What Are the Possible Values of Total Nuclear Spin and Its Components?

[BOAS]

Homework Statement

The nucleons in a nucleus have a net orbital angular momentum $L = 1$ and spin angular momentum $S = \frac{1}{2}$. What are the possible values of the total nuclear spin $J$, i.e. the total angular momentum of the nucleus, and what are the possible values of $J_z$ in each case?

Homework Equations

The Attempt at a Solution

I am very confused about how to solve this question.

I know that the total angular momentum $\vec{J} = \vec{L} + \vec{S}$ and that the lengths of these vectors are $|\vec{L}| = \sqrt{l(l+1)}$ and $\vec{S} = \sqrt{s(s+1)}$ respectively (in units of $\hbar$).

At a guess, I would say that the total nuclear spin can be 1/2 or 3/2, but I don't have a particularly convincing reason.

I'd really appreciate some help on what I need to understand to solve questions like this.

Thank you!

 

[kuruman]

At a guess, I would say that the total nuclear spin can be 1/2 or 3/2, but I don't have a particularly convincing reason.

That is correct, that's how angular momenta add in QM. What about the possible values of J_z?

 

[BOAS]

That is correct, that's how angular momenta add in QM. What about the possible values of J_z?

I think $J_z = m_j \hbar$ so the possible values should be $j_z = \frac{3}{2}, \frac{1}{2}, -\frac{3}{2}, -\frac{1}{2}$

 

[kuruman]

That is correct for J = 3/2. For J = 1/2 you have another pair of +1/2 and -1/2 values for a total of 6. Note that the number of states is
(2J₁+1) + (2J₂ + 1) = (2×3/2+ 1) + (2×1/2+1) = 4 + 2 = 6.
Counting the other (L + S) way, you get the same number of states
(2L + 1) (2S + 1) = (2×1 + 1)×(2×1/2 + 1) = 3×2 = 6
which is as it should be.