Delta function potential and Schrodinger Equation

[Seriously]

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 Schrödinger 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 Schrödinger 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 Schrödinger 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!

 

[Rach3]

Where you went wrong was in taking the second derivative of e^{-|x|}; it's actually a dirac delta, which cancels the other dirac deltas as you'd expect.

e^{-|x|} 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.

 

[Seriously]

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?