# Ehrenfest's Theorem

1. Aug 6, 2010

### calculus_jy

In proving the Ehrenfest Theorem
This is the typical first line:

$$\frac{d }{dt}<O> = \frac{\partial}{\partial t} <\psi|O|\psi> = <\dot{\psi}|O|\psi> + <\psi|O|\dot{\psi}>+<\psi|\dot{O}|\psi>$$

My question is how can the exact differential
$$\frac{d }{dt}<O>$$
be changed the partial differential
$$\frac{\partial}{\partial t} <\psi|O|\psi>$$
in the first equality. would it not be
$$\frac{d }{dt}<O>=\frac{\partial}{\partial x} <\psi|O|\psi> \frac{dx}{dt}+\frac{\partial}{\partial t} <\psi|O|\psi>$$

Have we assumed that $$\frac{dx}{dt}=0$$
If so why?

2. Aug 8, 2010

### facenian

I think that if we are working in abstrac state space the only variable is time since $|\psi(t)>$ is only function of time as well as the operator

Last edited: Aug 8, 2010