# Binding energies with muonic structures

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Mentor
TL;DR Summary
Could a hierarchical structure of protons, muons and positrons be stable without the weak interaction?
Summary: Could a hierarchical structure of protons, muons and positrons be stable without the weak interaction?

This discussion started here, I quote the relevant previous comments and then reply:
A positron could be in a stable orbital around a proton+2muon combination which itself would be stable similar to a hydrogen anion. But then the muons decay...
You're correct that this doesn't last any longer than the muons do, but I don't think this configuration is even stable. It is energetically favorable for it to rearrange to a muonic hydrogen atom and a (anti-)muonium atom. The H- ion is barely bound and both the muonic hydrogen atom and muonium atoms are deeply bound.
The binding energy of another electron is 0.75 eV, for the muon it should be 200 times that, or 150 eV, unless I miss some nonlinearity. The binding energy of muonium is just ~14 eV, similar to regular hydrogen. In addition you would miss most of the ~13.6 eV the positron has in the initial state.
We're getting off the track, but...

There is definitely a non-linearity (you can see it in the Hamiltonian), but you can see how the H- is bound semil-classically: one electron is near the proton forming an Hydrogen-atom like state, and the other is far away. The wavefunction is such that the expectation value of the distance between the far electron and the proton is slightly less than the expectation value of the distance between the far electron and the near electron, so you have a net attraction. (Pauli blocking is also important here, but I'll ignore it.) If I replace the electrons with negative muons, the first muon gets bound 200x deeper, but also has a wavefunction 200x closer (on average) so the second muon is bound ~200x deeper because of the mass, but only ~1/200 as deeply because muonic hydrogen is smaller. So I would expect binding of the outer muon to be of order eV (if it's bound at all).

But this is a detail. Even if you just scale everything up by 200, you can see the energetic instability. If you have (p mu- mu-) e+, the 1st muon is bound by 2.8 keV, the second by ~150 eV and the positron by 13.6 eV. If you rearrange to (p mu-) (mu- e+), the (p mu-) is bound by 2.8 keV, the (mu- e+) by half that, 1.4 keV, and if the two neutral "atoms" are bound at all, it will be of order eV. So the rearrangement is energetically favored by more than a kilovolt. Again, it's clearest semi-classically: the outer muon feels a much stronger attraction to the positron than it does to the neutral muonic hydrogen, so that's where it goes.

(mu- e+) is muonium, not true muonium - it is hydrogen-like where the muon replaces the (anti)proton, it has nearly the same binding energy as regular hydrogen, 13.6 eV. The binding energy decreases a tiny bit as the reduced mass decreases, but that is a ~0.5% effect.

I don't understand your scaling argument. If we shrink all wave functions by a factor 200 then all energies increase by a factor 200, this includes the difference between the potential energy the outer lepton has from proton and inner lepton.

fresh_42

Staff Emeritus
This might be better off in atomic physics.

Also, my x200 calculation in the previous post might have had one too many muons in it.

Let's start with the H- ion, since that's in many textbooks. The physics also quite beautiful. We start by choosing a candidate 1S wavefunction for both electrons scaled by what is conventionally called 1/ξ (this is the 1929 argument by Bethe, in modern language). The energy is scaled by ξ(ξ -2)/2, and the repulsion energy is now 5/8ξ instead of 5/8. So the total energy is ξ2 - 11/8ξ. This has an extremum at 11/16 (i.e. the ion is larger, and when you plug that in you get an energy where the electron is just slightly unbound (7/256 of a Hartree).

So we need to think of the two electrons separately. The "first" electron in H- ends up in a wavefunction similar to that in a hydrogen atom, no question about that. The "second" electron sees a dipole, not a point charge, so the field falls as 1/r3 and not 1/r2. But for now we are still working in terms of charges, not dipoles.

We attack this by introducing two ξ's and antisymmetrize the wavefunction. The integrals are horrible. I still have nightmares from graduate school. But what pops out is that one electron has more or less the usual wavefunction but the other is at a radius 4-5x farther out. Now you (barely) get a bound H- ion. Alas, the number is still not right - by a factor of 2 - i.e. the true 2nd electron in a hydride ion is bound twice as deeply as this predicts.

This tells us we only get a qualitative agreement by using unperturbed 1S wavefunctions. If instead you put the "second" electron in a wavefunction more appropriate for a dipole-monopole attraction, predictions get a lot closer.

This tells us something qualitative - the energy of the first electron will scale by 200 because of the muon mass, but the second will not: the dipole is 200x smaller. If it is bound at all (and I suspect that it is, barely), it's bound by eV, not keV.

OK, now to the question of whether (p mu- mu-) e+ is more stable than (p mu-) (mu- e+). A chemist might phrase this as whether the bond is covalent or ionic. Look at this from the point of view of the outer muon. It can either bind to a dipole far away - 4 or 5 Bohr radii - or to a positive charge nearby - about 1 Bohr radius. There's no question which is more strongly bound, so we know that (p mu- mu-) e+ will rearrange to (p mu-) (mu- e+) at its earliest opportunity.

