I read an unjustified treatment in a book, saying that in QED charge an not quantized by the gauge symmetry principle (which totally clear for me: Q the generator of $U(1)$ can be anything in $\mathbb{R}$) but for non-Abelian gauge symmetries the "charge" are quantized by virtue of this principle. Could someone give a hint (or reference) of the calculation showing that.

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    $\begingroup$ Huh, interesting question. I can't think of a reason offhand why non-Abelian-ness specifically would lead to quantization of the charge, but I will be interested to see what people come up with. $\endgroup$
    – David Z
    Feb 17, 2012 at 1:24
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    $\begingroup$ It is the property of non-Abelian gauge invariance that all matter fields interact with the gauge field with the same coupling (or couplings in case of a non-simple gauge group). Otherwise you cannot make the Lagrangian gauge invariant. This is the same as saying that the charges of all fields are determined solely by the structure of the gauge group and its representations up to an overall scale, which can be absorbed in the redefinition of the coupling. $\endgroup$ Feb 17, 2012 at 14:49
  • $\begingroup$ Ok thanks for the answer. I had already seen that, like the q + g -> q + g amplitude that requires the g of the fermion-gauge coupling to be the same as the g in the $A^{\mu}A^{\nu}\partial_{\mu}A_{\nu}$ but I was thinking about a quantization in the same way as magnetic monopoles quantize the electrical charge. $\endgroup$
    – toot
    Feb 17, 2012 at 17:45
  • $\begingroup$ This ref defines ladder operators based on the gauge generators which raise and lower isospin. I suspect that this works out similar to angular momentum ladder operators in elementary QM, where they lead to quantization of angular momentum. In the abelian case, the commutators would vanish so this would be doomed to fail. (Not posting as an answer because I'm not sure it's right, but it's something to look at...) $\endgroup$
    – twistor59
    Feb 17, 2012 at 18:02

2 Answers 2


First of all, note that the real Abelian Lie group $U(1)$ comes in two (multiplicatively written) versions:

  1. Compact $U(1)~\cong~e^{i\mathbb{R}}~\cong~S^1$, and

  2. Non-compact $U(1)$ $~\cong~e^{\mathbb{R}}\cong~\mathbb{R}_+\backslash\{0\}$.

Also note that in the physics literature, we often identifies charge operators with Lie algebra generators for a Cartan subalgebra (CSA) of the gauge Lie algebra.

Moreover, note that the choice of CSA generators is not unique, see also this answer. The ambiguity in the convention choice of charge operators is similar to the ambiguity in the convention choice of spin operators, see also this question. We shall from now on assume that we consistently stick to only one such possible convention.

Given a Lie algebra representation, the eigenvalues of the charge operator are called charges.

Now let us briefly sketch some lore and facts related to OP's question (v2).

  1. We observe in Nature that Abelian and non-Abelian charges are quantized, as accurately described by electric charge, electroweak hypercharge, electroweak isospin and color charges in the $U(1)\times SU(2)\times SU(3)~$ standard model.

  2. If there exist dual magnetic monopoles, then quantum theory provides a natural explanation for charge quantization. Namely, by playing with Wilson lines, the singlevaluedness of the wavefunction requires that charges are quantized (i.e., to take only discrete values), and that the gauge group is compact, as first explained by Dirac.

  3. It is a standard result in representation theory, that for a finite-dimensional representation of a compact Lie group, that the charges (i.e., the eigenvalues of the CSA generators) take values in a discrete weight lattice.

  4. If a gauge group contains both a compact and a non-compact direction, i.e. if its bilinear form$^1$ has indefinite signature, it is impossible to define a non-trivial positive-norm Hilbert subspace of physical, propagating, $A^a_{\mu}$ gauge field states.


$^1$ By a bilinear form is here meant a non-degenerate invariant/associative bilinear form on the Lie algebra. For a semisimple Lie algebra, we can use the Killing form.

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    $\begingroup$ @Qmechanics Can you please explain or refer to a reference about why charge operators are identified with Cartan generator? $\endgroup$
    – QGravity
    Feb 28, 2018 at 20:44
  • $\begingroup$ @Qmechanic In the Gell-mann basis of $\mathfrak{su(3)}$ in what sense are $\lambda_3$ and $\lambda_8$ charge generators of color? Can you expand on this? $\endgroup$
    – Craig
    Dec 19, 2018 at 4:06
  • $\begingroup$ Doesn't second quantisation already explain charge quantisation? We start with the conserved Noether charge of the classical field, and then second quantisation of the field makes the eigenvalue of this operator discrete. Then why do we need magnetic monopoles to explain the same thing? $\endgroup$
    – Ryder Rude
    Jul 26, 2022 at 5:37
  • $\begingroup$ Discreteness of charges does not imply that charges belongs to a lattice. $\endgroup$
    – Qmechanic
    Jul 26, 2022 at 6:00

The situation is similar with SU(2) spin quantization. Generators of the SU(2) are quantized, while U(1) this is not the case. Spin is quantized in 3D space, but in a 2D space it is continuous real number, with fractional quantum statistics intermediate between boson and fermion.

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    $\begingroup$ Interesting point related to braid groups and projective representations. $Spin(2)$ is here the compact $U(1)$. It is my understanding that so far, experimentally speaking, anyons in $2+1$ dimensional systems do not carry continuous, but only discrete fractional spin, which are therefore still quantized. $\endgroup$
    – Qmechanic
    Mar 3, 2012 at 19:37

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