I am trying to understand a problem in a subnuclear physics course. The question is the following:

The biggest difference between the low-lying states of $^{17}O$ and $^{19}O$ is that $^{19}O$ has two extra states with spin-parity $\frac{3}{2}^+$ and $\frac{9}{2}^+$. Show that these are the result of the configuration $(d_{5/2})^3$ and are therefore not expected in $^{17}O$.

What I have done:

I have used the shell model to confirm that the states with an unpaired neutron are $(d_{5/2})^1$ for $^{17}O$ and $(d_{5/2})^3$ for $^{19}O$, while the protons are always paired.

In the event of unpaired nucleons, the total spin can be all values satisfying $|j_p - j_n| < I < j_p + j_n$. But in this case of paired protons, how can there even be multiple states? And how can it be different between two configurations which both have $j = 5/2$?

I have a solution to the problem, that I don't understand. It goes like this, define $$ j_1 = j_2 =j_3 = 5/2. $$ Now, write down all integer values between $j_1 + j_2$ and $|j_1 - j_2|$. Set $j' = j_1 + j_2$. Now write down all the values between $j' + j_3$ and $|j' - j_3|$. We can see that in this second list we find the state that only exist for $^{19}O$.

I don't understand the solution. Any help would be much appreciated!


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