1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook