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Ehrenfest's Theorem

  1. Aug 6, 2010 #1
    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>

    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?
  2. jcsd
  3. Aug 8, 2010 #2
    I think that if we are working in abstrac state space the only variable is time since [itex]|\psi(t)>[/itex] is only function of time as well as the operator
    Last edited: Aug 8, 2010
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