[quarterwave]

Why is the muonic hydrogen required?

From the Bohr formula the Rydberg energy of the muonic hydrogen would be some 200 times larger than regular hydrogen. Anyone know how that plays a role?

It can't be sqrt(spring/mass) for vibration since the proton is anyway already some 2000x heavier than the electron. Unless spring somehow depended on the Rydberg energy, which is possible since the P.E-K.E would depend on mass via the K.E.

[alhw]

The decrease in potential energy upon vdW bond formation in reactions with typical hydrogen isotopes dominates the negligible gains in vibrational zero-point energy from reactants to products. With the muonic hydrogen substitution, the authors claim instead that the driving force for "bond formation" results from a decrease in zero-point energy which compensates for the expected losses in potential energy.

[gus_massa]

I have a few published articles in quantum chemistry, and I barely can understand your explanation. It feels right, but I think I need to take 30 minutes to try to understand the details. Can you explain this like I'm a graduate student with only 3 years in the university, please?

[alhw]

The authors of the article in topic provide a more coherent explanation than I could ever articulate:

>> Conventionally, the formation of chemical bonds is due to a decrease in potential energy (PE), often accompanied by small increases in vibrational zero point energy (ZPE). In principle, this basic mechanism can be completely reversed, wherein chemical bonds may even be formed by an increase in PE if there is a sufficiently compensating decrease in vibrational ZPE, giving rise to what has been coined “vibrational bonding” of molecules stabilized at saddle-point barriers on a potential energy surface (PES), far away from potential minima.

[gus_massa]

Thanks. But I think your previous comment has an interesting point about why muons are different than electrons. I'm not sure because I hadn't made the calculations, so any confirmation or refutation is welcome. Let's try:

When two normal molecules, with electrons, are close, they can form different kind of bonds. The weakest bond is the "van der Waals" bond. It's caused because the electrons of the molecules change their position slightly due to the presence of the other molecule.

In this experiment they only replace the electron of a hydrogen atom by a muon. The muons have much more mass than the electrons, so the radius of the orbit is much smaller. (They are quantum particles, so they don't have orbits, but please forgive this technical detail.) As the orbits are smaller, the displacement caused by the other molecule is smaller, so the van der Walls force is smaller.

In the normal (electron) case the van der Walls force cause the formation of the intermediate molecule. In this case (muon) the van der Waals forcé is so weak that other effects are more important.

[I left out the part about zero point energy. It's also interesting but this explanation is becoming larger than the article :) .]

[sp332]

It's the proton that is replaced, not the electron.

[gus_massa]

Ups! :( You are right and now I'm confused.

[quarterwave]

Thanks for both explanations.

[tjradcliffe]

This is muonium: anti-proton plus electron, not muonic hydrogen (proton plus muon). I made the same mistake: it's a very poorly written article.

[quarterwave]

Thanks for pointing this out.

The role of muon instead of proton could be: (i) to make the bond detectable as the muon decays, or (ii) 10X lighter mass of muonium compared to regular hydrogen leads to a dynamical regime with saddle points in P.E vs K.E (as pointed out earlier).