# Scattering States and Barriers

*Michael Fowler, University of Virginia*

### Streams of Particles

Our analysis of the time independent Schrödinger equation
using the spreadsheet limited us to real values of the wave function *ψ*(*x*). This is fine for
analyzing bound states in a potential, or standing waves in general, but cannot
be used, for example, to represent a stream of electrons being emitted by an
electron gun, such as in an old TV tube. The reason is that a real wavefunction *ψ*(*x*), in an
energetically allowed region, is made up of terms locally like cos*kx* and sin*kx*, multiplied in the full wave function by the time dependent
phase factor _{}, giving equal amplitudes of right moving waves _{} and left moving waves _{}. So if we are
interested in a system in which there are not equal numbers of particles moving
to the right and to the left, we must have a wave function such that even the *x*-dependent part is complex.

A simple example is a stream of particles of energy *E* moving from the left in one dimension
through a region of zero potential, encountering an upward step potential *V*_{0}, where *V*_{0} < *E*, at the origin *x* = 0,
so that classically the particles would climb the hill and continue to the
right. We shall represent the incoming
wave function by a plane wave,

_{}

It proves slightly more convenient to work with wave number *k* rather than particle momentum _{} in scattering problems
of this type. If we now think of the
classical picture of a particle approaching a hill (smoothing off the corners a
bit) that it definitely has enough energy to surmount, we would perhaps expect
that the wave function continues beyond *x *= 0 in the form

_{} for *x* > 0,

where *k*_{1} corresponds to the slower speed the particle will have after climbing the
hill.

Schrödinger’s equation requires that the wave function have
no discontinuities and no kinks (discontinuities in slope) so the *x* < 0 and *x* > 0 wave functions must match smoothly at the origin. For them
to have the same value, we see from above that *A* = *B*. For them to have the same slope we must have* kA* = *k*_{1}*B*. Unfortunately, the only way to satisfy both
these equations with our above wave functions is to take *k* = *k*_{1}—which
means there is no step potential at all!

*Question*: what is wrong with the above reasoning?

The answer is that we have been led astray by our mental
picture of the particles as little balls rolling along in a potential, with
enough energy to get up the hill, etc. Schrödinger’s equation is a *wave
equation*. Building intuition about
solutions should rely on experience with waves. We should be thinking about a light wave going from air into glass, for
example. If we do, we realize that at *any* interface *some of the light gets reflected*. This means that our expression for the wave function for *x* < 0 is incomplete, we need to add a *reflected* wave, giving

_{}

If we now match the wave function and its derivative at the origin, we find

_{}

Recalling that the square of the wave function denotes probability, it is easy to check that the fraction of the wave that is reflected

_{}

Evidently, the fraction of the wave transmitted

_{}.

*Question*: isn’t the amount transmitted just given
by *B*^{2}/*A*^{2}?

The answer is no. The
ratio *B*^{2}/*A*^{2} gives the relative
probability of finding a particle in some small region in the transmitted
stream relative to that in the incoming stream, but the particles in the
transmitted stream are moving more slowly, by a factor *k*_{1}/*k.* This means that just comparing the densities
of particles in the transmitted and incoming streams is not enough. The physically significant quantity is the *probability current* flowing past a given
point, and this is the product of the density *and* the speed. Therefore,
the transmission coefficient is *B*^{2}*k*_{1}/*A*^{2}*k*.

*Exercise*: prove that even a step *down* gives rise to some reflection.

### Barriers

If a plane wave coming in from the left encounters a step at
the origin of height *V*_{0} > *E*, the incoming energy, there
will be total reflection, but with an exponentially decaying wave penetrating
some distance into the step. Suppose now
we replace the step with a barrier,

*V* = 0 for *x* < 0

*V* = *V*_{0} for 0
< *x* < *L*

*V* = 0 for *L* < *x*.

In this situation, the wave function will still decay exponentially
into the barrier (assuming the barrier is thick compared to the exponential
decay length), but on reaching the far end at *x* = *L*, a plane wave
solution is again allowed, so there is a nonzero probability of finding the
particle beyond the barrier, moving with its original speed. This phenomenon is
called *tunneling*, since in the
classical picture the particle doesn’t have enough energy to get over the top
of the barrier.

### Alpha decay

A good example of tunneling, and one which helped establish
the validity of quantum ideas at the nuclear level, is alpha decay. Certain large unstable nuclei decay
radioactively by emitting an alpha-particle, a tightly bound state of two
protons and two neutrons. It is thought
that alpha- particles may exist, at least as long lived resonances, inside the
nucleus. For such a particle, the strong
but short ranged nuclear force creates a spherical finite depth well having a
steep wall more or less coinciding with the surface of the nucleus. However, we must also include the
electrostatic repulsion between the alpha-particle and the rest of the nucleus,
a potential _{} outside the
nucleus. This means that, as seen from
inside the nucleus, the wall at the surface may not be a step but a barrier, in
the sense we used the word above, a step up followed by a slide down the
electrostatic curve. Therefore, an alpha-particle bouncing around inside the
nucleus may have enough energy to tunnel through to the outside world.

It is evident that the more energetic the alpha-particle is, the thinner the
barrier it faces. Since the wave function decays exponentially in the barrier,
this can make a huge difference in tunneling rates. It is not difficult to find
the energy with which the alpha-particle hits the nuclear wall, because this
will be the same energy with which it escapes. Therefore, if we measure the
energy of an emitted alpha, since we think we know the shape of the barrier
pretty well, we should be able, at least numerically, to predict the tunneling
rate. The only other thing we need to know is how many times per second alpha's
bounce off the wall. The size of the nucleus is of order 10^{-14} meters,
if we assume an alpha moves at, say, 10^{7} meters per second,
it will bang into the wall 10^{21} times per second. This is a bit
handwaving, but all alpha-radioactive nuclei are pretty much the same size, so
perhaps it's safe to assume this will be about the same for all of them. If we
do that, we get impressive agreement with experiment over a huge range of
lifetimes. polonium^{212} emits alpha's with energy 8.95 MeV, and lasts
3 10^{-7} seconds, thorium^{232} emits 4.05 MeV alpha's, and
lasts 1.4 10^{10 }years. These can both be understood in terms of
essentially the same barrier being tunneled through at the different heights
corresponding to the alpha energy. (French, *QM*, page 407).

*Exercise*: assume the nucleus has a
charge of 90e, and a radius of 10^{-14} meters. Estimate the height of
the barrier at its maximum, and the width of barrier an alpha must tunnel
through for polonium and thorium, discussed above. (Of course, the tunneling
rate is not the same as for a rectangular barrier-one must include the
variation of the decay length with the changing barrier height. This is the
main part of the so-called WKB approximation, see any book on quantum
mechanics.)