Prove that $i\hbar\dot{\hat{\Omega}}=[\hat{\Omega},\hat{H}]$ , where $\hat{\Omega}(\hat{x},\hat{p})$ is Taylor expandable How do I show that $i\hbar\dot{\hat{\Omega}}=[\hat{\Omega},\hat{H}]$ , where $\hat{\Omega}(\hat{x},\hat{p})$ is Taylor expandable, $\hat{H}(\hat{x},\hat{p})$ is the Hamiltonian of the system and $\hat{x}, \hat{p}$ are the positional and momentum operators respectively.
I have already proved that $i\hbar\dot{\hat{x}}=[\hat{x},\hat{H}]$ and $i\hbar\dot{\hat{p}}=[\hat{p},\hat{H}]$ and also the classical analogue using Poisson bracket that $\dot{\Omega}=\{\Omega, H\}_{PB}$. I am stuck after this point, and have no clue how to proceed with the question.
Any help regarding this is appreciated.
 A: Assuming a time-independent Hamiltonian $H$  (the time-dependent case works just the same, but with a heavier formalism), Heisenberg's equations of motion are solved by :
\begin{align}
x(t) = e^{-iHt/\hbar}x(0)e^{iHt/\hbar} \\
p(t) = e^{-iHt/\hbar}p(0)e^{iHt/\hbar} 
\end{align}
Then, we have :
\begin{align}
\Omega(t) &= \Omega(x(t),p(t)) \\
&= \Omega\left(e^{iHt/\hbar}x(0)e^{-iHt/\hbar},e^{iHt/\hbar}p(0)e^{-iHt/\hbar} 
\right) \\
&=e^{iHt/\hbar}\Omega(0)e^{-iHt/\hbar}
\end{align}
So, taking the time derivative, we get :
$$i\hbar\dot \Omega (t) = [\Omega,H]$$
A: So you mentioned that you already have $i\hbar\dot{x}=[x,H]$ and $i\hbar\dot{p}=[p,H]$  (I am omitting the hats above the operators for convinience). Now there exists a "product rule" for the commutator, with which I mean that for Operators $A_1,A_2,H$ we have $[A_1A_2,H]=A_1[A_2,H]+[A_1,H]A_2$ which you can see by just writing out the commutators. This can of course be extended to arbitrary products by induction: $[\prod_{k=1}^{n}A_n,H]=\sum_{k=1}^{n}(\prod_{j=1}^{k-1}A_j )[A_k,H] (\prod_{l=k+1}^{n}A_l)$ where the "empty product" is supposed to just be the identity, i.e. for example $\prod_{j=1}^{0}A_j=1$. The proof by induction is not very difficult. You can also see the sum as a telescoping sum when you write the commutator in the middle as a difference and obtain the result.
Well, now look at the terms in the Taylor expansion of $\Omega$: They are of the form $x^np^m$ with constants in front. This is just a product of $n+m$ operators so we can use our "product rule": $[x^np^m,H]=\sum_{k=1}^{n}(\prod_{j=1}^{k-1}x^j )[x,H] (\prod_{l=k+1}^{n}x^l)p^m+x^n\sum_{k=1}^{m}(\prod_{j=1}^{k-1}p^j )[p,H] (\prod_{l=k+1}^{m}p^l)$. Here we have two sums because I split the product first in the factors $x^n$ and $p^m$ before applying the rule.
Okay, we are almost finished. Just plug in $i\hbar\dot{x}=[x,H]$ and $i\hbar\dot{p}=[p,H]$ for the commutators and then use the product rule of the time derivative "backwards". That means: $[x^np^m,H]=i\hbar \sum_{k=1}^{n}(\prod_{j=1}^{k-1}x^j )\dot{x} (\prod_{l=k+1}^{n}x^l)p^m+  i\hbar x^n\sum_{k=1}^{m}(\prod_{j=1}^{k-1}p^j )\dot{p}  (\prod_{l=k+1}^{m}p^l)=i\hbar \frac{d}{dt} x^np^m$. Now we have shown the statement for one term in the Taylor expansion and because of linearity it extends to the whole Taylor series.
Hopefully that was helpful!
