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?


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.


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