- #1
tommy01
- 40
- 0
Hi there,...
For a derivation of the Ehrenfestequations i found the following commutator relations for the Hamilton-Operator in a book:
[tex]H = \frac{p_{op}^2}{2m} + V(r,t)[/tex]
and the momentum-operator [tex]p_{op} = - i \hbar \nabla[/tex] respectively the position-operator [tex]r[/tex] in position space:
[tex][H,p_{op}] = -i \hbar (V(r,t) \nabla - \nabla V(r,t)) = i \hbar \nabla V(r,t)[/tex]
and
[tex][H,r] = \frac{\hbar^2}{2m} (\Delta r - r \Delta) = - \frac{\hbar^2}{m} \nabla[/tex]
so i don't understand how to get these results. it looks nearly like the use of the product rule but the signs don't match and especially in equation two: how does the laplace-operator become a nabla?
sorry if this problem is obviously but i dont't see it.
thanks and greetings.
tommy
For a derivation of the Ehrenfestequations i found the following commutator relations for the Hamilton-Operator in a book:
[tex]H = \frac{p_{op}^2}{2m} + V(r,t)[/tex]
and the momentum-operator [tex]p_{op} = - i \hbar \nabla[/tex] respectively the position-operator [tex]r[/tex] in position space:
[tex][H,p_{op}] = -i \hbar (V(r,t) \nabla - \nabla V(r,t)) = i \hbar \nabla V(r,t)[/tex]
and
[tex][H,r] = \frac{\hbar^2}{2m} (\Delta r - r \Delta) = - \frac{\hbar^2}{m} \nabla[/tex]
so i don't understand how to get these results. it looks nearly like the use of the product rule but the signs don't match and especially in equation two: how does the laplace-operator become a nabla?
sorry if this problem is obviously but i dont't see it.
thanks and greetings.
tommy