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Bound states for a Spherically Symmetric Schrodinger equation

  1. Dec 11, 2006 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m moves in three dimesions in a potential energy field
    V(r) = -V0 r< R
    0 if r> R

    where r is the distance from the origin. Its eigenfunctions psi(r) are governed by

    [tex] \frac{\hbar^2}{2m} \nabla^2 \psi + V(r) \psi = E \psi [/tex]
    ALL in spherical coords.

    Consider a spherically symmertic eigenfunction with no angular dependance of the form

    [tex] \psi(r) = \frac{u(r)}{r} [/tex]

    Solve for u(r) in teh regions r< R and r > R and yb imposiing boundary conditions, find the eigenfunction of a bound state with energy [itex] E = \hbar^2 \alpha^2 / 2m [/itex]

    Show taht there is one bound state of this kind if the depth of the weel obeys
    [tex] \frac{\hbar^2 \pi^2}{8mR^2} < V_{0} < \frac{9\hbar^2 \pi^2}{8 mR^2} [/tex]

    2. Relevant equations
    Ok i found te solution of the wavefunction to be
    [tex] C \sin (k_{0}r) /r [/tex] if r < R
    [tex] A e^{\alpha r}/ r [/tex] if r > R
    The solutions are such because the solutions are found a bound state that is E <= V0. Also the solutions are spherically symmetric.

    where [tex] k_{0} = \sqrt{\frac{2m}{\hbar^2} (V_{0} + E)} [/tex]

    3. The attempt at a solution
    Furthermore i found that
    [tex] k_{0} \cot k_{0} R = -\alpha [/tex]
    [tex] k_{0}^2 + \alpha^2 = \frac{2m}{\hbar^2} V_{0} [/tex]

    How would i prove the condition for V0?? Would i do this graphically assuming different values for R???

    Thanks for your help!!!
  2. jcsd
  3. Dec 11, 2006 #2


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  4. Dec 11, 2006 #3
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