# Delta-function potentials, scattering states

1. Sep 26, 2009

### AntiStrange

I'm reading through the Introduction to Quantum Mechanics book by Griffiths (2nd edition)
and it is describing delta-function potential wells.
When it describes how to find bound states the energy is E < 0 (negative) in the region x < 0 (negative).

It says the general solution is:
$$\psi (x) = Ae^{-\kappa x} + Be^{\kappa x}$$
It explains that as x goes to -∞ the first term blows up. Which makes sense because it will go to infinity, while the second term only goes to zero.

However, they then look at the situation when E > 0, x < 0
again the general solution is:
$$\psi (x) = Ae^{-\kappa x} + Be^{\kappa x}$$

but this time they say that neither term blows up. I don't understand why that is. Shouldn't the first term still go to infinity as x goes to -∞ ?

2. Sep 27, 2009

### kanato

No, for E > 0, x < 0 the general solution is
$$\psi(x) = Ae^{-ikx} + Be^{ikx}$$
which oscillates but does not exponentially grow or decay (for real k).