2
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ $\left\{ \hat{O}_{A},\hat{O}_{B}\right\} =i\alpha_p\left[ \hat{O}_{A},\hat{O}_{B}\right]$. $\endgroup$ – AccidentalFourierTransform Nov 18 '17 at 15:17
  • $\begingroup$ I edited my question. Now, there are two independent parameters $\endgroup$ – Paramore Nov 18 '17 at 15:32
3
$\begingroup$
  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.

$\endgroup$
  • $\begingroup$ 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 '17 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. $\endgroup$ – AccidentalFourierTransform Nov 18 '17 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 '17 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 '17 at 10:11
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Nov 19 '17 at 11:28
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I don't think so. Sign of commutator is absolutely independent from sign of Poisson bracket $\endgroup$ – Paramore Nov 18 '17 at 15:20
  • 1
    $\begingroup$ 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 '17 at 15:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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