Spin (for example of an electron) is described as an intrinsic form of angular moment, and we often say the electron has therefore magnetic moment, due to this angular moment. Well, I suppose when we say "magnetic moment" we actually mean magnetic dipole moment. In that case, is that dipole ideal? That is, are quadrupole, octupole, etc, components of magnetic moment zero for an elementary particle with spin? If not, why (since an elementary particle is as close as I think we can get to the definition of an ideal dipole)?


1 Answer 1


Is an electron an ideal magnetic dipole?

Theoretically, yes. I'll explain the theoretical prediction. If any experiment ever disagrees with this prediction, please let me know. Please let everybody know!

The theoretical answer is based on the Wigner-Eckart theorem, as explained in an existing answer to the related question Why do spin-$\frac{1}{2}$ nuclei have zero electric quadrupole moment? The answer I'm posting here is really just a more pedagogically-worded version of that existing answer, tailored for this new question.

An electron has intrinsic angular momentum equal to $\hbar/2$. To the degree that an electron can be localized at a point, this is a statement about how a single-electron state transforms under rotations about that point. Namely, it transforms as a spin-$1/2$ representation of the (covering group of the) rotation group. The electron can also have non-intrinsic (also called "orbital") angular momentum that depends on its motion through space, and that can have associated higher multipole moments, like it does in excited states of a hydrogen atom. The question asks about the electron's intrinsic properties, though, and for that, we can treat the electron as being localized at a point.

Multipole moments, by definition, also transform in specific ways under the rotation group: a monopole moment transforms as a spin-$0$ representation, a dipole moment transforms as a spin-$1$ representation, a quadrupole moment transforms as a spin-$2$ representation, and so on. Again, this is by definition.

Consider any multipole observable $M$ that transforms as a spin-$j$ representation, and let $|e\rangle$ be a single-electron state. The expectation value of $M$ in this state is proportional to $\langle e|M|e\rangle$, the inner product of $|e\rangle$ with $M|e\rangle$. If we rotate everything, including all states and all observables, then the theory's predictions should be unchanged. In particular, $\langle e|M|e\rangle$ should be unchanged. This is possible only if $|e\rangle$ and $M|e\rangle$ both transform the same way, so that their inner product remains unchanged under arbitrary rotations. The state $|e\rangle$ has spin $1/2$ and the observable $M$ has spin $j$, so the state $M|e\rangle$ is a linear combination of spins $j+ 1/2$ and $j-1/2$. We need $|e\rangle$ and $M|e\rangle$ to transform the same way if we want their inner product $\langle e|M|e\rangle$ to be invariant, so we must have either $j+ 1/2=1/2$ or $j-1/2=1/2$, which implies $j\in\{0,1\}$. In other words, an electron can have a monopole moment (and we know that it does have an electric monopole moment) and a dipole moment (and we know that it does have a magnetic dipole moment), but it cannot have any higher multiple moments, neither electric nor magnetic.

By the way, this argument is not limited to elementary particles. It also works for composite particles, such as nuclei (which is what the linked question is about). A composite particle may have various excited states with different angular momenta, so a generic state of the composite particle may be a superposition of several different intrinsic angular momenta, allowing it to have several different intrinsic multipole moments in such a state. But in its ground state, it transforms as a specific representation of the (covering group of the) rotation group, so the preceding argument still applies to a composite particle in its ground state. In this context, the only special thing about elementary particles is that they're always in their ground state. I suppose that's one way to define what "elementary particle" means.

  • $\begingroup$ Thanks for the answer! Your last paragraph is exactly the reason I assumed this might be a different question to that of the nucleus, as that is a more complicated system, and thanks for the insights on that too :) $\endgroup$ Commented Dec 28, 2020 at 11:22

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