# Trying to understand how nucleon number relates to spin of nucleus

Nuclei with even number of protons and of neutrons have a net spin of zero.

This I understand, as to have the lowest energy state, by the Pauli exclusion principle, all protons will pair wth one of each pair being spin up and th other spin down; and all neutrons will pair as well.

The next 'rule' is

Nuclei with odd number of protons and neutrons will have integer spin of 1,2,3....

In all of the sources I have read, I have not seen spin of 0 mentioned here. This is my first point of confusion. Could it not be that the net proton and net neutron aline with opposite spins to give 0? Perhaps the answer lies in considering energy- maybe the nucleus is in a lower energy state if the spins of the net proton an neutron are aligned? But then surely a spin-zero excited nucleus CAN exist.

And when considering energy, I do not see how we can achieve nuclear spin states of 2, 3 etc. This would imply that there are some protons/neutrons which could pair with another of the same nucleon, but are not doing so? Why would this be energetically favourable, and a spin 0 nucleus is not?

Final 'rule':

A nucleus with odd number of protons or neutrons but the other nucleon has an even number has integer spin: 1/2, 3/2, 5/2...

Again, my confusion with this rule is that for spin magnitudes greater than 1/2, there must be nucleons which could be pairing but are not?

• – user139561
Dec 20, 2016 at 12:57
• – user4552
Jul 5, 2019 at 21:50

Nuclear angular momentum includes not only nucleon spin but also nucleon orbital angular momentum. The rules developed for building a nucleus in the nuclear shell model therefore bear a lot of resemblance to Hund's rules for filling electron shells, although the pairing interaction is stronger among nucleons (for which there are two species) than among electrons.

Here's a hand-waving way to think of it:

1. As you say, even-even nuclei like to relax to spin and parity $J^P = 0^+$.

2. An even-odd or odd-even nucleus must have half-integer $J$, because of the spin of the unpaired nucleon. If the extra nucleon also carries orbital angular momentum — for instance, if the extra nucleon is in a $p$-shell or a $d$-shell — the orbital angular momentum must be added to the total spin of the nucleus. Here are a couple of examples where you can think of a single extra nucleon orbiting a $0^+$ core.

3. An odd-odd nucleus must have integer $J$, but not necessarily $J=0$. One model is an even-even core orbited by a deuteron.

Spin is a funky quantum version of angular momentum that comes in units of $\hbar.$ There are two ways that you can space things by $\hbar$: you can either have integer spin, $\{\dots,-2\hbar, -\hbar, 0, \hbar, 2\hbar, \dots\},$ or you can have half-integer spin, $\{\dots,-\frac32\hbar, -\frac12\hbar, \frac12\hbar, \frac32\hbar,\dots\}.$

These amounts are components of angular momentum along an arbitrary axis. Let me give you a quick quantum-angular-momentum refresher. So we choose this arbitrary $z$-axis and measure $L_z = \hbar m$ for either integer or half-integer $m$, and then we are uncertain about $L_x$ and $L_y$ by the uncertainty principle. However it turns out that the total-angular-momentum operator $L^2 = L_x^2 + L_y^2 + L_z^2$ commutes with $L_z$ and therefore we don't have an uncertainty relation there, we can simultaneously know both $L^2 = \hbar^2 \ell (\ell + 1)$ and $L_z = \hbar ~m.$ This bounds $|m| \lt \ell.$ Both $m$ and $\ell$ will either be integers or half-integers together, if I recall correctly.

Now for quarks, and protons and neutrons are both made up of three quarks and similarly have spin-1/2, meaning $\ell = 1/2.$ This has a nice interpretation which turns out to be wrong: if you're thinking that it's as easy as "two quarks spin one way and the other spins the other way" it turns out that's overly simplistic and this is an open mystery.

If there is an even amount of half-integer spin total, either from (even number of protons + even number of neutrons) or from (odd number of protons + odd number of neutrons), then the nucleus has an even number of half-integer spins and has an integer spin in general. The argument is really simple, just line up all of the spins so that they all point up, you find that if there are N neutrons and P protons then there is a total spin in the z-direction of (N + P)/2, which is an integer. "But they are not all spinning up!" you object. Well that's fine, start flipping some of them down: each time they will go from $+\hbar/2$ to $-\hbar/2$ and they will therefore reduce the spin in the $z$ direction by $\hbar$, changing the integer by another integer. So you don't need to worry about anything else, every configuration has an integer spin upwards.

You express a worry that this spin component cannot be 0 but in many cases it is, even when you have an odd + odd combination. For example, check out the isotopes of cobalt, atomic number 27. Cobalt is kinda funky and therefore has some awesome MRI uses; its natural form is all 59Co and it has nucelar spin 7/2 in its most common case. But it also has an isotope of 54Co which we can confidently say would have spin-zero. It's odd + odd and all the spins should pair up oppositely.

Similarly if $N+P$ is odd then you have a half-integer spin up when they're all spinning up, and as you make some of them spin down they change the spin by a full $\hbar$ and therefore you map half-integers to half-integers still. So again, there is no change. And indeed as you're surmising, the spins are simply unpaired to create a net spin one way or the other.

Now as @rob says the nucleus also has internal angular momentum due to the nuclear shells, so it may be a little more complicated in general to work out the total angular momentum of the nucleus. However I noticed that you are specifically asking about the spin angular momentum and this is a plausible way to describe it. Furthermore the nuclear shells force certain spins to be paired, otherwise a nucleon would have to "jump" to the next shell which requires a ton of energy.

• Nice answer. My only point of clarification would be that the "intrinsic spin" of a composite system like a nucleus includes all the angular momentum of its constituents, both their own intrinsic spins and their orbital angular momenta; you can't distinguish between the two unless some symmetry in the Hamiltonian separates them. …
– rob
Dec 20, 2016 at 16:49
• … For instance, there are a couple dozen different ways to combine two protons and two neutrons into a $J^P=0^+$ composite state, and all of them contribute to the ground-state wavefunction for the helium nucleus. Even for nucleons: proton spin has contributions from valence quark spin, virtual gluon spin, virtual quark-antiquark ("sea quark") spin, and the orbital angular momenta of all these constituents.
– rob
Dec 20, 2016 at 16:49
• Ah, I didn't realize that, but it makes a lot of sense. Dec 20, 2016 at 17:16