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.

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    $\begingroup$ Related. $\endgroup$
    – rob
    Nov 18, 2022 at 0:21

2 Answers 2


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.

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    $\begingroup$ And if you turn on relativistic corrections, the hyperfine ground state of the hydrogen atom is a $F=0$ singlet with zero magnetic moment at zero field. However, this state has a big magnetic polarizability; if you put it in an appreciable magnetic field (breaking rotation invariance!), the electron spin alignment will determine the magnetic moment of the whole atom (the anomalous Zeeman effect). $\endgroup$
    – Buzz
    Nov 18, 2022 at 0:29
  • $\begingroup$ Rob, Good point about $F$. I was being sloppy thinking $J$ would be more familiar as "total" angular momentum. $\endgroup$ Nov 18, 2022 at 0:50

After some thought, I think I have mostly resolved my confusion.

  • The Wigner-Eckart theorem tells us that a spinless particle in a non-degenerate energy eigenstate (i.e. a stationary state) cannot have an intrinsic dipole moment.
  • The theorem does not prevent spinless composite particles with degenerate states from having a dipole moment.
    • A spinless particle can have an induced dipole moment if it has weakly separated states whose degeneracy is broken by an external field.
    • A spinless particle can have an permanent dipole moment if it has strongly separated degenerate states whose lifetime is longer the measurement timescale, and sometimes longer than the age of the universe.

As is discussed in Electric dipole moments revisited, some confusion is due to the slightly differing uses of the term "dipole moment" in different contexts, e.g. "intrinsic", "induced", "permanent". Except for the Higgs boson, all known spinless particles are composite particles whose constituent spins and orbital angular momenta sum to zero. Whether such a composite particle has a dipole moment depends on its quantum state, the time-scale over which the dipole moment is observed, whether constituent magnetic dipole moments cancel, or if geometric structure creates an electric dipole moment.

This question of spinless particle dipole moments is also an example of a more general problem: the experimental observation of states that violate the symmetries of the system's Hamiltonian. As early as 1927, Friedrich Hund recognized the paradox created by the well-established existence of parity-violating chiral molecules. A key point in the resolution of this paradox is noted in the answer to "Are broken symmetry states non-stationary?":

  • Non-degenerate stationary states are invariant under the symmetries of the Hamiltonian, but broken symmetry non-stationary states can exist that are linear combinations of degenerate stationary states.

There is, however, still not a complete consensus on how to reconcile molecular chemistry with quantum mechanics.

Let's discuss a few examples.

Non-degenerate energy eigenstates: Pseudoscalar Meson Magnetic Moments

Pions and kaons have no magnetic moments because they are all non-degenerate energy eigenstates, so the Wigner-Eckart theorem applies. Their structure is such that all their constituent magnetic moments cancel. This is also why the ground (parahydrogen) state of the hydrogen $\mathrm{H_2}$ molecule has no magnetic moment.

Weakly separated degenerate states: Hydrogen Atom Magnetic Moment

As discussed by Feynman and also in this answer, the $H(1s)$, the ground-state hydrogen atom has four combinations of electron spin ($S$) and proton spin ($I$): $$ \vert S_3 I_3 \rangle = \vert\! \uparrow \uparrow \rangle,\,\vert\! \uparrow \downarrow \rangle,\,\vert\! \downarrow \uparrow \rangle,\,\vert\! \downarrow \downarrow \rangle $$

These are not, however, stationary states. The four energy eigenstates are the ground-state singlet $\vert j=0, m \rangle = \vert 0, 0 \rangle$ with spin state

$$ \vert S_3 I_3 \rangle = \frac{\vert\! \uparrow \downarrow \rangle -\vert\! \downarrow \uparrow \rangle}{\sqrt{2}} $$

and three degenerate triplet states $\vert j=1, m \rangle = \left(\vert 1, 1 \rangle, \,\vert 1, 0 \rangle,\,\vert 1, -1 \rangle\right)$ with spin states

$$ \vert S_3 I_3 \rangle = \vert\! \uparrow \uparrow \rangle,\;\frac{\vert\! \uparrow \downarrow \rangle + \vert\!\downarrow \uparrow \rangle}{\sqrt{2}},\;\vert\! \downarrow \downarrow \rangle $$

The spin-spin interactions shift the singlet state energy down and the triplet up, with singlet-triplet splitting of $\Delta E_{s-t} =5.874\,\mu\mathrm{eV}$, corresponding to the famous 21 cm (1420 MHz) hydrogen line.

