Understanding Ortho- and Para- Water: Explaining the Physics behind Ka+Kc+v3

[DanAbnormal]

I was wondering if anyone was familiar with what I am stuck on.

I understand the origins of ortho/para hydrogen, and how it is extended to the water molecule.

For 2 hydrogens:
If total nuclear spin = 0 -> Singlet state (para)
of total nuclear spin = 1 -> Triplet state (ortho)

This gives rise to the 3:1 ortho:para ratio for molecular hydrogen (and water) etc etc.

Now I am working with spectroscopy of water, and I have quantum numbers describing energy states. The numbers are:

J,Ka,Kc,v1,v2,v3

The J,Ka,Kc numbers being the standard asymmetric top q-numbers describing rotation.
J= total ang. mom.
and Ka, Kc the projections on the A and C axes, respectively.

the v1,v2,v3 numbers correspond to symmetric stretch, symmetric bend and asymmetric stretch vibration modes, respectively.

THE QUESTION: I have read in countless papers (without explanation) that ortho and para states can be distinguished like so:

If Mod(Ka+Kc+v3,2)=0 -> para
or if =1 -> ortho

Or equivalently, if Ka+Kc+v3 is even -> para
or if odd -> ortho

I don't understand why this is so. Can someone explain the physics behind this, I really don't get it... Am I missing something?

Thanks
Dan

 

[DrDu]

First some questions: Do you understand how ortho and para are defined for D2?
Do you know how symmetry arguments lead to the restrictions of J for ortho and para H2?

 

[DanAbnormal]

First some questions: Do you understand how ortho and para are defined for D2?
Do you know how symmetry arguments lead to the restrictions of J for ortho and para H2?

I do not know the answer to the first question.
But for the second question, I understand the following:

The total wavefunction is a superposition of individual states corresponding to rotation, vibration, nuclear and electron.

Both the rotational state and nuclear spin state can be symmetric or antisymmetric, but the overall wavefunction must be antisymmetric therefore by choosing one of them to be either symmetric/antisymmetric collapses the other into an antisymmetric/symmetric state (opposite).

Since we have 2 spin-1/2 nuclei, this gives rise to four spin states: comprising of a triplet and a singlet.

This sort of makes sense to me as there is a 3:1 ratio between nuclear spin states.

Im not sure about the symmetry argument though. The singlet state is antisymmetric yes?
I don't fully understand how this extends to constraining J, but I am guessing the constraint is on whether J takes odd or even values?
If so, can you explain?

Thanks

 

[DrDu]

Yes, as hydrogen nucleus is a fermion, the total wavefunction has to be antisymmetric under nuclei exchange. The total wavefunction of the molecule is in lowest approximation (sufficient for symmetry considerations) a product of an electronic wavefunction, a vibrational and a rotational wavefunction and of the nuclear spin wavefunctio, the latter being symmetric for s=1 and antisymmetric for s=0.
The electronic ground state wavefunction is symmetric and the vibrational coordinate, too. Hence it is only the J of the rotational wavefunction which is restricted by symmetry.
In H2O the situation is more complicated, as both the asymmetric stretch co-ordinate nu_3 is antisymmetric and also the rotational wavefunction for some combination of J, K_A and K_C (which aren't true quantum numbers but refer to some iealized limiting geometries of the molecule, see: http://www.pci.tu-bs.de/aggericke/PC4e/Kap_III/Asymmetrischer_Kreisel.htm).
Hence the vibrational wavefunction for nu_3 with an even number of quanta is even and with an odd number is odd. It has to be combined with fitting rotational wavefunctions, which are to be specified in terms of J, K_A and K_C (although I don't know about the details).