# CG coefficient in Wigner Eckhart Theorem-how to find

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1. Apr 25, 2016

### lonewolf219

1. The problem statement, all variables and given/known data
The Clebsh Gordon coefficient is < 11;00 |11> = < l_1,m_1;l_2m_2 | kq >
2. Relevant equations
My professor determined the CG to be 1

3. The attempt at a solution
How do we look up these numbers in the table? What is the direct product basis? 1x1? What is JM and what is m_1 and m_2... very confused. Any help greatly appreciated

2. Apr 29, 2016

### blue_leaf77

Suppose you have two angular momenta vector operator $\mathbf{L}_1$ and $\mathbf{L}_2$ acting on different vector spaces. The eigenket of $L_{1z}$ in the first vector space is denoted by $|l_1,m_1\rangle$, while the eigenket of $L_{2z}$ in the second space is denoted by $|l_2,m_2\rangle$. These kets can also serve as the bases in their respective vector spaces.

But now you want to combine the two spaces together through the tensor product. In this new space, the two angular momenta add together $\mathbf{J} = \mathbf{L}_1 + \mathbf{L}_2$. One of many possible bases in this composite space is formed by directly pairing the basis kets from its two composition spaces: $|l_1,m_1\rangle |l_2,m_2\rangle$. This form of basis kets is also an eigenket of both $L_{1z}$ and $L_{2z}$ (more precisely $L_{1z} \otimes \mathbf{1}$ and $\mathbf{1} \otimes L_{2z}$). But you can also find a different basis by forming certain linear combination of the old basis $|l_1,m_1\rangle |l_2,m_2\rangle$, namely
$$|J,m_j\rangle = \sum_{m_1} \sum_{m_2} C_{m_1m_2} |l_1,m_1\rangle |l_2,m_2\rangle$$
such that $|J,m_j\rangle$ is an eigenket of both $J^2$ and $J_z$ operators. The coefficient $C_{m_1m_2}$ is called the Clebsch-Gordan coefficient.