# Addition of three angular momenta

To add three mutually commuting angular momenta $$\mathbf{J}=\mathbf{J}_1+\mathbf{J}_2+\mathbf{J}_3$$, we may follow this method: add $$\mathbf{J}_1$$ and $$\mathbf{J}_2$$ to obtain $$\mathbf{J}_{12}=\mathbf{J}_1+\mathbf{J}_2$$, then add $$\mathbf{J}_{12}$$ and $$\mathbf{J}_3$$ to obtain $$\mathbf{J}=\mathbf{J}_{12}+\mathbf{J}_3$$.

Denote the eigenstates of $$\mathbf{J}^2_k$$ and $$J_{k,z}$$ as $$|j_k,m_k\rangle, k=1,2,3$$. Then the joint basis that diagonalizes the CSCO $$\{ \mathbf{J}^2_k, J_{k,z}| k=1,2,3\}$$ is the direct product of the corresponding eigenstates, which can be written as $$|j_1j_2j_3;m_1m_2m_3\rangle$$.

Now, I want to considerd the CSCO $$\{ \mathbf{J}^2_1,\mathbf{J}^2_2,\mathbf{J}^2_3,\mathbf{J}^2_{12}, \mathbf{J}^2,J_z\}$$ with joint eigenstates $$|j_1j_2j_3;j_{12},j,m\rangle$$. My goal is to express these eigenstates in terms of $$|j_1j_2j_3;m_1m_2m_3\rangle$$:

For the coupling of $$\mathbf{J}_1$$ and $$\mathbf{J}_2$$, we have $$|j_1j_2;j_{12},m_{12}\rangle= \sum_{m_1}\sum_{m_2} \langle j_1j_2;m_1m_2\mid j_1j_2;j_{12}m_{12}\rangle|j_1j_2;m_1m_2\rangle,$$ where $$m_{12}= m_1+m_2$$ and $$|j_1-j_2|\le j_{12} \le j_1+j_2$$. My notes then proceed with saying that by adding $$\mathbf{J}_{12}$$ and $$\mathbf{J}_3$$, the eigenstate $$|j_1j_2j_3;j_{12},j,m\rangle$$ is given by $$\sum_{m_{12}}\sum_{m_3} \langle j_1j_2;m_1m_2\mid j_1j_2;j_{12}m_2\rangle\langle j_{12}j_3; m_{12}m_3\mid j_1j_2j_3;j_{12},j,m\rangle|j_1j_2j_3;m_1m_2m_3\rangle,$$ where $$m = m_{12}+m_3$$ and $$|j_{12}-j_3|\le j\le j_{12}+j_3$$.

I don't see where this last expression is coming from. When adding $$\mathbf{J}_1$$ and $$\mathbf{J}_2$$, one could use the resolution of the identity in the old (uncoupled) basis to get the identity above, but how can the last expression be derived?

• Related. Might start with this one. Dec 23 '20 at 20:10

## 1 Answer

In triple coupling you couple every $$j_{12}$$ that occurs in $$j_1\otimes j_2$$ with $$j_3$$ so for each $$j_{12}$$ you need to consider $$j_{12}\otimes j_3$$ which gives the range of final $$j$$ as indicated.

Note that different $$j_{12}$$ in the decomposition $$j_1\otimes j_2$$ can give the same final $$j$$, so that a specific value of $$j$$ can occur more than once.

An example of this would be $$\textstyle\frac{1}{2}\otimes\frac{1}{2}\otimes\frac{1}{2}=\frac{1}{2}\oplus\frac{1}{2}\oplus\frac{3}{2}. \tag{1}$$

Start with $$\textstyle\frac{1}{2}\otimes\frac{1}{2}=0\oplus 1$$ and couple those to $$\frac{1}{2}$$. It is easily seen that one of the $$\frac{1}{2}$$ in Eq(1) arises from $$j_{12}=0$$ coupled with $$\frac{1}{2}$$ while the other arises from $$j_{12}=1$$ coupled with $$\frac{1}{2}$$.

In your last expression it is assumed you have first constructed states with the correct $$j_{12}$$ value, and then coupled those with $$j_3$$ to get the final $$j$$. The $$j_{12}$$ label functions as an intermediate label to distinguish distinct states with the same $$j_3m_3$$ values but distinct “$$j_{12}$$” paths.