where we must remember that
is a function of
,
not just a constant; the dispersion relation
determines all the key physical properties of the wave such as
phase velocity
and
group (physical) velocity
.
direction, giving
the 1-dimensional version
with
and
.
has a mean wavenumber
and a variance
(so that
is the standard deviation of k).
or
where
If we now let
so that
, we have
=
Completing the square,
, giving
where
The definite integral has the value
(look it up in a table of integrals!) giving
or
where
That is, the rms width of the wave packet about its
initial mean of
is
and the product of the x and k widths obeys the
uncertainty relation
at t = 0.
=
=
where
.
Again the definite integral equals
, giving
or
=
.
We have now fully described
.
has a different k
and therefore progapates at a different velocity
=
=
.
Thus they all move away from x = 0 at a different rate
and become spread out or dispersed
[hence the name ``dispersion relation''
for
]
relative to their average position
[the centre of the wave packet]
at
.
The width of the wave packet,
, therefore increases
with time from its minimum value
at t = 0.
The time dependence can be calculated with some effort
(not shown here);
the result is
The normalization constant A will decrease with time
(as the spatial extent of the wave packet increases)
in order to maintain
.
Thus the probability of finding the particle within dx of
its mean position
steadily decreases with time as the wave packet disperses.

nm
(roughly atomic dimensions)
in a gaussian wave packet. For simplicity we will let
;
that is, the electron is (on average) at rest.
If the electron is free
(as we have assumed throughout this treatment)
then its wave packet will expand to
times its initial size
in a time
s.
m,
the time required for it to disperse until
is
times longer:
ns.
s.
(That is,
years!)

This raises the question: What do the ``wiggles'' represent? When the particle is (on average) at rest, its wave packet is just a ``bump'' that spreads out with time; when it is moving, it acquires all these oscillations of phase with a wavelength satisfying de Broglie's formula. Is it really ``there'' at the peaks and ``not there'' at the points where the function crosses the axis? No. Except for the overall ``envelope'' it is just as ``there'' at one point as at another. This is a direct consequence of using the complex exponential form (rather than a cosine) for the travelling wave. Although the plots above show only the real part, there is an imaginary part that is a maximum when the real part is zero and vice versa so that the absolute magnitude is always (except for the overall ``envelope'') the same.
In that case, what is the point of even having these ``wiggles?'' Well, although no experiment can measure the absolute phase of a wavefunction, the relative phase of two probability amplitudes being added together is what causes interference, which is the key to all observable quantum mechanical phenomena.
It is also worth remembering that by adding together two travelling waves propagating in opposite directions it is possible to make a standing wave, whose wavefunction really is a real oscillatory function for which the particle is actually never found at positions where the amplitude is zero.