Say we have a classical Hamiltonian for a conservative system given by $H=\omega xp$, $\omega$ being a constant, $x$ and $p$ are the position and its conjugate momentum respectively.

What should its quantum mechanical counterpart be, in coordinate representation?

I saw somewhere that first we need to symmetrize the classical operator. So \begin{align} H=\omega xp \rightarrow \frac{\omega}{2}(xp + px) \end{align} And only then we can transform the classical observables into operators. So then in QM, we shall have \begin{align} &\hat{H}=\frac{\omega}{2}(\hat{x}\hat{p}+\hat{p}\hat{x})\\ \Rightarrow &\hat{H}\equiv-i\hbar\omega(x\frac{\partial}{\partial x}+\frac{1}{2}) \end{align} in coordinate representation.

My questions are-

  • Am I correct?

  • Why am I correct, i.e, why do we have to symmetrize a product of classical observables before raising them as operators in quantum mechanics?

  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/726069/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Sep 10, 2022 at 4:45
  • 1
    $\begingroup$ Conf. the link for an in - depth answer (especially important wanting to keep symmetries of the original hamiltonian when quantizing), but for a simple example that could result from not symmetrizing - your classical H could in principle have been written $H = \omega x p + \omega x(xp - px)$ as classicaly the term in parenthesis is zero. When quantizing you get the arbitrary $\hat H = \omega\hat X \hat P + i\hbar \omega\hat X $. (I dont understand your expression after the rightarrow, that is not how the X - and P operator look in the x - basis ) $\endgroup$
    – user330563
    Commented Sep 10, 2022 at 9:18
  • $\begingroup$ @user330563 thanks! I was looking for this. (After the right arrow) I have just simplified the operator expression as we do by taking a test wave function:: $(\hat{x}\hat{p}+\hat{p}\hat{x})\psi(x)=-i\hbar(x\partial_{x}{\psi(x)}+\partial_{x}{(x\psi(x)}))$. Also, I didn't get what you mean by "Conf. the link". Can you please elaborate? $\endgroup$ Commented Sep 10, 2022 at 13:50
  • $\begingroup$ OK, I just meant the link to the duplicate question $\endgroup$
    – user330563
    Commented Sep 10, 2022 at 15:46