# Are $\mathrm U(1)$ charges quantised?

Consider a theory with a $\mathrm U(1)$ symmetry, i.e., such that there exists a unitary operator $U\equiv\mathrm e^{iQ}$ that commutes with the $S$ matrix (or the Hamiltonian). The hermitian operator $Q$ represents a conserved charge, which we may refer to as the electric charge (or baryonic charge, etc.).

The states of the theory are classified according to unitary representations of the symmetry group, which in this case contains a $\mathrm U(1)$ factor. Now, the unitary representations of this group are of the form $z\mapsto z^n$ with $z\in\mathbb C-\{0\}$ and $n\in\mathbb Z$, which means that states are labelled according to $$Q|n,\dots\rangle=n|n,\dots\rangle$$ where "$\dots$" refers to other labels. From this I would conclude that electric charge, or any other $\mathrm U(1)$ charge, is always quantised. There exists a minimal charge, say $q$, such that the charge of any other state is $nq$ for some $n\in\mathbb Z$. This seems to be in agreement with what we observe experimentally.

Now comes my question: I would have expected that we should allow for projective representations rather than regular ones. This means that we may now allow $z\mapsto z^n$ with $n\in\mathbb R$, i.e., the quantum number is no longer quantised. Any charge should be observed rather than only those a scalar multiple of some minimal one. This does not seem to agree with what we observe experimentally. Why is this? Why must we only consider regular representations rather than projective ones? Should or should not $\mathrm U(1)$ charges be quantised?

• U(1) symmetry does not explain quantization of charge. Commented Aug 23, 2017 at 14:55
• @Thomas thank you for your comment. I guess that the answer to the title is nevative then: $\mathrm U(1)$ charges are not quantised, right? Is it because the representations are projective, as I said in the OP, or for some other reason? Commented Aug 23, 2017 at 14:59
• In nature charge is obviously quantized. In the SM there is no real argument for charge quantization, but the charge assignments are restricted by anomaly cancellation. In GUTs you can get charge quantization by embedding the U(1) into a larger group. Commented Aug 23, 2017 at 17:01

The unitary operator $U~=~e^{i\theta}$ has as its general argument a phase. The angle $\theta$ is not directly the charge.

Let us think of the Aharonov-Bohm effect. This is a phase induced on a charged particle that passes over a very long solenoid with a magnetic field $\vec B$ inside. The phase induced is $$\psi~\rightarrow~e^{ie/\hbar\oint{\vec A}\cdot d\vec x}$$ where the loop integration is around the solenoid. Now Stoke's rule tells us that $$e/\hbar\oint{\vec A}\cdot d\vec x~=~e/\hbar\iint\nabla\times{\vec A}\cdot d\vec a ~=~-e/\hbar\iint{\vec B}\cdot d\vec a.$$ The vector $\vec a$ is a normal pointing out of the opening of the solenoid and the last integral is then a Gauss' law result that gives the effective magnetic monopole charge $g$. To eliminate the influence of the solenoid, or Dirac string, we sent this phase to $\theta~=~-2\pi N$ so that $$eg~=~2\pi N\hbar.$$ This is a Bohr-Sommerfeld quantization condition on the electric and magnetic charge.

This is the only charge quantization predicted, which suggests there should be a magnetic charge. There are some field theories with this magnetic charge, which induces a topology on the theory. Inflationary cosmology damps the occurrence of magnetic monopole charges so while they might still exists they are extremely sparse.

• Thanks! I've accepted D.'s answer because it was more along the lines of what I was looking for, i.e., QFT rather than non-relativistic QM. Commented Aug 23, 2017 at 16:47

For any Lie group $$G$$, the projective unitary representations of $$G$$ correspond to unitary representations of its universal cover $$\widetilde{G}$$. The universal cover of $$\mathrm{U}(1)$$ is the real line $$\mathbb{R}$$. The unitary representations of $$\mathbb{R}$$ are all direct sums of irreducible representations of this form:

$$\rho(x) = \exp(iqx) \qquad x \in \mathbb{R}$$

where the charge $$q$$ is an arbitrary real number.

So yes: if we allow projective unitary representations of $$\mathrm{U}(1)$$ then the charge does not need to be quantized.

Nice question, AccidentalFourierTransform!

To be specific, I'll consider here the case of the electric charge.

There are two models which can explain the quantization of the electric charge.

1. When the electric charge generator is one of the generators of a broken large simple group, (such as in the Georgi–Glashow model) then the electric charge is represented by a matrix acting on a vector wave function. This transformation is not projective, thus in order to have a proper action, the charge needs to be quantized.

2. In the Kaluza-Klein scenario. here the charge is actually the velocity in the fifth dimension. If the fifth dimension is a circle, then a quantization of the electric charge will take place similar to the quantization of the momentum of a particle moving on a circle.