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Solutions of the quantum SHO

  1. Aug 12, 2007 #1
    The even solutions of an SHO are:

    [tex]h^+(y) = \sum_{s = 0}^{\infty}a_s y^{2s} [/tex]

    where a is given by the recursion

    [tex] a_{s+1} = a_s \left( \frac{4s + 1 - \epsilon}{2(s+1)(2s+1)} \right)[/tex]

    The solutions are square integrable iff

    a_n = 0 so that the polynomial is finite.

    What I do not understand is why my book (Robinett) says

    h_0(y) = 1 and not 0 when a_0 = 0 for n = 0?
  2. jcsd
  3. Aug 12, 2007 #2


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    Who's "n" and what does it have to do with what you wrote ?
  4. Aug 12, 2007 #3
    So, there should be one solution for every n because

    a_n = 0 --> epsilon = (4n +1).

    So, I guess n is the degree (maybe +/- 1 of the polynomial). It is also the energy level.
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