A question about Dirac Delta Potential Well solution

1. Aug 19, 2013

Positron137

In Griffith's Introduction to Quantum Mechanics, on page 56, he says that for scattering states
(E > 0), the general solution for the Dirac delta potential function V(x) = -aδ(x) (once plugged into the Schrodinger Equation), is the following: ψ(x) = Ae^(ikx) + Be^(-ikx), where k = (√2mE)/h. After that, he states that in the general solution for ψ(x) (stated above), both terms do NOT blow up in the section of the well where x < 0. But this doesn't make sense, because earlier, when he was demonstrating bound states (E < 0) , he stated that the second term, Be^(-ikx), blows up at infinity when x < 0. But here, for scattering states, he states that NEITHER term blows up as x < 0, which seems contradictory. Could anyone explain why this is true (why neither term blows up for a scattering state, when x < 0)? Thanks!

2. Aug 19, 2013

jeppetrost

The difference, I believe is in the "i" (of the beholder BWAAHAHAHA). But seriously. The scattering states have an i, thus are oscillatory, and the bound states don't have an i, and hence are 'regular' exponentials, which blow up at one of the infinities (+ or -).

3. Aug 19, 2013

Positron137

Ah ok. Thanks! LOL I was getting confused. So the reason why it doesn't "blow up" as we would expect it to is because for complex exponentials, as x -> infinity, e^(ikx) and e^(-ikx) don't blow up? Actually, that kinda makes sense because e^ix is like going in a circle in the complex plane. Thanks for the clarification!

4. Aug 19, 2013