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Quantum mechanics: what does d<x>/dt mean physically?

  1. Jul 17, 2012 #1
    1. The problem statement, all variables and given/known data

    I am reading the Griffths book on quantum mechanics. In the first chapter on Momentum (Sect 1.5) what does d<x>/dt mean in the physical sense?

    2. Relevant equations

    NA

    3. The attempt at a solution

    If the expectation value <x> is already determined by the schrodinger equation,

    (i) thus <x> must be in the form of a number and hence d<x>/dt must equal to zero always. So calculating d<x>/dt is not relevant.

    (ii) unless we are saying <x> changes in time but this will contradict my above statement..... since the expectation is fixed once the wave function of a particle is known)

    (iii) If this rate of change is about when the initial measurement is made and <x> spikes to one of the eigenvalues and then 'smears off' into the original wave function form, then my question is (from a mathematics perspective) how does the equation allow for a change from a non continuous form (when ψ collapses and <x> spikes) and then into a continuous form (when ψ goes back to its original form and <x> becomes the expectation value of ALL possible eigenvalues)

    Thanks very much for any assistance!

    gennes77
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 17, 2012 #2
    [itex]\langle x \rangle[/itex] is necessarily not a function of position, but it can still be a function of time for time-dependent wavefunctions.
     
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