How can you tell if a Gauss-type orbital is an eigenfunction of Hamiltionan $H$? For example:

$$GTO = N z^2 \exp\left(-\alpha r^2\right)$$

I know it is and eigenfunction of $L_z$ and not $L_x$ and $L_y$. But is it not an eigenfunction of the Hamiltonian $H$ and $L^2$? If I know it is one of $H$, I know it will be one of $L^2$ as well since they commute.

  • $\begingroup$ No, don't jump to such decisions, two commuting operators have a common set of eigenfunctions, but that set doesn't compulsorily comprise your function. $\endgroup$ – Sofia Jan 15 '15 at 17:56
  • $\begingroup$ Hmm good point! But you can make it so that they have a common set of eigenfunctions (which commute to the second operator) derived from the vectorspace created by the eigenfunctions of the first operator right? But that aside, how do you derive that a GTO is not an eigenfunction of H to start with? $\endgroup$ – Upsellum Jan 15 '15 at 18:16
  • $\begingroup$ Do you know $H$? How does it look like in your problem? I read in the site that you indicate, that you have to do with molecules, but no Hamiltonian is indicated there. $\endgroup$ – Sofia Jan 15 '15 at 18:30
  • $\begingroup$ Oh I'm sorry, $H$ is the operator connected with the electronic Schrödinger equation. It works on the wavefunction of an electron and gives forth the oribital energy. Normally the wavefunction is a real orbital such as $2px$ which is an eigenfunction of $H$. The problem with the GTO's is that they apparently are not eigenfunctions of $H$. That's why I would like to know how you can deduce they are not an eigenfunction. (But if you're not familiar with quantum chemistry don't mind my question. No offense ofcourse! Just saving your time :) ) $\endgroup$ – Upsellum Jan 15 '15 at 18:40
  • $\begingroup$ I just reread that last part and it still sounds rude which is totally not my aim. I'm even gratefull you replied so far. $\endgroup$ – Upsellum Jan 15 '15 at 18:48

Your Hamiltonian operator should be something like $$\hat{H}=\frac{-\hbar^2}{2m}\nabla^2+V(r).$$ All you have to do is act with it on your GTO function, ie calculate the derivatives and check if the result is of the form $a\cdot GTO$ where $a$ is a number. If it is of this form, then it is an eigenfuction with eigenvalue $a$ and if not, then it is not an eigenfunction.

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  • $\begingroup$ It is indeed of that shape. I also have the potential energy operator $V(r)$ and the laplacian $∇2$ specified. I will try what you suggested. Thanks! $\endgroup$ – Upsellum Jan 15 '15 at 19:16

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