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Let's consider Poisson bracket

$$\left\{ A,B\right\} =\alpha_{p} \left( \frac{\partial A}{\partial p_{k}}\frac{\partial B}{\partial q^{k}}-\frac{% \partial A}{\partial q^{k}}\frac{\partial B}{\partial p_{k}}\right) \tag{1}$$

where $\alpha _{p}=\pm 1;$ then

$$\frac{dA}{dt}=\frac{1}{% \alpha_{p}}\left\{ H,A\right\} \tag{2}$$ we know relation between Poisson bracket and commutator

$$\left\{ \hat{O}_{A},\hat{O}_{B}\right\} =i \alpha_{c}\left[ \hat{O}_{A},\hat{O}_{B}\right] \tag{3}$$

then Heisenberg equation

$$\frac{d\hat{A}}{dt}=\frac{i \alpha_{c}% }{\alpha_{p} }\left[ \hat{H},\hat{A}\right] \tag{4}$$

for momentum and coordinate

$$\frac{d\hat{p}}{dt}=\frac{i \alpha_{c}}{\alpha_{p} }\left[ \hat{H},\hat{p}\right] \qquad\text{and} \qquad\frac{d\hat{q}}{dt}=\frac{i \alpha_{c}}{\alpha_{p} }\left[ \hat{H},\hat{q}% \right] \tag{5}$$

finally we get

$$\frac{i \alpha_{c}}{\alpha_{p} }\frac{\partial \psi }{\partial t}=H\psi\tag{6} $$

so there are two independent parameters $\alpha_{c}$ and $\alpha_{p} $ I can choose $\alpha_{c}=1$ and $\alpha_{p}=-1$.

Is that computation correct?

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    $\begingroup$ $\left\{ \hat{O}_{A},\hat{O}_{B}\right\} =i\alpha_p\left[ \hat{O}_{A},\hat{O}_{B}\right]$. $\endgroup$ Nov 18, 2017 at 15:17
  • $\begingroup$ I edited my question. Now, there are two independent parameters $\endgroup$
    – Paramore
    Nov 18, 2017 at 15:32

2 Answers 2

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  1. With the standard CCR $$[\hat{q},\hat{p}]~=~i\hbar{\bf 1}, \tag{A}$$ and with OP's convention (1) for the Poisson bracket $$\{p,q\}_{PB}~=~\alpha_p,\tag{B}$$ the operator $\leftrightarrow$ function correspondence reads
    $$ [\hat{f},\hat{g}]\quad\longleftrightarrow\quad\frac{\hbar}{i\alpha_p}\{f,g\}_{PB} +{\cal O}(\hbar^2) .\tag{C}$$ Note that the traditional physics convention is $\alpha_p=-1$. See e.g. this related Phys.SE post.

  2. Now let's address OP's title question. The time-dependent Schrödinger equation (TDSE) $$ i\hbar \frac{d |\psi \rangle}{d t}~=~\hat{H}|\psi \rangle, \tag{D}$$ and the Heisenberg equation $$ \frac{d\hat{f}}{dt}~=~\frac{i}{\hbar}[\hat{H},\hat{f}] + \frac{\partial \hat{f}}{\partial t}, \tag{E}$$ are independent of the Poisson bracket convention (1); while Hamilton's/Liouville's equation $$ \frac{df}{dt}~=~\frac{1}{\alpha_p}\{H,f\}_{PB} + \frac{\partial f}{\partial t}, \tag{F}$$ does depend on the Poisson bracket convention (1).

  3. For discussions of non-standard sign conventions in the TDSE, see e.g. this & this Phys.SE posts.

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  • $\begingroup$ you write there same parameter $\alpha_{p}$ but it could be any free parameter. see my edited answer. There is second parameter $\alpha_{c}$ . but you get $\alpha_{p}=\alpha_{c}$ why? $\endgroup$
    – Paramore
    Nov 18, 2017 at 15:48
  • $\begingroup$ @Qmechanic A slight generalisation: If you define $[A,B]=\alpha_c(AB-BA)$, then the Heisenberg equation depends on $\alpha_c$. Moreover, the relation between $\{\cdot,\cdot\}$ and $[\cdot,\cdot]$ becomes $\alpha_p\{\cdot,\cdot\}=\alpha_c[\cdot,\cdot]$. Finally, the Sch. equation is both independent of $\alpha_p$ and $\alpha_c$. This $\alpha_c$ is not the same as in the OP, but I believe this is what they had in mind. $\endgroup$ Nov 18, 2017 at 17:57
  • $\begingroup$ @AccidentalFourierTransform: Good point. That would bring the conventional choices in the definitions of $[\cdot,\cdot]$ and $\{\cdot,\cdot\}_{PB}$ on equal footing. $\endgroup$
    – Qmechanic
    Nov 18, 2017 at 17:59
  • $\begingroup$ Your derivation is based on that fact that $[\hat{q},\hat{p}]~=~i\hbar{\bf 1}. \tag{A}$ But do you have any argument why it can't be $[\hat{q},\hat{p}]~= - ~i\hbar{\bf 1}? \tag{A}$ $\endgroup$
    – Paramore
    Nov 19, 2017 at 10:11
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Nov 19, 2017 at 11:28
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Because you add a sign to the Poisson brackets, you should do the same to the commutator for the relation

$$\left\{\hat{O}_{A},\hat{O}_{B}\right\}\leftrightarrow i\left[\hat{O}_{A},\hat{O}_{B}\right]$$

to still hold. Thus you just define the Poisson brackets/commutator differently, keeping the explicit equations the same. There is no point of doing this.

EDIT 1: The time evolution of a quantity in analytical mechanics is unique

$$\frac{dA}{dt}=\left\{A,H\right\}$$

with the regular definition of the Poisson brackets. Also Ehrenfest's theorem is unique

$$\frac{d\left<\hat{A}\right>}{dt}=\frac{1}{i\hbar}\left<\left[\hat{A},\hat{H}\right]\right>$$

with the regular definition of the commutator. This leads us to the usual correspondence

$$\left\{A,B\right\}\leftrightarrow\frac{1}{i\hbar}\left[\hat{A},\hat{B}\right]$$

If for some reason you want to change the usual definitions and write

$$\left\{A,B\right\}_{\rm OP}=\alpha_{\rm PB}\left\{A,B\right\}$$

$$\left[\hat{A},\hat{B}\right]_{\rm OP}=\alpha_{\rm C}\left[\hat{A},\hat{B}\right]$$

then all the above equations change accordingly, but that's only because you defined something differently. It is only a cosmetic change.

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  • $\begingroup$ I don't think so. Sign of commutator is absolutely independent from sign of Poisson bracket $\endgroup$
    – Paramore
    Nov 18, 2017 at 15:20
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    $\begingroup$ So you should change the relation instead, as @AccidentalFourierTransform already stated in the comment. You have to do either one of the two. QM and Analytical Mechanics are independent, and this is just a correspondence between the two. $\endgroup$
    – eranreches
    Nov 18, 2017 at 15:22

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