130_notes.dvi

9.3 The Delta Function Potential*.

Take a simple,attractive delta function potentialand look for the bound states.

V(x) =−aV 0 δ(x)

These will have energy less than zero so the solutions are

ψ(x) =

{

Aeκx x < 0

Ae−κx x > 0

where

κ=

√

− 2 mE

̄h^2

.

There are only two regions, above and below the delta function. We don’t need to worry about the
one point atx= 0 – the two solutions will match there. We have already made the wave function
continuous atx= 0 by using the same coefficient,A, for the solution in both regions.

0

x

Energy

0

x

E

V(x)

e

κx e−κ

We now need to meet the boundary condition on the first derivative atx= 0. Recall that the delta
function causes a known discontinuity in the first derivative (See section 8.7.2).

dψ

dx

∣

∣

∣

∣

+ǫ

−

dψ

dx

∣

∣

∣

∣

−ǫ

=−

2 maV 0

̄h^2

ψ(0)

−κ−κ=−

2 maV 0

̄h^2

κ=

maV 0

̄h^2