Clebsch–Gordan coefficients mathematical proof for bigger $\Delta m$ make the sign between vertical jump Suppose $j_1=2$ and $j_2=1$,where states were represented by $|jm>$.
Where we have $|33>$ as the top states in the beginning.
We applied $j_-$ to $|33>$ and obtain the $|32>$.
The vertical jump of $j$, i.e. $|32>\rightarrow|22>$ could be calculated out by normality and orthogonal up to a sign($\pm$).
However, my professor told me that the sign could be determined by the $\Delta m$ and someone from 50s proved it(I didn't remember the exact name). 
My question was that: Could you give me the mathematical proof of the argument, and was there some new way to derive it?
 A: In your example, 
\begin{align}
\vert 33\rangle & = \vert 22\rangle \vert 11\rangle \\
\vert 32\rangle &=\sqrt{\frac{2}{3}}\vert 21\rangle\vert 11\rangle 
+\sqrt{\frac{1}{3}}\vert 22\rangle \vert 10\rangle\, .
\end{align}
The only other possible state with $m=2$ must be $\vert 22\rangle$ and it must be orthogonal to $\vert 32\rangle$ since they are eigenstates of $\vec L\cdot\vec L$ with different eigenvalues.
By orthogonality we thus have
$$
\vert 22\rangle = \pm \left(-\sqrt{\frac{1}{3}}\vert 21\rangle\vert 11\rangle +\sqrt{\frac{2}{3}}\vert 22\rangle \vert 10\rangle\right)
$$
The sign in front is purely conventional but usually chosen using the Condon-Shortley phase convention, which states that the coefficient of $\vert j_1j_1\rangle\vert j_2m_2\rangle$ in the highest state $\vert JJ\rangle$ should be positive.  Thus, here we should keep the $+$ sign.
This is not the only convention but it is by far the most common one.  It is quite possible that for some specific application a more specialized convention might be more useful.  However, most of the "modern" tables use Condon-Shortley.  Mathematica has a built-in CG calculator which also uses this convention.
As far as I know, most "classics" like Varshalovich, Rose or Edmonds also use the Condon-Shortley convention and, if there was a derivation, these references would supply it.
