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Traveling and Standing Waves in Quantum Mechanics

  1. Oct 9, 2009 #1
    1. The problem statement, all variables and given/known data
    A wave in quantum mechanics is represented by Aei(kx-[tex]\omega[/tex]t). Show that a standing wave looks like 2iAe-i[tex]\omega[/tex]tsin(kx) by subtracting two waves moving in opposite directions. (Hint: make the k negative in one of the waves)


    2. Relevant equations
    As far as I know, the 2 equations in the problem


    3. The attempt at a solution
    Honestly, I have no idea where to start. Any help would be greatly appreciated.
     
  2. jcsd
  3. Oct 9, 2009 #2

    Doc Al

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    Staff: Mentor

    Just follow the hint. How would you represent the wave moving in the opposite direction?
     
  4. Oct 9, 2009 #3
    That's the thing... I'm not sure. If I had to guess, I would say it would look something like:

    Aei(kx-[tex]\omega[/tex]t) - Bei(-kx-[tex]\omega[/tex]t)

    But how do I relate it to sine?
     
  5. Oct 9, 2009 #4

    Doc Al

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    Good, but use A for the second wave instead of B.
    Hint: Look up Euler's formula.
     
  6. Oct 9, 2009 #5
    Thanks for the hint about Euler's formula... it really helped. So from the original equation, I now have:

    A[cos(kx-wt)+i sin(kx-wt)]-A[cos(-kx-wt)+i sin(-kx-wt)]

    I distributed the A term throughout, and found that the cosine terms canceled. I was left with:

    Aisin(kx-wt)-Aisin(-kx-wt)

    which could be written as

    Aisin(kx-wt)+Aisin(kx+wt)

    But now I have to somehow re-introduce an exp function and take the 'wt' term from the sine. Hmmmm....
     
  7. Oct 9, 2009 #6

    Doc Al

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    Another hint: ea+b = eaeb
     
  8. Oct 9, 2009 #7
    OK... so I found that the original equation can be written as follows:
    Aeikxe-iwt
    Using Euler's formula on the eikx gives:
    Ae-iwt (cos(ikx)+i sin(ikx))

    I know that the cosine term will eventually cancel when I subtract. But the complex number in the sine bothers me. Won't that end up being a hyperbolic function? I carried the math through and came up with this:

    2iAe-iwt (sin(ikx))

    I'm a lot closer, but I don't know how to deal with the complex number in the trig function
     
    Last edited: Oct 9, 2009
  9. Oct 10, 2009 #8

    Doc Al

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    Not exactly. Euler's formula says:
    eix = cosx + isinx
     
  10. Oct 10, 2009 #9
    Oh, I'm an idiot! Of course the complex number isn't in the sine!

    Thanks Doc Al for all of your help! I'll never forget about Euler's formula again! :)
     
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