Uncertainty principle##\Delta H\Delta Q##, where Q is time-independent

[Foracle]

TL;DR

Why is $\Delta H\Delta Q \ge |\frac{ħ}{2}$d<Q>/dt$|$

Let Q be a time-independent operator.
$[H,Q] = iħ[\frac{d}{dt},Q]$
Since Q is time-independent, $[H,Q]=0$

And from the uncertainty principle :
$\Delta H\Delta Q \ge |<\Psi|\frac{1}{2i}[H,Q]|\Psi>|$
From $[H,Q] = 0$, I concluded that $\Delta H\Delta Q \ge 0$

But by evaluating d<Q>/dt, it can be found that $\Delta H\Delta Q \ge |\frac{ħ}{2}$d<Q>/dt$|$

I know that the right answer is the latter, but I just want to know why $\Delta H\Delta Q \ge 0$ is wrong.

 

[Haborix]

You go wrong when you equate the Hamiltonian with the time derivative. Your identification implies there is no difference between the Hamiltonians describing different systems or the time-evolution of the wavefunctions for those systems.