Neither of two spinless $H(1s)$ atom energy eigenstates $$ \vert S_3 I_3 \rangle = \frac{\vert\! \uparrow \downarrow \rangle -\vert\! \downarrow \uparrow \rangle}{\sqrt{2}}, \qquad \frac{\vert\! \uparrow \downarrow \rangle +\vert\! \downarrow \uparrow \rangle}{\sqrt{2}} $$ have a magnetic dipole moment since they both have equal amplitudes of $\vert\! \uparrow \downarrow \rangle$ and $\vert\! \downarrow \uparrow \rangle$ electron and proton spins. If we brought a hypersensitive magnetometer close to an isolated static hydrogen atom we would detect no dipole magnetic field.

When an external magnetic field is applied, however, the degenerate spinless states undergo Zeeman splitting that is linear with the magnetic field strength $B$. The splitting between the two spinless states is $\Delta E_B \rightarrow 2 (\mu_e-\mu_p) B$ for $\Delta E_B >> \Delta E_{s-t}$, as expected for a particle with magnetic moment $\mu'=\mu_e-\mu_p$, where $\mu_e$ and $\mu_p$ are the magnetic moments of the electron and proton. (In principle, such splitting could occur for the pion as well, but it would require a magnetic field $\gtrsim {m_\pi^2 c^2}/{\hbar e} \sim 3\times 10^{14}\,\mathrm{T}$, which isn't even reached on the surface of magnetars.)

Weakly separated degenerate states: Ammonia Electric Dipole Moment

Not only should electric dipole moments not exist for spinless energy eigenstates, they are more generally forbidden for any electromagnetic stationary state because they violate parity ($P$) and time reversal ($T$). Many polar molecules exist, however, with the tetrahedral ammonia ($\mathrm{NH_3}$) molecule being one classic example.

For any isolated ammonia molecule rotating around its symmetry axis, there two electric dipole states, $|\mathrm{L}\rangle$ & $|\mathrm{R}\rangle$, corresponding to when the electric dipole is aligned or anti-aligned with the molecules rotational angular momentum. These states exist even if the total molecular angular momentum (rotational+orbital+nuclear+electronic) is zero, as long as the rotational component is non-zero to define a "handedness".

As discussed by Feynman and elsewhere, this can be modelled by a simple two-well potential with a finite barrier between the two wells. The molecule has two energy eigenstates with opposite parities: $$ \vert + \rangle = \frac{\vert\! L \rangle +\vert\! R \rangle}{\sqrt{2}},\qquad\qquad \vert - \rangle = \frac{\vert\! L \rangle -\vert\! R \rangle}{\sqrt{2}} $$ Neither of these stationary states has a electric dipole moment, but an external electric field removes the $|\mathrm{L}\rangle$/$|\mathrm{R}\rangle$ degeneracy, causing Stark splitting that is linear in electric field strength, corresponding to an electric dipole moment of 1.42 debye. This electric field can be due to other molecules, which is why molecular polarity is such a useful concept in chemistry. Similarly, a water molecule has an electric dipole moment of 1.8546 debye, despite violating time reversal.

Strongly separated degenerate states: Chiral Molecules

Many stable right-handed or left-handed molecules (enantiomers) exist despite violating the parity symmetry of their Hamiltonian. (We are no longer talking just about spinless particles, but discussing the general point that long-lived states can violate the symmetries of their Hamiltonian.) In principle, these molecules have parity conserving energy eigenstates, but inversion energy barrier is so large that the tunnelling times between the chiral states is so long - sometimes even longer than the age of the universe - that they never relax into a energy eigenstate as happens with the hydrogen atom or the ammonia molecule. Such a state is "permanent", but not strictly speaking "stationary".

As Woolley noted, there is sort of a chiral uncertainty principle: an enantiomer can have a well-defined structure or a well-defined parity, but not both at the same time.


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