Hamiltonian in classical mechanics is $$H=wxp $$ $x=$ position, $p=$ momentum coordinate.

Find the corresponding Hamiltonian in quantum mechanics!

  • 2
    $\begingroup$ This question is ill-posed - there is no unique map between classical observables and quantum mechanical observables, you have to choose an ordering convention or something equivalent, and even then there's still trouble with naive approaches to quantization. Please be more specific about what you actually want to achieve. $\endgroup$
    – ACuriousMind
    Commented Jul 7, 2016 at 18:07

1 Answer 1


The naive thing to do would be to just replace $x$ by $\hat{x}$ and $p$ by $-i\hbar \nabla$. Note however that $(\hat{x}\hat{p})^{\dagger} = \hat{p}\hat{x} \neq \hat{x}\hat{p}$, i.e. the operator is not Hermitian due to the noncommutativity of $\hat{x}$ and $\hat{p}$, which is a problem of course.

So what can we do to fix this? One possible way is to symmetrize:

$$\hat{H} = w\left(\frac{\hat{x}\hat{p}+\hat{p}\hat{x}}{2}\right).$$

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    $\begingroup$ Notice that $(\hat{p} f)(x)= -i\partial_x f(x)$ is the representation of the momentum operator on the position basis, not the momentum operator itself. $\endgroup$
    – gented
    Commented Jul 6, 2016 at 13:17
  • $\begingroup$ @GennaroTedesco, how does that distinction help the answer? What other representation could possible be relevant here? $\endgroup$ Commented Jul 6, 2016 at 14:08
  • $\begingroup$ @JánLalinský well I don't see why the monentum basis wouldn't work just as well, but I think the answer is fine as it stands. $\endgroup$
    – anon01
    Commented Jul 6, 2016 at 14:17
  • $\begingroup$ @ConfusinglyCuriousTheThird, I think you're right, in this case the momentum representation would be the same. $\endgroup$ Commented Jul 7, 2016 at 21:05

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