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When writing the quantum Hamiltonian from a classical hamiltonian, we cannot simply replace position and momentum variables with their respective operators. We have to make sure that the resultant Hamiltonian is also hermitian.

In R.Shankar book he mentions that this is ambiguous but usually symmetrization is used. So a product r.p in the Hamiltonian becomes: \begin{equation} r.p \rightarrow \frac{\hat{r}\hat{p} + \hat{p}\hat{r}}{2}. \end{equation}

My question was why can't we use: \begin{equation} r.p \rightarrow i(\hat{r}\hat{p} - \hat{p}\hat{r}) \end{equation} which would also keep the Hamiltonian hermitian. What are the physical consequences of choosing such a method of quantising the Hamiltonian?

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    $\begingroup$ It doesn't reduce to $r \cdot p$ in the quasi-classical limit maybe, is it not zero in that limit? $\endgroup$
    – bolbteppa
    Commented Dec 2, 2021 at 11:01
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    $\begingroup$ oh yeah makes sense. Thanks! $\endgroup$
    – KaV
    Commented Dec 2, 2021 at 11:24
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    $\begingroup$ @bolbteppa's answer is true in this context. But once you go to operators that are more than quadratic in $r$ and $p$, the ambiguities grow and there are multiple quantum operators with the same classical limit. $\endgroup$
    – Quantum Mechanic
    Commented Dec 2, 2021 at 14:25
  • $\begingroup$ @QuantumMechanic, it would be interesting to see an explicit example as an answer. $\endgroup$
    – Cham
    Commented Dec 2, 2021 at 17:53
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    $\begingroup$ Linked. $\endgroup$
    – Cosmas Zachos
    Commented Dec 2, 2021 at 17:54

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$\hat{r}\hat{p} - \hat{p}\hat{r} = [\hat{r}, \hat{p}] = i \hbar$ in 1D. So your second hamiltonian would be a constant, which is different from the first one

For more information on commutators you can check https://en.wikibooks.org/wiki/Quantum_Mechanics/Operators_and_Commutators

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