- #1
jfy4
- 649
- 3
Hi,
This thread looks like GR/SR, but it has grounds QM and maybe only stays in that realm, which is what I'm asking
I was looking at some everyday non-relativistic quantum mechanics and I spotted something I thought was interesting. Consider the time evolution of an observable
[tex]
\frac{d \hat{A}}{dt}=\frac{i}{\hbar}[H,\hat{A}]
[/tex]
and next consider a form of that for the position operator
[tex]
\frac{d^2 \hat{x}}{dt^2}=\frac{i}{\hbar}[H,\hat{v}]
[/tex]
This seems to give an expression for the acceleration. My question is whether there is anything deeper going on here. Is this expression trying to say that "the extent to which the velocity field doesn't commute with the Hamiltonian of the system is the acceleration"? More specifically, can this be generalized to
[tex]
\frac{d^2 \mathbf{x} }{dt^2}=\mathcal{L}_{H}\,\mathbf{u}?
[/tex]
Thanks,
This thread looks like GR/SR, but it has grounds QM and maybe only stays in that realm, which is what I'm asking
I was looking at some everyday non-relativistic quantum mechanics and I spotted something I thought was interesting. Consider the time evolution of an observable
[tex]
\frac{d \hat{A}}{dt}=\frac{i}{\hbar}[H,\hat{A}]
[/tex]
and next consider a form of that for the position operator
[tex]
\frac{d^2 \hat{x}}{dt^2}=\frac{i}{\hbar}[H,\hat{v}]
[/tex]
This seems to give an expression for the acceleration. My question is whether there is anything deeper going on here. Is this expression trying to say that "the extent to which the velocity field doesn't commute with the Hamiltonian of the system is the acceleration"? More specifically, can this be generalized to
[tex]
\frac{d^2 \mathbf{x} }{dt^2}=\mathcal{L}_{H}\,\mathbf{u}?
[/tex]
Thanks,