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In John Preskill's review of monopoles he states

Nowadays, we have another way of understanding why electric charge is quantized. Charge is quantized if the electromagnetic U(l)em gauge group is compact. But U(I)em is automatically compact in a unified gauge theory in which U(l)em is embedded in a nonabelian semisimple group. [Note that the standard Weinberg-Salam-Glashow (35) model is not "unified" according to this criterion.]

The implication of the third sentence is that, in some circumstances, the $U(1)_{em}$ gauge group may not be compact. How could this be? Since $U(1)$ as a differentiable manifold is diffeomorphic to $S^1$ isn't it automatically always compact?

The following paragraph:

In other words, in a unified gauge theory, the electric charge operator obeys nontrivial commutation relations with other operators in the theory. Just as the angular momentum algebra requires the eigenvalues of $J_z$ to be integer multiples of $\frac{\hbar}{2}$, the commutation relations satisfied by the electric charge operator require its eigenvalues to be integer multiples of a fundamental unit. This conclusion holds even if the symmetries generated by the charges that fail to commute with electric charge are spontaneously broken.

is OK, but I don't follow what that has to do with the compactness of $U(1)$.

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1 Answer 1

up vote 19 down vote accepted

By the "noncompact $U(1)$ group", we mean a group that is isomorphic to $({\mathbb R},+)$. In other words, the elements of $U(1)$ are formally $\exp(i\phi)$ but the identification $\phi\sim \phi+2\pi k$ isn't imposed. When it's not imposed, it also means that the dual variable ("momentum") to $\phi$, the charge, isn't quantized. One may allow fields with arbitrary continuous charges $Q$ that transform by the factor $\exp(iQ\phi)$.

It's still legitimate to call this a version of a $U(1)$ group because the Lie algebra of the group is still the same, ${\mathfrak u}(1)$.

In the second part of the question, where I am not 100% sure what you don't understand about the quote, you probably want to explain why compactness is related to quantization? It's because the charge $Q$ is what determines how the phase $\phi$ of a complex field is changing under gauge transformations. If we say that the gauge transformation multiplying fields by $\exp(iQ\phi)$ is equivalent for $\phi$ and $\phi+2\pi$, it's equivalent to saying that $Q$ is integer-valued because the identity $\exp(iQ\phi)=\exp(iQ(\phi+2\pi))$ holds iff $Q\in{\mathbb Z}$. It's the same logic as the quantization of momentum on compact spaces or angular momentum from wave functions that depend on the spherical coordinates.

He is explaining that the embedding of the $Q$ into a non-Abelian group pretty much implies that $Q$ is embedded into an $SU(2)$ group inside the non-Abelian group, and then the $Q$ is quantized for the same mathematical reason why $J_z$ is quantized. I would only repeat his explanation because it seems utterly complete and comprehensible to me.

Note that the quantization of $Q$ holds even if the $SU(2)$ is spontaneously broken to a $U(1)$. After all, we see such a thing in the electroweak theory. The group theory still works for the spontaneously broken $SU(2)$ group.

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Ah thanks! your first sentence explains everything. It was the whole idea of using an "unwrapped" $U(1)$ and still calling it $U(1)$ that I wasn't familiar with. When I read $U(1)$ I automatically think of a little circle. But yes, I see the Lie algebra is the same whether or not it's wrapped. –  twistor59 Feb 4 '13 at 10:46
Could you clarify the statement that "$exp(iQ\phi)=exp(iQ(\phi+2\pi))$ holds IFF $Q\in \mathbb{Z}$" Naively I would have thought that $Q\in\mathbb{Z}$ would imply $exp(iQ\phi)=exp(iQ(\phi+2\pi))$ even if $\phi+2\pi$ is not identified with $\phi$. Is a more correct statement something like : if $Q\in \mathbb{Z}$ then $\phi + 2\pi$ and $\phi$ cannot be distinguished using physical processes involving these quantized charges ? –  symanzik138 Jul 21 '14 at 20:16
Sorry, I don't understand what your confusion could possibly be, @symanzik138. Your assertion is also correct when formulated correctly but I probably needed a different, "converse" statement. I said that an operator identity $\exp(2\pi i Q)=1$ holds if the spectrum of $Q$ is a subset of integers. If there is no non-integer, the equation holds, if there is a non-integer, it fails, therefore "iff". It's an equivalence. If there were two correct statements in maths, then they are equally correct and one cannot be "more correct" - at most, it may be "more understandable to you". –  Luboš Motl Jul 22 '14 at 6:37

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