I Correlation between Symmetry number & Total wavefunction

[JohnnyGui]

TL;DR Summary

If the classical symmetry number must come from quantum mechanics and has the same correction value, is there somehow a deep correlation between a molecule's physical structure and its allowed rotational quantum states based on the total wavefunction being symmetrical/anti-symmetrical? And how can this correlation be explained?

Some rotational quantum states are not allowed for a rotating particle. At quantum level, these "forbidden" quantum states is based on the requirement of the total wavefunction being either symmetrical or anti-symmetrical, depending on whether the particle is a fermion or boson. The particle's rotational partition function therefore only sums up the quantum states with the allowed $J$ values. Source

In the classical regime, the exclusion of these "forbidden" quantum states is done by using the so-called symmetry number $\sigma$ instead, which is based on the number of physical orientations of a molecule that are indistinguishable because of its physical symmetrical structure. Explanation on Page 2

The first source is stating, starting from Page 6, that the symmetry number is of classical mechanical origin but at the same time, must come from quantum mechanics. It then proceeds to show with calculations that they have the same correction value for excluding the forbidden quantum states.

If the classical symmetry number must come from quantum mechanics and has the same correction value, is there somehow a deep correlation between a molecule's physical structure and its allowed rotational quantum states based on the total wavefunction being symmetrical/anti-symmetrical? And how can this correlation be explained?

 

[DrClaude]

If the classical symmetry number must come from quantum mechanics and has the same correction value, is there somehow a deep correlation between a molecule's physical structure and its allowed rotational quantum states based on the total wavefunction being symmetrical/anti-symmetrical? And how can this correlation be explained?

It is simply that rotation by π is the same as the inversion or mirror symmetry when two nuclei are identical.

 

[JohnnyGui]

It is simply that rotation by π is the same as the inversion or mirror symmetry when two nuclei are identical.

But whether the total wavefunction is symmetric/anti-symmetric depends on the spins of the nuclei, not on whether the nuclei are identical or not. It goes even further by the fact that the total wavefunction of e.g. dihydrogen can be symmetrical even if the two atoms have opposed spins.

So it makes me question how a correction based on physical structural symmetry is somehow linked to a correction based on wavefunction symmetry derived from nuclear spins (as well as other wavefunctions such as rotation)

 

[PeterDonis]

whether the total wavefunction is symmetric/anti-symmetric depends on the spins of the nuclei

Yes.

not on whether the nuclei are identical or not.

No. The wave function of a system consisting of two non-identical nuclei (say helium and iron) does not have to be symmetric or antisymmetric.

Of course if you use a more detailed model where, instead of nuclei, you view the system as composed of protons and neutrons, the total wave function will have to be antisymmetric under exchange of a pair of protons or a pair of neutrons. But it won't have to be antisymmetric under exchange of a proton and a neutron.

 

[JohnnyGui]

But it won't have to be antisymmetric under exchange of a proton and a neutron.

Does "won't have to" imply that it is still possible to stay antisymmetric? Because in that case, the classical symmetry number would still not have a clear link (to me) since it relies on the physical indenticality of particles.

 

[PeterDonis]

Does "won't have to" imply that it is still possible to stay antisymmetric?

I don't know what you mean by "stay antisymmetric".

 

[JohnnyGui]

I don't know what you mean by "stay antisymmetric".

I understood that specific phrase "won't have to be" as in "not obliged to be" which made me think that it is also possible for such a model to be antisymmetric under exchange of a proton and neutron as well, just as in the exchange of a proton pair or neutron pair.
But I assume you meant that "it has to" be symmetric under exchange of a proton and a neutron?

Which in that case shows that not only nuclear spin, but also the identicality of particles determine whether a total wavefunction should be symmetrical or antisymmetrical?

 

[PeterDonis]

I understood that specific phrase "won't have to be" as in "not obliged to be" which made me think that it is possible for such a model to be antisymmetric under exchange of a proton and neutron as well

I don't think there are wave functions that are antisymmetric (or, for that matter, symmetric) under the exchange of a proton or [Edit: and] a neutron. That would require that the two states (before and after exchange) were physically indistinguishable, and they're not.

assume you meant that "it has to" be symmetric under exchange of a proton and a neutron?

No. See above.

 

[JohnnyGui]

I don't think there are wave functions that are antisymmetric (or, for that matter, symmetric) under the exchange of a proton or a neutron.

"proton or neutron" as in exchange of a proton pair "or" neutron pair? If yes, could you please elaborate how this quote is compatible with this:

the total wave function will have to be antisymmetric under exchange of a pair of protons or a pair of neutrons

I am a bit confused.

 

[PeterDonis]

"proton or neutron" as in exchange of a proton pair "or" neutron pair?

I should have said proton and neutron--i.e., exchanging one proton with one neutron. Sorry for the typo. I have edited my previous post to clarify.