Can someone please provide an insightful solution or verification of my solution to this question:
Compute the commutator $$[p^2,q^2]$$ (in the one variable case), and compare it to quantization $$\widehat{\{p^2, q^2\}}$$ of the Poisson bracket.
Attempted Solution:
Recall that $$\hat{p} = - i\hbar\frac{d}{dq}$$ and that let $$\hat{q} = \hat{q}$$, hence the expression $$[\hat{p}^2,\hat{q}^2]$$ gives us:
$$ [\hat{p}^2,\hat{q}^2] = \hat{p}^2\hat{q}^2 - \hat{q}^2\hat{p}^2 = (i^2\hbar^2\frac{d^2}{dq^2}q^2) - (i^2\hbar^2 q^2\frac{d^2}{dq^2}) $$ $$\qquad = (- \hbar^2\frac{d^2}{dq^2}q^2) - (- \hbar^2 q^2\frac{d^2}{dq^2})= (- \hbar^2\frac{d^2}{dq^2}q^2) + (\hbar^2 q^2\frac{d^2}{dq^2}) $$ $$\qquad (- \hbar^2\frac{d^2}{dq^2}q^2) + (\hbar^2 q^2\frac{d^2}{dq^2}) = (-2\hbar^2) + (- 2\hbar^2 q\frac{d}{dq}) = (-2i\hbar\hat{p}\hat{q}) - (2\hbar^2)$$
Now divide by $\hbar$ and take the limit as $\hbar \to 0$, we get the following: $\frac{1}{\hbar}[\hat{p}^2,\hat{q}^2] = -2i\hat{p}\hat{q}$, and now multiply both side by $i$ to get the desired expression that: $\frac{i}{\hbar}[\hat{p}^2,\hat{q}^2] = \widehat{\{p^2,q^2\}}$.
This would have been a bland "check my homework" question, if it did not fail to specify the "quantization" $$\widehat{\{p^2, q^2\}} = -4 \widehat{qp}.$$
For every quantization prescription, there is a different answer. For quadratic functions, all answers differ by an $\hbar$ correction, (which I think you eliminated, dangerously, by taking the $\hbar\to 0$ limit).
In any case, for Weyl quantization, you indeed have $$ -2 (\hat q \hat p + \hat p \hat q), $$ which is the exact expression determined by the commutator, $$ -[\hat q ^2, \hat p^2]=-\{\hat q,\{\hat p,[\hat q,\hat p]\}\}= -2i\hbar (\hat q \hat p + \hat p \hat q), $$ where, of course, now, the curly brace denotes the operator anticommutator, not the PB!