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Homework Help: Discrete or continious spectrum in QM

  1. Nov 17, 2009 #1
    1. The problem statement, all variables and given/known data

    Here is the question: how can we know that if we have discrete or continuous spectrum just by looking at the potential graph?

    Specifically, let`s consider the potential V(x)=-F*x (F:const) . After we solve, we can conclude wavefunctin is airy function, and so both continious and discrete spectrum. But, without sloving, how can we decide?
  2. jcsd
  3. Nov 17, 2009 #2


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    Gold Member

    There are special cases when we can guess what the spectrum would be. For example, when the particle travels in the central field potential it can be shown that: when the energy [tex]E[/tex] of the particle is positive the spectrum is continuous and when [tex]E<0[/tex] the spectrum is discrete.
    For example, if we have repulsive potential energy of the form [tex]U \sim 1/r[/tex], the total energy [tex]E[/tex] of the particle is positive. (Really,
    [tex]E=\frac{1}{2m}\int {\psi^* \hat{\mathbf{P}}^2}\psi dV + \int {\psi^*U \psi}dV[/tex]
    which is positive since [tex]U[/tex] is positive and the eigenvalues of [tex]\hat{\mathbf{P}}^2}[/tex] are positive numbers too).
    So, in this potential field we have [tex]E>0[/tex] and continuous spectrum.
    If [tex]U \sim -1/r[/tex] we have –°oulomb attraction and to possibilities: [tex]E>0[/tex] (continuous spectrum, ionized electron) and [tex]E<0[/tex] (discrete spectrum).
    Finally, in case of 6-12 potential we have continuous spectrum for [tex]E>0[/tex] (dissociated molecule) and discrete spectrum for [tex]E<0[/tex].
    Such examples show that we can guess what the spectrum is without doing anything with Schrödinger equation.
  4. Nov 17, 2009 #3
    The way that was explained to me is if there are 2 turning points, then it is discrete. If there is only 1 turning point, it is continuous. I, unfortunately, was never able to ask my professor whether something that had three (or more) turning points would be continuous or not.
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