Total angular momentum of deuteron 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?
 A: 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.
A: As I understand it, by experimental measurements, it is known that the deuteron has:


*

*an "even" spatial parity

*and total nuclear spin = 1 (as it's a boson)


And as parity is characterized by (-1) to the power l, any odd 
  l
 value like 
  l
  =
  1
 gives us an "odd" parity, which is inconsistent with our measurement, so options (b) and (c) must be eliminated.
Further, 
  l
  =
  0
 and 
  l
  =
  2
 options (which of course give "even" parity), when combined with spin 
  s
  =
  1
, give correct total spin = 1. So deuteron's ground state is a superposition of 
  l
  =
  0
 and 
  l
  =
  2
 states.
Note, that we have not talked about isospin yet, which ensures that the total wave function describing deuteron(which is a system of two fermions) is anti-symmetric, as required by Pauli-exclusion principle.
A: For the (c) case, you have parallel spins so total spin is $S = 1$. Then we have orbital angular momentum (quantum number) $L = 1$. Our total nuclear spin will then have any of the values $I = |L-S|, ..., L+S$. This means $I = 0, 1, 2$. So, this combination of $L$ and $S$ does give us a total $I$ of $1$, even though $I=1$ is not the only possibility! The thing you need to understand is that if we say "a certain combination of $L$ and $S$ gives us $I =1$", we don't necessarily mean that $I=1$ is the only possible result from that combination.
A: Best place to start is with assuming that the deuteron is in an l = 0 state. The reason for assuming this is because orbital angular momentum zero is the lowest-energy option for positive parity.
Next, we have to deal with spin. The spins of the proton and the neutron are aligned in the deuteron (i.e. the proton and neutron both have spin projection "plus 1/2" if you like). This is because there is a component in the nucleon-nucleon interaction which takes on its greatest value when the spins are aligned. 
So: now we have orbital angular momentum zero and spin 1. This gives us a total angular momentum of 1. 
A: $ 3 \otimes 3 = 1 \oplus 3 \oplus 5$
Where $3$ means "triplet", or a $l=1$ state ($m\in [-1, 0, 1]$). When you combine 2 triplet states, the eigenstate of total angular momentum can be singlet ($l=0$), triplet ($l=1$)--this is the one we're interest in, and "5"-which is $l=2$.
