i am kind of stupid, so, for most text book with algebra example, i am easy to lost in the middle.

Thus, now, i am going to present a detail calculation based on Recursion Relations.

we just need equation and few observations to calculate all. i like to use the J- relation:

$K(j,-m-1) C_{m_1 m_2}^{j m}= K(j_1,m_1) C_{m_1+1 m_2}^{j m+1}+ K(j_2,m_2) C_{m_1 m_2+1}^ {j m+1}$

$K(j,m) = \sqrt{ j(j+1) - m(m+1)}$

$C_{m_1 m_2}^{j m}$ is the coefficient.

Notice that the relation is only on fixed j, thus, we will have our $m_1 m_2$ plane with fixed j, so, we have many planes from $j = j_1+j_2$ down to $j = |j_1-j_2|$.

We have 2 observations:

1. $C_{j_1, j_2}^{j_1+j_2 , j_1+j_2} = 1$ which is the maximum state. the minimum state also equal 1.
2. For $m \ne m_1+m_2$ the coefficient is ZERO.

Thus, on the $j = j_1 + j_2$ plane. the right-upper corner is 1. then using the relation, we can have all element down and left. and then, we can have all element on the plane.

the problem comes when we consider $j = j_1 + j_2 -1$ plane. no relation is working! and no book tells us how to find it!

Lets take an example, a super easy one, $j_1 = 1/2 , j_2 = 1/2$. possible $j = 0, 1$, so we have 2 planes.

The j = 1 plane is no big deal.

but the j = 0 plane, there are only 2 coefficient. and we can just related them and know they are different only a sign. and we have to use the orthonormal condition to find out the value.

See? i really doubt is there somebody really do the actually calculation. J.J.Sakuarai just skip the j = l-1/2 case. he cheats!

when going to higher j1+j2 case, we have w=to use the J- relation to evaluate all coefficient. the way is start from the lower left corner, and use the J- relation to find out the relationship between each lower diagonal coefficients. then, since all lower diagonal coefficients have same m value, thus, the sum of them should be normalized. then, we have our base line and use the J+ to find the rest.

i will add graph and another example. say, j1 = 3, j2 = 1.