Rotation matrices in Schwinger's oscillator model of angular momentum

Question

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 $hat D(R)|jmrangle$:

I employ $hat D^{-1}hat D=1$ and ignore the denominator to write
begin{align}
 hat D(R)|jmrangle&=   hat D  bigg[big(hat a^dagger_+big)^{j+m}big(hat a^dagger_-big)^{j-m}|0,0ranglebigg]
&=   hat D bigg[1^{j+m}timesbig(hat a^dagger_+big)^{j+m}times1^{j+m}timesbig(hat a^dagger_-big)^{j-m}times1^{j-m}|0,0ranglebigg] 
&=   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}|0,0ranglebigg] 
&=     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}}_{*}|0,0rangle  
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.

Cosmas Zachos · Accepted Answer

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),
$$
bbox[yellow]{e^{-ibeta J_y/hbar} = e^{-ibeta j_y/hbar} otimes e^{-ibeta j_y/hbar} otimes ...otimes e^{-ibeta j_y/hbar} }, 
$$
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
$$
e^{-ibeta j_y/hbar} a_+^dagger e^{ibeta j_y/hbar} otimes e^{-ibeta j_y/hbar} a_+^dagger e^{ibeta j_y/hbar} otimes e^{-ibeta j_y/hbar} a_+^dagger e^{ibeta j_y/hbar} otimes  ... otimes e^{-ibeta j_y/hbar} a_-^dagger e^{ibeta j_y/hbar}  
$$
a total of 2j tensor factors, which acts on the vacuum. This amounts to each tensor factor transforming as
$$
a_+^dagger mapsto e^{-ibeta j_y/hbar} a_+^dagger e^{ibeta j_y/hbar}  = a_+^dagger cos(beta/2) + a_-^dagger sin(beta/2) 
a_-^dagger mapsto e^{-ibeta j_y/hbar} a_-^dagger e^{ibeta j_y/hbar}  = a_-^dagger cos(beta/2) - a_+^dagger  sin(beta/2) ,
$$
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,
$$
|1,0rangle=  a_+^dagger   a_-^dagger |0rangle,
$$
so that, acting on the left with this rotation yields
$$
Bigl  (e^{-ibeta j_y/hbar} otimes e^{-ibeta j_y/hbar} Bigr ) Bigl (a_+^dagger otimes    a_-^daggerBigr  )|0rangle  
=Bigl (a_+^dagger cos(beta/2) + a_-^dagger sin(beta/2) Bigr )Bigl ( a_-^dagger cos(beta/2) - a_+^dagger  sin(beta/2) Bigr )  |0rangle  
= bigl (sin beta ~ ((a_-^dagger)^2 -(a_+^dagger) ^2 )/2 + cosbeta ~ a_+^dagger a_-^daggerbigr )|0rangle 
= cosbeta ~|1,0rangle + frac{sinbeta }{sqrt 2}|1,-1rangle -  frac{sinbeta }{sqrt 2}|1,1rangle ,
$$
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,
$$
bbox[yellow]{e^{-ibeta J_y/hbar}= e^{-{betaover 2} (J_+-J_-)/hbar}  
equiv  e^{frac{beta}{2} (a_-^dagger a_+ - a_+^dagger a_-)}   },
$$
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 $a_-^dagger$.

chungls123 · Answer

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.

Rotation matrices in Schwinger's oscillator model of angular momentum

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