Expectation of a commutation relation Is there any significance to: $\langle[H,\hat{O}]\rangle =0$ (which can easily be shown) where $H$ is the Hamiltonian, $\hat{O}$ is an arbitrary operator? Thanks.
 A: Yes.  The commutator of an operator with the hamiltonian is related to the rate of change of the expectation value of the operator.
$$\frac{\partial}{\partial t} \langle \Psi \mid \hat{O} \mid \Psi \rangle = \langle \frac{\partial}{\partial t} \Psi \mid \hat{O} \mid \Psi \rangle +  \langle \Psi \mid \frac{\partial}{\partial t} \hat{O} \mid \Psi \rangle + \langle \Psi \mid  \hat{O} \mid \frac{\partial}{\partial t} \Psi \rangle$$
applying Schrodinger's equation
$$i\hbar \frac{\partial}{\partial t} \mid \Psi \rangle = H\mid\Psi\rangle$$
and its conjugate, we get
$$\frac{\partial}{\partial t} \langle \Psi \mid \hat{O} \mid \Psi \rangle = \frac{i}{\hbar} \left(\langle \Psi \mid H \hat{O}\mid\Psi\rangle   - \langle \Psi \mid \hat{O}H\mid\Psi\rangle\right) + \langle \Psi \mid \frac{\partial}{\partial t}\hat{O}\mid\Psi\rangle $$
where we've exploited that fact that $H$ is Hermitian.  Combining the first two terms gives
$$\frac{\partial}{\partial t} \langle \Psi \mid \hat{O} \mid \Psi \rangle = \frac{i}{\hbar}\langle \Psi \mid [H,\hat{O}] \mid \Psi \rangle + \langle \Psi \mid \frac{\partial}{\partial t}\hat{O}\mid\Psi\rangle$$
Specifically, if the commutator is zero, as you specified, and the operator is time-independent, then its expectation value does not change.
As Ron suggested, we can get a little more from this analysis.  The generalized uncertainty principle states
$$\sigma_H\sigma_O \geq |\langle \Psi \mid \frac{1}{2i}[H,O]\mid\Psi\rangle|$$
Using the above relation  and assuming the operator is time independent, we get
$$\sigma_H\sigma_O \geq \frac{\hbar}{2}\left|\frac{\textrm{d}\langle\hat{O}\rangle}{\textrm{d}t}\right|$$
or
$$\sigma_H\frac{\sigma_O}{\left|\frac{\textrm{d}\langle\hat{O}\rangle}{\textrm{d}t}\right|} \geq \frac{\hbar}{2}$$
$\sigma_O$ represents the spread in $\hat{O}$.  If $\langle \hat{O}\rangle$ changes by $\sigma_O$, that indicates a significant change.  So $\frac{\sigma_O}{\left|\frac{\textrm{d}\langle\hat{O}\rangle}{\textrm{d}t}\right|}$ represents a time scale on which $\langle \hat{O}\rangle$ changes a significant amount.  If we call that $\Delta t$ and call the standard deviation in the energy $\Delta E$, we have
$$\Delta E \Delta t \geq \frac{\hbar}{2}$$
