# Quantum mechanics and atomic bonding

I'm learning quantum mechanics in high school this year, and I have several doubts. I've done my research on various websites but my understanding is still fuzzy. I understand that when I punch a wall or sit on a chair I don"t go through it due to quantum degeneracy pressure. I've studied the Pauli exclusion principle, but here"s my first question:

1. What exactly is quantum degeneracy pressure? I've read answers that say "Atomic orbitals resist squishing", and the like, but I don"t understand why.

2. If orbitals resist squishing, why do covalent bonds form?

From the Wikipedia picture of a H2 bond it seems like the s orbitals of the individual atoms were squished from a circle(2-D) to an ellipse. How do we know when this squishing results in a lower energy state and when it doesn't? If the wall was made of atoms that didn't have an octet and my hand too would my hand bond with the door when I punched it?

-

The degeneracy pressure does not directly apply here, I think. It will be important for neutron stars. The particles that matter is made up with (electrons, protons, neutrons) all have spin 1/2. That means that they are fermions.

The wavefunction, which contains all the information about the system, has so change its sign when two particles are exchanged. Since particles of the same kind are indistinguishable, that means two particles cannot be in the exact same quantum state (something like ”spot”). That means that even if you compress something, the neutrons will still not fit into a single spot. That will give you a neutron star, where the degeneracy pressure works against its collapse.

The fact that you do not fall through the chair should just be the electromagnetic repulsion. The electron shells of various atoms repel each other. When you from a covalent bond, the wavefunctions of both atoms overlap. Since the electrons are indistinguishable, that will cause them to spread between the two nuclei.

If you look at the overlapping wavefunction, you find eventually that there are combinations with more and with less energy. See the $\sigma$ and the $\sigma^*$ orbitals in $\mathrm H_2$. Both electrons will go into the $\sigma$ molecular orbital which has lower energy. Therefore, it will form a covalent bound. In $\mathrm{He}_2$, there two more electrons, which just have to occupy the $\sigma^*$ since there is no other molecular orbital around. Since that has an energy which is too high, it is anti-binding. The total energy in the bond is no less than for the individual atoms, so there will not be any bonding.

Your hand and the wall is not likely to bond as a macroscopic bond, since they are just that big. But I could imagine that individual atoms or molecules bind to some of the wall.

-
Wikipedia says "It is quantum degeneracy pressure and not electrostatic repulsion as previously assumed that makes solid matter impervious." Also, as the s shell is spherical, the centre of negative and positive charges will be the same in say, hydrogen, and so why should atoms repel when they come close? I think I got the covalent bond, thanks. – user42991 Mar 21 '14 at 8:26
That is interesting, do you have a link? The spherical shape only applies if there is nothing there (the system is spherical). As soon as you introduce another atom, it will break the symmetry. Then the atoms will start to have a little dipole moment. – Martin Ueding Mar 21 '14 at 10:35
Line 3: en.wikipedia.org/wiki/Electron_degeneracy_pressure Also, when you say " As soon as you introduce another atom, it will break the symmetry" Is this due to instantaneous dipoles(Vaan der Waals)? – user42991 Mar 21 '14 at 12:33
What I mean is that the hamiltonian $\hat H$ will not be rotationally symmetric any more with two atoms. That will break the symmetry of the wavefunctions. The wavefunction is not centered around the nucleus any more, therefore you have an electric dipole moment. I think this is Van-der-Walls-Forces in essence. – Martin Ueding Mar 21 '14 at 12:42
Ok. We have a graph in our textbook which I think represents the Lennard-Jones potential. So for a given set of atoms, there may or may not be a lower energy state that includes orbital overlapping(the potential well, a covalent bond) but once you get too close there is always repulsion due to the Pauli exclusion principle? That"s what I gather from en.wikipedia.org/wiki/Lennard-Jones_potential – user42991 Mar 21 '14 at 12:49