# Capactiy of an orbtial to hold muons and electrons

In a normal atom, there is a limit of 2 electrons per obital due to the Pauli-Exclusion principle. I have seen people talking about replacing an electron with a muon, but since muons and electrons are distinguishable, would this then allow 2 electrons and 2 muons to occupy the same orbital, or would only 2 leptons be allowed to have the same set of quantum numbers?

For example, would a Beryllium atom with 2 muons and 2 electrons use only the 1S orbital, or would the 2S be used as well? Spectroscopic methods should be able to distinguish between the two.

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## 2 Answers

The atomic orbital description of an atom is only an approximation for any atom with more than one electron. Atomic orbitals only exist when the potential is centrally symmetric. With a centrally symmetric potential the wavefunction of the atom factors into separate atomic orbitals so we can talk about $1s$, $2s$, $2p$ etc orbitals.

However, when there is more than one electron in the atom the repulsion between the electrons spoils the central symmetry. The end result is that the atomic orbitals get mixed up and the wavefunction can no longer be split into separate atomic orbitals. In a multi-electron atom you can't uniquely assign each pair of electrons to an atomic orbital.

The reason we still talk about the atomic orbitals in multi-electron atoms is that even though it's just an approximation, actually it's not a bad approximation. When you're calculating the wavefunction of the atom you can average out the positions of all the electrons, and the average then gives you a central potential that you can use to calculate separate orbitals, though these orbitals are obviously different from the ones you get for a one electron atom.

The point of all this is that it's not obvious to me that you could usefully average out the electrons and muons together because their masses are so different. I suspect that what would happen is that you'd average the two muons and separately average the two electrons. The averaging would still give you a central potential, but what you'd end up with is two distinct orbitals with the two muons in one and the two electrons in the other. Both orbitals would look like a $1s$ orbital, but the muon orbital would lie much closer to the nucleus that the electron orbital. In fact I'd guess the electron orbital would be very similar to the $1s$ orbital in a helium atom: the muons would partialy screen the nucleus so the electrons would just see a +2 nuclear charge. Your wavefunction would end up looking like $\Psi = 1s_{muon}1s_{electron}$.

But note that writing the wavefunction as two $1s$ type orbitals would be an approximation that ignored the electron-electron and electron-muon correlations.

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If you could produce Hg with 1 muon captured / molecule, I suppose it would chemically look very similar to gold then! –  Nick Oct 5 '12 at 14:22
@Nick, if you are interested in the method that John talks about in the second paragraph the phrase you want is "mean field model". –  dmckee Oct 5 '12 at 14:32
If you're (like me) a humble chemist you'll know this as the Hartree Fock method or the Self Consistent Field (SCF) method. –  John Rennie Oct 5 '12 at 15:07
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They are distinguishable so of course the number of states doubles if you double the number of species, i.e. if you use both electrons and muons.

So with 2 muons and 2 electrons, 2 muons would first sit to the 1s orbitals very close to the nucleus, resembling a "small helium". The electrons are 200+ times lighter so their orbitals are 200+ times larger. (In other words, the Bohr radius for the muon is 200+ times smaller than that for the electron.) The electrons would effectively view the nucleus' charge reduced by the charge of the two muons: the "small muonic helium atom" would act as a nucleus for the electrons. But otherwise the electrons would sit into their 1s states, resembling the ordinary helium.

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