I am reading through Wu-Ki Tung's Group Theory in Physics and I met a problem when going through the part of the Wigner-Eckart theorem for irreducible tensor operators.
In the 4.3 part of the book, the author defines irreducible tensor operators as operators having the following transforming property($U$ is a general representation of the group $G$, while $D^{\mu}$ is the $\mu$'s irreducible matrix representation) : $$
U(g) O_{i}^{\mu} U(g)^{-1}=O_{j}^{\mu} D^{\mu}(g)_{i}^{j}
$$
and then he shows that the set of vectors $U(g) O_{i}^{\mu}\left|e_{j}^{v}\right\rangle$ ($|e_{j}^{v}\rangle$ is a irreducible set of $\nu$'s irreducible representation, corresponding to some physical states) transform
according to the direct product representation $D^{\mu \times \nu}$:$$
\begin{aligned}
U(g) O_{i}^{\mu}\left|e_{j}^{v}\right\rangle &=U(g) O_{i}^{\mu} U(g)^{-1} U(g)\left|e_{j}^{v}\right\rangle \\
&=O_{k}^{\mu}\left|e_{l}^{v}\right\rangle D^{\mu}(g)^{k}{ }_{i} D^{v}(g)^{l}{ }_{j}
\end{aligned}
$$so we can express this set of vectors using the irreducible vectors of $U^{\mu}\times U^{\nu}:\left|w_{\alpha l}^{\lambda}\right\rangle$as follow (the coefficients of RHS are the Clebsch-Gordan coefficients,$\alpha$ distinguish different sets of the same irreducible representation) :$$
O_{i}^{\mu}\left|e_{j}^{v}\right\rangle=\sum_{\alpha, \lambda, l}\left|w_{\alpha l}^{\lambda}\right\rangle\langle\alpha, \lambda, l(\mu, v) i, j\rangle
$$
then, using the orthogonality of irreducible sets of different representations, the author states that we can proof the following formula for the matrix element $\left\langle e_{\lambda}^{l}\left|O_{i}^{\mu}\right| e_{j}^{v}\right\rangle$(the formula (4.3-4) of the book):$$
\left\langle e_{\lambda}^{l}\left|O_{i}^{\mu}\right| e_{j}^{v}\right\rangle=\sum_{\alpha}\langle\alpha, \lambda, l(\mu, v) i, j\rangle\left\langle\lambda\left|O^{\mu}\right| v\right\rangle_{\alpha}
$$where $\left\langle\lambda\left|O^{\mu}\right| v\right\rangle_{\alpha} \equiv n_{\lambda}^{-1} \sum_{k}\left\langle e_{\lambda}^{k} \mid w_{\alpha k}^{\lambda}\right\rangle$ is called the reduced matrix element, where $n_{\lambda}$ is the dimension of the $\lambda$'s irreducible representation.
then the author states that the through this formula the $i$– $j$– and $l$– dependence of the matrix element are all in the Clebsch-Gordan coefficients which can be determined through the representation theory.
My first question is, how can we derive the last formula through the process described above? I have tried but I can't get the right result:$$\left\langle e_{\lambda}^{l}\left|O_{i}^{\mu}\right| e_{j}^{v}\right\rangle=\sum_{\alpha, \lambda^{\prime}, l^{\prime}}\langle e_{\lambda}^{l}\left|w_{\alpha l^{\prime}}^{\lambda^{\prime}}\right\rangle\langle\alpha, \lambda^{\prime}, l^{\prime}(\mu, v) i, j\rangle \\=\sum_{\alpha , l^{\prime}}\langle e_{\lambda}^{l}\left|w_{\alpha l^{\prime}}^{\lambda}\right\rangle\langle\alpha, \lambda, l^{\prime}(\mu, v) i, j\rangle$$the second equality is due to the orthogonality of different irreducible sets.
My second question is, where does the $\mu$,$\nu$dependence of the reduced matrix $\left\langle\lambda\left|O^{\mu}\right| v\right\rangle_{\alpha} \equiv n_{\lambda}^{-1} \sum_{k}\left\langle e_{\lambda}^{k} \mid w_{\alpha k}^{\lambda}\right\rangle$come from? I think as follow: if we want to look up the Clebsch-Gordan coefficients from some pulished books, we must construct the set $|w_{\alpha l}^{\lambda}\rangle$ from $ O_{i}^{\mu}\left|e_{j}^{v}\right\rangle$ in a specific way.
Best Answer
Thank you for your clarification on $\alpha$, so I can provide the answer more accurately.
The trick to deriving the last equation is to consider the product $\langle e^{l}_{\lambda}|O^{\mu}_i|e^{\nu}_{j}\rangle$ with \begin{align} \langle e^{l}_{\lambda}|O^{\mu}_i|e^{\nu}_{j}\rangle & = \langle e^{l}_{\lambda}|U(g^{-1})U(g)O^{\mu}_i|e^{\nu}_{j}\rangle \\ & = \langle e^{\color{red}{k}}_{\lambda}|(D^{\lambda}(g)^{\color{red}{k}}_{l})^*\sum_{\alpha,\lambda',l'}{\bigg(D^{\lambda'}(g)^{\color{red}{q}}_{l'}|\omega^{\lambda'}_{\alpha,{\color{red}{q}}}\rangle \langle \alpha,\lambda',l'(\mu,\nu)i,j \rangle \bigg)} \\ & = \sum_{\alpha,\lambda',l'}{\langle \alpha,\lambda',l'(\mu,\nu)i,j \rangle \bigg((D^{\lambda}(g)^{\color{red}{k}}_l)^*D^{\lambda'}(g)^{\color{red}{q}}_{l'}\bigg)\langle e^{\color{red}{k}}_{\lambda}|\omega^{\lambda'}_{\alpha,{\color{red}{q}}}\rangle} \end{align} where red labels are summed over as Einstein's convention. Now we integrate both sides over $g$, and use the orthogonality relation to come up with \begin{align} \langle e^{l}_{\lambda}|O^{\mu}_i|e^{\nu}_{j}\rangle & = \sum_{\alpha,\lambda',l'}{\langle \alpha,\lambda',l'(\mu,\nu)i,j \rangle{1 \over 2\lambda+1}\delta_{\lambda\lambda'}\delta_{{\color{red}{k}}{\color{red}{q}}}\delta_{ll'}\langle e^{\color{red}{k}}_{\lambda}|\omega^{\lambda'}_{\alpha,{\color{red}{q}}}\rangle} \\ & = \sum_{\alpha}{\langle \alpha,\lambda,l(\mu,\nu)i,j \rangle{1 \over 2\lambda+1}\sum_{k}{\langle e^{k}_{\lambda}|\omega^{\lambda}_{\alpha,k}\rangle}} \end{align}
This answers your first question, and about the second one, your thought is correct, the set of vectors $\big\{|\omega^{\lambda}_{\alpha,k} \rangle\big|\lambda=|\mu-\nu|,\cdots,\mu+\nu, \ \text{and} \ |k| \leq \lambda\big\}$ are constructed from $\big\{O^{\mu}_i|e^{\nu}_{j}\rangle\big||i| \leq \mu, \ \text{and} \ |j| \leq \nu\big\}$. Therefore, the dependence of $\mu$ and $\nu$ are not absent from the reduced matrix.