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I have read these questions:

Are contravariant basis vectors and basis 1-forms identical?

Where John Rennie's answer says that electrons do have an electric dipole moment and we imagine that in math with a cloud of virtual particles.

Now that cloud has a shape and that is what he is referring to to be spherical.

Now the electron has a magnetic dipole moment too, and it is related to the spin.

And this question:

How do you fit a dipole in an electron?

Classically a non-pointlike spinning charged object possesses a magnetic dipole moment due to the fact that charged particles in the object are spinning around some axis. In contrast, the electron has a dipole moment that arises from its intrinsic spin angular momentum. As you point out, the electron has no internal structure, so the spin does not refer to actual physical spinning. The dipole moment has spatial dimensions outside of the point where the electron exists because it arises from the quantum spin property of the electron, it's not itself a property of the electron.

Now none of these talk about how the electron can have a dipole moment, when it is a point particle.

Usually a permanent magnet has two poles.

The word dipole comes from the words two poles. Magnets do have two poles, we can call them north and south, but there is no magnetic monopole.

Now is that because like the electric dipole is in math made up of a cloud of virtual particles, and the magnetic dipole is made up of a cloud of virtual particles? And do those virtual particles have two poles, north and south?

Question:

  1. Now how can a point particle, that has no spatial extension have two poles, a north and south, at the same point, and still have magnetic dipole moment?

  2. Is the magnetic dipole moment the same way in math made up of a cloud of virtual particles (like the electric dipole moment)?

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    $\begingroup$ think of it as going to the limit of d-->0 where d is the length of the dipole. It is allowed mathematically, and it is the same problem as with the electron spin, as nothing is spinning if you have a point. It is allowed to go to a point, and particle physics uses it as a postulate, that the particles in the standard model table are points with no extent $\endgroup$ – anna v Jul 19 '18 at 9:11
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See Wikipedia's "Neutron magnetic moment" section: "Anomalous magnetic moments and meson physics" where electron magnetic moment is explained better than it is on their: "Magnetic moment of an electron" page:

The physical picture is that the effective magnetic moment of the electron results from the contributions of the "bare" electron, which is the Dirac particle, and the cloud of "virtual," short-lived electron–positron pairs and photons that surround this particle as a consequence of QED. The small effects of these quantum mechanical fluctuations can be theoretically computed using Feynman diagrams with loops.$^{[51]}$

[51] "Revised value of the eighth-order QED contribution to the anomalous magnetic moment of the electron" (Feb 20 2008), by T. Aoyama, M. Hayakawa, T. Kinoshita, and M. Nio:

See Page 6: "The amplitude of the magnetic moment contribution of a diagram is obtained by applying Feynman-Dyson rules of QED in the momentum space." and appendix B on page 38.

See also page 223: "Concepts of Modern Physics" (.PDF) Sixth Edition by Arthur Beise:

The magnetic moment of the orbital electron in a hydrogen atom depends on its angular momentum L. Hence both the magnitude of L and its orientation with respect to the field determine the extent of the magnetic contribution to the total energy of the atom when it is in a magnetic field. The magnetic moment of a current loop has the magnitude

$$\mu = IA$$

where $I$ is the current and $A$ the area it encloses. An electron that makes $\mathcal f$ rev/s in a circular orbit of radius $r$ is equivalent to a current of $- e \mathcal f$ (since the electronic charge is $-e$), and its magnetic moment is therefore

$$\mu = -e \mathcal f \pi r^2$$

Because the linear speed $v$ of the electron is $2 \pi \mathcal f r$ its angular momentum is

$$L = mvr = 2 \pi m \mathcal f r^2$$

Comparing the formulas for magnetic moment $\mu$ and angular momentum $L$ shows that

$$\text{Electron magnetic moment}\qquad\qquad\mu = - \left( \frac {e}{2m} \right) L \qquad\qquad\qquad\qquad\qquad\qquad\qquad (6.39)$$

Figure 6.16

Figure 6.16 (a) Magnetic moment of a current loop enclosing area A. (b) Magnetic moment of an orbiting electron of angular momentum L.

for an orbital electron (Fig. 6.16). The quantity ($-e/2m$), which involves only the charge and mass of the electron, is called its gyromagnetic ratio. The minus sign means that $\mu$ is in the opposite direction to $L$ and is a consequence of the negative charge of the electron. While the above expression for the magnetic moment of an orbital electron has been obtained by a classical calculation, quantum mechanics yields the same result. The magnetic potential energy of an atom in a magnetic field is therefore

$$\qquad\qquad\qquad\qquad\qquad\qquad\qquad U_m = \left( \frac {e}{2m} \right) LB \, \cos \, \theta\qquad\qquad\qquad\qquad\qquad (6.40)$$

which depends on both B and $\theta$.

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