In this case we only have two regions separated by a point discontinuity in the potential at x=0. Region s ends at x=0 and Region t begins at x=0. Region c is just the point x=0.
The left-incident boundary conditions at 
set the
form of the wave function in Region t to a wave of the
form 
, where 
.
The general form of the solutions in Region s are oscillatory
with wave vector 
. The most convenient form of
such solutions for
matching the boundary conditions at x=0 is a combination of sine and
cosine centered at x=0, giving
Matching boundary conditions at x=0 gives the solution,
A and B determine our full solution. To determine s and r, we now expand the sines and cosines in complex exponentials,
where for completeness, we have also included the present form of the transmitted solution.
Given the raw solution 26, the next step in the solution
process is to set the overall normalization of the solution. To
extract proper quantum amplitudes, we must have a unit incoming
current. The we achieve by multiplying the entire raw solution
through by 
resulting in
This form now has a unit incoming current. In addition, we have the
reflected beam expressed as a simple factor times the unit
reflected beam 
. The one difficulty with
this form is that the transmitted beam is not also written in terms of
a unit current beam because the wave vector in the transmitted part
does not match the wave vector appearing in the
normalizing square-root 
. To rectify this
situation, we rearrange the factors in 
, taking care to do
absolutely nothing to change the final value of this part of the wave
function, which is set entirely by Schrödinger's equation, our
boundary conditions and the normalization to unit incoming current,
Our final wave function, after some minor algebra of gathering factors and clearing fractions, is
The quantum amplitudes for reflection and transmission are thus
