We have already studied
the bound states in such a potential. We then used a shifted energy origin
so that the potential was zero at the bottom of the well. Here
we take the the *top* fo the well to be *U*=0.
This is the same strategy we adopted in 1d when we studied
bound states in finite square wells
and then scattering from
finite square wells.

As we have seen, classically
the effect of entering a region with *U*=-*U*_{0}
is much like refraction. The component of velocity parallel to the
interface's __normal__ is increased (as required by conservation of energy) while
the component of velocity parallel to the interface is
unchanged (as there is no force in that direction). The angle of
incidence () and the angle of refraction
() end up related as in Snell's Law:

We seek to relate the impact parameter (*b*) to the
scattering ange (). Remember that on the
previous page we found the relationship between the
counter-clockwise __reflection__
angle (here called _{0}) and the
impact parameter. The effect of refraction is identical to that
of reflection except the normal of the exit point has been
rotated clockwise by the angle:

-2

Thus the result of refraction is our previous result
minus (because it's *clockwise*) the rotation of the
normals.

While classically (and even in the WKB approximation) there is
no reflection, we know from our previous work
that reflection will accompany refraction in quantum mechanics.
Thus we can have immediate reflection or we can have one or more
reflections inside the potential well. That is to say
the path of the "particle" may include *N*=0,1,2,...
internal chords. The net result of an *N* internal chords, is to
rotate the exit normal from the entrance normal by:

*N*(-2)

Thus, we have the relationship between the scattering angle and the impact parameter:

We plot some of these results in the case *n*=1.33
(which happens to be the refractive index for visible light
in water).

Note in particular the curves for *N*=2, (and 3).
We see that the scattering angle has a minimum (maximum)
at an angle equivalent to =2.40
(2.27). (*N*=2 scatters light in the range:
[138°, 180°], *N*=3 scatters light in the range:
[0, 130°]. Between the bows we have Alexander's dark band.)
Since the angle is extremal, the derivate of
with respect to *b* will be zero there, and hence the
differential cross-section will be infinite at that point since:

What's happening is that particles from a fair range of *b*
near the extremal, end up scattering to pretty much the same
degree. So particles from a "finite" range of *b* are being
concentrated into an "infinitesimal" range of ,
and the result is an infinite cross-section. This is called
a *rainbow*. [It explains why a rainbow is bright; rainbows
are colored because *n* depends on the the light's color, so
the different colors of light end up having slightly different
rainbow angles.] For water we see that there is a rainbow
(the "primary" rainbow) from scattering
with two internal chords: 42° from the anti-solar point
[i.e., nearly backscattered off the raindrops].
Scattering with three internal chords produces a rainbow
(the "secondary" bow, often
not seen and always dimmer than the primary bow---it also has its
colors reversed from the primary bow) 50° from the anti-solar point
(i.e., beyond the primary bow). The below shows the extremal paths of the
rays for *N*=2,3.

We can cobble together a "classical" cross-section by taking
our previous result for the probability of reflection, and apply
it every time the particle touches the interface. Here is the result
for *n*=1.33:

Note that most of the scattering comes from *N*=1,
i.e., pure refraction. Nevertheless the infinite cross-sections
due to the rainbows for *N*=2,3 are quite visible.
Here is a blow-up of the rainbow region:

Note (most noticable from the green *N*=2 curve) that
the "rainbow" has a "double valued function"...two different
rays end up at the same angle. In classical mechanics we would just
add these scattering probabilities. In wave mechanics, different paths
means different phases...we should expect interference in the rainbow
region between the rays that have taken different paths to get to the
same scattering angle.

In making the connection between particles and QM, note that
*n* is a strong function of energy. Thus at low energy
*n* is large and reflection becomes quite likely; whereas at high energy
*n*1 and the cross-section declines.

In quantum mechanics we proceed much as on the previous page. Outside the well we have solutions of the form:

*J*_{m}(*k'r'*) cos(*m*) and
*Y*_{m}(*k'r'*) cos(*m*)

where *E'*=*k'*^{2}. We choose to
write our solution a as sum of Hankel functions:

[*H*_{m}^{(2)}(*k'r'*) +
exp(2*i*_{m})
*H*_{m}^{(1)}(*k'r'*) ] cos(*m*)

Inside the well the solution must be regular at the origin and hence a *J*:

*J*_{m}(*q'r'*) cos(*m*)

where *E'*+*U*_{0}'=*q'*^{2}
(i.e., the particle has more momentum when inside the well).

