While fundamental point particles with zero spin cannot have a dipole moment because spin is their only directional property, this does not seem to be true for spinless composite "particles".
An obvious example is the spinless ground state hydrogen atom which has a large magnetic moment due to its electron.
I realize I don't fully understand statements such
"the dipole moment of any spin-zero state must vanish exactly" in answers such as What is the magnetic dipole moment of a pion? or Is there an experimental value for the magnetic moment of the Kaons?
Do pseudoscalar mesons have zero magnetic moments because they have zero spin and the Wigner-Eckart theorem applies, or because they have a particular quark model internal structure due to Quantum ChromoDynamics (QCD)?
Consider, for example, a $K^+$ ($u\bar{s}$) meson.
It's spin-flavour wave function is
$$K^+=-\frac{1}{\sqrt{2}}\left(u\uparrow\bar{s}\downarrow-u\downarrow\bar{s}\uparrow\right)$$
and we can see that the average spins for $u$ and $\bar{s}$ quarks are both zero separately, so their intrinsic quark magnetic moments independently cancel out. This does not happen with the proton and electron in a hydrogen atom.
Of course a meson is very different from a hydrogen atom. The meson's binding energy is $\sim10^7$ larger then that of hydrogen and the binding QCD force doesn't care about a quark's flavour or electric charge. A meson is orders-of-magnitude small than an atom. Changing the quark-antiquark spins from an antiparallel spin-0 state to a parallel spin-1 state spin inside a meson increases the meson mass by hundreds of MeV, compared to $\sim 6\times10^{-6}$ eV for changing a hydrogen atom from spin 0 to spin 1.
So I guess my confusion is about how the Wigner-Eckart theorem applies to composite systems, and what is the essential difference between a composite spinless meson that cannot have a dipole moment and a composite spinless atom that does have one?
Edit Note: I originally incorrectly used $J$ for the atom's total angular momentum instead of the correct F, which is why it is mentioned in Rob's answer.
Best Answer
I believe the difference is that atoms can be modeled in the non-relativistic limit where total, orbital, and spin angular momenta $J,L,S$ for the electronic sector, and the nuclear spin $I$, are all commuting observables. (Note that hydrogen’s electronic ground state is $^{2S+1}L_J={}^2S_{1/2}$. The hyperfine transition whose micro-eV energy you give usually gets the quantum numbers $F=1\to0$, rather than your $J$.) Because the operators for $S$ and $I$ both commute with the Hamiltonian, they are good quantum numbers for energy eigenstates and you can sensibly talk about a magnetic moment proportional to one or the other.
In relativistic systems, including the smallest nuclei, this approximation is no good. The two nucleons in the deuteron, for example, are in a mixture of $^3S_1$ and $^3D_1$ angular momentum states, so we can’t assign a direction to the nucleon spins in polarized deuterium.