Number of coupled states of the total angular momentum

Question

Let $L$ be the angular momentum, $S$ the spin and $l$ and $s$ the respective quantum numbers.

Then, there are $(2l+1)(2s+1)$ states $|l;m_l;s;m_s\rangle$ for fixed $l$ and $s$.

Let $J = L + S$ be the total angular momentum and $j$ the resprective quantum number.

I know that $|l-s| \leq j \leq l + s$ holds, but how to show that there are as many coupled states $|l;s;j;m_j\rangle$ as uncoupled states?

Answer 1

If $j$ ranges from $\left|{l-s}\right|$ to $l+s$ and there are $2j+1$ states for every value of $j$, then the total number of states is (assuming $l>s$)

$$\sum_{j=l-s}^{l+s} (2j+1). $$

This is equal to the average term times the number of terms. There are $2s+1$ terms and the average is $2l+1$, so the total number of states is $(2l+1)(2s+1)$. The calculation is the same for the case where $l<s$.