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

enter image description here

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.

  • 4
    $\begingroup$ 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


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 $a_-^\dagger$.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.