TeethWhitener
Gold Member
This tells us something qualitative - the energy of the first electron will scale by 200 because of the muon mass, but the second will not: the dipole is 200x smaller. If it is bound at all (and I suspect that it is, barely), it's bound by eV, not keV.
I’m curious as to what this means for the proton-muon-electron system. If the proton-muon-muon system is bound on the eV scale, does this mean that p-##\mu##-e is bound on the meV scale? Since the 1s wavefunction is spherically symmetric, this kind of looks like an electron-neutron interaction (which I’ve always assumed was basically nonexistent, but which I’ve admittedly not thought too deeply about). The meV scale is definitely experimentally accessible—though maybe not in the types of experiments that would generate p-##\mu##-e systems. Is there any work in this area?

Staff Emeritus
The total binding is of a larger scale. If you are talking about removing one particle, you need to say which one.

Gold Member
If you are talking about removing one particle, you need to say which one.
The electron.

Mentor
Atomic physics is better, I agree, moved.
This tells us something qualitative - the energy of the first electron will scale by 200 because of the muon mass, but the second will not: the dipole is 200x smaller.
The dipole is smaller but we are also a factor 200 closer. The second muon won't have 4-5 times the Bohr radius, it will have 4-5 times the "muonic Bohr radius". Reduce the dipole moment by a factor 200, reduce the distance by a factor 200 and forces from the dipole moment and forces from individual charges have the same scaling again.

More formally: If we treat the proton as having infinite mass the Hamiltonian should be invariant under rescaling the masses and distances by a factor 200 and 1/200 and the energies by a factor 2002.

An electron would have an extremely weak bond to muonic hydrogen, sure, but that's not what we consider here.
OK, now to the question of whether (p mu- mu-) e+ is more stable than (p mu-) (mu- e+). A chemist might phrase this as whether the bond is covalent or ionic. Look at this from the point of view of the outer muon. It can either bind to a dipole far away - 4 or 5 Bohr radii - or to a positive charge nearby - about 1 Bohr radius. There's no question which is more strongly bound, so we know that (p mu- mu-) e+ will rearrange to (p mu-) (mu- e+) at its earliest opportunity.
I don't see the analogy to chemistry here. We don't consider the place of an electron, we consider the place of the heavy particles (i.e. what would be a nucleus for chemistry). Chemists don't consider nuclear fusion.

Treating the system as "what if the muon moves" would only work if the positron would be much heavier than the muon. Otherwise it is the positron that is mobile.

Let's assemble both step by step:
• p mu releases R*2002 with the regular Rydberg energy R, adding another muon gives an energy X where we disagree on how much, but we both expect X>0. Adding a positron to this system will give about R, neglecting effects from the finite size of the pseudo-nucleus.
• p mu releases R*2002, same as before. (mu e) releases R*200/201, slightly less than regular antihydrogen due to the lower reduced mass. We are missing X. The only thing left that could release more energy is some weird sort of molecule-state between the very compact (p mu) and the hydrogen-like (mu e). But - as you mentioned - the dipole of the first is very small. For that combination to release eV-scale energies the muon from (mu e) must get very close to (p mu), much closer than the size of the positron orbital. If that happens we get the other state, ((p mu mu) e).

Staff Emeritus
I don't see the analogy to chemistry here. We don't consider the place of an electron

We most certainly consider where the electrons want to go. The question here is exactly analogous to bonding - does the electron move to the other heavy guy?

Mentor
There is no "other heavy guy" in (p mu mu). If (p mu) and mu are separate then it is obvious where the electron is, but that is not the question considered here.

Staff Emeritus
This is turning into a semantic argument. I don't see any problem with (p mu mu) e+ being considered covalent (or at least more covalent) and (p mu) (mu e) being considered ionic.

Relative stability is important. Going back to the H- ion, if you just use 1S wavefunctions, you find it is bound by -3/8, but unfortunately neutral hydrogen and the other electron at infinity is bound by -1/2, so H- is calculated to be unstable. I have seen no argument that {(p mu mu) e+} is energetically stable to (p mu) (mu e) decay, and I have presented some arguments that it probably isn't.

I am still not convinced that (p mu mu) is even bound. I argued that to get it bound at all requires a fairly detailed calculate. There are things that weren't considered that might be important for the muon case. For example, in H-, the two electron spins are antiparallel. Flipping the spin of one is sufficient to unbind the system. So there is a magnetic attraction between the electrons that weakens the electric repulsion. Not so for the muons (where it is 40,000x smaller) For the electrons, we can ignore the proton size: it's less than a part per thousand effect. For muons it;s larger - the few percent level. That's another effect.

Mentor
I don't see any problem with (p mu mu) e+ being considered covalent (or at least more covalent) and (p mu) (mu e) being considered ionic.
I don't think the second one can exist as stable bound state.
I have seen no argument that {(p mu mu) e+} is energetically stable to (p mu) (mu e) decay, and I have presented some arguments that it probably isn't.
See post 6, I get the impression you overlooked that.
For example, in H-, the two electron spins are antiparallel. Flipping the spin of one is sufficient to unbind the system. So there is a magnetic attraction between the electrons that weakens the electric repulsion. Not so for the muons (where it is 40,000x smaller) For the electrons, we can ignore the proton size: it's less than a part per thousand effect. For muons it;s larger - the few percent level. That's another effect.
These are interesting points, and I fear they are beyond what we can solve here.

Staff Emeritus