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Show that the radial component of the wave function satisfies the radial equation

  1. Nov 14, 2011 #1
    1. The problem statement, all variables and given/known data
    Show that radial components of the continuum electron wave function
    satisfies the radial equation:

    [itex]{\left[\frac{-\hbar }{2m}\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial }{\partial r}\right)+\frac{\hbar ^2l(l+1)}{2m
    r^2}-\frac{Z e^2}{r}\right]R=E R}[/itex]

    where:
    E=(k*hbar)2/2m
    R=Rkl(r)

    2. Relevant equations
    so, the radial component R is given as:

    [itex]{R_{\text{kl}}(r)=\frac{C_{\text{kl}}}{(2l+1)!}(2{kr})^l\text{Exp}[-\text{ikr}]*F(i/k+l+1,2l+1;2\text{ikr})}[/itex]

    3. The attempt at a solution
    First I have rewritten R as:

    Rkl(r)= C rl Exp[-ikr] F[i/k + l + 1,2l+1;2ikr]

    where in C is everything that does not depend on r.

    Great, I was thinking this will be pretty straightforward to show.

    My reasoning was to insert R in the above radial equation, take those derivatives and at the end I will be able the factorize R out, so that I'm left with something like
    HR=AR where A is the energy.

    It did not work out like that :smile:
    After taking the derivatives I'm left with a mess and I don't see how to factorize R out.
    I mean, derivatives of Exp[-ikr] give me back the same function (multiplied by constant) so I can factorize that out but rl and F give me problems after derivation.
    Also I was looking at the properties of these confluent hypergeometric functions F
    and tried to adjust them back to original function but without success.

    Can someone say whether I'm on the right track trying to solve the problem like that or
    have I completely missed the point ?
     
  2. jcsd
  3. Nov 14, 2011 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Interesting problem. I would first try to relabel the confluent hypergeom function as u(r) and plug the solution into the original ODE. Then I'd be led to a second order ODE for u(r) which should have exactly the same solution with the confluent hypergeom function I'm given.
     
  4. Nov 15, 2011 #3
    I'm not sure if I understood you well, but this is what I have tried:

    Now R is given as:

    Rkl(r)= C rl Exp[-ikr] u[r]

    And a second order ODE that u[r] obeys is:

    [itex]z\frac{d^{2}u}{dz^{^{2}}}+(b-z)\frac{du}{dz}-au=0[/itex]

    where u is u[r]~F[a,b,z]=F[i/k+l+1,2l+1,2ikr]

    So, I should insert R the into radial equation and with the help of this ODE for u[r] somehow simplify the result, right ?
    Ok, I focused here only on the result I get when acting on R with this part of the radial equation
    [itex]\frac{1}{r^{2}}\frac{\partial }{\partial r}(r^{2}\frac{\partial }{\partial r})[/itex]

    the result after acting on R with this part is:
    [itex]{c e^{-i k r} r^{-2+l} \left(\left(l+l^2-2 i k l r-k r (2 i+k r)\right) u[r]+r \left(2 (1+l-i k r) u'[r]+r u''[r]\right)\right)}[/itex]

    Now, I have tried to bring this expression(or a part of it) to form of ODE for u[r].
    But when I adjust the coefficient that multiplies u''[r] so that I have 2ikr*u''[r] , which is required by the ODE for u[r] then the coefficient for u'[r] is not in a proper form anymore.
    And the rest of the equation is also of no help since these are the only u''[r] and u'[r] terms.
    So I'm stuck again :)
     
    Last edited: Nov 15, 2011
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