# Spin of Nucleus

N has 7 protons and 7 neutrons per atom. So the spin is S=1 for one N-atom. Why is this? I know that they do pairs in the shell model. So I only need to look at the spin of a single proton and a single neutron in N. If both spins are parallel it is 1 or -1. But why is 0 not an option for this (antiparallel spin ex. S = $$\frac{1}{2}$$ for proton and S=$$-\frac{1}{2}$$ for neutron)?

Therefore the spin of $$N_2$$ must be 0 or 2, right? 1 isnt possible because 0 isn't an option for a single N-atom as far as I understand.

I just do not understand the detail. I need to tell if the angular momentum quantum number of $$N_2$$ can be even, uneven or both if the wave function of the nucleus is antisymmetric. So I need to tell first if the wavefunction is symmetric or not.

• From Blatt & Weiskopf: "Since nuclei are built up of neutrons and protons, each possesses an angular momentum I which is the combined effect of the intrinsic spin of the constituents and of the angular momentum of the orbital motion within the nucleus" Nov 20 '20 at 16:23

In the nuclear shell model, $$^{14}$$N is seen as an inert $$^{12}$$C core, with 2 extra nucleons. The ground state for the strong nuclear force is going to put these nucleons in the same spatial state (a symmetric S state) and the same spin state (which for spin 1/2 particles is the spin-1 triplet state). Anti-symmetrization of the wave function is accomplished in the isospin sector via the singlet isospin wave function:

$$|I=0, I_3=0\rangle = \frac 1 {\sqrt 2}[|p\rangle|n\rangle - |n\rangle|p\rangle]$$

It's important to note that individual nucleons do not have a definite "proton" or "neutron" identity with isospin. It is exactly the same as a spin 1/2 particle not having a definite spin.

So we can treat each $$^{14}$$N as indistinguishable vector bosons.

You can work out the Clebsch-Gordan coefficients for combining two vectors, but the general structure is:

$${\bf 3} \otimes {\bf 3} = {\bf 5}_S \oplus {\bf 3}_A \oplus {\bf 1}_S$$

which means the spin-2 and spin-0 multiplets are symmetric under interchange and the spin-1 combination is antisymmetric. For example:

$$|J=1,J_z=0\rangle = \frac 1 {\sqrt 2}[|1,+1\rangle_1|1,-1\rangle_2 - |1,-1\rangle_1|1,+1\rangle_2]$$

while:

$$|J=0,J_z=0\rangle = \frac 1 {\sqrt 3}[|1,+1\rangle_1|1,-1\rangle_2 + |1,-1\rangle_1|1,+1\rangle_2 - |1,0\rangle_1|1,0\rangle_2 ]$$

The molecular wave function then needs to have the same symmetry as the spin wave function to make the overall wave function symmetric.