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Delta function potential and Schrodinger Equation

  1. Oct 21, 2006 #1
    I have a time dependent wavefunction for inside a delta function potential well: V(x) = -a delta(x).

    It reads
    Psi(x,t) = (sqrt(m*a)/hbar) * exp(-m*a*abs(x)/hbar^2) * exp(-iEt/hbar)

    I'm supposed to stick this back into the time dependent Schrodinger Equation and solve for E.

    Taking my Psi(x,t), I found the second derivative with respect to x, and also found the time derivative. Then I plugged directly back into the time dependent Schrodinger Equation, with V(x) given as above. The problem is that I can't seem to make the delta function go away. I get
    -(m*a^2)/(2*hbar^2) - a*delta(x) = E

    How do I get the delta to go away? I can't just say x=0, because then the delta function is infinity.

    I also tried: Plugging directly back into the time dependent Schrodinger Equation, I integrate both sides from -e to +e, where is some really small distance around x=0. Instead of using dPsi/dt, I write dPsi/dt in terms of d^2Psi/dx^2. Then I integrate both sides. The only problem with doing this, is when I integrate d^2Psi/dx^2 dx, this becomes dPsi/dx evaluated from -e to +e -- and because of the absolute value of x in Psi, this is zero. Help!!!
  2. jcsd
  3. Oct 22, 2006 #2
    Where you went wrong was in taking the second derivative of [tex]e^{-|x|}[/tex]; it's actually a dirac delta, which cancels the other dirac deltas as you'd expect.

    [tex]e^{-|x|}[/tex] is continuous and non-differentiable; the first derivative is has a jump discontinuity like the step function; and the derivative of that has a singularity like the dirac delta. (What you actually get is the sum of a dirac delta and a differentiable function.) A useful fun exercise for delta functions is problem 6 from Robert Jaffe.
    Last edited by a moderator: Oct 22, 2006
  4. Oct 22, 2006 #3
    Thank you for the reply, Rach3, I really appreciate it.

    I see now that I'm not differentiating exp(-abs(x)) correctly, let alone the second derivative. But when I try to follow your instructions, I find that I'm still not comfortable with differentiating this.

    Phi = exp(-abs(x))
    dPhi/dx = - dabs(x)/dx exp(-abs(x))
    where dabs(x)/dx = -1 for x<0, undef for x=0, +1 for x>0

    Is the second derivative d2abs(x)/dx2 exp(abs(x))?
    Or is it (-(dabs(x)/dx)^2 + d2abs(x)/dx2 ) exp(abs(x)) ?
    Or none of the above, lol?
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