Energetics of bonding.

This post is related to things we will cover in the future.
added: You can use this:
$e^2/(4 \pi \epsilon_o) = 1.44 \:eV nm$
Also, here is another useful relationship:
$\hbar^2/m = .076 \:eV nm^2$   

This post is about the energetics of chemical bonding.  Why would 2 nitrogen atoms in the air prefer to be in the form of N2? Or why would two hydrogen atoms prefer to be paired, to form H2, rather than remaining separate? Additionally, what determines the distance between the two nuclei in a molecule like N2 or H2? Perhaps we can learn about the fundamental origin of bonding and of bond length by looking at the simplest case, H2+. I think studying that will help us understand the energetics of sp2 bonding as well.

H2+ involves two protons and one electron. I believe we can write the wave equation for that as:
$\frac{-\hbar^2}{2m} \bigtriangledown^2 \psi(\vec{r}) +[ \frac{-e^2}{4 \pi \epsilon_{o} r} + \frac{-e^2}{4 \pi \epsilon_{o} \:(\vec{r}-b \hat{x})}] \:\psi(\vec{r}) = E  \:\psi(\vec{r})$,
where the first term is the kinetic energy term for our one electron, and the 2nd term is the potential energy of the electron-proton interactions (two protons). b is the distance between the two protons. We don't know what that should be, but we can vary b to find out how to get the lowest energy.

Finding the exact ground state wave-function for the above wave equation is very challenging, but suppose we try an intuitive molecular state of the form,
$\psi_m( \: \vec{r}) = \frac{c_m}{\sqrt{2}} ( \psi_{1s} (r) + \psi_{1s}(\:\vec{r}-b \hat{x}))$
where b is, again. the distance between the two protons. Note that $c_m$ is a function of b, is close to 1 for most values of b, and is unit-less. It has to be evaluated by a normalization integral for each value of b. (added 2-26-18)
$\psi_{1s} (\vec{r}) =  \frac{1}{\sqrt{\pi a^3}} e^{-r/a}$
is the atomic ground state, with a fixed at .053 nm (the Bohr radius).
So then $\psi_{1s}(\:\vec{r}-b \hat{x}))=  \frac{1}{\sqrt{\pi a^3}} e^{-\sqrt{(x-b)^2 + y^2 +z^2)}/a}$

So then the total potential energy of this system of three objects would consist of:
1) the proton-proton repulsion, which is simply $U_{p-p} =  \frac{+e^2}{4 \pi \epsilon_{o}b}$ and,
2) the expectation value of the electron-proton interaction potential energy. That involves an integral of the wave function squared over 3d space, right?  I think the integrand for that may be:
$\psi_m (\vec{r}) [ \frac{-e^2}{4 \pi \epsilon_{o} r} + \frac{-e^2}{4 \pi \epsilon_{o} \:(\vec{r}-b \hat{x})}] \:\psi_m (\vec{r})$

So there is a positive potential energy associated with the proton-proton repulsion, and a negative potential energy associated with electron proton attraction. So we wonder: which one is bigger? How much bigger would it have to be to justify bonding? How do they depend on b, the separation between the two protons (and the parameter that appears in both the repulsive and attractive potential energy terms)?

Bottom line: Could someone calculate this for a few values of b and see what you get. Like say b = 1.5a, 2a and 2.5a, where a is the fixed Bohr radius, 0.053 nm.
Better: Just input a as a fixed value, .053 nm, and calculate for b values of:
0.06 nm, 0.08 nm,  0.10 nm, 0.12nm,  0.14 nm and 0.16 nm.
See what you get with that.