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With regard to the Wigner-Eckart Theorem the following is stated: The following is an outline of the proof in a text I am using:

"Consider the action of a tensor-operator component on an angular-momentum state $$T^{(k)}_{q}| \alpha' j' m' \rangle$$ where $\alpha'$ represents other quantum numbers that do not represent angular dependence of the state. $T^{(k)}_{q}$ transforms via the rotation matrix in the same way as the angular-momentum ket $| k q \rangle$. Thus, the state $T_{q}^{(k)} | \alpha' j' m' \rangle$ transforms as the composite state $| k q \rangle |j' m' \rangle$. We can consider the usual angular-momentum addition relation $$|k q;j' m' \rangle = \sum_{k'}\sum_{q'} |k' q' \rangle \langle k' q' | j' m' ; k q \rangle\tag{*}$$ and write in analogy to it the same superposition $$T^{(k)}_{q}| \alpha' j' m' \rangle = \sum_{k' q'} | \tilde{\alpha} k' q ' \rangle \langle k' q' |j' m'; k q \rangle$$ where $\tilde{\alpha}$ is some set of transformed radial quantum numbers since the states in the relations transform equivalently. Now we can operate from the left with $\langle \alpha j m |$, we find the matrix element: $$\langle \alpha j m | T_{q}^{(k)}| \alpha' j' m' \rangle = \sum \langle \alpha j m| \tilde{\alpha} k' q'| j' m' ; k q \rangle = \langle \alpha j m | \tilde{\alpha} j m \rangle \langle j m | j' m'; k q \rangle. "$$

Question: Can anyone see how $T^{(k)}_{q}| \alpha' j' m' \rangle = \sum_{k' q'} | \tilde{\alpha} k' q ' \rangle \langle k' q' |j' m'; k q \rangle$ is in analogy with (*)? Instead of $|k' q' \rangle$ we have $| \tilde{\alpha k' q' \rangle}$ but then we kept $\langle k' q' |j' m'; k q \rangle$? How is this form decided on in analogy with $(*)$?

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  • $\begingroup$ $\uparrow$ Which text? $\endgroup$ – Qmechanic May 22 '17 at 17:18
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The quick and dirty argument starts with $$ T^{(k)}_{q}| \alpha' j' m' \rangle = \sum_{k' q'} | \tilde{\alpha} k' q ' \rangle \langle k' q' |j' m'; k q \rangle\, . $$ Act on it with - say - $\hat L_+$: \begin{align} \hat L_+ T^{(k)}_{q}| \alpha' j' m' \rangle&= \left(\hat L_+ T^{(k)}_{q}-T^{(k)}_q \hat L_+ + T^{(k)}_q \hat L_+ \right) | \alpha' j' m' \rangle\, ,\\ &=[\hat L_+,T^{(k)}_{q}]| \alpha' j' m' \rangle+ T^{(k)}_q \left[\hat L_+| \alpha' j' m' \rangle\right]\, ,\\ &= \sqrt{(k-q)(k+q+1)}T^{(k)}_{q+1}| \alpha' j' m' \rangle\\ &\quad + T^{(k)}_q\sqrt{(j'-m')(j'+m'+1)}| \alpha' j' m'+1 \rangle \tag{1} \end{align} where the next-to-last line follows because $T^{(k)}_{q}$ is the component of a tensor operator.

Notice this is exactly the same action as \begin{align} \hat L_+ \left(\vert\alpha kq\rangle| \alpha' j' m' \rangle\right) &=\left[\hat L_+\vert\alpha kq\rangle\right]\vert \alpha' j' m' \rangle + \vert\alpha kq\rangle \left[\hat L_+| \alpha' j' m' \rangle\right]\, ,\\ &=\sqrt{(k-q)(k+q+1)}\vert\alpha k,q+1\rangle\vert\alpha'\,j'm'\rangle\\ &\qquad + \sqrt{(j'-m')(j'+m'+1)}\vert \alpha kq\rangle\vert\alpha' j',m'+1\rangle \tag{2} \end{align} which means Eq(1) has the same combination rules as $\vert\alpha kq\rangle| \alpha' j' m' \rangle$, i.e. can reorganized as suggested using CG technology.

Basically, if you compare (2) and (1), you realize that the action of $\hat L_+$ in (1) is the same as the action of $\hat L_+$ in (2). Knowing that the left hand side of (2) can be expanded over $\vert \alpha JM\rangle$ states with resulting overlaps that are CG coefficients, you expand (1) over the same set of $\vert \alpha JM\rangle$ states and the expansion coefficients will be identical to those of (2), i.e. will be a CG.

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  • $\begingroup$ I understand your working up to the conclusion: "Notice this is..." I'm not sure what you mean here? Isn't $$\hat{L}_{+}(| \alpha k q \rangle |\alpha' j' m'+1 \rangle) = \sqrt{(k-q)(k+q+1)}\hbar | \alpha k q+1 \rangle|\alpha j'm'+1\rangle + \sqrt{(j'-m'-1)(j'+m'+2)}\hbar | \alpha k q \rangle | \alpha' j' m'+2 \rangle$$? Could you elaborate on this and how it justifies the form of $T^{(k)}_{q}|\alpha'j'm'\rangle$? $\endgroup$ – user101311 May 22 '17 at 20:08
  • $\begingroup$ @Moses fixed several typos. $\endgroup$ – ZeroTheHero May 22 '17 at 21:22
  • $\begingroup$ Okay thanks a lot. One thing, by "the left hand side can be expanded over $| \alpha J M \rangle$ do you mean $(\sum_{JM} | \alpha J M \rangle \langle \alpha J M |)\hat{L}_{+}(| \alpha k q \rangle | \alpha' j' m')$"? $\endgroup$ – user101311 May 22 '17 at 21:38
  • $\begingroup$ I meant more $\vert \alpha k q\rangle \alpha' j'm'\rangle$. The strategy is to show that $T^{(k)}_q\vert \alpha'j'm'\rangle$ generates the same relations as $\vert \alpha k q\rangle \alpha' j'm'\rangle$. If you can expand $\vert \alpha k q\rangle \alpha' j'm'\rangle$ using CGs, you can also expand $T^{(k)}_q\vert \alpha'j'm'\rangle$ using CGs. $\endgroup$ – ZeroTheHero May 22 '17 at 21:47
  • $\begingroup$ Oh okay, would this do the trick $\hat{L}_{+}(\sum_{_{JM}}|\alpha'' JM\rangle\langle \alpha'' JM|)(|\alpha k q \rangle | \alpha' j' m \rangle)$ (apologies if I did ping you for this already, not sure if I did)? $\endgroup$ – user101311 May 23 '17 at 15:28

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