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1D infinite potential well with origin shift

  1. Feb 15, 2006 #1
    Hi guys, I am solving the 1D infinite potential well for a particle, but in this case instead of the potential being 0 from -a to a, its shifted to 0 to 2a. I have calculated that the even parity solution is zero.
    My question is, I have calculated that k=n*Pi/(2*a) by applying the boundary conditions, what values of n is this for? Previously it was n is even for odd parity solution, but here as there is only an odd parity solution, surely there are still all the energy levels, so n is all integers, 1,2,3 etc?

    Thanks very much! If you need more detail please dont ignore, just ask! :-)

    Gareth
     
  2. jcsd
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