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Propagation of a Free Classic Particle and its Quantum Wave Packet Representation

The wave packet that I use is a simple complex sinusoid embedded in various envelopes (see "Starting Wave Packet Envelope Shapes"). Its propagating function changes are most accurately computed by solving the time dependent Schrodinger equation (TDSE)

`(d^2psi)/(dx^2)=iℏ(dpsi)/dt`

using finite element analysis (FEA). Another computation method is to use the fact that we know that the energy of a free particle is related to its momentum and we can add its time dependence as `exp[-iℏk^2/(2mass)*t]` and that is the method I'll use here.
I give the learner the choice of 3 wave packet envelope initial shapes:
1: Gaussian (standard bell curve used in statistics)
2: Triangular
3: Square
4: Semicircle
The evolution of the Gaussian is the slowest but it eventually gets spread out as required by the Uncertainty Principle; The Triangle, which has a slope discontinuity, remains intact for a while but it definitely loses its resemblance to its original shape. The Square, which has value discontinuities as well as slope dicontinuities, loses its identity fastest. Note that both of these latter two packets tend toward a gaussian shape. To compute the propagation of these packets, I first choose a set of constants, ck, that I can use to represent the original shape of the packet:

`psi_0(x)=sum_-oo^ooc_kexp(ikx)`

where

`c_k+int_-oo^oo"envelope"(x)exp(i(k-k_0)xdx`

Having obtained `c_k` and `psi_0[x]` it is simple to give it the time dependence associated with a free particle of kinetic energy (ℏk)^2/(2*mass) so that psi becomes:

`psi(x,t)=sum_(k=-oo)^(k=oo)c_kexp(ikx-i(ℏk^2)/(2mass)t)`

After summing over all k, this expresses a complex wave that moves in the +x direction at speed `ℏ/(mass)*k_0` Since the values of k are continuous, the starting values of the above shapes cannot be accurately approximated by the sum over values of k. The reason is that the k dependent functions are not orthogonal on the region in which envelope(x) is non-zero. If they were orthogonal, then we could express the shapes as a Fourier series and the fidelity to the envelope would be much better.