I Can We Always Determine Eigenfunctions for an Asymmetric Top?

[kelly0303]

Hello! I read that for a symmetric top (oblate or prolate) we can find the exact eigenfunctions (in terms of Winger matrices) and eigenstates, but we can't do it in general for an asymmetric top. I am not sure I understand why. The Hamiltonian for an asymmetric top, for a given J, can be written in terms of $J^2$, $J_z$, $J^+$ and $J^-$ (where the operators are in the intrinsic frame). So for a given J, we can build the Hamiltonian (2J+1 x 2J+1 matrix), which will mix different values of K. But we can diagonalize this Hamiltonian (which is initially in the basis of the symmetric top wavefunction), and from there extract the energies and eigenfunctions (which will be linear combinations of the symmetric top wavefunction). Why is this not true in general? Can't we always get the eigenfunctions of an asymmetric top in this way? Thank you!

 

[Twigg]

we can't do it in general for an asymmetric top

I think in this case "we can't find eigenfunctions" translates to "we could find them numerically if we really wanted to, but they're ugly". It really just means that there's no conventional set of special functions in the major textbooks that describes asymmetric top eigenfunctions. In contrast, for the symmetric top we have the Wigner D matrices which have all those fancy algebraic properties. Saying "we can't find the asymmetric top eigenfunctions" is really more of a cosmetic statement than a factual one. Smells like theorist-speak to me

 

[kelly0303]

I think in this case "we can't find eigenfunctions" translates to "we could find them numerically if we really wanted to, but they're ugly". It really just means that there's no conventional set of special functions in the major textbooks that describes asymmetric top eigenfunctions. In contrast, for the symmetric top we have the Wigner D matrices which have all those fancy algebraic properties. Saying "we can't find the asymmetric top eigenfunctions" is really more of a cosmetic statement than a factual one. Smells like theorist-speak to me

Thank you! But I am still confused. Assuming we know the 3 moments of inertia, we can write the wavefunction of an asymmetric top as a linear combination of wavefunctions of symmetric tops (by diagonalizing the Hamiltonian in a J subspace, as I mentioned in the original post). So for example for $J=1$ the wavefunction would be of the form:

$$aD_{-1}^1+bD_{0}^1+cD_{1}^1$$

where D are the Wigner matrices and a, b and c are constants that depend on the moments of inertia. So the wavefunction has a clear analytical solution. I am not sure why we would need numerical methods for this. Also one can still take advantage of the Wigner matrix properties appearing there (of course not as easily as in a pure symmetric top case).

 

[Twigg]

I feel like you're right, since $J^2$ should still commute with the asymmetric top hamiltonian. The coefficients a, b, and c will change in time according to the time-dependent schrodinger equation, and those may need to be solved numerically (maybe it has analytic solutions, I really don't know off the top of my head).

Edit: I misunderstood what you meant initially. Now I understand that you mean you could solve for the eigenfunctions of the time-independent schrodinger equation for a, b, and c that make the stationary states of the asymmetric top. Yes this makes perfect sense to me! Nice work!