An electron in orbit is a good approximation of a magnetic dipole so n the introduction of chapter 8 in the book: "Quantum Physics of Atoms, Molecules, Solids, Nuclei and Particles",

a classical derivation of the magnetic dipole moment is presented:


$\mu$ being the dipole moment of the electron, $i$ is the current and $A$ is the area of the circular orbit of the electron.


with $e$ the charge of the electron and $T$ the time it takes to complete one orbit so:

$i=\frac{ev}{2\pi r}\tag3$

Where $r$ is the radius of the orbit. The dipole moment is then:

$\mu=\frac{ev}{2\pi r}\pi r^2=\frac{evr}{2\pi}=\frac{eL}{2m}\tag4$

Where $m$ is the mass of the electron and $L=mvr$ is the angular momentum of the electron.

Question 1: should I be convinced by the book that a quantum mechanical derivation of the magnetic dipole would yield the same relation, given the fact that the ratio between the angular momentum and the dipole moment is independent of the details of the orbit, i.e. the radius, shape, orbital frequency, and why?

I can't see how this is at all a good argument since quantum mechanics is definitely not always an "averaging out" of sorts of classical mechanical results, as the double slit experiment or quantization of energy show.

Question 2: How does one go about calculating the dipole moment quantum mechanically?

I imagine one could create a "dipole moment operator" which would just be the angular momentum operator multiplied by the constant $\frac{e}{2m}$. Then a total angular momentum eigenstate would be a "dipole moment magnitude eigenstate"? Since no more than one component of the angular momentum can be know exactly, this "dipole moment magnitude eigenstate" could not also be a "dipole moment direction eigenstate".

I'm very surprised that I can't seem to find good information about this anywhere, it really bugs me when books make these leaps of faith so I hope somebody can clarify this for me.


Quantum mechanically, you would calculate the dipole moment of the atom by using perturbation theory to computing the change in the ground state energy under the introduction of a small magnetic field and then setting $$ \delta E_0 = -\boldsymbol{\mu}\cdot {\bf B}. $$ This leads to the Lande g-factor, which is the multiplicative correction to the classical $$ \boldsymbol{\mu}= \frac{|e|}{2m_e} {\bf L}. $$

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  • $\begingroup$ For a hydrogen molecule this g-factor is one, is this not because in a total angular momentum eigenstate, the electron will have a definite dipole moment because in a hydrogen atom the ratio between angular momentum and dipolemoment is just a constant? $\endgroup$ – Stijn Boshoven Apr 20 at 16:56

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