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I first recall the Dirac's quantization rule, derived under the hypothesis that there would exit somewhere a magnetic charge: $\frac{gq}{4\pi} = \frac{n\hbar}{2} $ with $n$ natural.

I am wondering how the quantization of electric charge can be deduced from it. The quantization of the product $gq$ is certainly not enough; what else is demanded?

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The quantization of gq is certainly enough, since if you make q small, the unit of g rises to infinity. If there are arbitrarily small charges, the smallest charge monopole is infinitely charged. If there is a fixed charge monopole, the unit of charge is the inverse of the magnetic charge. What more is there to say? –  Ron Maimon May 12 '12 at 17:31
    
Your answer suffices in a universe with only two particles: a magnetic charge $g$ and an electric charge $q$. But if you now consider a bigger universe, with other particles, I have the feeling that yes, there is more to say... –  Isaac May 12 '12 at 20:10
    
@Isaac: Nowhere in my answer is assumed that there are only two types of particles. On the other hand, if we have established charge quantization via two types of particles, having more types of particles around means weakly more restrictions (weakly, because they might not be new restrictions), but at least they cannot undo the previous conditions, only tighten them. So the charge quantization remains in place. –  Qmechanic May 12 '12 at 20:24

1 Answer 1

i) First of all, the Dirac quantization rule

$$\tag{1} \frac{qg}{2\pi\hbar} ~\in~ \mathbb{Z} $$

for magnetic monopoles can be generalized to the Dirac-Zwanziger-Schwinger quantization condition

$$\tag{2} \frac{q_1g_2-q_2g_1}{2\pi\hbar} ~\in~ \mathbb{Z} $$

for dyons. (In a slight misuse of terminology, we shall in the following also include purely electrically charged particles and pure magnetic monopoles into the definition of dyons.)

II) Let $\Gamma=\{(q,g)\}$ denote the set of electric and magnetic charges for dyons. It is natural to think of $\Gamma$ as a subset of the plane $\mathbb{R}^2$. The left-hand side of (2) has a geometric meaning as a signed area spanned by two vectors $(q_1,g_1)$ and $(q_2,g_2)$.

III) Now assume that $\Gamma\backslash\{(0,0)\} $ is non-empty, i.e. there exists a dyon $(q_1,g_1)\neq(0,0)$ to begin with. What points $(q_2,g_2)\in\Gamma$ of $\mathbb{R}^2$ would not conflict with condition (2)? The answer is a set of equidistant discrete lines parallel to the vector $(q_1,g_1)$.

IV) Now assume that $\Gamma$ contains at least two linearly independent vectors $(q_1,g_1)$ and $(q_2,g_2)$. What points $(q_3,g_3)\in\Gamma$ of $\mathbb{R}^2$ would not conflict with condition (2)? The answer is a discrete grid/lattice of intersection points, namely precisely where the corresponding two sets of equidistant discrete parallel lines from section III meet. In other words, the charges are quantized.

V) As a special case, if there exist at least one purely electrically charged particle and at least one pure magnetic monopole, we are in the situation described in section IV, and hence the charges must be quantized.

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