Need help with proof for expectation value relation.

AI Thread Summary
The discussion centers on proving the relation \hbar \frac{d}{dt}\langle L\rangle = \langle N \rangle, where L represents angular momentum and N represents torque. The user initially attempted to use Ehrenfest's theorem to express the equation but struggled with the commutator term. After working through the calculations involving the commutation of the angular momentum operator and the Hamiltonian, the user realized that the angular momentum operator is time-independent, leading to the term vanishing. Ultimately, this insight allowed the user to correctly derive the expected result. The problem was resolved by recognizing the properties of the operators involved.
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Homework Statement


I have to prove the following:
<br /> \hbar \frac{d}{dt}\langle L\rangle = \langle N \rangle<br />

Edit: L = Angular Momentum & N = Torque

Homework Equations


I used Ehrenfest's theorem, and I've got the equation in the following form:

<br /> \frac{1}{i} \left(\left[L,H\right]\right) + \hbar \left\langle \frac{\partial L}{\partial t}\right\rangle<br />

The Attempt at a Solution



I pretty much need to prove the commutator term vanishes, but I'm not sure if it does. I've done the following with the commutation:

<br /> i\hbar\frac{d}{dt}(\mathbf{r}\times\mathbf{p}) - i\hbar(\mathbf{r}\times\mathbf{p})\frac{d}{dt}<br />

<br /> -i^2\hbar^2\frac{d}{dt}(\mathbf{r}\times\mathbf{\nabla}_r) - -(i^2)\hbar^2(\mathbf{r}\times\mathbf{\nabla}_r)\frac{d}{dt}<br />

<br /> \hbar^2\frac{d}{dt}(\mathbf{r}\times\mathbf{\nabla}_r) - \hbar^2(\mathbf{r}\times\mathbf{\nabla}_r)\frac{d}{dt}<br />

<br /> \hbar^2\frac{d}{dt}\left(\epsilon_{ijk}\mathbf{r}^j\mathbf{\nabla}^k\right) - \hbar^2\left(\epsilon_{ijk}\mathbf{r}^j\mathbf{\nabla}^k\right) \frac{d}{dt}<br />

Can I switch the order of the derivatives/operators in such a way that I get the following:
<br /> \hbar^2\frac{d}{dt}\left(\epsilon_{ijk}\mathbf{r}^j\mathbf{\nabla}^k\right) - \hbar^2\frac{d}{dt}\left(\epsilon_{ijk}\mathbf{r}^j\mathbf{\nabla}^k\right) = 0<br />
 
Last edited:
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Never mind, figured it out. I forgot that the angular momentum operator is time-independent, so it disappears. Then I took the commutation of [L,H] a little differently then the way I did it above, and I got the answer.
 
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