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Bound states in finite spherical well

  1. May 13, 2015 #1

    Maylis

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    1. The problem statement, all variables and given/known data
    upload_2015-5-13_21-11-58.png

    2. Relevant equations


    3. The attempt at a solution
    for ##r \le a## and ##l = 0##, the radial equation is

    $$- \frac {\hbar^{2}}{2m} \frac {d^{2}u}{dr^{2}} - V_{0} = Eu $$
    $$- \frac {\hbar^{2}}{2m} \frac {d^{2}u}{dr^{2}} - [V_{0} + E]u = 0$$
    call ##k^{2} = \frac {2m}{\hbar^{2}}[V_{0} + E]##
    The solution to this differential equation comes out to be
    $$u(r) = A sin(kr) + B cos(kr)$$
    Since ##u = R(r)r##, as ##r \rightarrow 0##, then ##\frac {cos(kr)}{r} \rightarrow \infty##, therefore ##B = 0##
    Leaving ##u(r) = Asin(kr)##

    For ##r > a##
    $$- \frac {\hbar^{2}}{2m} \frac {d^{2}u}{dr^{2}} = Eu$$
    ##\kappa^{2} = - \frac {2mE}{\hbar^{2}}##
    therefore, ##u(r) = Ce^{-\kappa r} + De^{\kappa r}##
    As ##r \rightarrow \infty##, then the second term blows up, so ##D = 0##

    At the ##r = a##, the ##u(r)## must be the same for both sides. Therefore,
    ##A sin(ka) = Ce^{- \kappa a}##
    Also, the function must be continuous, so the derivative must be equal at ##r = a##
    ## kA cos(ka) = - \kappa C e^{-\kappa a}##
    Dividing through,
    $$\frac {1}{k} tan(ka) = - \frac {1}{\kappa}$$

    From here, how am I supposed to find the bound states? I have always been confused about bound and unbound states.
     
  2. jcsd
  3. May 13, 2015 #2
    Hi Maylis,
    Try to expand tan(ka)
    Note that the ground state have E~0
    but no zero, we can give it a great approximation !
     
  4. May 13, 2015 #3

    PeroK

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    Bound states are defined by E < 0, which you assumed to get the solution outside the well. So, this is the solution for bound states you have.

    To solve the transcendental equation in ##tan(ka)## you need a bit of jiggery-pokery. See the earlier chapter on the 1D finite well for how it's done. Let ##z = ka## etc.
     
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