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Linear operator

  1. Sep 23, 2006 #1
    let be the linear operator: (Hermitian ??)

    [tex] L = -i(x\frac{d}{dx}+1/2) [/tex]

    then the "eigenfunctions" are [tex] y_{n} (x)=Ax^{i\lambda _{n} -1/2 [/tex]

    then my question is how would we get the energies imposing boundary conditions? (for example y(0)=Y(L)=0 wher L is a positive integer )....:rofl: :rofl:
  2. jcsd
  3. Sep 25, 2006 #2


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    Science Advisor

    Energies? That is certainly not a Hamiltonian. In any case, since it is a first order operator, you can't satisfy two boundary conditions, only one.
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