JJ Thompson in 1904. I'm guessing that my actual question is also something that has been answered a long time ago too.

I see. You mean something like an atomic configuration. In order to give some meaning to $\vec{J}$, this quantity must appear naturally in some "atomic" calculation. Take as an example a Hydrogen atom in some $|n,l,m\rangle$ state and give us a situation where its electromagnetic field matters. Then we will see some analogy with your "dion".

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