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?

  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.

  • $\begingroup$ 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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