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I read chapter 6.12/6.13 in J. D. Jackson's Classical Electrodynamics about the magnetic monopole and a certain detail is confusing me.

First in a semiclassical consideration of a magnetic monopole, the fact that change of angular momentum must occur in integral multiples of $\hbar$ is used to show that magnetic and electric charge must have discrete values.

Then a simplified discussion is given of Dirac's original argument that leads to the same quantisation condition for electric and magnetic charge. But in this presentation - when I understand it correctly - the quantisation of angular momentum is not used. Instead, single-valuedness of wave functions is used, together with gauge invariance.

So I am not sure, whether quantisation of angular momentum is really needed to find the quantisation condition?

Of course when not, the quantisation condition for electric/magnetic charge would explain the quantisation of angular momentum, wouldn't it?

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  1. Well, the point is that the quantization of angular momentum arises semiclassically from the WKB approximation by imposing that the wave function should be single-valued. Here we use an azimuthal angle and its corresponding angular momentum $(\varphi,L_z)$ as canonical variables for simplicity rather than the 3D position and momentum, cf. the geometric setup mentioned in Jackson. The single-valuedness condition becomes a periodicity condition $$ \psi(\varphi+2\pi)~=~\psi(\varphi).$$ This leads to the Bohr-Sommerfeld quantization rule $$ \oint \!L_z ~\mathrm{d}\varphi~\in~h\mathbb{Z} ,$$ which in turn leads to the quantization condition for the angular momentum $L_z$.

  2. Hence both (i) the above semiclassical argument for angular momentum quantization and (ii) Dirac's magnetic monopole quantization argument rely on the single-valuedness of the wavefunction.

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  • $\begingroup$ Btw, do you think that single-valuedness is essential for modern qm to somehow "explain" quantization vs. just adding it to the mathematical framework (non-commutable operators). $\endgroup$ – Gerard Aug 15 '16 at 11:53
  • $\begingroup$ @Gerard: Good question. Let me record for later that this Phys.SE question can be viewed as a special case of what you ask in above comment. $\endgroup$ – Qmechanic Aug 15 '16 at 15:08

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