• The general formula for the magnetic moment of a charge configuration is defined as $$\vec{\mu} = \frac{1}{2} \int \vec{r} \times \vec{J} \,d^3r$$

  • For an electron it's said that the correct equation relating it's spin and magnetic moment is is $$\vec{\mu} =g\frac{q}{2m}\vec{S}$$

  • It's said that the above equation cannot be justified classically and is a quantum mechanical phenomenon.

What is the definition of magnetic moment used in the quantum mechanical equation $$\vec{\mu} =g\frac{q}{2m}\vec{S}$$


1 Answer 1


The magnetic moment $\vec{\mu}$ of a charge configuration is defined by the torque $\vec{\tau}$ it feels when being in an external magnetic field $\vec{B}$ $$\vec{\tau}=\vec{\mu}\times\vec{B}$$ or equivalently by its potential energy $U$ when being in this external field $\vec{B}$ (see Magnetic Moment - Effects of an external magnetic field) $$U=-\vec{\mu}\cdot\vec{B}.$$

This definition is used both for classical and for quantum-mechanical systems.

The difference begins when you want to get a relation between magnetic moment $\vec{\mu}$ and angular momentum. For orbital momentum $\vec{L}$ you have (both classically and quantum-mechanically) $$\vec{\mu} = g\frac{q}{2m}\vec{L}\quad\text{, with } g=1$$ which can be derived theoretically and confirmed experimentally (by measuring the torque or energy).

But for spin angular momentum $\vec{S}$ of the electron experiments show $$\vec{\mu} = g\frac{q}{2m}\vec{S}\quad\text{, with } g=2.0023$$ which is roughly double the size than for orbital momentum. This cannot be understood by classical mechanics. But from Pauli's equation (i.e. with non-relativistic quantum mechanics) or from Dirac's equation (i.e. with relativistic quantum mechanics) you can derive this formula with $g=2$. And with the full theory of quantum electrodynamics you can even derive it with the exact value $g=2.0023$ (see $g$-factor).

  • $\begingroup$ The factor $g$ is actually not exactly 2 but 2.00231930436. This value has been found both experimentally and theoretically. $\endgroup$
    – md2perpe
    Jan 23 at 22:20
  • $\begingroup$ @md2perpe You are right. I have improved the answer for this. $\endgroup$ Jan 23 at 22:58
  • $\begingroup$ You can understand the factor of 2 using just non relativistic quantum mechanics $\endgroup$
    – Mauricio
    Jan 23 at 23:33
  • $\begingroup$ @Mauricio You are right. I've reread the derivation from Pauli's equation $\endgroup$ Jan 24 at 1:56
  • $\begingroup$ @Thomas Fritsch, thank you. I've seen on many places the definition of magnetic moment to be as $\vec{\mu} = \frac{1}{2} \int \vec{r} \times \vec{J} \,d^3r$ ,that's why I got confused. $\endgroup$
    – Kashmiri
    Jan 24 at 4:32

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