Take the 2-minute tour ×
Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers and students. It's 100% free, no registration required.

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

2p static

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

orbitals bonding

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:

2p animated

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.

share|improve this question
4  
When you construct such molecular orbitals as a linear combination of atomic orbitals, you are simply choosing the AO as a basis set to project the Schrödinger equation onto. You are not actually adding them up, neither with nor without their phase. Each MO will then feature a phase of its own, depending on its energy. –  F'x Feb 18 '13 at 12:46
    
@F'x - you should consider making this an answer. –  Richard Terrett Feb 19 '13 at 1:49
    
@RichardTerrett I should find time to expand it… or someone should, and make it an answer :) –  F'x Feb 19 '13 at 8:27
add comment

2 Answers

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.