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Homework Help: Probability current

  1. Oct 11, 2007 #1


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

    I have a wave function,


    All I am required to do is calc. the prob. current. This I have done using the usual

    [tex]j = \frac{\hbar}{m}.|A|^{2}.\frac{\partial \varphi}{\partial z}[/tex]

    where [tex]\varphi}[/tex] is the imaginary component of the exponent of e.

    It comes (of course) to [tex]j = \frac{\hbar}{m}.|A|^{2}.k_{0}[/tex].

    But, doing it the long way (I just thought I would), with

    [tex]j = \frac{\hbar}{2mi}.(\psi*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi*}{\partial x})[/tex]

    I end up with the same result, but [tex]e^{-2ax^{2}}[/tex] attached. Since taking [tex]\psi*[/tex] only changes the sign of the imaginary component, this seems inevitable... anyone know what I'm doing wrong?

    Cheers :-)
  2. jcsd
  3. Oct 11, 2007 #2


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    Apparently A from the general expression you quoted

    [tex] j=\frac{\hbar}{m}|A|^{2}\frac{\partial \phi}{\partial x} [/tex]

    includes the real exponential...
  4. Oct 11, 2007 #3


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    ???? Where did you see this equation? It does not look right to me. are you sure this is not only for plane waves??
  5. Oct 14, 2007 #4


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    I'm thinking the simpler equation can only apply where the exponent of e in the wave function is wholly imaginary.

    Using the equation

    [tex]j = \frac{\hbar}{2mi}\left( \psi^{*} \frac{\partial\psi}{\partial x} - \psi \frac{\partial\psi^{*}}{\partial x} \right)[/tex]

    I obtain

    [tex] \frac{\hbar}{m}|A|^{2}e^{-2ax^{2}} k_{0}[/tex]

    I presume that's correct. In which case, I don't think the [tex]e^{-2ax^{2}}[/tex] should be absorbed into A. The probability current is a function of x.

    Last edited: Oct 14, 2007
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