Probability calculation of the measurement of the total angular momentum for an electron Hello fellow scientists,
I also posted this on the physics stack exchange, please tell me if it off topic for this one-I don't see any reason why it should be-.  
I will start with an example:

An electron has an orbital angular momentum of $l=2$ and $m_l=2$. Say its spin is $m_s=-1/2$ . What is the probability of the measurement of the total angular momentum to equal $j=3/2$

I know that the total angular momentum can have values that range from $j_\min =\lvert l-s\rvert$ to $j_\max = l+s$ but I don't get how the probability calculation works math-wise. I am also aware that I have to construct the probability using $J_-$ and $J_+$ operators but my bibliography is of poor quality.
Any help would be greatly appreciated.

The question is related to math only because of funny quantum-mechanical angular momentum calculus.
So it is said that an electron has orbital angular momentum of $2$ and we know that electron always has spin $1/2$. Now we construct the multiplet of total angular momentum eigenstates (in the following LHS are total angular momentum eigenstates, RHS are tensor products of orbital angular momentum eigenstates by spin eigenstates):
$$|5/2,5/2\rangle =|2,2\rangle\otimes|1/2,1/2\rangle $$
$$|5/2,3/2\rangle =\alpha|2,1\rangle\otimes|1/2,1/2\rangle+\beta|2,2\rangle\otimes|1/2,-1/2\rangle$$
$$|3/2,3/2\rangle =\gamma|2,1\rangle\otimes|1/2,1/2\rangle+\delta|2,2\rangle\otimes|1/2,-1/2\rangle$$
Our final goal is to find $\delta$ modulo squared, $|\delta|^2$, which is the probability we are looking for.
We start from the top state of the multiplet $|5/2,5/2\rangle$ and apply lowering operator
$$J_{-}=L_{-}+S_{-}$$
to both sides; we have
$$J_{-}|j,m\rangle=\hbar\sqrt{j(j+1)-m(m-1)}|j,m-1\rangle$$
the same is for $L_{-}$ and $S_{-}$, ex
$$L_{-}|l,m_l\rangle=\hbar\sqrt{l(l+1)-m_l(m_l-1)}|l,m_l-1\rangle$$
so
$$J_{-}|5/2,5/2\rangle=\hbar\sqrt{5/2\cdot 7/2-5/2\cdot 3/2}|5/2,3/2\rangle=\hbar\sqrt{5}|5/2,3/2\rangle$$
$$(L_{-}+S_{-})|2,2\rangle\otimes|1/2,1/2\rangle=(L_{-}|2,2\rangle)\otimes|1/2,1/2\rangle+|2,2\rangle\otimes(S_{-}|1/2,1/2\rangle)$$
$$=\hbar \cdot 2|2,1\rangle\otimes|1/2,1/2\rangle+\hbar \cdot |2,2\rangle\otimes|1/2,-1/2\rangle$$
so $\alpha=2/\sqrt{5}$, $\beta=1/\sqrt{5}$.
To find $\gamma$ and $\delta$ it is enough to note that the states $|5/2,3/2\rangle$ and $|3/2,3/2\rangle$ are orthonormal, and we can choose (up to common phase which does not matter here) $\gamma=-1/\sqrt{5}$, $\delta=2/\sqrt{5}$, and the final answer is
$$P=|\delta|^2=\frac{4}{5}$$
