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Particle in a Box Question

  1. Aug 11, 2008 #1
    What are the eigenstates of a particle in a box whose bounds are [tex]x = -a/2[/tex] and [tex]x = a/2[/tex]?

    Well, the eigenstates where [tex]x = 0, a[/tex] are just

    [tex]\varphi_n = \sqrt{\frac{2}{a}} \sin \frac{n \pi x}{a}[/tex],

    so why wouldn't the eigenstates just be

    [tex]\varphi_n = \sqrt{\frac{2}{a}} \sin \frac{n \pi (x+a/2)}{a} = \sqrt{\frac{2}{a}}\sin \left(\frac{n \pi x}{a} + \frac{n \pi}{2}}\right)[/tex]

  2. jcsd
  3. Aug 11, 2008 #2


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    No reason at all; those are the eigenstates!
  4. Aug 11, 2008 #3
    Yes, I just figured out the reason... I was slightly afraid of the negative sign that resulted in my answer for the Sin x solution, but it doesn't matter... negative signs don't have any effect on eigenstates.
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