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I was following Nakahara's book Geometry, Topology and Physics, specifically subsection 10.5.2. I do understand how he obtains the equality,

$$ \Delta \varphi = \int \mathrm d\varphi = \int \limits^{2\pi}_{0} 2g \, \mathrm d\phi = 4\pi g$$

and that therefore this has to be a multiple of $2\pi$ for $t_{NS}$ to be uniquely defined. But I don't see directly how, $$\Delta \varphi / 2\pi = 2g \in \mathbb{Z},$$ is equivalent to

$$ \frac{eg}{\hbar} = 2\pi n ~(n\in \mathbb{Z}).$$

Can somebody tell me what I am missing?

JamalS
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Dylan_VM
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2 Answers2

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Please note that the transition function $t_{NS}$ defined on the equator $U_N \cap U_S \approx S^1$ of the bundle $P(S^2, U(1))$ is given by the formula:

$$ t_{NS}(\phi) = \exp[i\varphi(\phi)] \qquad (\varphi:S^1\to \mathbb R) \tag{10.90}$$

and it is periodic in the azimuthal angle $\phi\ $: $\ t_{NS}(\phi +2\pi) \equiv t_{NS}(\phi).$ Observe that the transition function represents a phase in the quantum wavefunction. Since a vector $A_\mu$ couples to an infinitesimal element of the worldline $\textrm dx^\mu$ via terms such as $(1+\frac{ie}\hbar A_\mu\textrm dx^\mu),$ the said phase is expressed as: $t_{NS}=\exp[\frac{ie}\hbar\int A_\mu\textrm dx^\mu].$ For a closed worldline $\Gamma =\partial\Sigma$ representing a full rotation around the equator, this phase is determined by

$$ \Delta\varphi = \frac{e}\hbar \oint_\Gamma A_\mu\textrm dx^\mu = \frac{e}\hbar \int_\Sigma (\vec\nabla \times \vec{A})\cdot \textrm d\vec\Sigma = \frac{e}\hbar \int_\Sigma \vec B\cdot \textrm d\vec\Sigma = \frac{eg}\hbar \,. $$

Since you understand that $\Delta \varphi \in 2\pi \mathbb Z,$ it follows that you also understand $\frac{eg}\hbar \in 2\pi \mathbb Z.$

Nanashi No Gombe
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It is just a matter of convention. Nakahara uses the gauge covariant derivative $D_\mu \phi = ( \partial_\mu + \mathcal{A}_\mu ) \phi$ in the chapter. This is equivalent to the common definition $D_\mu \phi = ( \partial_\mu + i e/\hbar A_\mu ) \phi$ if $$\mathcal{A}_\mu = i e/\hbar A_\mu \,.$$ Nakahara's definition of the magnetic charge $g$ is $$4 \pi i g = \int \mathcal{F}\,.$$ If you define your magnetic charge as $$\tilde g = \int F\,,$$ you find that $g = \frac{e \tilde g}{4 \pi \hbar}$, and $$2g = \frac{e \tilde g}{2 \pi \hbar} \in \mathbb{Z} \,.$$

Friedrich
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