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Differential equation, quantum physics

  1. Aug 4, 2007 #1
    1. The problem statement, all variables and given/known data

    Solve for u(r) in the region r>R (i.e r goes towards infinity). Show that psi(r)=(A/r)e^-alpha*r

    2. Relevant equations

    [tex]\frac{\(d^2u(r)}{dr^2}=\alpha^2u(r)[/tex]

    [tex]\(u(r)=\psi(r)*r[/tex]

    3. The attempt at a solution

    [tex]\frac{\(d^2u(r)}{dr^2}=\alpha^2u(r)\Longleftrightarrow\frac{\(d^2\psi(r)*r}{dr^2}-\alpha^2\psi(r)*r=0\Longleftrightarrow\frac{\(d^2\psi(r)}{dr^2}+\frac{\(2}{r}\psi(r)-\alpha^2\psi(r)=0\Longrightarrow\mbox{characteristical equation}\Longrightarrow[/tex]

    (z^2) + (2/r)z - alpha^2 = 0

    which has the roots z = (-1/r) +/- ((1/r^2) + alpha^2 )^(1/2)

    This doesn't seem correct. What have I done wrong? (BTW: sorry for not using LATEX everywhere; I didn't have enough time)
     
  2. jcsd
  3. Aug 4, 2007 #2

    Gokul43201

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    Wouldn't it be simpler to just solve for the DE in u(r) first?
     
  4. Aug 5, 2007 #3

    dextercioby

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    Sides, characteristic equations cannot be used for ODE-s z with variable coefficients.
     
  5. Aug 5, 2007 #4
    Yes, but I cannot see how to get the 1/r term. I only seem to get Ae^-alpha*r (after the boundary condition has been imposed).
     
  6. Aug 5, 2007 #5
    This was actually something I was thinking of. Makes sense.
     
  7. Aug 5, 2007 #6

    Gokul43201

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    Is this a solution for u(r) or psi(r)?
     
  8. Aug 5, 2007 #7
    The former I think. Now I didn't substitute psi(r)*r for u(r) as I did in the OP.
     
    Last edited: Aug 5, 2007
  9. Aug 5, 2007 #8

    nrqed

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    That's right. So you found the solution for u(r) with the approprioate boundary conditions. To get psi, just divide u(r) by r!! [itex] \psi(r) = \frac{u(r)}{r} [/itex] and you are done.
     
  10. Aug 5, 2007 #9
    It's that simple?! I thought I had to somehow include the r term in the derivation (d^2)/dr^2 :blushing: Well, thanks then!
     
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