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Wave Function: Normalization Constant

  1. Aug 5, 2015 #1
    1. The problem statement, all variables and given/known data
    Consider a free particle, initially with a well defined momentum ##p_0##, whose wave function is well approximated by a plane wave. At ##t=0##, the particle is localized in a region ##-\frac{a}{2}\leq x \leq\frac{a}{2}##, so that its wave function is

    ##\psi(x)=\begin{cases} Ae^{-ip_0x/\hbar} & if -\frac{a}{2}\leq x \leq\frac{a}{2} \\0 & \text{otherwise} \end{cases}##

    Find the normalization constant ##A## and sketch ##Re(\psi(x))##, ##Im(\psi(x))## and ##|\psi(x)|^2##.

    2. Relevant equations


    3. The attempt at a solution
    So here's what I done:

    ##A^2\int_{-\frac{a}{2}}^\frac{a}{2} e^{-ip_0x/\hbar}dx=1##

    ##A^2.-\frac{\hbar}{ip_0}.e^{-ip_0x/\hbar}=1##

    ##A^2=-\frac{ip_0}{\hbar}.\frac{1}{e^{-ip_0a/2\hbar}-e^{-ip_0a/2\hbar}}##

    Is this the correct method? Also I have no idea how to sketch the function asked. Any help would be greatly appreciated.
     
  2. jcsd
  3. Aug 5, 2015 #2

    George Jones

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    How did you get this?
     
  4. Aug 5, 2015 #3
    I took it from ##P(t)=\int_{-\frac{a}{2}}^\frac{a}{2}|\psi(x,t)|^2=1## subbed my wave function into that then. Is this method wrong?
     
  5. Aug 5, 2015 #4

    George Jones

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    If ##\psi \left( x \right) =Ae^{-ip_0x/\hbar}##, what is ##\left| \psi \left( x \right) \right|^2##?
     
  6. Aug 5, 2015 #5
    ##|\psi(x)|^2=A^2e^{-2ip_0z/\hbar}##?
     
  7. Aug 5, 2015 #6

    George Jones

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    Not quite. If ##Z## is a complex number, are ##Z^2## and ##\left|Z\right|^2## the same?
     
  8. Aug 5, 2015 #7
    No, ##|Z|^2## is the modulus squared ie. ##(\sqrt{a^2+b^2})^2##
     
  9. Aug 5, 2015 #8

    George Jones

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    Yes. What if the complex value ##Z## is expressed in polar notation?
     
  10. Aug 5, 2015 #9
    Is it ##cos\theta +isin\theta##?
     
  11. Aug 5, 2015 #10

    George Jones

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    Well, ##e^{i\theta} = \cos \theta + i \sin \theta##. What is ##\left| e^{i\theta} \right|^2## for real ##\theta##?
     
  12. Aug 5, 2015 #11
    ##sin2\theta + 1##?
     
  13. Aug 5, 2015 #12

    George Jones

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    No. You're making this more difficult than it actually is. Forget about a and b and theta. Can you express ##\left| Z \right|^2## in terms of ##Z## and the complex conjugate of ##Z##?
     
  14. Aug 5, 2015 #13
    Remember, also. The complex conjugate of ##\psi##, and ##|\psi|^2=\psi^{*}\psi##.

    Chris
     
  15. Aug 5, 2015 #14
    Oh ##|\psi|^2=z\bar{z}##
     
  16. Aug 5, 2015 #15

    George Jones

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    Yes! And if ##z = e^{i\theta}##, then ##\bar{z} = ##?
     
  17. Aug 5, 2015 #16
    Is it ##\bar{z}=e^{-i\theta}##?
     
    Last edited: Aug 5, 2015
  18. Aug 5, 2015 #17

    George Jones

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    When complex conjugating, what is done to every ##i##?
     
  19. Aug 5, 2015 #18
    Sorry I was supposed to put a minus in that I'll edit
     
  20. Aug 5, 2015 #19

    George Jones

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    And ##z \bar{z} =##?
     
  21. Aug 5, 2015 #20
    They ##i\theta## and ##-i\theta## cancel each other out so ##e^0=1##
     
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