How do the orbitals in atoms synchronize their phases during the formation of a bond?

Question

Suppose I have two hydrogen atoms, each with one electron in their 1s orbital. Then we can represent the wave functions of these two as say $|\Psi\rangle_1$ and $|\Psi\rangle_2$ for each of the two atoms respectively. Since we have two identical atoms, $|\Psi\rangle_2$ is just $|\Psi\rangle_1$ translated to a different point in space with some phase difference, so $|\Psi\rangle_2 = e^{i\theta}|\Psi\rangle_1'$ where the $'$ denotes that we've shifted the wavefunction in space.

We know that if the two atoms have wave functions in phase with each other ($\theta = 0$) then when they come together the orbitals will combine with each other and a $\sigma$-bond will be formed. Equally if the wave functions are completely out of phase with each other then the anti-bonding orbital of the combined system would have 2 electrons while the bonding orbital would have 0 and since the anti-bonding orbital has higher energy this transformation is not energetically favourable so no reaction occurs.

I'm curious about what happens when the phase is not equal to either $0$ or $\pi$, which is a case that happens with probability $1$. Clearly the atoms do end up reacting in real life so the phases must get synchronized somehow before the reaction takes place. I'd like to ask how exactly this happens.

Answer 1

The expression $|\Psi\rangle_2 = e^{i\theta}|\Psi\rangle_1'$, for $\theta \neq k \pi$ ($k \in \mathbf{Z}$) is likely to reach to a complex function. This situation is unlikely under an attracting potential because the position tends to be more precise (requires a real wave function) at the expense of the momentum (which requires a complex wave function), even if the exact wavefunction is unknown.

On the other hand if we consider a translation operator $\hat{T}$ acting on a $s$ orbital of one hydrogen atom of a site $i$ transferring the orbital to another site $j$ in a chain of $n > 0$ hydrogen atoms having $a$ as interatomic distance in a $x$ axis $\mathbf{R} = l a \hat{e_x}$ with $l=1,2,...,n$. The case $n = 2$ is the hydrogen molecule $H_2$. Generally, $\hat{T}$ and the Hamiltonian $\hat{H}$ do not commute : $[\hat{H},\hat{T}] \neq 0$ this means it is "unlikely" that they share the same eigenvectors because of the repulsion $V_{ee}=+\sum_{ij}\frac{1}{|r_i-r_j|}$.

We can consider the case $V_{ee}$ can be ignored so that $[\hat{H},\hat{T}] = 0$, the operator $\hat{T}$ gives $\hat{T}|\Psi\rangle_1 = e^{i\mathbf{k}\mathbf{R}} |\Psi\rangle_1$, if $k=\frac{2\pi}{a}$, therefore $\hat{T}|\Psi\rangle_1 = e^{i 2\pi l} |\Psi\rangle_1 = |\Psi\rangle_2 $. Here the phase shift is due to the translation operator and can only be a positive integer (1). A similar result compared to the expression above, but there is no anti-bonding.

This result is due to the fact that $\hat{T}$ is a generator of bonding orbitals, transferring the wave function from one atom to another. The state described above is similar to a singlet state, where $V_{ee}$ is ignored because there is an exchange interaction (attractive) $V_{x}$ which lowers significantly the impact of $V_{ee}$, for $n=2$, $\Psi_B \propto (|\Psi\rangle_1 + |\Psi\rangle_2)$. Now for the antibonding is due to the fact that $V_{x}$ is only a pairwise interaction, it means the exchange do not act on more than two states, this anti-bonding state has the impact of $V_{ee}$ and $[\hat{H},\hat{T}] \neq 0$, the state is not an eigenvectors of $\hat{T}$, the only possible choice is $\Psi_A \propto (|\Psi\rangle_1 - |\Psi\rangle_2)$.

I did not answer the question directly by I hope it helps.