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Proofs with current density and wavefunctions

  1. Apr 8, 2014 #1
    1. The problem statement, all variables and given/known data

    So I was able to find a problem that was kind of similar to a homework problem that I am working on. Unfortunately, I'm not quite sure what is going on partially within the problem.

    In the problem they state that [itex]\phi[/itex]=[itex]\phi[/itex]*, but it does not state why. I was wondering if someone could explain this to me?

    Also, later on in the problem e[itex]^{-i\phi}[/itex]e[itex]^{i\phi}[/itex]+e[itex]^{i\phi}[/itex]e[itex]^{-i\phi}[/itex] cancels out along with the 1/2i. I'm not quite sure how this happened either. I have tried to review the euler's equation to understand where this comes from because I'm supposing it is being canceled out based on its properties.

    I have attached the proof to this forum.


    I would greatly appreciate some help because the problem I am working on involves proving that the integral of the current density over a surface is equal to the momentum operator/m. I understand conceptually what this means but I need a push in the write direction with the math behind it.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Apr 8, 2014 #2
    You seem to be missing a factor of [itex]i[/itex] when you plug back in. The [itex]i[/itex] factors come from the [itex]\frac{\partial}{\partial x}[/itex].

    Also,

    [itex]e^{i\phi}e^{-i\phi} = e^{0} = 1[/itex].

    This comes from the properties of exponentials, but you can show this using Euler's formula as well. Note [itex]\left(a + ib\right)\left(a-ib\right) = a^2 + b^2[/itex], and [itex]\sin\left(-\phi\right) = -\sin\phi[/itex].
     
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