# CG coefficient in Wigner Eckhart Theorem-how to find

## Homework Statement

The Clebsh Gordon coefficient is < 11;00 |11> = < l_1,m_1;l_2m_2 | kq >

## Homework Equations

My professor determined the CG to be 1

## 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

## Answers and Replies

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blue_leaf77
Science Advisor
Homework Helper
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.