While we know that any physical state obeys the TISE in each of the
three regions I, II and III, we must be careful in how we connect our
solutions across the interfaces at x=-a and x=a. Because
is a physically measurable quantity we expect it
and thus 
always to be continuous. It is not necessary to
add this as an additional physical requirement, because this
continuity is a consequence of the TISE. Recall that the finite
square well potential represents an idealization of a smooth potential
which happens to vary extremely rapidly at the interfaces between the
regions. In all physical cases where the potential is ``smooth''
(continuous with all of its derivatives continuous, a so called
``
'' function), the solutions to the TISE will also be
smooth. All we must then determine is the behavior of the solutions
of the TISE, as we make the walls of the well steeper and steeper.
The best way to predict this behavior of the wavefunctions is to apply
the mean value theorem
of calculus. The mean value theorem states that for any function
whose derivative
exists over the range between two points 
and 
, there exists a
third
point c between 
and 
for which
To apply this theorem in our case, we take
in (2). In
any case where 
is continuous (although perhaps very steep) for
any two points 
straddling the interface at x=-a,
for some point c in the range 
.
Because 
remains finite in the range 
, as we take
the limit 
and 
, the left hand side of the
above equation will
approach zero and thus
, and we have
proven that the derivative 
remains continuous even if the
potential possesses a finite discontinuity.
A similar argument may be applied to
to show that whenever 
is finite, the wavefunction
itself is also continuous. Taken together, these two facts give our
second boundary condition:
This tells us how to connect solutions across the interface at x=-a and continue our solution from region I into region II.