Our job is to match the logarithmic derivative
() at the well boundary (*r'*=1). Thus:

Consider the
previously worked case of
*U*_{0}'=50. Recall that we had three bound states
for *m*=0, two bound states for *m*=1,2, one
bound state for *m*=3,4, and no other bound states
for larger *m*. The highest energy *m*=0
bound state was extremely weakly bound. We plot below the
phase shifts (*m*=0-5) for this problem as function
of particle momentum *k'*:

The phase shifts are positive for this attractive potential whereas they were negative for the reflecting (repulsive) potential.

Note that at high energy, all phase shifts approach the same value, which as given in the WKB approximation is:

=½*U*_{0}'/*k'*

Note that at low energy, all the phase shifts (seemely except
*m*=0 which we address below), approach a multiple of
, which means the same as =0:
no scattering. If |*B*| is large (as it usually is at low energy
[small *k'*]) the above phase-shift formula reduces to the phase shift formula for
the fully reflective well. Thus, exactly as on the previous page,
we expect the *m*th phase shift to go to "0" like
*k'*^{2m}. The *k'*=0 value of
the phase shift is *j*, where
*j* is the number of bound states with that value
of *m*. (The highest energy bound state has "sucked in"
*j* of phase [e.g., it has
*j*-1 nodes]. Higher energy states [e.g., *E*=0]
start with this phase which is reduced at high energy
(where *n*1) to zero.)

The *m*=0 state is different in this case because
there is a very weakly bound state an thus
*J'*_{m}(*q'*) is quite small and hence
one must go to extremely low values of *k'* before
|*B*| become large. A non "0" phase shift means
significant scattering. The below plots of differential cross-sections
show larger, low energy *s* scattering than we saw in the
fully reflective case.

It is helpful to define the total cross-section:

The total cross-section is related to the fraction of
the particles experiencing *any* scattering in
the sample. We plot below the total cross-section
as a function of *k'*

(Note the scale change in the "dull" region: *k'*>10.)
Note the sharp peak (*resonance*) near *k'*=2.4
(and weaker peaks at 4.5, 6, ...).
Notice that the *m*=5
phase shift reaches maximum scattering (90°) near
*k'*=2.4. Consider three differenctial cross-section
plots at *k'*=2.3 (red), 2.4 (green), and 2.5 (blue).

Notice that the "resonant" *k'*=2.4 cross-section is
much different from the others. If we display the *m*=5
wavefunction [cos(5)]^{2}, I can
maybe convince you that the difference is due to the appearance
and then disappearance of the *m*=5 wavefunction.

The below is a plot of the solution to Schrödinger's equation
for *k'*=2.4 *m*=5.

Note that there is a perfectly good classically allowed region
between *r'*=0.7 and 1.0 which is connected through a disallowed
region between 1 and 2.1 to the large *r'* allowed region.
Tunneling should come to mind...
Classically we can have a particle bound between 0.7 and 1, but tunneling
allows the probability to escape. To have a standing wave solution
to Schrödinger's equation we need to feed a little flux into
the well to make up for what tunnels out. The oscillation in the
large *r'* allowed region is the incoming and outgoing flux
from/to what is nearly a bound state. If we shift the energy slightly
the near bound state condition disappears. Our Schrödinger's equation
solution (10x) now looks like:

Note particularly the inside/outside ratio is now much reduced.

There is nothing special about *m*=5; there are other
resonances for larger *m* [i.e.,
(*m*,*k'*)=(6,4.5), (7,6), ...]. The "sharpness" of the resonance
[i.e., the small range of energy in which it is obvious] is degraded
by reducing the size of the disallowed region, which happens at higher
energy. Thus *m*=5 is the first and hence sharpest of these
resonances.

Here are solme plots of the probability density for our
*U*_{0}'=50 potential. First at *k'*=1
with a blow-up of the central region.

followed by *k'*=10.

Finally, here is a set of cross-sections with increasing *k'* but with
*U*_{0}' also increased so that *n*=1.33. Note the
main *N*=1 pure refraction peak, and the much small "rainbows" filled
with interference.

*k'*=30:

*k'*=100:

*k'*=300:

Here is a repeat of the classical equivalent: