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Edit: I think what I was looking for wasn't very well understood, so I'm reformulating my question to make it clearer. Hope this helps.

In the book of "Angular Momentum: An Illustrated Guide to Rotational Symmetries for Physical Systems" (p. 309), the Racah definition for the commutators of spherical operators is given by:

\begin{align} [J_0,T_{kq}]&=[J_z,T_{kq}]=qT_{kq} \\ [J_{\pm 1},T_{kq}]&=\sqrt{(k\mp q)(k\pm q+1)}T_{kq\pm1} \end{align}

Using the commutator for spherical operators provided by Sakurai in Modern Quantum Mechanics (p. 239):

$$[J\cdot n, V_{q}^{(k)}]=\sum_{q'}V_{q'}^{(k)}\langle k,q'|J\cdot n|k,q\rangle$$

And by taking $n=z$ and $n=x\pm i y$ respectively, as well as applying the well known eigenvalue equations: \begin{align} J_z|j,m\rangle & =\hbar m |j,m\rangle\delta_{m,m'} \\ J_\pm|j,m\rangle & =\hbar\sqrt{(j\mp m)(j\pm m+1)} \delta_{j,j'}\delta_{m',m\pm 1} \end{align} We can easily deduce the Racah commutators given at the beginning (for a detailed proof, see p. 21 of: http://bohr.physics.berkeley.edu/classes/221/1011/notes/wigeck.pdf).

Now, I've been working on an exercise which asks to generalize the above commutators for the case $k=1$, and show that by expressing the generators of rotations $J$ in the spherical basis $J_q$, the commutators with a spherical operator can be rewritten as:

$$[J_{q},V_{q'}] =(-1)^{q}\hbar\sqrt{2}\langle 1,q+q';1,-q|1,q'\rangle V_{q+q'}$$

However, I don't see how to combine both commutators for into this single relation or how to derive this expression in the first place. While there's a similar equation in the book of "Angular Momentum", the author just states it like that and so far I haven't found any detailed derivation of this last commutator.

Over the past week, I tried several methods to derive this expression, but most of them ended in me already knowing the general expression and proving particular cases. Nevertheless, I'm looking for a proof in the same spirit as the ones I did for the first two commutators.

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Example:

I tried the following. Let's start with the basics and set $k=1$ Recalling the commutator in (Sakurai):

$$[J\cdot n, V_{q}]=\sum_{q'}V_{q'}\langle 1,q'|J\cdot n|1,q\rangle$$

Now taking $n=(n_0,n_{+1},n_{-1})$, where $(n_0,n_{+1},n_{-1})$ are the components of the spherical base:

$$[J_{q'}, V_{q}]=\sum_{q''}V_{q''}\langle 1,q''|J_{q'}|1,q\rangle$$

Using the Wigner-Eckart theorem, with $j=j'$:

$$\langle j,m'|T_q^{(k)}|j,m\rangle=\langle j,m;k,q|j,m\rangle \frac{\langle j||T^{(k)}||j\rangle}{\sqrt{2j'+1}}$$

With some algebra (see https://www.ipnl.in2p3.fr/perso/bennaceur/emr.pdf), we can prove that the reduced matrix element is:

$$\langle j||T^{(k)}||j'\rangle=\hbar \sqrt{j(2j+1)(j+1)}$$

Reintroducing into the W-E theorem:

$$\langle j,m'|T_q^{(k)}|j,m\rangle=\langle j,m;k,q|j,m\rangle \frac{\hbar \sqrt{j(2j+1)(j+1)}}{\sqrt{2j+1}}$$

Substituting now in the commutator given in (Sakurai), with $j=j'=1$:

$$[J_q, V_{q'}]=\sum_{q'}V_{q'}\langle 1,q'|J_q|1,q\rangle$$

$$[J_q, V_{q'}]=\sum_{q'}V_{q'}\langle 1,q;1,q|1,q'\rangle\hbar \sqrt{2}$$

Which should somehow reduce to:

$$[J_{q},V_{q'}] =(-1)^{q}\hbar\sqrt{2}\langle 1,q+q';1,-q|1,q'\rangle V_{q+q'}$$

But I can't find a way to do it, assuming my procedure is correct.

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    $\begingroup$ This is just the relation between $3j$ and CG coefficients. You can check that the square root factor is actually proportional to the appropriate CG. $\endgroup$ – ZeroTheHero Oct 3 '18 at 23:37
  • $\begingroup$ Thanks, I know that for the last commutator the two lines are connected by the relation between the $3j$ and CG coefficients. However, I wanted to know how did the author derived the last commutator (how did the $3j$ or the CG coefficients arose in the first place). $\endgroup$ – Charlie Oct 3 '18 at 23:40
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    $\begingroup$ Compare $\sqrt{(j\mp m)(j\pm m+ 1)}$ with analytical expression for CG... The more general expression lies in the Wigner-Eckart theorem. $\endgroup$ – ZeroTheHero Oct 3 '18 at 23:49
  • $\begingroup$ That seems like a good idea, I'll try it in a while and comment if I succeed to finding it. $\endgroup$ – Charlie Oct 4 '18 at 1:48
  • $\begingroup$ Well I tried doing it but the general expression of the C-G coefficients (which has sums of products of permutations) is extremedly hard to reduce without computer, so I ended just comparing individual cases again. Not sure if this approach is correct or there's a formal demonstration. $\endgroup$ – Charlie Oct 4 '18 at 3:30

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