# How to understand the formation of covalent bond from the viewpoint of wavefunction interaction?

The formation of the covalent bond is usually understood from the electron sharing, like explained by the following figure:

However, this simple physical picture seemingly hinders many things. In quantum mechanics or chemistry, the electron is represented fully by the wavefunction.

Therefore, how can I understand the formation of the covalent bond from electron/wavefunction interaction?

I have gained some initial ideas when two wavefunctions are touched, such as:

• Deformation;
• Electrostatic attraction;
• Exchange;
• Polarization;
• Charge transfer.

These processes/concepts play an important role when two wavefunctions interacted. (I'm sorry I can't find concrete references for these concepts or processes.) In these processes, the attraction interaction (bonding) will compete with the repulsion interaction (antibonding) to find the ground state.

I post this question to look a full understanding of the formation of the covalent bond from electron wavefunction interaction.

• When you calculate the full system, let said, two atoms, you will have only one wavefunction for the whole system, not two "interacting" wavefunctions (one for each atom). By the way, the wavefunctions do not interact one with other. The bond is "defined" starting from the electronic density that is calculated from the probability density using the wavefunction.
– Camps
Oct 28 '20 at 13:43
• Basically, when two atoms are very far apart, the wavefunction of the total system can be taken as a product of two wavefunctions to a high degree of accuracy. In other words, the system hamiltonian is the sum of the hamiltonians for each atom, with no coupling terms. So, what you're asking is basically how good the atom-only basis is at representing the case when there is a strong interaction between two atoms. The answer is that it will be very bad. You can attempt to say what physical effects lead to this breakdown, as you've suggested, but this is never unique and always contentious. Oct 28 '20 at 18:10
• @jheindel you should submit that as an answer Oct 29 '20 at 16:10
• @Camps Great! you should submit that as an answer Oct 29 '20 at 16:11
• This is a good question, and probably the best result will be to get multiple answers explaining different ways of thinking about it. Oct 29 '20 at 16:12

There is more than one way to answer this question well. I'll give one answer here.

Let's treat the two atoms (that will form a covalent bond) as two localized election states (orbitals). The detailed shape isn't important. The Hamiltonian of this system could be written:

$$H = V (n_{\uparrow,1} n_{\downarrow,1}+ n_{\uparrow,2} n_{\downarrow,2}) - t \sum \limits_{s=\pm1/2} (c^\dagger_{s,2} c_{s,1} + c^\dagger_{s,1} c_{s,2})$$

Here $$n_{s,i}$$ is the total number of electrons on site $$i$$ (0,1, or 2) and the $$c^\dagger_{s,1} c_{s,2}$$ terms 'hop' an electron with spin $$s$$ from site 1 to 2 (and vice versa).

Basically here you have $$V$$ representing an electron-electron repulsion (they don't want to be on the same site) and $$t$$ representing a tunneling between the two sites. This tunneling is roughly proportional to the wavefunction overlap between the site 1 and 2:

$$t \approx \int d^3r \psi_1 \psi_2$$

Now, when the two atoms are far away, that overlap will be small and $$t\to 0$$. In this limit we just have two isolated atoms. If we put two electrons in the system, the ground state would have one on each site with any combination of spins.

As you bring the two atoms closer together, the tunneling becomes finite and it breaks the degeneracy, lowering the energy of a pair of opposite aligned spins because they can tunnel back and forth and pick up a $$-t^2/V$$ decrease in energy (from second order degenerate perturbation theory.

This example is essentially the Hubbard model.

• Very nice ideas. When the competition between kinetic energy and interaction energy is minimized, the covalent bond is formed?
– Jack
Oct 30 '20 at 0:57