Angular momentum coupling-calculation of Clebsch–Gordan coefficients I am facing problem in calculating the value of given Clebsch–Gordan coefficients representing the coupled angular momenta of two-particle system. For example
$$\begin{pmatrix}2 & 1 & 2 \\ 1 & -1 & 0\end{pmatrix}$$
In the book it is first expanded in three parts like
$$\begin{pmatrix}2 & 1 & 2 \\ 0 & 0 & 0\end{pmatrix},\begin{pmatrix}2 & 1 & 2 \\ 1 & -1 & 0\end{pmatrix},\begin{pmatrix}2 & 1 & 2 \\ -1 & 1 & 0\end{pmatrix}$$
I am really very much confused that which symmetry property should use?
I know here orthonormality condition of coefficient applied here. But why here in the second row values are first $\begin{pmatrix}0 & 0 & 0\end{pmatrix}$ then $\begin{pmatrix}1 & -1 & 0\end{pmatrix}$ and then $\begin{pmatrix}-1 & 1 & 0\end{pmatrix}$? Please help! I will be thankful to you. 
 A: If you look on the Wikipedia entry for the Wigner 3-j symbols: 
http://en.wikipedia.org/wiki/Wigner_3-j_symbols
You will see that those symbols are related combining two spin states to get a third state:
$ \begin{pmatrix}
  j_1 \, j_2 \, j_3\\
  m_1 \, m_2 \, m_3
\end{pmatrix} \equiv \frac{(-1)^{j_1-j_2-m_3}}{\sqrt{2j_3+1}} \langle j_1 m_1 j_2 m_2 | j_3 \, {-m_3} \rangle.$
Ignoring any coefficients, the interpretation is that the first two columns are the states that are added and the third column is the resulting state.
This tells us a few things.  First, it tells us that the sum of the bottom row must equal 0.  This is just conservation of angular momentum (in the z-direction).  Thus, $m_1 + m_2 = m_3$.
So, back to your original question, the original 3-j symbol can be expanded into a sum of other symbols.  Physically, this means: given two particles of fixed total spin (in this case, spin 1 and spin 2), how can I add them together to get an effective state of total spin 2?
Focusing only on the bottom row, which represents the z-direction angular momenta of the particles, the first entry can be 2, 1, 0, -1, -2 (the possible z-direction spins of a spin-2 particle) and the second entry in the second row can only be 1, 0, -1 (it's a spin-1 particle).  But these two must add up to 0.  So, given those choices, there are only three combinations that work: 0 + 0 = 0, 1 + -1 = 0, -1 + 1 = 0.  This is why only those three symbols are listed: all others vanish.
The physical interpretation is: given a spin-2 particle and a spin-1 particle, I can combine  them together to form an effective spin-2 particle with 0 angular momenta in the z-direction.  To do this, I only need 3 terms in the sum, and the value of the 3-j symbols gives you the coefficients of each term in the sum.
