Bonding and anti-bonding orbitals in the light of time-dependent Schrödinger equation?

Question

In organic chemistry, people draw 2p orbitals like this:

and then they explain how the orbitals combine to non-bonding (π*) or bonding (π) molecular orbitals, like this:

depending on whether the orbitals overlap out-of-phase (blue + white) or in-phase (blue + blue, or white + white).

But in reality, the orbitals are time-dependent solutions of the time-dependent Schrödinger equation, so they are not really like the first image above, but more like this:

So, what does it mean, when the orbitals overlap in-phase or out-of-phase? As far as I understand, the phase difference of the orbitals of two random atoms could be anything from 0 to 2π, and the cases of perfect overlap (0 phase difference) or perfect anti-overlap (π, or half-cycle, phase difference) should be quite improbable.

Answer 1

As F'x has indicated, the sum of atomic orbitals $\phi_k$ (AO) that form the molecular orbital $\psi_j$ (MO) $$ \psi_j = \sum_k c_{jk} \phi_k $$ is just a solution (an approximate solution) to a self-consistent variational procedure to solve the time-independent Schrödinger equation, obtained after choosing a particularly well suited basis to represent the solution. They approximately solve this equation: $${\sf F}\psi_j = \epsilon_j \psi_j$$ where ${\sf F}$ is the Fock operator or if you wish, an effective one-electron energy operator and $\epsilon_j$ is an effective "one-electron" energy that gives a good approximation to the ionization potential from that orbital.

Then you have to construct your overall n-electron wave function $\Psi({\bf x}_1,\ldots,{\bf x}_n)$, which in the simplest case (the one that follows from the minimal molecular orbital theory) is a Slater determinant. You can notice that for this wave function the coefficients $c_{jk}$ are not unique. Following the general properties of determinants, by a similarity transformation of the basis you can get a new set of coefficients $c'_{jk}$ that will not solve the Fock equation, but after which you obtain the same overall electronic wave function.

In everything that I have said so far, there are no time-dependent phases: we only have one "real" (observable) energy, an eigenvalue of the time-independent Schrödinger equations for the electronic Hamiltonian $$ {\sf H}_{elec} \Psi({\bf x}_1,\ldots,{\bf x}_n) = E \Psi({\bf x}_1,\ldots,{\bf x}_n)$$ for a fixed geometry. Now is where you can put the time-dependent Schrödinger equation into work. In principle you could create electronic superposition states combining $\Psi({\bf x}_1,\ldots,{\bf x}_n)$ of different energy. These are the ones that would evolve in a manner similar to what you have assumed, with time-dependent evolving phases.

Answer 2

It is important to note that the phases of the MOs are irrelevant. Assuming that the Hamiltonian can be solved with a Slater determinant $\Psi_0=|\phi_1(1)\dots\phi_N(N)|$ of MOs $\phi_i$ with eigenvalue $E:= \sum \epsilon_i$, where $\epsilon_i$ are the eigenvalues of the Fock operator, then your time-dependent solution will be $$ \Psi(t) = e^{-\frac{Et}{\Im \hbar}} \Psi_0.$$ Now you see that no phase can be uniquely attributed to any one MO. Canonically the MOs are chosen to be real-valued.