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Singular potential problem

  1. Aug 10, 2007 #1
    My book says that the solution to a singular potential at zero problem has the form
    \psi(x) &=& A exp(Kx) for x< 0 \\
    & =& Aexp(-Kx) for x > 0 \end{array}

    How do you get that from the general solution of the Schrodinger equation
    [tex]\hbar^2/2m d^2 \psi(x)/dx^2 = E \psi(x) = -|E|\psi(x) [/tex]
    which is
    psi(x) &=& A exp(Kx) + B exp(-Kx) for x< 0 \\
    & &= Cexp(-Kx) + Dexp(Kx)for x > 0

    by "imposing the condition that the wavefunction be square integrable and continuous at x = 0".

    Obviously the second condition gives you A + B = C + D but I do not see how that helps reduce to a single coefficient.

    And why are my equation arrays not working? :(
    Last edited: Aug 11, 2007
  2. jcsd
  3. Aug 11, 2007 #2


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    the condition that it be square integrable means the we must set
  4. Aug 11, 2007 #3
    How does square integability imply that?
  5. Aug 11, 2007 #4


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    As a quick test, you want to evaluate each of those (modulus squared) integrals over their respective domains. Ask yourself which ones will converge and which will diverge (for non-zero coefficients)?
    Last edited: Aug 11, 2007
  6. Aug 11, 2007 #5
    I see. Thanks.
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