Are protons and neutrons affected by the Pauli Exclusion Principle?

I'm very confused about the Pauli exclusion principle. Wikipedia states it as "two identical fermions cannot occupy the same quantum state in a quantum system". I understand this for electrons that for each energy level in an atom there are two possible electrons that may occupy this energy state but with opposite spin numbers.

What about for protons and neutrons?

Protons and neutrons are both fermions, so why in a nucleus can multiple protons and neutrons simultaneously exist. I understand that neutrons and protons are not identical fermions but considering them individually, suppose in a nucleus with X protons, are the energies of individual protons different from one another (and similarly for neutrons in the nucleus)?

Apologies, I'm not very familiar with quantum theory or the maths involved. I super confused about how the exclusion principle works for protons and neutrons. The only explanations I've been able to find consider 2 protons and state that they can have different spin. What happens when we consider more than 2 protons/neutrons?

• Have a look at: physics.stackexchange.com/q/320858 – flippiefanus Sep 5 '17 at 5:27
• I'd add to what JR said that the wavefunction of identical indistinguishable fermions is described by the en.wikipedia.org/wiki/Slater_determinant So it's just the mathematical nature of fermions' descriptions (vs bosons) that gives rise to this kind of result. – John Forkosh Sep 5 '17 at 6:42
• The Pauli exclusion principle for Neutrons is what keeps a neutron star from collapsing. – Arthur Sep 5 '17 at 7:47
• The wavefunction of multiple fermions is in general a superposition of Slater determinants. – Robin Ekman Sep 5 '17 at 16:10

To a reasonable approximation the protons and neutrons in a nucleus occupy nuclear orbitals in the same way that electrons occupy atomic orbitals. This description of the nucleus is known as the shell model. The exclusion principle applies to all fermions, including protons and neutrons, so the protons and neutrons pair up two per orbital, just as electrons do. Note that the protons and neutrons have their own separate sets of orbitals.

I say to a reasonable approximation because neither nuclear orbitals nor atomic orbitals really exist. The atomic orbitals we all know and love, the $1s$, $2s$, etc, appear in an approximation known as the mean field. However the electron-electron pair repulsion mixes up the atomic orbitals so strictly speaking they don't exist as individual separate orbitals. This effect is small enough to be ignored (mostly) in atoms, but in nuclei the nucleons are so close that the nuclear orbitals are heavily mixed. That means we have to accept that the shell model may be a good qualitative description, but we have to be cautious about pushing it further than that.

• The only comment I would add to this is that if it weren't for Pauli exclusion, adding a new neutron would almost always make a nucleus more stable, as free neutrons only weakly decay into protons and their leading term would be to lower energy by the strong force. The shape of the valley of stability therefore owes a lot to this fact that neutrons have to be inserted in states with higher angular momentum in the nucleus. – CR Drost Sep 5 '17 at 5:44
• This raises the question, what are they orbitting? I can no longer remember my Quantum Physics lectures :( – Pureferret Sep 5 '17 at 10:07
• @Pureferret: They don't orbit in the classical sense. The wavefunction that describes the nucleons is centred on the overall centre of mass of all the nucleons involved. – John Rennie Sep 5 '17 at 10:17
• – dmckee Sep 5 '17 at 14:48
• @JohnRennie is that not how orbiting ultimately works (asked as the most ignorant on the subject) ? – deg Sep 6 '17 at 5:50

Neutrons are certainly distinguishable from protons, and both separately satisfy the Pauli exclusion principle, i.e. the exclusion is on identical protons by themselves, and on identical neutrons by themselves.

The nuclear force is largely independent of the electric charge and affects neutrons and protons in approximately the same wave. As a result, both species live in a common potential knows as Wood-Saxon (or inverted Fermi) potential. This is the average potential felt by one nucleon and generated by all other nucleons, irrespective of species.

Whereas neutrons are not subject to a Coulomb force, protons are so, as a result of the Coulomb repulsion, the energy levels of protons are typically higher than those of the neutrons, and the proton part of the potential also has a " tunnelling lip". This is well illustrated in the figure below, which sketches the filling of (not realistic) nuclear levels for a nucleus with 6 protons and 6 neutrons. Note that the nuclear potential is spherical, so the split in the middle is meant to cleanly separate the radial shape of the proton and neutron potentials. The nuclear radius $r_0 A^{1/3}$ is very nearly the same for both species.

Note that in addition to the above two interactions, there is also a very strong inverted nuclear spin-orbit interaction, which splits the spherical symmetry and produces energy levels with strong $j$-dependence, where $j$ is the total angular momentum of a nucleon.