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In several textbooks and papers, like this one for example, a claim like the following is made:

The EDM of a system $\vec{d}$ must be parallel (or antiparallel) to the average angular momentum of the system $\hbar\langle\vec{J}\rangle$.

What is the motivation for making such an assertion? It seems not to make much sense if you consider, for example, two opposite electric charges held apart by a short, rigid massless rod (the usual classical picture of an electric dipole). The system can have zero angular momentum, but still has a nonzero electric dipole moment.

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  • $\begingroup$ See answer to my question here: physics.stackexchange.com/a/467662/128186 The answer is related to irreducible subspaces. Namely, the assumption is that the system is in a state of fixed total $|\boldsymbol{J}|$. When this is the case (to my surprise) expectation values of all vector operators are parallel. I think this is just one of the symmetries of rotations that arises when you restrict yourself to a certain total angular momentum subspace. If the system is in a superposition of multiple total angular momenta then it now become possible for vector expectations to be non-parallel. $\endgroup$
    – Jagerber48
    Commented Jun 12, 2020 at 1:30
  • $\begingroup$ @jgerber Ok, so this isn't a general statement, but rather one that only applies to systems with fixed total angular momentum? That makes more sense, but then why is it being treated as a general assumption in these papers, being stated without any motivation? After all, not every system we study has fixed total angular momentum. $\endgroup$
    – probably_someone
    Commented Jun 12, 2020 at 1:32
  • $\begingroup$ As far as I understand, what you say is correct. If the system has a superposition of total angular momenta then the statement will not hold. Why do they assume it in these papers? My guess (without thinking harder) is that the assumption holds for the systems they consider. Atoms, molecules, nuclei, and particles are all typically found in states of fixed angular momentum. This is especially true if they systems are cooled as they must be for some EDM experiments. For example, the ground state for these systems have fixed angular momenta as do the typical basis states. $\endgroup$
    – Jagerber48
    Commented Jun 12, 2020 at 1:50
  • $\begingroup$ Does this answer your question? Why must the electron's electric dipole moment (EDM) always be aligned with the spin? $\endgroup$
    – Emilio Pisanty
    Commented Sep 21, 2023 at 20:22
  • $\begingroup$ I am closing this (for now) as a duplicate of Why must the electron's electric dipole moment (EDM) always be aligned with the spin?. The arguments there are focused on the spin-1/2 case, but the proof there covers the assertion in your quote, so long as you understand $\vec d$ to be an expectation value. (If you don't, then $\vec d$ and $\vec J$ can cease to be proportional as operators if $j$ is large enough. But then the quote makes no sense.) If you feel the linked answer doesn't cover your question, you should sharpen this post to explain how. $\endgroup$
    – Emilio Pisanty
    Commented Sep 21, 2023 at 20:25

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They are probably talking about the quantum mechanical expectation value, which has to be along the angular momentum vector. If the angular momentum is zero, the expectation value of any vector would be zero.

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  • $\begingroup$ This doesn't really answer the question, though: why does the (expectation value of the) dipole moment vector have to be along the (expectation value of the) angular momentum vector? $\endgroup$
    – probably_someone
    Commented Jun 12, 2020 at 1:11
  • $\begingroup$ Anyway, it's not true in general that the angular momentum being zero means that the expectation value of any vector is zero. Take, for example, a spin-0 particle whose wavefunction is spherically-symmetric about the origin at $t=0$, and moves rigidly in some straight line at constant velocity afterward. If you calculate the expectation value of the angular momentum at $t=0$, you will see that it is zero (each volume element is precisely cancelled by its reflection about the origin), but the state nevertheless has nonzero linear momentum. $\endgroup$
    – probably_someone
    Commented Jun 12, 2020 at 1:15
  • $\begingroup$ The assertion (that if $j=0$ then the expectation value of any vector is zero) is correct. @probably_someone - your proposed counter-example does not work, as the wavefunction is not spherically symmetric at $t=0$ if it is about to set off at any velocity, since that would give it momentum (and therefore a phase gradient parallel to the momentum, thereby breaking spherical symmetry, so the angular momentum is nonzero). $\endgroup$
    – Emilio Pisanty
    Commented Sep 21, 2023 at 20:30

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