Undergrad Angular momentum of an Odd-Odd nucleus (in its ground state)

[JD_PM]

TL;DR

Issue explaining why sometimes the table just gives one of the possible values for the total AM of an O/O nucleus.

What I know is the following:

The total angular momentum of the nucleus is just the total sum of the angular momentum of each nucleon.

If the nucleons are even the total angular momentum in the ground state will simply be $0+$.

If the odd number of nucleons is close to one of the magic numbers then the shell model can explain what total angular momentum we'll get.

There are 3 cases to consider:

1) E/E nucleus (40-Ca for instance): Both proton and neutron numbers are even so the total angular momentum of the nucleus (I have read in other sources they call it Isospin) in the ground state will be $0+$.

2) E/O nucleus (41-Ca for instance): The proton number is even. Thus the contribution by protons to the final angular momentum is none because $j_p = 0$. What about neutrons? The unpaired neutron will be located in the 1f7/2 shell, then $j_n = 7/2-$. Then the final angular momentum of 41-Ca in its GS will be $I = 7/2-$.

Here comes my issue:

3) O/O nucleus (14-N for instance): here both proton and neutron numbers are odd so we expect that both contribute to the final angular momentum of 14-N.

Proton contribution: The unpaired proton will be located in the 1p1/2 shell, then $j_p = 1/2-$.

Neutron contribution: The unpaired neutron will be located in the 1p1/2 shell, then $j_n = 1/2-$.

Now it is about adding both AMs up to get $I$:

$$I = 1+, 0+$$

But in the table just 1+ is given. Why?

 

[snorkack]

Summary: Issue explaining why sometimes the table just gives one of the possible values for the total AM of an O/O nucleus.

Now it is about adding both AMs up to get $I$:

$$I = 1+, 0+$$

But in the table just 1+ is given. Why?

Because the states of different angular momenta are not, in general, of equal energy.
Therefore only one of the different states is the ground state. The others must be excited states.
If the table must be that of ground states then it has to give just one, and namely the one which is the ground state.
Though this sometimes causes problems of relevance. Notoriously so with Ta-180.

 

[JD_PM]

Therefore only one of the different states is the ground state.

I see, but how can we predict theoretically which of the states is the GS?

In my example how can we know which is the GS? (I = 1+, 0+; the table says 1+ but I guess that is an experimental result).