The TISE in this region is precisely the
same as it was in region I. Using the same definition for 
(4) , the general solution in region III is of
the form 
. To ease
the algebra we again centered the solution about
the boundary x=a. To further reduce the algebra, we introduce our
final ``trick.''
Taking linear combinations we define new constants
and 
, so that
Again, one may arrive at 13 directly. In a
classically forbidden region
(where we define 
's instead of k's), the solutions are
always sinusoidal and direct substitution verifies that
(13) solves (3).
Matching the boundary conditions with this form is straight forward, we find
The solution in region III as propagated from region I through region II then reads
