# Why the proportionality factor of 'Quantum' Poisson brackets is imaginary?

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.

• Is the question effectively Why the proportionality factor is imaginary? May 4, 2020 at 10:19

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. $$\Big\{\hat A_1,\hat B_1\Big\}_{\text{QM}}= i\alpha \Big[\hat A_1,\hat B_1\Big]_-\quad\quad\quad \alpha\in\Re$$