Talk:Dihydrogen cation

actually technically, teh simplest molecular ion is He⁺.Nergaal (talk) 07:01, 23 September 2008 (UTC)

That's not technically a molecule, but if you count He+, why not H+? :) --Itub (talk) 08:30, 23 September 2008 (UTC)

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Two notes: 1) It may be worth specifying that the Schrodinger Equation for H2+ can be solved exactly only in the Born-Oppenheimer (i.e., clamped nuclei) approximation. The full, electrons+nuclei equation is NOT exactly soluble. 2) It is said that its solution "is included as an example in most quantum chemistry textbooks." In fact I think most modern texts do not discuss the details of the solution of this equation... —Preceding unsigned comment added by L0rents (talk • contribs) 17:24, 7 May 2009 (UTC)

Good points. Regarding the second, I think the article could use clarification, but this system is indeed discussed in every quantum chemistry textbook that I've seen. What is not included is all the details of the analytical solution. Just some historic mention, and then it used as a simple example of a LCAO approximation. --Itub (talk) 11:38, 19 May 2009 (UTC)

I checked in more details some references and I'd like to point out a couple of things.

First, concerning my previous point 1), it is perhaps misleading to say that the clamped-nuclei Schrödinger equation (SE) for H2+ can be "exactly solved". By separation of variables the original 3-dimensional SE is reduced to the solution of three coupled 1-dimensional differential equations. The equation in the variable corresponding to the symmetry around the inter-nuclear axis can be solved straight away but the other two equations are non-trivial and have to be solved numerically (and the numerical algorithm of solution is non-trivial). What *is* true is that with today's computers and using efficient solution strategies one can get "virtually exact" solutions (say, with 15-digit accuracy) in the flap of an eye, so in this sense the problem is solved.

In the article it is said "The analytical solutions for the energy eigenvalues are a generalization of the Lambert W function [...]". I don't believe this is true as the Lambert-W-function expression refers to a 1-dimensional model of the H2+ ion, not to the physically-real one which, of course, lives in 3D space.

As I said, the energy eigenvalues of this latter can be found only numerically (or by various series-expansion expressions, if you wish, but even these come with several caveats).

What I find very suggestive is that even the solution of the SE of such a (supposedly) extremely simple system is in fact rather complicated, even in the Born-Oppenheimer approximation. It is a testament to the immense difficult of the solution of the many-particle SE.

3D H2+ problem is indeed licked

Hello, I have added a theoretical discussion for explanation. I am glad to have included references to the work of Cizek, Paldus and also Dalgarno. For the 1-D version of H2+, the eigenenergy is indeed a standard W function for the homonuclear case. I can assure you that or the 3-D version, one needs a GENERALIZATION as shown in the paper co-authored with J. Grotendorst. Have a look at it. The treatment is indeed in 3D. The authors obtain values for the eigenenergy to within arbitrary accuracy which is independent of choice of basis. The problem can now be down entirely within a computer algebra system. —Preceding unsigned comment added by 171.71.55.235 (talk) 22:59, 26 August 2009 (UTC)

I should also add that the work by Cizek et all is one incredible piece of work: perhaps the biggest paper to ever appear in Phys. Rev. A (at least that I know of). They provided asymptotic series expansions with quite a few terms for the 10 lowest discrete states of H2+. As you can see, quite a bit of analytical work has been applied here and it's probably high time to mention it. If there are any questions, or need for clarifications or desire additional material, please let me know. - Tony —Preceding unsigned comment added by 98.234.179.52 (talk) 02:38, 27 August 2009 (UTC)

Asymptotic expansion wrong?

I'm confused by the asymptotic expansion. If it was correct, the total energy would have expansion (in atomic units) Etot = Eg/u + 1/R = -1/2 + 1/R - 9/4R^4 + ..., which is not consistent with the figure (or with bonding)? Smeuuh (talk) 08:19, 24 August 2023 (UTC)

It looks like that was added back in August 2009, by the Tony mentioned above. Perhaps this is user:TonyMath and if so Tony may care to comment. ref arxiv version: http://arxiv.org/abs/physics/0607081 Graeme Bartlett (talk) 10:37, 24 August 2023 (UTC)

Yes, I put those energies. Why do you say, it is not consistent with the figure? As R tends to infinity, Etot becomes -1/2 a.u. i.e. the ground state of a hydrogen atom, which is correct. However, you cannot use that formula for R->0 or even near the bonding. When R becomes small, subdominant exponential terms kick in and you are in a different regime. The formula were taken from the paper by Cizek et al. I have rechecked it, namely eq. (1) with their Table XI. Note that the internuclear 1/R term is not asymptotic and that will ensure that all energies diverge as R -> 0. I think you might be troubled by the 1/R^4 term which only applies as R-> infinity NOT R->0. Having said all this, I verified if the quoted total energy for the ground state include this internuclear term 1/R. I checked that indeed for the ground state this is a perfect cancellation of the 1/R terms which means E_tot = E_{g/u}. They are definitely in the graph because as R->0, the energy is that of a united atom and the energy graphs clearly have an additive 1/R term i.e. the energies diverge positively as R->0. So I believe, your concerns are addressed: the reported asymptotic formula and the graph of energies are in agreement. I have added a few words of explanation and an extra reference.TonyMath (talk) 12:42, 15 October 2023 (UTC)