Rotation matrices in Schwinger's oscillator model of angular momentum I am Section 3.9 in Sakurai's Modern QM, 3rd ed (which is Section 3.8 in 2nd ed.)  I am trying to obtain the given form for $\hat D(R)|jm\rangle$:

I employ $\hat D^{-1}\hat D=1$ and ignore the denominator to write
\begin{align}
 \hat D(R)|jm\rangle&=   \hat D  \bigg[\big(\hat a^\dagger_+\big)^{j+m}\big(\hat a^\dagger_-\big)^{j-m}|0,0\rangle\bigg]\\
&=   \hat D \bigg[1^{j+m}\times\big(\hat a^\dagger_+\big)^{j+m}\times1^{j+m}\times\big(\hat a^\dagger_-\big)^{j-m}\times1^{j-m}|0,0\rangle\bigg]\\ 
&=   \hat D \bigg[\big[\hat D^{-1} \hat D \big]^{j+m}\big(\hat a^\dagger_+\big)^{j+m}\big[\hat D^{-1} \hat D \big]^{j+m}\big(\hat a^\dagger_-\big)^{j-m}\big[\hat D^{-1} \hat D \big]^{j-m}|0,0\rangle\bigg]\\ 
&=     \underbrace{\big(\hat D^{-1} \big)^{j+m-1}}_{*} \big(\hat D \,\hat a^\dagger_+\,\hat D^{-1} \big)^{j+m}\underbrace{\big(\hat D \big)^{2m}}_{*}\big(\hat D  \,\hat a^\dagger_-\,\hat D^{-1} \big)^{j-m}\underbrace{\big(\hat D \big)^{j-m}}_{*}|0,0\rangle \\ 
\end{align}
Among the three indicated $*$ terms, I have one extra factor of $\hat D$ so that I will obtain the expression given in Sakurai.  However, I need to show that $\hat D$ commutes with $\hat a_\pm^\dagger$ or else that it commutes with $\hat D \,\hat a^\dagger_\pm\,\hat D^{-1}$.  What would be the easiest way to show this?  I think it will be unnecessarily involved to find an expression for $\hat J_y$ in terms of the Schwinger oscillator operators.
 A: Your state (3.8.18) is a fully-symmetrized tensor (Kronecker) product of 2j oscillators, or spin doublets (spin 1/2s) arrayed to yield a spin j object, in this ingenious Jordan (Schwinger) construction.
So, by construction, (recalling this),
$$
\bbox[yellow]{e^{-i\beta J_y/\hbar} = e^{-i\beta j_y/\hbar} \otimes e^{-i\beta j_y/\hbar} \otimes ...\otimes e^{-i\beta j_y/\hbar} }, 
$$
where $j_y=\sigma_y/2$ for each tensor factor. That is to say, each of the 2j tensor factors only acts on its doublet/oscillator subspace and ignores all others. I am skipping the vacuum, a singlet, since it is rotationally invariant; also in this language.
So, sandwiching the product of oscillators in (3.8.18) by this rotation operator on the left and its inverse on the right, amounts to
$$
e^{-i\beta j_y/\hbar} a_+^\dagger e^{i\beta j_y/\hbar} \otimes e^{-i\beta j_y/\hbar} a_+^\dagger e^{i\beta j_y/\hbar} \otimes e^{-i\beta j_y/\hbar} a_+^\dagger e^{i\beta j_y/\hbar} \otimes  ... \otimes e^{-i\beta j_y/\hbar} a_-^\dagger e^{i\beta j_y/\hbar}  
$$
a total of 2j tensor factors, which acts on the vacuum. This amounts to each tensor factor transforming as
$$
a_+^\dagger \mapsto e^{-i\beta j_y/\hbar} a_+^\dagger e^{i\beta j_y/\hbar}  = a_+^\dagger \cos(\beta/2) + a_-^\dagger \sin(\beta/2) \\
a_-^\dagger \mapsto e^{-i\beta j_y/\hbar} a_-^\dagger e^{i\beta j_y/\hbar}  = a_-^\dagger \cos(\beta/2) - a_+^\dagger  \sin(\beta/2) ,
$$
by the well-known reduction of Pauli vector exponentials. This is the expression following (3.8.20) in your display.
To test-drive this with a simple tractable example, consider j=1,
$$
|1,0\rangle=  a_+^\dagger   a_-^\dagger |0\rangle,
$$
so that, acting on the left with this rotation yields
$$
\Bigl  (e^{-i\beta j_y/\hbar} \otimes e^{-i\beta j_y/\hbar} \Bigr ) \Bigl (a_+^\dagger \otimes    a_-^\dagger\Bigr  )|0\rangle  \\
=\Bigl (a_+^\dagger \cos(\beta/2) + a_-^\dagger \sin(\beta/2) \Bigr )\Bigl ( a_-^\dagger \cos(\beta/2) - a_+^\dagger  \sin(\beta/2) \Bigr )  |0\rangle \\ 
= \bigl (\sin \beta ~ ((a_-^\dagger)^2 -(a_+^\dagger) ^2 )/2 + \cos\beta ~ a_+^\dagger a_-^\dagger\bigr )|0\rangle \\
= \cos\beta ~|1,0\rangle + \frac{\sin\beta }{\sqrt 2}|1,-1\rangle -  \frac{\sin\beta }{\sqrt 2}|1,1\rangle ,
$$
the coefficients  yielding the $d^1_{0,m}$s.
NB If you really wish to eschew the above symmetric tensor product structure, simply recall that, by Schwinger's definition,
$$
\bbox[yellow]{e^{-i\beta J_y/\hbar}= e^{-{\beta\over 2} (J_+-J_-)/\hbar}  
\equiv  e^{\frac{\beta}{2} (a_-^\dagger a_+ - a_+^\dagger a_-)}   },
$$
so you braid this operator past each of your 2j oscillators, all the way to the right where it trivializes to 1 operating on the vacuum. You will, of course, find the same result provided above!
$$
e^{\frac{\beta}{2} (a_-^\dagger a_+ - a_+^\dagger a_-)} a_+^\dagger e^{-\frac{\beta}{2} (a_-^\dagger a_+ - a_+^\dagger a_-)} =  a_+^\dagger \cos(\beta/2) + a_-^\dagger \sin(\beta/2) ,
$$
and the orthogonal form for $a_-^\dagger$.
A: You may use the following formula:
$$(ABA^{-1})^{m} (ACA^{-1})^{n}=(ABA^{-1})(ABA^{-1})...(ABA^{-1})(ACA^{-1})(ACA^{-1})...(ACA^{-1})=AB^{m}C^{n}A^{-1}$$
where $A,B,C$ are operators and $m,n$ are some positive integers.
