6
$\begingroup$
  • Vibrational wavefunctions corresponding to different vibrational levels of the same normal mode are orthogonal because they are eigenfunctions of the harmonic oscillator Hamiltonian, which is hermitian.
  • Displacement vectors corresponding to different normal modes are orthogonal because they are eigenvectors of the Hessian matrix, which is symmetric.

QUESTION: given the above, how can one show that vibrational wavefunctions corresponding to different normal modes are orthogonal?

$\endgroup$
  • 4
    $\begingroup$ @A.K. but then how can you tell what is chemistry and what is physics? One might say that all of theoretical chemistry is basically physics... $\endgroup$ – GingerBadger Jul 10 '18 at 15:16
  • 10
    $\begingroup$ I don't know if agree with this being moved. While it certainly could be asked on physics, I think it also falls well within the normal bounds of computational chemistry. $\endgroup$ – Tyberius Jul 10 '18 at 15:19
  • 9
    $\begingroup$ This question absolutely fits under the category of chemistry. There is absolutely no reason to redirect computational chemistry questions and would only serve to hurt our community by prohibiting this field of study. $\endgroup$ – LordStryker Jul 10 '18 at 16:37
3
$\begingroup$

given the above, how can one show that vibrational wavefunctions corresponding to different normal modes are orthogonal?

The Hamiltonian for a system with two (or more) normal modes can be written as a tensor product of (Hermitian) Hamiltonians for each normal mode. For example for two modes (stretching and bending):

$$ H_{\rm{total}} = H_{\rm{stretching}}\otimes H_{\rm{bending}} $$

The tensor product of two Hermitian matrices is also Hermitian. Therefore the wavefunctions are orthogonal for the same reason that you figured out on your own for the single mode case.

More generally, all Hamiltonians (Hermitian operators) have orthogonal wavefunctions (eigenfunctions) for distinct energy levels (eigenvalues). So wavefunctions corresponding to different energy levels are orthogonal as a consequence of the Schroedinger equation which is based on a Hermitian object called the Hamiltonian, and is the basis for the existence of all distinct energy states (vibrational or not) in quantum chemistry.

$\endgroup$
  • $\begingroup$ Thank you. But then why are the vibrational wavefunction necessarily eigenfunctions of a single total Hamiltonian? Because (1) one almost always assumes that vibrational are independent or, at most, interacting via an averaged self-consistent field, which isn't true in $H_{total}$, (2) anharmonicity effects are commonly neglected. $\endgroup$ – GingerBadger Jul 11 '18 at 10:26
  • $\begingroup$ @munchnone: In your question you say it works because the Hamiltonian is the harmonic oscillator Hamiltonian, but actually it works for anharmonic vibrations such as the Morse/Long-range potential as well. You ask why the wavefunctions are eigenfunctions of a single total Hamiltonian? Actually it is a postulate of quantum mechanics that the universe can be described by $H|\psi_n\rangle = E_n|\psi_n\rangle$ where $H$ is the Hamiltonian of the universe. Then if you want to look at smaller parts of the universe (like a water molecule) you consider a sub-matrix of the full $H$. $\endgroup$ – user1271772 Jul 11 '18 at 11:22
  • $\begingroup$ Thanks. So is what you mean is to consider the total harmonic Hamiltonian matrix, which is a tensor product of the Hamiltonian matrices for each normal mode because the total harmonic Hamiltonian operator is the sum of Hamiltonian operators for each normal mode? Note that this would not be the total Hamiltonian, because it contains higher order terms that couple all normal modes. $\endgroup$ – GingerBadger Jul 11 '18 at 11:56
  • $\begingroup$ Is there an easier way to think about this? E.g. if we have $f_1(x_1)$ and $f_2(x_2)$, where $x_1$ and $x_2$ are orthogonal directions in space, then I assume it is always true that $f_1(x_1)$ is orthogonal to $f_2(x_2)$? What is a good way to show it? $\endgroup$ – GingerBadger Jul 11 '18 at 12:00
  • $\begingroup$ @munchnone: No the total Hamiltonian (harmonic or anharmonic) is Hermitian, because all Hamiltonians have to be Hermitian. In your question you say the harmonic Hamiltonian is Hermitian and that's why wavefunctions are orthogonal, but actually all Hamiltonians are Hermitian (harmonic or anharmonic doesn't matter) $\endgroup$ – user1271772 Jul 11 '18 at 12:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.