Average Acceleration for a Particle in an Infinite Well?

A classical free particle has a specific momentum p and kinetic energy pp/2m. In an ideal gas with no potential P(p)=P(-p) so p average=0 at each x. A quantum free particle has a wavefunction exp(ipx), a specific momentum of p and a kinetic energy of pp/2m. A special feature seems to be that exp(ipx) is associated with a length proportional to 1/p which repeats itself. We argue that this length is associated with the motion p in one dimension together with the two dimensionality. In other words, motion is not a simple translation. If a particle is placed in an infinite well of 1 dimension and length L, then low bound states have a special length (wavelength) of the order of the box. In other words one does not know at which point the particle is located or whether it has already reflected. Thus we argue that from a probabilistic point of view one must include p and -p together. These have different probabilities exp(ipx) and exp(-ipx) because this is a dynamic probability. The interference of the two removes one dimension exposing the periodic cos(px) like nature. If one calculates pp/2m average using  -1/2m d/dx d/dx W / W (where W is the wavefunction) one obtains pp/2m (the classical result) because both p and -p have the same magnitude. The dynamic probability, however, does not have the same value for each. Thus p average = -i dW/dx / W is not zero and this result still holds if one multiplies by W*W. In other words there at each x there seems to be more weight to p or to -p. This suggests that an average acceleration may appear in fact:  d/dx { -dW/dx / W ) =   d/dx d/dx W / W - (dW/dx / W)(dW/dx / W).

    The quantum bound scenario is a statistical one with an inherent physical (we argue) kind of motion linked to a wavelength. If a particle is knocked out of bound state it has a wavefunction exp(ipx) and a momentum p. It is no longer in a hybrid inteferring statistical state. The modulus of exp(ipx) is 1 at each x as well. Furthermore  P(p)=P(-p) = a*(p)a(p)=a*(-p)a(-p) which is not the same result as exp(ipx) versus exp(-ipx) for the dynamical probability. Given P(p)=P(-p) and no knowledge of x one might guess that one has a constant spatial density classical scenario, but this is not the case. As a result a special type of measurement is needed to discern W*(x)W(x) it seems, but this picture of humps and low probability regions is also in keeping with an average acceleration we argue.