Next: Potential scattering
Up: Scattering states in one
Previous: Scattering states in one
  Contents
The simplest case is the one where the potential
is zero throughout: the two
independent solutions of the Schrödinger
equation then are plane waves
 |
|
|
(114) |
with positive energy
.
We denote both solutions as
with
either positive or negative. They are plane waves with fixed wave vector
and therefore fixed momentum
.
We can have no physically meaningful solutions with negative energy
because in this case the wave function would become infinite either for
or
.
A problem arises, however, because
can not be
normalized over the whole
-axis according to
 |
|
|
(115) |
because this integral is infinite: the probability density, i.e. the square
is
constant, i.e.
everywhere.
In particular, we have for the mean square deviation of the momentum and the position
There is no uncertainty in the momentum of the particle, but there is maximum uncertainty
in its position: the wave function
describes a particle with fixed momentum
which is completely delocalized (spread) over the
-axis.
A practical solution is to consider a large, but finite interval
instead of the total
-axis, and to normalize the wave functions according to
 |
|
|
(117) |
Then, the boundary conditions at
have to be specified. Again, a convenient (but not necessary
the only) choice are periodic boundary conditions: we bend the interval
into a ring
such that the points
coincide. Demanding continuity of
, i.e.
, we obtain a quantization condition for the possible
-values,
 |
|
|
(118) |
i.e. a discrete set of possible
values and therewith discrete energies
.
To each energy
there are two linearly independent plane waves with wave vectors
and
. One says the energy value
is two-fold degenerate.
If
is very large, the possible values for
are still discrete but
subsequent
-values get very close to each other.
Next: Potential scattering
Up: Scattering states in one
Previous: Scattering states in one
  Contents
Tobias Brandes
2004-02-04