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Excited state

  1. Nov 14, 2006 #1
    A particle in a potnetial V(x) has a definite energy

    [tex] E = - \frac{\hbar^2 \alpha^2}{2m} [/tex]

    and its eigenfunction is given by

    [tex] \psi(x) = Nx \exp(-alpha x) [/tex] if 0 <= x < infinity
    zero elsewhere

    Prove that [tex] V(x) = -alpha \hbar^2/mx [/tex] 0 <= x < infinity
    infinity elsewhere

    Given that the eigenfunction describes the ground state of the aprticle in teh potnetial V(x) roughly sketch the the eigenfunction whifchcdescribes the first excited state

    ground state ook like sketch 1

    no zeroes
    i think the first excited state will look like sketch
    it has to have 1 zero...
     

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  2. jcsd
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