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

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.

- First consider an electron that is initially confined to a region
of a size
*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*. - If the same electron is confined much more loosely
to a region of a size
*m*, the time required for it to disperse until is times longer:*ns*. - The same electron initially confined to a 1
*mm*sized wave packet will take 0.0172*s*to disperse to a wave packet 1.414*mm*in size; and so on. - A one-gram marble localized to within 0.1
*mm*will delocalize spontaneously (the physical meaning of dispersion) to 0.1414*mm*only after*s*. (That is,*years!*) - The centre of a wave packet with a finite
*k*_{o}moves with time at the**group velocity**, as expected for the mean position of a particle. It simultaneously*broadens*just like the (on average) stationary particle; this must be so in order to preserve Galilean invariance, which is still applicable as long as the velocities are nonrelativistic.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*of a wavefunction, the**absolute**phase**relative****interference**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.

Jess H. Brewer - Last modified: Mon Nov 23 17:33:28 PST 2015