Classically, the downward step in potential at *x*=0
corresponds to an infinite force in the *x* direction located
just on the *x*=0 line. Everywhere except *x*=0 the potential
is flat, and so we are almost always in a no-force (i.e., "free particle") situation.
Particles move with the speed required for their energy.
On the left hand side:

**p**^{2}=2*mE*

and on the right hand side we have a faster speed:

**p**^{2}=2*m*(*E*+*U*_{0})

Since there is no force in the *y* direction, the momentum in the
*y* direction: *p*_{y} is conserved, even
at the potential step. Thus
when a particle makes the energy jump at *x*=0,
*p*_{x} must increase to supply the increase in
kinetic energy. Since *p*_{x} is enlarged and
*p*_{y} is conserved, the direction of the particle
changes. We find:

We see that the particle is Snell's Law "refracted" as if with an
"index-of-refraction" *n*. Newton used such a particle-based
explanation for the behavior of light. We note some oddities:

- Moving into a more "optically dense" material
(i.e.,
*n*>1, like from air to glass) the particle increases in speed. For light, we find the speed of light is slower in glass than in air. - The "index-of-refraction"
*n*depends on the particle's energy: higher energy corresponds to smaller*n*. For visible light in most materials, higher energy means larger*n* - A step up in potential (e.g.,
*U*_{0}<0) corresponds to*n*<1. For light that would be like a transition from glass to air. - Classically we have no reflection. As I commented before, even in the WKB approximation there is no reflection; reflection is a result of a large change in potential in a wavelength or so of the wave.

Schrödinger's equation for our problem is:

Thus:

with solutions:

We note that Schrödinger's equation does not require, say,
*q*_{x}, to be a real number, only that
*q*_{x}^{2}+*q*_{y}^{2}
have a given value. If *q*_{x} is imaginary, we can
have exponentially decaying solutions like: exp(-|*q*_{x}|*x*).

Note that since we no longer have a finite sized box we've lost our length scale. We are forced to use dimensioned coordinates.

We now consider a solution that includes incident, reflected, and refracted waves:

As usual we must match and
along the boundary *x*=0. Thus we end up considering
the values of for **r**=(0,*y*)
and produce functions like:

exp(*ik*_{y}*y*) and exp(*iq*_{y}*y*)

If *k*_{y} does not equal *q*_{y} we would have
different wavelengths on each side of the boundary so while we could match
at a point, there would be no hope of matching all along the boundary. Thus
we immediately produce the "refraction" condition that:

*k*_{y}=*q*_{y}

We now proceed to match and
and derive the reflected amplitude *R*:

Since:

We can simplify our expressions for *R*:

We note that quantum mechanical reflection is exactly the same as the reflection
of TE polarized light. Note the obvious check that
here is no reflection if *n*=1. At normal incidence (=0)
*R*=(1-*n*)/(1+*n*). At grazing incidence
(=90°) *R*=-1. If *n*<1
(i.e., *U*_{0}<0) we can have total internal reflection
for sin>*n*. As we show below, we have produced
an imaginary *q*_{x}, so the "transmitted" wave does not propagate,
rather it decays away for positive *x*: