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When trying to understand the correspondence principle, I found a proof in this section (of this book) about why the quantum Poisson brackets ($\{\,,\,\}_{\text{QM}}$) must be proportional to the commutator ($[\,,\,]_{-}$). But, I'm stuck in the last step:

$$\Big[\hat A_1,\hat B_1\Big]_{-}\Big\{\hat A_2,\hat B_2\Big\}_{\text{QM}}=\Big\{\hat A_1,\hat B_1\Big\}_{\text{QM}}\Big[\hat A_2,\hat B_2\Big]_{-}$$

Since $\hat A_i,\hat B_i$ are almost arbitrary chosen operators, this result suggests that:

$$\Big\{\hat A,\hat B\Big\}_{\text{QM}}=i\alpha\Big[\hat A,\hat B\Big]_{-}$$

My question is, why is that suggestion so straightforward?

I thought it could be obtained by choosing $\hat A_1$ and $\hat B_1$ such that $\dfrac{\{\hat A_1,\hat B_1\}_{\text{QM}}}{[\hat A_1,\hat B_1]_{-}}=i\alpha$, but that is redundant.

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    $\begingroup$ Is the question effectively Why the proportionality factor is imaginary? $\endgroup$
    – Qmechanic
    May 4, 2020 at 10:19

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Notice that this equation $$\Big[\hat A_1,\hat B_1\Big]_{-}\Big\{\hat A_2,\hat B_2\Big\}_{\text{QM}}=\Big\{\hat A_1,\hat B_1\Big\}_{\text{QM}}\Big[\hat A_2,\hat B_2\Big]_{-}$$ suggests that $\Big\{\hat A_1,\hat B_1\Big\}_{\text{QM}}\propto \Big[\hat A_1,\hat B_1\Big]_-$. That I think you understand. But next thing that must be noticed that $\hat{A}$ and $\hat{B}$ are hermitian operators as they represent the observables. So its poisson bracket must also be hermitian. Take an example of $\dot{q}=\{q,H\}$. $q$ and $H$ are observable along with $\dot{q}$. This is evident that the poisson bracket of two observable must be observable. However, commutation is an anti-hermitian for two hermitian operator. $$[\hat{A},\hat{B}]_-^{\dagger}=-[\hat{A},\hat{B}]_-$$ So the proportionality constant between poisson bracket and the commutator must be a purely imaginary quantity. $$\Big\{\hat A_1,\hat B_1\Big\}_{\text{QM}}= i\alpha \Big[\hat A_1,\hat B_1\Big]_-\quad\quad\quad \alpha\in\Re$$

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The quantum PB is assumed to take 2 Hermitian operators into a Hermitian operator, while the commutator clearly takes 2 Hermitian operators into an anti-Hermitian operator. If they are supposed to be proportional, this can only work if the proportionality factor is imaginary.

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