Addition of orbital angular momentum and spin

1. The problem statement, all variables and given/known data



Consider an electron with spin [itex]\frac{1}{2}[/itex] and orbital angular momentum l=1. Write down all possible total angular momentum states as a combination of the product states [itex] | l=1 , m_l > | s = \frac{1}{2} , m_s > [/itex]



2. Relevant equations



Lowering operator : [itex] J_- |j, m> = \sqrt{(j + m)(j - m + 1)} |j, m-1> [/itex]



3. The attempt at a solution



Since total angular momentum [itex] | l-s | <= j <= (l+s) [/itex]

and its z-component [itex] -j <= m_j <= +j [/itex]

I know that the possible [itex] |j, m_j > [/itex] states are:



[itex] | \frac{1}{2} , \frac{-1}{2} > [/itex]

[itex] | \frac{1}{2} , \frac{1}{2} > [/itex]

[itex] | \frac{3}{2} , \frac{-3}{2} > [/itex]

[itex] | \frac{3}{2} , \frac{-1}{2} > [/itex]

[itex] | \frac{3}{2} , \frac{1}{2} > [/itex]

[itex] | \frac{3}{2} , \frac{3}{2} > [/itex]



As for finding the product states, I know that:

[itex] | \frac{3}{2} , \frac{3}{2} > = |1, 1> | \frac{1}{2} , \frac{1}{2} > [/itex]

as this is the maximal spin state. I can then find [itex] | \frac{3}{2} , \frac{1}{2} > [/itex], [itex] | \frac{3}{2} , \frac{-1}{2} > [/itex] and [itex] | \frac{3}{2} , \frac{-3}{2} > [/itex] using the lowering operator above. I don't know how I can use this information to find [itex] | \frac{1}{2} , \frac{1}{2} > [/itex] and [itex] | \frac{1}{2} , \frac{-1}{2} > [/itex] however.





Thanks.