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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.

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Which state is being used to calculate the expectation value? – David Z Sep 2 '11 at 23:18
Any eigenstate of $H$.:) – bra-ket Sep 2 '11 at 23:28
up vote 4 down vote accepted

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

$$\sigma_H\sigma_O \geq \frac{\hbar}{2}\left|\frac{\textrm{d}\langle\hat{O}\rangle}{\textrm{d}t}\right|$$


$$\sigma_H\frac{\sigma_O}{\left|\frac{\textrm{d}\langle\hat{O}\rangle}{\textrm{d}t}\right|} \geq \frac{\hbar}{2}$$

$\sigma_O$ 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

$$\Delta E \Delta t \geq \frac{\hbar}{2}$$

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It is perhaps nice with regard to this operator to mention the L.I Mandelshtam- I.E Tamm interpretation of the time-energy uncertainty principle which is mentioned in Wikipedia's uncertainty principle article. – Ron Maimon Sep 3 '11 at 2:38
@Ron Sure. I'll add that. – Mark Eichenlaub Sep 3 '11 at 3:03
@Mark and Ron:Thanks guys! – bra-ket Sep 3 '11 at 7:46

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