# Why doesn't the ${\rm H}_2$ molecule have a permanent dipole while the neutral ${\rm H}\,{\rm\small I}$ has one?

According to the National Radio Astronomy Observatory, the $21\,{\rm cm}$ line of hydrogen can be emitted by the neutral hydrogen atom due to the magnetic interaction between the spins of the proton and electron. However, it says the hydrogen molecule ${\rm H}_2$ does not posses a permanent dipole, and thus does not emit a spectral line at radio frequencies. Why is that the interaction between the spins of the proton and electrons no longer manifests itself when we are talking about the ${\rm H}_2$ molecule?

$H_2$ contains 2 electrons in the same ground-state orbital; by Pauli exclusion, one must be spin-up and the other must be spin-down. The 21cm line is generated in a normal hydrogen atom when an electron's spin flips from being aligned with the proton to being anti-aligned with the proton. In $H_2$, an electron's spin cannot flip because it would then be occupying the same state as the other electron. Even if it could, the differing shape of the orbital would produce a line at a significantly different wavelength than 21cm.

The two protons, since they're not in the same orbital, can exhibit a similar type of transition as the 21cm line (i.e. proton spin flips from aligned to anti-aligned), but since the protons are highly localized, massive, and far apart, any spin-spin coupling between them is insignificant. Indeed, when measured, this transition has a frequency of just 72 kHz, corresponding to a wavelength of over 4 km.

• Does the same argument also hold for the proton spins in the hydrogen molecule? So that they can only be anti-parallel? And no flipping to parallel allowed that could produce a photon emission? – freecharly Jan 22 '18 at 15:45
• The difference is that the protons are highly localized and very far apart from each other, so any sort of proton spin-spin interaction can only result in a tiny difference in energy. Namely, this transition has a frequency of only 72 kHz, which corresponds to a wavelength of 4.17 km. Source: ac.els-cdn.com/000926149385520X/… – probably_someone Jan 22 '18 at 18:35
• Thank you very much for your interesting explanation! So that would be in the longwave radio band. – freecharly Jan 22 '18 at 18:45
• Indeed it would. The issue is that, since the energy for the transition is so low, the half-life for spontaneous transitions is extremely long. So you'd either need a huge (even by astronomical standards) amount of molecular hydrogen gas or very contrived experimental circumstances to actually see it, which is why it's usually not mentioned. – probably_someone Jan 22 '18 at 18:49

The neutral $\mathrm{H}$ atom is a permanent magnetic dipole because the electron and proton have different g-factors, so they cannot completely cancel even when their magnetic dipoles are anti-aligned.

In the $\mathrm{H}_2$ molecule, on the other hand, the magnetic moments can completely cancel. That's not to say that they do in practice - I wouldn't be surprised if there were some extremely long wavelength transition that corresponds to flipping the $\mathrm{H}_2$ nuclear spins aligned vs. anti-aligned. The thing to keep in mind, though, is that the lower the energy difference the longer the life-time of the excited state. The decay that produces the $21\operatorname{cm}$ line, for example, has a half-life of about $11$ million years. That we see it, at all, is primarily a factor of the quantity of hydrogen in the universe. I would expect a nuclear spin flip transition to be even lower energy by at least an order of magnitude.

All of that said, @probably_someone's explanation is correct: the electron cloud in $\mathrm{H}_2$ has a zero spin ground state, and the exclusion principle requires that any non-spin zero state must also be a spatially excited state, too.

In detail, when the molecule is in its ground state the spins are in a total spin zero state, $$|s=0\rangle = \frac{|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle}{\sqrt{2}},$$ and the anti-symmetry of the spins allows the spatial wave-functions to be symmetric, occupying the same lowest energy level. When the spin state has total spin $1$, one of \begin{align} |s=1\rangle & = |\uparrow\uparrow\rangle, \\ & = |\downarrow\downarrow\rangle, \ \mathrm{or} \\ & = \frac{|\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle}{\sqrt{2}}, \end{align} then the spatial wave function has to be anti-symmetric, making the energy of the state significantly higher.