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I guess this is more of a chemistry question, but whatever. I think it's interesting.

Suppose you had two bare atomic nuclei. For concreteness, lets assume the nuclei are the same with atomic number $Z$. Lets bring in a single electron and focus on the ground states of the nuclei.

When the nuclei are far apart, the ground states are degenerate. When we bring the nuclei together, the ground state splits into the bonding and anti-bonding orbitals. Let $\Delta E$ represent some measure of the energy difference between the bonding and anti-bonding orbitals.

From intuition, I would expect $\Delta E$ to increase with decreasing internuclear distance $R$. What happens as $R$ shrinks to zero?

I expect the bonding orbital to become the ground state of an "atom" with charge $2Z$. Is that correct? More importantly, what happens to the anti-bonding orbital?

This isn't an exercise in the Born-Oppenheimer Approximation. I magically hold the nuclei at a distance $R$, so their repulsion doesn't matter. Also, electron-electron repulsion doesn't matter because I only introduce one electron.

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I think that this is fine as a physics questions (it is straight ahead quantum mechanics) despite the chemical terminology. However, for future reference there is a Chemistry.SE beta site for question which are unequivocally chemical in nature. – dmckee Jul 12 '12 at 3:32
I would second @dmckee's comment. I also consider this on topic here, but I also consider it on topic over at Chem.SE. If you don't get any useful answers here, consider asking it over there. – Colin McFaul Jul 12 '12 at 3:53

The split into the bonding and anti-bonding orbitals comes from the LCAO (linear combination of atomic orbitals) approximation, and this approximation breaks down long before the two nuclei merge. I'm not sure it makes sense to ask what becomes of the antibonding orbital when the nuclei get close.

Later: I looked up the LCAO approach in my venerable (1978!) copy of Atkins' book Molecular Quantum Mechanics. The energy of the two states is given by:

$$ E = \frac{\alpha \pm \beta}{1 \pm S} $$


$$ S = \int d\tau_1 \phi_1 \phi_2 $$ $$ \alpha = \int d\tau_1 \phi_1 \hat{H} \phi_1 $$ $$ \beta = \int d\tau_1 \phi_1 \hat{H} \phi_2 $$

Assuming the approximation remains valid up to zero separation the energy will diverge to (I think) infinity. I say "I think" because it actually ends up as zero divided by zero at zero separation.

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Although you are right that the approximation is wrong at these short distances, the idea makes sense as an adiabatic flow, where you can trace each state over time as you slowly change a parameter. This only fails when two eigenvalues collide, and there is no (real) eigenvalue collision (involving these two levels) except at infinite separation (where the two levels become degenerate). – Ron Maimon Jul 12 '12 at 8:18
The new answer is not correct, even in the LCAO approach. First, as I said in my answer, the adiabatic process links the 2Z atom to the separated atom correctly, and shuffles the LCAO antibound state to the n=2 l=1 state. The approximation obviously is invalid at small separations, and also the LCAO answer is not infinity. The overlaps you give just reduce the quantum mechanics problem to a two state system, and at any separation you can diagonalize the matrix without divergence, the matrix elements are finite at all separations. – Ron Maimon Jul 12 '12 at 9:22

The binding orbital is parity invariant to reflections around the center between the two nuclei, while the anti-binding orbital gets a minus sign under this reflection, from the fact that it is mixing the two ground states with a minus sign phase. This property is preserved as you adiabatically bring the nuclei close, and you can trace the level you get using only this fact.

The spin of the electron is decoupled and irrelevant, the binding ground state is necessarily adiabatically linked to the 2Z atom bound state, and the anti-binding state must become the lowest parity odd state. The obvious candidate is the p-orbital (n=2 l=1) which is aligned along the axis of separation of the two nuclei (as the separation becomes infinitesimal). In order for this to not be the endstate of the adiabatic flow of eigenstates, there must be another eigenstate which becomes degenerate with the anti-binding configuration. This is not going to happen, because the states which come from the excited states of atoms (at infinite separation) are always higher in energy than the anti-binding state until the separation gets to zero (this is intuitively clear, although it would be hard to prove). At exact collision, you get a degeneracy of the antibinding state with 2 excited states, since there are 3 degenerate p orbitals at n=2. This is the first degeneracy that develops in the system, which is clear from going the other way.

If you split the nuclear charge of 2Z into two nearby Z nuclei, you add a dipole moment to the nucleus. This dipole moment on average doesn't affect the n=2 S-orbital at first order in perturbations, only at second order. But at first order, this perturbation splits the n=2 p levels so that the one where the p-lobes are aligned with the separation of the nuclei is lower energy. This is the state that's turning into the antibinding.

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I know this looks qualitative, but it isn't--- this is the right answer, there is no chance of mistake. – Ron Maimon Jul 12 '12 at 17:24
Thanks for the reply. No chance of mistake? A bold claim, but you're right in that I was thinking this looks qualitative. How do you know the anti-bonding state doesn't merge into the continuum? Does the anti-bonding state avoid crossing the other excited states? – ChickenGod Jul 18 '12 at 8:22
Furthermore, how does it make any sense for the anti-bonding orbital to become a 2p level? Take hydrogen, whose ground state is -13.6eV. Then when you bring two hydrogen atoms together, you expect the anti-bonding energy to increase monotonically. However, the energy of the 2p level of a Z=2 atom is also -13.6eV, not higher than the ground state of the hydrogen atom. – ChickenGod Jul 18 '12 at 8:30
@ChickenGod: It's not rigorous, but I'll try to make you certain. The reason is that there are no level crossings during the process, and the qualitative perturbation theory of the n=2 states matches what you expect from parity and shape of the endstate. The anti-binding energy doesn't increase monotonically, there's no reason why it should. The reason you know there is no level crossing is because at the first degeneracy, the real antibound state with it's node sheet between the atoms has to be degenerate with a mixed higher n state with a node that makes spheres around the nuclei. Can't be. – Ron Maimon Jul 18 '12 at 9:45

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