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 complete system of commuting observables (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 consider 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?


1 Answer 1


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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