- #1
calculus_jy
- 56
- 0
In proving the Ehrenfest Theorem
This is the typical first line:
[tex]\frac{d }{dt}<O> = \frac{\partial}{\partial t} <\psi|O|\psi> = <\dot{\psi}|O|\psi> + <\psi|O|\dot{\psi}>+<\psi|\dot{O}|\psi>
[/tex]
My question is how can the exact differential
[tex] \frac{d }{dt}<O>[/tex]
be changed the partial differential
[tex] \frac{\partial}{\partial t} <\psi|O|\psi> [/tex]
in the first equality. would it not be
[tex] \frac{d }{dt}<O>=\frac{\partial}{\partial x} <\psi|O|\psi> \frac{dx}{dt}+\frac{\partial}{\partial t} <\psi|O|\psi>[/tex]
Have we assumed that [tex] \frac{dx}{dt}=0[/tex]
If so why?
This is the typical first line:
[tex]\frac{d }{dt}<O> = \frac{\partial}{\partial t} <\psi|O|\psi> = <\dot{\psi}|O|\psi> + <\psi|O|\dot{\psi}>+<\psi|\dot{O}|\psi>
[/tex]
My question is how can the exact differential
[tex] \frac{d }{dt}<O>[/tex]
be changed the partial differential
[tex] \frac{\partial}{\partial t} <\psi|O|\psi> [/tex]
in the first equality. would it not be
[tex] \frac{d }{dt}<O>=\frac{\partial}{\partial x} <\psi|O|\psi> \frac{dx}{dt}+\frac{\partial}{\partial t} <\psi|O|\psi>[/tex]
Have we assumed that [tex] \frac{dx}{dt}=0[/tex]
If so why?