Some time ago, (a few years back) I heard that the orbitals of a single water molecule were solved analytically (or was it just numerically). What do they look like?

I am interested in the shape (and equations) of those orbitals. There should be an exact solution around and that is what I am looking for (no LCAO or similar please).

  • $\begingroup$ @Jan yes, I meant molecular orbitals, but analytical solution. However, it seems that there is none (by 2015). I'll fix the title. $\endgroup$ – Juha Oct 8 '15 at 18:40
  • $\begingroup$ Check them here $\endgroup$ – Jan Oct 8 '15 at 19:02

If I am interpreting this question correctly, you mean experimentally? You may be thinking of this Nature Materials paper

I'm not sure that solved would be the correct word, but the frontier orbitals of water were certainly imaged with STM.

  • $\begingroup$ Wau, cool. I didn't know about the experimental part, but that's really cool. However, I meant the exact analytical solution for the orbitals. In the nature paper you linked they are comparing the experimental results to theoretical ones. These theoretical ones I am after. There should be analytical calculation starting from Schrödingers equation somewhere. In this paper they use DFT because of the NaCl surface and its an approximate solution (although it probably is pretty close to exact one). $\endgroup$ – Juha Oct 8 '15 at 5:41
  • 5
    $\begingroup$ All theoretical calculations are approximate. The exact analytical solutions are available for H-like ions and $\ce{H2^+}$. $\endgroup$ – Ivan Neretin Oct 8 '15 at 8:54
  • $\begingroup$ Oh, I see what you meant Juha. Ivan is right. There is a nice list of QM systems with analytical solutions here: en.wikipedia.org/wiki/… ... Is there a reason why DFT is insufficient for you? $\endgroup$ – Aaron Oct 8 '15 at 9:47
  • $\begingroup$ Exact calculations are difficult I admit. I just wondered how far in the exact calculations they were. I even thought helium was solved. Ok, the answer to my question is that there is no exact solution, yet. The experimental result and DFT is then perfectly fine. $\endgroup$ – Juha Oct 8 '15 at 18:35
  • 1
    $\begingroup$ @Juha The problem is that the $\hat{V}_\mathrm{el-el}$ operator does not permit an analytical solution; if I recall my quantum chemistry class correctly, it has been proven to be unsolveable. $\endgroup$ – Jan Oct 8 '15 at 19:00

First things first, orbitals are not real: neither atomic, nor molecular. Except for, probably, one-electron atoms (also known as hydrogen-like atoms), where one might still argue that to get the orbitals you need to treat the nucleus as a classical point particle that just creates a spherically symmetric electric potential which can be already considered as an approximation. For many-electron systems, one have to necessarily make the so-called mean-field approximation (plus the Born-Oppenheimer approximation prior to that for the case of molecules).

Second, in principle, no further approximation is done in LCAO-MO procedure; it is a purely algebraic trick of expanding a function over a complete set. Such expansion is in principle exact, but the problem is that a complete set is infinite, so in real-world calculations we have to truncate the linear combination at some point which indeed is yet another approximation.

Thirdly, analytical expressions for orbitals are known for one-electron systems only (such as $\ce{H}$ atom, or $\ce{H2+}$ molecular ion) since the electronic Schrödinger equation was solved analytically only for such systems.

And finally, strictly speaking, it is electron density that was imaged with STM, not orbitals. The fact that it closely resembles the density corresponding to orbitals we get in our calculations justifies the quantum chemistry as a whole: at the end of the day, with all the above mentioned approximations we still get something that is close the reality!

  • 1
    $\begingroup$ What do you mean by non-relativistic approximation only? Dirac equation for electron in Coulomb potential is solvable exactly, so you can have an orbital for this case too (with all its caveats, but still true). $\endgroup$ – Ruslan Nov 23 '16 at 19:50
  • $\begingroup$ @Ruslan, honestly speaking, now I have no idea what I meant by saying so. You're right, of course: Dirac equation can be solved for a one-electron system resulting in a set of 4-component wave functions, which are also called orbitals. May be I just meant that these orbitals are different from their usual non-relativistic counterparts, i.e. 2-component wave functions we get when solve the Schrödinger equation. But anaway, currently, this part of answer is wrong, as you noticed. $\endgroup$ – Wildcat Nov 23 '16 at 21:16
  • $\begingroup$ OK, I removed that wrong statement plus provided a link to Wikipedia article on hydrogen-like atom that mentions, among other things, 4-component solutions of Dirac equation for that system. $\endgroup$ – Wildcat Nov 23 '16 at 21:31

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.