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I previously asked a question about how it is possible for different fields in a gauge theory to have different $U(1)$ charges. I think the issue is that I do not actually know what the $U(1)$ irreps and their associated generators are.

Comparing with the relationship between $SU(n)$ and $su(n)$ as written in the linked question above, are the generators of the different $U(1)$ irreps just different real numbers, given that $e^{ict}$, where $c \in \mathbb{R}$, is an element of a $U(1)$ representation for any real number $t$ as the generator?

tomdodd4598
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1 Answers1

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An irreducible unitary complex $\mathrm{U}(1)$ representation must be 1-dimensional by Schur's lemma, since all $\mathrm{U}(1)$ elements commute with each other and so are multiples of the identity, and each one-dimensional subspace is an invariant subspace of multiples of the identity.

So irreps of $\mathrm{U}(1)$ are given by homomorphisms $\mathrm{U}(1)\to\mathrm{U}(1)$. If we think of $\mathrm{U}(1)$ as the unit circle in the complex plane and parametrize its elements as $\mathrm{e}^{\mathrm{i}\phi}, \phi\in[0,2\pi)$, then all such homomorphisms are given by $$ \rho_n : \mathrm{U}(1)\to\mathrm{U}(1), \mathrm{e}^{\mathrm{i}\phi}\mapsto \mathrm{e}^{\mathrm{i}n\phi},$$ for some $n\in\mathbb{Z}$ (can you figure out what goes wrong for non-integral $n$? Consider what happens at $\rho_n(1) = \rho_n(\mathrm{e}^{\mathrm{i}\pi}\mathrm{e}^{\mathrm{i}\pi}) = \rho_n(\mathrm{e}^{\mathrm{i}\pi})\rho_n(\mathrm{e}^{\mathrm{i}\pi})$). The generator is then $\mathrm{i}n$ and so the irreps of $\mathrm{U}(1)$ are classified by a single integral number $n\in\mathbb{Z}$.

In terms of fields/particles/objects and their charges, we usually pick the smallest ("elementary") charge to be the representation $n=\pm1$ (the sign choice is arbitrary), and then all other stuff transforms in the representation labeled by the multiple of that elementary charge they have.

ACuriousMind
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  • I suppose the problem is that I don't see what goes wrong for non-integral $n$. It might be something subtle or just because I don't know enough group/representation theory. – tomdodd4598 Dec 03 '21 at 23:59
  • @turbodiesel4598 I wrote down an equation that must hold for $\rho_n$ to be a group homomorphism (that I actually had a typo in, corrected now). What happens for non-integral $n$? (Further hint: The l.h.s. is $\rho_n(\mathrm{e}^{\mathrm{i}2\pi}) = \rho_n(1) = 1$) – ACuriousMind Dec 04 '21 at 00:06
  • Ah, I see now, thanks... $e^{i 2 \pi n}$ is only $1$ for integral $n$, and so the definition of a representation only holds for integral $n$. Very interesting! Right, I think I have one more question to post on this topic before I can feel better about it :P – tomdodd4598 Dec 04 '21 at 00:17
  • Never mind - my last question was going to be why charge quantisation is an issue given this result, but there's already some good answers out there. – tomdodd4598 Dec 04 '21 at 01:34