# The AntiBonding Orbital with Shrinking Interatomic Distance

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:

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

where:

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

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