Currently trying to solve the following problem:
Consider an operator $O_x$ for $x = 1$ to $2$, transforming according to the spin $1/2$ representation as follows:
$$ [J_a, O_x] = O_y[\sigma_a]_{yx} / 2 $$
where $\sigma_a$ are the Pauli matrices. Given
$$ <3/2, -1/2, \alpha|O_1|1, -1, \beta> \ = \ A$$
find
$$<3/2, -3/2, \alpha|O_2|1, -1, \beta>$$
The author gave the following example in text:
Supposing $\ <1/2, 1/2, \alpha|r_3|1/2, 1/2, \beta> = A \ \ $ find $\ \ <1/2, 1/2, \alpha|r_1| 1/2, -1/2, \beta>$
Solution:
For a tensor operator being a set of operators $O^s _m$ such that $[J_a, O^s _l] = O^s _m [J^s _a]_{ml}$, we make a change of basis as follows:
since $[J_3, r_3] = 0$, the $r_3$ operator carries m=0, so we assign $r^0 = r_3$. We can find $r^{\pm 1}$ by using the commutation relations for the spin 1 raising and lowering operators: $[J^{\pm}, r_3] = r_{\pm 1} = \mp (r_1 \pm ir_2)/\sqrt{2}$.
Since $r^0 = r_3, \ <1/2, 1/2, \alpha|r^0|1/2, 1/2, \beta> = A$ and also from the new tensor basis, $r_1 = \frac{1}{\sqrt{2}} (-r^{+1} + r^{-1})$, thus
$$ <1/2, 1/2, \alpha|r_1|1/2, -1/2, \beta> \ = \ <1/2, 1/2, \alpha|\frac{1}{\sqrt{2}} (-r^{+1} + r^{-1})|1/2, -1/2, \beta> \\ = -\frac{1}{\sqrt{2}} <1/2, 1/2, \alpha|r^{+1}|1/2, -1/2, \beta>$$
Now, from the highest weight deconstruction, we know that $|3/2, 3/2> \ = \ r^{+1} |1/2, 1/2, \beta>$ is a 3/2, 3/2 state because it is the highest weight that can get as a product of an $r^{l}$ operator acting on an $|1/2, m>$ state. Furthermore,
$$ \tag{*}|3/2, 1/2> \ = \ \sqrt{\frac{2}{3}} J^- |3/2, 3/2> \ = \ \sqrt{\frac{2}{3}} r^0 |1/2, 1/2, \beta> + \sqrt{\frac{1}{3}} r^{+1} |1/2, -1/2, \beta>$$
Since we know $0 \ = \ <1/2, 1/2, \alpha| 3/2, 1/2> $, then
$$ 0 \ = \ <1/2, 1/2, \alpha| 3/2, 1/2> \ = \ \sqrt{\frac{2}{3}} <1/2, 1/2, \alpha|r^0|1/2, 1/2, \beta> + \sqrt{\frac{1}{3}} <1/2, 1/2, \alpha|r^{+1}|1/2, -1/2, \beta> $$
so finally,
$$<1/2, 1/2, \alpha|r^{+1}|1/2, -1/2, \beta> = -\sqrt{2}<1/2, 1/2, \alpha|r^{0}|1/2, 1/2, \beta> = -\sqrt{2} \ A \\$$
$$\implies <1/2, 1/2, \alpha|r_3|1/2, 1/2, \beta> \ = \ <1/2, 1/2, \alpha|r_1| 1/2, -1/2, \beta> \ = \ A $$
I have lots of questions here, I don't really understand what's going on. Firstly, in the example we went over, why do the new tensor operators seem to act like raising and lowering operators? I don't see how to obtain equation (*). Could someone explain why we performed the change of basis at all? It seems that the tensor operators don't transform under the same representation as the entire representation that the kets live in. I think that the worked example is the case where states live in the tensor product space of $1/2 \otimes 1$.
I see that there's some sort of spherical symmetricness going on, but why did the example utilize the spin-1 raising and lowering operators specifically? Also, what does an inner product such as $<1/2, 1/2, \alpha|1/2, 1/2>$ mean? I know that the $\alpha$ is there to imply that within the Hilbert space, there could be other characteristics of the quantum state not governed by (in this case) SU(2). How is it evaluated, if it's different than $<1/2, 1/2|1/2, 1/2>$?
For the problem I'm trying to solve, where do I begin? I'm not seeing what this chapter of my book is wanting me to understand about these tensor operators we're using. I believe that there's some relation between how the operators acting on kets transform and the original basis. Perhaps the tensor operator acting on a ket altogether transforms like an individual ket? I'm not sure.
Edit: I'd like to add that I do see where the Wigner Eckert theorem itself applies; it's when we get to the element $-\frac{1}{\sqrt{2}} <1/2, 1/2, \alpha|r^{+1}|1/2, -1/2, \beta>$. I believe the application here is that we can split up a matrix element that would require a lot of different configurations to integrate (or in this case sum) over into a product of two different matrix elements that by themselves don't have as many terms needed sum together (it's just a clebsch gordan coefficient and whatever the value of the other matrix element is, depending on the explicit eigenbasis).
