I'm studying for my nuclear physics exam and the book we use is Introductory Nuclear Physics by K.S. Krane. In the chapter on Basic Nuclear Structure, we research the deuteron. However, when discussing the total angular momentum of the deuteron, something confuses me (p84 if someone has the book). I'll copy the paragraph:
Spin and Parity
The total angular momentum $I$ of the deuteron should have three components: the individual spins $s_n$ and $s_p$ of the neutron and proton (each equal to $\frac{1}{2}$), and the orbital angular momentum $l$ of the nucleons as they move about their common center of mass:
$$I = s_n + s_p + l $$
[…]The measured spin of the deuteron is $I = 1$ […]. Since the neutron and proton spins can be either parallel (for a total of $1$) or antiparallel (for a total of zero), there are four ways to couple $s_n$, $s_p$, and $l$ to get a total of $1$:
(a) $s_n$ and $s_p$ parallel with $l = 0$
(b) $s_n$ and $s_p$ antiparallel with $l = 1$
(c) $s_n$ and $s_p$ parallel with $l = 1$
(d) $s_n$ and $s_p$ parallel with $l = 2$
Unfortunately, this is not explained deeper and options (b) and (c) are shown to be invalid because the measured parity is $1$, whereas those options have parity $-1$.
Now, (a), (b) and (d) seem okay with me to get total angular momentum $1$, but I don't understand how we can get to $1$ in (c): $I = 1$ if $l=1$ and $s_n + s_p = 1$. This gives either $0$ or $2$? Or can $I$ take all values between $l-s_n-s_p$ and $l+s_n+s_p$ rather than just those two extremes?
Best Answer
I think this is just standard angular momentum coupling, not anything specific to nuclear physics. In (c), you're coupling spin 1 to spin 1. There are 9 states, which can be broken down into multiplets with spins 0, 1, and 2.