I was wondering as to why would two unpaired single electron radicals form a bond. So, I tried to compute the added ground state energies of two particles in two distinct, yet identical boxes, The energy of ground states would be $E_1=\frac{\hbar^2\pi^2}{2mL^2}$ each, so the total energy of the system would be $E=\frac{\hbar^2\pi^2}{mL^2}$. But when I compute the ground state energies of putting both the particles in the same box. (I am keeping the box length the same, just trying to look at it like an approximate model). Then I have a ground state energy of $E=\frac{5\hbar^2\pi^2}{2mL^2}$. So, the ground state energy of having two electrons in a box is higher than the combined energy of the system having one electron each in a separate but identical box. How does one explain bonding to himself. Is there any literature on it, or am I going wrong somewhere? I would appreciate if I was directed towards some literature on it.

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    $\begingroup$ An electron in a molecule (say, H2) doesn't experience the same potential as an electron in an atom (hydrogen atom in this case), because there are now two nuclei to deal with, as well as a second electron. $\endgroup$
    – orthocresol
    Nov 5 '17 at 0:39
  • $\begingroup$ @orthocresol: Is there some QM model which explains this in a better way?? $\endgroup$ Nov 5 '17 at 0:44
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    $\begingroup$ I don't know how much simple models can accurately reproduce molecules - maybe they can to some extent, but I genuinely don't know. The "real" explanation is really quite involved and I don't entirely understand it myself, but you might want to take a look at this page which seems to be a good overview of the topic and has some key references. users.csbsju.edu/~frioux/h2-virial/virialh2.htm $\endgroup$
    – orthocresol
    Nov 5 '17 at 0:48
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    $\begingroup$ Regardless of what model you use, though, you should probably put both electrons in the lowest energy state (one can be spin up and the other spin down. The standard QM particle in a box doesn't include spin). In your particle in box treatment, that results in the same energy for bound vs. unbound, which is not a huge surprise - if you assume the system doesn't change at all upon bond formation, that is equivalent to saying that both atoms have zero interaction with each other. Clearly, it is this change in the system that leads to bonding. $\endgroup$
    – orthocresol
    Nov 5 '17 at 0:50
  • $\begingroup$ @orthocresol Thanks a lot for the article, I'll take a look at it and then I can post questions about the parts I don't completely understand :) $\endgroup$ Nov 5 '17 at 1:52

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