Orthogonality condition for angular momentum eigenstates in the coupled and uncoupled basis

Question

I've been studying the addition of angular momentum in quantum mechanics but I can't understand the orthogonality condition for the coupled and uncoupled basis.

I know they are related by the Clebsch-Gordan coefficients $$|j_{1}j_{2}jm\rangle = \sum_{j_{1}m_{1}j_{2}m_{2}}\langle j_{1}m_{1};j_{2}m_{2}|j_{1}j_{2}jm\rangle |j_{1}m_{1};j_{2}m_{2}\rangle\tag{1}$$

where $j_{1},j_{2},m_{1},m_{2}$ correspond to the quantum numbers of each of the angular momentum in the system.

But what does $$\langle j_{1}^{'}j_{2}^{'}j^{'}m^{'}|j_{1}j_{2}jm\rangle\tag{2}$$ gives?

I think the result should be something proportional to $$\delta_{j_{1}^{'}j_{1}}\delta_{m_{1}^{'}m_{1}}\delta_{j_{2}^{'}j_{2}}\delta_{m_{2}^{'}m_{2}}\tag{3}$$ in order to the operator $\hat{j}^{2}$ be diagonal in the coupled basis but I'm not sure about it.

It's this true? It's this the whole answer?

Answer 1

The best way to think about this is to shorten $ \vert j_1m_1\rangle\vert j_2m_2\rangle=\vert j_1m_1;j_2m_2\rangle$ and observe that the $\{\vert j_1m_1\rangle\vert j_2 m_2\rangle\}$ span your space of states, with $$ \hat 1 = \sum_{m_1m_2} \vert j_1m_1; j_2m_2\rangle \langle j_1m_1;j_2m_2\vert $$ so that a state $\vert jm\rangle$ is just $$ \vert jm\rangle=\sum_{m_1m_2} \vert j_1m_1; j_2m_2\rangle \langle j_1m_1;j_2m_2\vert jm\rangle\, . $$ This way, another state $\vert j'm'\rangle$ would still be expanded in terms of $\vert j_1m_1\rangle\vert j_2 m_2\rangle$ and $\langle j'm'\vert jm\rangle=\delta_{jj'}\delta_{mm'}$

If $\vert j'm'\rangle$ does not live in the space spanned by $\{\vert j_1m_1\rangle\vert j_2 m_2\rangle\}$ but in some other space $\{\vert j'_1m'_1\rangle\vert j'_2 m'_2\rangle\}$ then indeed the fuller orthogonality relation is $$ \langle j'm'\vert jm\rangle=\delta_{jj'}\delta_{mm'}\delta_{j_1j'_1} \delta_{j_2j'_2}\, . $$