± exp[-*x*]

for
*x*>*a* and

± exp[+*x*]

for
*x*<-*a* as 0
as |*x*| to produce
normalizable wavefunctions.

Our solutions
for |*x*|<*a* are:

Inside the well the potential is zero so:

*k'*^{2}=*E'*

For |*x*|>*a* we have:

where: '^{2}=*U*_{0}'-*E'*

Our matching points are *x'*=±½
(i.e., *x*=±*a*). The wavefunction and its derivative must
be continuous at the matching point. The second derivative of
will have a step at *x*=±*a* to match the step in potential as
required by Schrödinger's equation:

Thus we require:

Here the expression indicates that
we are to use the top item (__cos__) for *odd* *n* and the
bottom item (__sin__) for *even* *n*. The expression:
½^{-} indicates evaluation a little less ("-") than ½
i.e., in the classically allowed region
and ½^{+} indicates evaluation a little more ("+") than ½,
i.e., in the classically disallowed region.
(Technically what's going on here is limits from the left hand side and
limits from the right hand side. We are making sure that our formulae
for agree at the turning point.) *A* and *B* are unknown
constants having to do with the normalization of the wavefunction in each
region.

Now we match the derivative of :

We can eliminate the unknowns *A* and *B* by dividing the
second equation by the first.

For odd *n* we have:

For even *n* we have:

The quantity formed by dividing the derivative of by : is known as the logarithmic derivative since it is the derivative of ln(). Thus we would say the above two expressions require the continuity of the logarithmic derivative across the turning points.

The above two equations are nonlinear equations we can solve
for *k'*, and from *k'* we can find *E'*. The solutions,
of course, depend on *U*_{0}'. Once we know *k'*,
we can solve the -match equations for *A/B*.
The overall scale of is set by the normalization
requirement:

We are seeking the energy levels for various possible values
of *U*_{0}'. Here are the results for
*U*_{0}'=120:

The match, so carefully made in the above work, is hard to
see. In the below we show the *n*=4 wavefunction and blow up
one matching region. Notice that on the left of *x'*=½
the curvature (second derivative) of is positive
whereas on the right of *x'*=½ the curvature is smaller
and negative. The jump in curvature is required by Schrödinger's equation.
A line with the matching slope is shown in blue.

Here are the results for
*U*_{0}'=100:

If we reduce *U*_{0}' to 80, the
*n*=4 state is lost:

Notice that as *U*_{0}' is reduced, all the states move
a bit lower in energy.
For example, the third state was at *E'*=61.3 for *U*_{0}'=120
and moved to 55.6 for *U*_{0}'=80. With *U*_{0}'=,
third state has *E'*=(3)^{2}=88.8. Note how the probability of being found in
the classically disallowed region grows as *E'* nears
*U*_{0}'. The smaller the energy "deficit"
in the classically disallowed region, the more likely it is
to find a particle in this "disallowed" region. As we said earlier, if *U*_{0}'=
the wavefunction must be exactly zero in the disallowed region.

In the below plot we display the loss of states as *U*_{0}' is reduced
from infinity (i.e., 1/*U*_{0}' increases from 0). On the *y*-axis
*k'*/ is plotted; Recall: *E'*=*k'*^{2}.
For the infinite square well
*k'*=*n*, so for 1/*U*_{0}'=0 the *n*^{th}
level's *k'*/ is *n* and and states exist for each whole number
*n*. As *U*_{0}' is reduced each level's *k'* is reduced
until *k'*/=*n*-1 at which point that level disappears.
Thus for *U*_{0}'=100 (1/*U*_{0}'=.01) we have but 4 levels left,
and the fourth is about to disappear.

The *n*=2 level disappears at *U*_{0}'=^{2}
(near 1/*U*_{0}'=0.1), the ground state remains for even
arbitrarily small *U*_{0}'.