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I Intuition for why d<p>/dt = -dV(<x>)/dx

  1. Dec 4, 2018 #1
    Is there any good physical or graphical intuition for why ##\frac{d \langle p \rangle}{dt} = -\frac{\partial V(\langle x \rangle)}{\partial x}##? Classically this is apparently true.

    Thanks.
     
  2. jcsd
  3. Dec 4, 2018 #2

    PeroK

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    In classical physics potential is defined so that ##F = -\frac{\partial V}{\partial x}##.

    Your equation is, however, not correct. It should be:

    ##\frac{d \langle p \rangle}{dt} = -\langle \frac{\partial V(x)}{\partial x} \rangle##
     
  4. Dec 4, 2018 #3
    I think ##\frac{d \langle p \rangle}{dt} = -\langle \frac{\partial V(x)}{\partial x} \rangle## is the Ehrenfest theorem. The Wikipedia page however notes that were quantum expectation values to be consistent with Newtonian mechanics, we'd have ##F = -\frac{\partial V}{\partial x}##. Unless I'm misunderstanding something.
     
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