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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 $\hat D(R)|jm\rangle$:

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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.

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    $\begingroup$ I believe what is being used here is not multiplication by identities but instead the transformation rule for operators and states under rotations. $\endgroup$
    – secavara
    Jan 15, 2021 at 20:04
  • $\begingroup$ You are wildly misreading the formula. ${\mathcal D} a_+^\dagger {\mathcal D}^{-1}=a_+^\dagger \cos(\beta/2)+ a_-^\dagger\sin (\beta/2)$ and the orthogonal form for $a_-^\dagger$. I don't believe you grasp Schwinger's (actually Jordan's) picture. $\endgroup$ Jan 15, 2021 at 20:57
  • $\begingroup$ Hot tip: You are rotating states in a tensor product of 2j spin-1/2 doublets. To satisfy yourself you understand how your aspirational coproduct construction should work, test drive it for j =1, ie, just compose two doublets. Trying the simplest possible example to appreciate the language is a technique I've watched Wigner utilize to devastating effect, destroying colloquium speakers by 2x2 matrix counterexamples. His complex formulas didn't start life complex: they abstracted simple cases. Do you want an illustration here? $\endgroup$ Jan 15, 2021 at 22:31
  • $\begingroup$ Linked. $\endgroup$ Jan 16, 2021 at 1:10
  • $\begingroup$ I think so. Since $\hat D$ is an operator on a Hilbert space, it doesn't make sense to think of the same $\hat D$ acting on a $|j,m\rangle$ state as $\hat D$ acting on a $\hat a^\dagger\hat a|0\rangle$ state from a different Hilbert space. I was failing to separate the abstract operator from its representation. $\endgroup$ Jan 21, 2021 at 4:24

2 Answers 2

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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 $a_-^\dagger$.

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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.

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