- #1
jonnaraev
- 3
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Hello, first post here.
I am preparing for my Introductory Quantum Mechanics course, and in the exam questions, we are asked to use Ehrenfest's theorem to show that
[tex] \frac{d}{dt}\langle \vec{r}\cdot \vec{p} \rangle = \langle 2T-\vec{r}\cdot \nabla V \rangle [/tex]
Now, from other results:
[tex] \frac{d}{dt}\langle \vec{r} \rangle = \frac{1}{m} \langle \vec{p} \rangle [/tex]
[tex] \frac{d}{dt}\langle \vec{p} \rangle = \langle -\nabla V \rangle [/tex]
it can be solved using a product rule of differentiation:
[tex] \frac{d}{dt}\langle \vec{r}\cdot \vec{p} \rangle =\frac{d}{dt}\langle \vec{r} \rangle \cdot \vec{p} + \vec{r} \cdot \frac{d}{dt}\langle \vec{p} \rangle [/tex]
But I severely doubt this is the right way, or even mathematically sound. So my question is, is the above equation sound, or should I be calculating the commutators with the Hamiltonian, as per Ehrenfest's?
I am preparing for my Introductory Quantum Mechanics course, and in the exam questions, we are asked to use Ehrenfest's theorem to show that
[tex] \frac{d}{dt}\langle \vec{r}\cdot \vec{p} \rangle = \langle 2T-\vec{r}\cdot \nabla V \rangle [/tex]
Now, from other results:
[tex] \frac{d}{dt}\langle \vec{r} \rangle = \frac{1}{m} \langle \vec{p} \rangle [/tex]
[tex] \frac{d}{dt}\langle \vec{p} \rangle = \langle -\nabla V \rangle [/tex]
it can be solved using a product rule of differentiation:
[tex] \frac{d}{dt}\langle \vec{r}\cdot \vec{p} \rangle =\frac{d}{dt}\langle \vec{r} \rangle \cdot \vec{p} + \vec{r} \cdot \frac{d}{dt}\langle \vec{p} \rangle [/tex]
But I severely doubt this is the right way, or even mathematically sound. So my question is, is the above equation sound, or should I be calculating the commutators with the Hamiltonian, as per Ehrenfest's?
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