Originally Posted By Zhukov:
I've been watching some videos on the Schrodinger equation and the solutions for electron orbitals. I think many people are familiar with , say, the 2p_(x,y,z) orbitals:

https://www.sas.upenn.edu/~milester/courses/chem101/AQMChem101/AQMImages2/image57.gif

One thing that I was thinking is this: Take Neon, which is the lowest atom that has all the 2p orbitals populated. If you spatially superimpose the three 2p_* orbitals, you invariably overlap the regions of probability where an electron can be found. That brings up two questions:
1) To what degree do electrons interact since there is a non-zero probability that any of the 6 electrons could occupy the same space. It seems like there are electromagnetic repulsion would be observed to prevent collisions.
2) Can electrons "swap" states, ie a 2p_x electron swap with a 2p_y electron (ignoring the spin for simplicity)? Would that be detectable?

Originally Posted By eugene_stoner:


I'm going to answer your question in extreme detail as I don't think that many other people on this site would be able to.

I don't know how familiar you are with the Schrodinger equation, but there are a few things about it that we can start by discussing. The Schrodinger equation should be thought of as the quantum equivalent of Newton's law F=ma in the sense that it is assumed to be a fundamental law of nature and it yields the dynamical behavior of the system. For a set of masses and a set of forces acting on those masses, Newton's law will give you a set of differential equations that, when solved, will yield the positions of the masses as a function of time. Likewise, the Schrodinger equation describes a set of particles with interactions and forces between them and, when solved, will yield the probability distribution of finding any of the particles anywhere at any given time.

Newton's laws are actually quite cumbersome to use and may be recast into something known as Hamilton's equation Hamiltonian Mechanics wherein the central object is known as the Hamiltonian (denoted by H) and is essentially just the energy of all the particles in the system.


The Schrodinger equation also has a Hamiltonian (also denoted H), but the key difference is that the Schrodinger equation, being quantum in nature, has promoted the Hamiltonian to an operator in the sense of linear algebra. This means that the Hamiltonian operates on some vector (known as the wavefunction). The eigenvalues of this Hamiltonian operator are the different energies of the system. Any observable quantity (energy, angular momentum, spin, etc.) is promoted to an operator in quantum mechanics with exactly the same formalism.

The piece of the Schrodinger equation that yields these energy states may be written out as follows: H |psi> = E |psi> where E is the scalar energy of a state, and |psi> is the wavefunction. If you solve this equation for a Hydrogen atom, the Hamiltonian takes the following form:

H = ( -D^2 / 2m ) + ( e^2 / r )

where the ( -D^2 / 2m ) part is the kinetic energy operator, and the ( e^2 / r ) is the potential energy that a single electron feels a distance of r away from a positive charge (i.e. the proton in the nucleus). If you solve this equation, you will get that the proper wavefunctions (i.e. the eigenstates) are the orbitals that you describe. Due to a hidden symmetry in the problem (known as an SO(4) symmetry), it turns out that the energy only depends on a quantum number known as n such that the energy of each state is given by:

E_n ~ -1 / n^2

For each value of n however, there are several states with different angular momentum that all have the same energy (these are known as degenerate states). The angular momentum is denoted by a quantum number known as l and it is traditional to refer to it via letters (l = 0 is S, l = 1 is P, l = 2 is D, etc.). Within each value of l there is actually another level of degeneracy protected by something called SO(3) symmetry whereby there are ( 2 l + 1) degenerate states that are each assigned a quantum number known as m. In your original question, you reference the l = 1 is P state which has a degeneracy of ( 2 l + 1 = 3) which describes three states labeled by m = -1, 0, or +1. These three states are the px, py, pz orbitals and we can now begin to address your question:


1)  Strictly speaking, as we discussed above, these orbitals denoted by the n, l, and m quantum number are only valid for a single electron experiencing the Columb potential from some positive charge. Once you start to add in multiple electrons and include terms to the Hamiltonian that describe electron-electron Colomb repulsion, there is no longer an exact solution to the problem and, strictly speaking, the px, py, pz orbitals are no longer true eigenstates of the system. Generally though, the electron-electron interaction may be weak and can be either ignored, or renormalized into the problem such that the n, l, and m quantum numbers are still approximately good and that we may still gain insight in treating the problem through them.

There is actually a fourth quantum number in this problem known as spin (denoted by the quantum number s) and it may take on the values of s = -1/2 or + 1/2 per value of m. Since both values of spin are degenerate in energy in this problem (as we have not included any terms int he Hamiltonian which couple the spin and orbital sectors), this essentially means that each P-orbital can hold two and only two electrons.

So, to answer the question, in the limit wherein the P-orbitals are exact solutions, the electrons cannot "collide" via electrostatic repulsion because they literally do not feel any Coulomb force between themselves as there is no such term in the Hamiltonian. The details of how the interaction works once you try to add it into the Hamiltonian depends on the particular system in question.

2) The electrons do actually "feel" each other in this problem via the quantum degeneracy "force" (also known as the Pauli exclusion principle). This means that it a system has an electron in one state (i.e. with one set of quantum numbers n, l, m, and s ), the system cannot have a second electron in the same state with the same quantum numbers. This means that your hopping question depends very strongly on how many electrons are in the P-orbitals. For example, if the px orbital has two electrons in it (i.e. the electrons have quantum numbers (n = 1, l = 1, m = -1, s  = -1/2) and (n = 1, l = 1, m = -1, s  = +1/2)) then another electron in say the py orbital (with say n = 1, l = 1, m = 0, s  = -1/2) cannot sawap in or hop into the px orbital. Although it is not a true "force" the Pauli exclusion principle effectively makes it so that the electrons in this example feel an infinitely strong repulsion force between the px and py orbitals.

If, on the other hand, there was only a single electron in the problem, then Epstein didn't kill himself, so there is no reason why it could not jump from px to py to pz as it would not feel any Pauli exclusion pressure and all three of these orbitals are degenerate and at the same energy. In fact, for fixed n and l quantum numbers (for example) there would be a 1/3 chance of finding the single electron in either px, py, or pz when you observe it. It would essentially be in all three of the orbitals at once until you perform a measurement to collapse its wavefunction into a specific orbital (i.e. a specific value of m). After you did that, if your experiment was not sensitive to spin, you would furthermore have a 1/2 chance of finding the particle in the s = -1/2 (spin down) or the s = +1/2 (spin up) state.

Originally Posted By redgineer:
Just got back from deer hunting, and was surprised to come across this on arfcom. I'm just a chemical engineer with a master's, and school was a while ago. That being said...I'll do my best to explain in simple terms. eugene_stoner can correct me if I'm wrong.

The orbital shapes will change based on the valence, also known as hybridization. VSEPR theory is what I learned in inorganic chem. This is easily observed when looking at the geometry of polyatomic molecules. For example, methane is tetrahedral, while carbon dioxide is linear. Those shapes maximize the space between orbitals for each molecule.

For neon, the orbitals will have a teardrop shape that comes to a point at the center. This eliminates overlap between px, py, pz orbitals. I don't think I've ever seen p orbitals represented as spheres until reading this thread.

Originally Posted By Zhukov:
I'm familiar with sp, sp2, and sp3 hybrids. It explains 1s2, 2p4 in the case of carbon, but what about 1s2, 2p6?

ETA: I'm actually sitting in my hunting blind right now. 2 mediocre does about 15 minutes ago, morning else yet.