Compact supported square shape packet

Next: Compact supported triangular shape Up: Examples (with figures) Previous: Minimising wave packet

Compact supported square shape packet

This wave packet has constant probability density in an interval of length L and zero probability density outside this interval. The simplest expression which gives constant $\rho (x)$ is just a wave function which is itself constant [7].

$\begin{eqnarray} \psi_{Square} ( x; L ) &=& \frac {1} { \sqrt {L} } {\rm Wind... ...a, b )$\space ; \cr 0, & if $x \not\in ( a, b )$\space . \cr } .\end{eqnarray}$

Its Fourier transform is:

$\begin{equation} \varphi_{Square} ( k; L ) = \frac {1} { \sqrt L } \frac {L} {... ...t{2 \pi} } \cdot \frac {\sin { L/2 \cdot k } } { L/2 \cdot k } .\end{equation}$

The position indeterminacy value can be calculated easily using (20) and give

$\begin{equation} \Delta x = \frac {L} { \sqrt {12} } .\end{equation}$

The integrals for $\Delta k$ does not converge, however,

$\begin{equation} \Delta k = \infty .\end{equation}$

The properties of the square initial state are summarised as follows:

It has compact support in x , the particle is confined to the [ - L / 2, L / 2 ] interval.
In k it has infinite support and it is very poorly localised because $\varrho (k)$ decreases only slowly as . (Cf. Fig. 2).
$\Delta k$ is infinite. This means that $E_{kin}$ is infinite which shows that this state is unphysical because it would be necessary to perform an infinite amount of work to prepare this wave function. It is natural, however, because the square wave function has jumps and so does not fulfil the continuity requirement of the wave functions. Hence the square wave packet can only be approximated in real physical scenarios.

It is easier to calculate $\psi_{Square}(x,t;L)$ with the convolution formula (18) than with the Fourier transform formula (16):

$\begin{equation} \psi_{Square}(x,t;L) = \psi_{Square} (x,t=0;L) * P_{Free}(x,t)... ...int_{-L/2}^{L/2} { \exp { i \frac { (y-x)^2 } { 2 t } } dy } .\end{equation}$

These kinds of integrals can be expressed by the so called E(x) function [8]:

$\begin{eqnarray} \psi_{Square}(x,t;L) &=& \frac {1} { \sqrt{ 2 L } } \exp { -i... ...\ E(x) &=& \sqrt{ \frac{2}{\pi} } \int_0^x { e ^ { i t^2 } dt }\end{eqnarray}$

It is important to note that $\psi_{Square}(x,t;L)$ is not any more a compact supported function for any time $t \neq 0$ . This means that if we switch off the potential that was used to squeeze the wave function into the finite interval [-L/2,L/2] the wave function spreads into the whole $(-\infty,\infty)$ space during an infinitesimal dt time. Hence we could find the particle at any far distance from its original position after an infinitesimal amount of time. This would involve a motion with infinite speed. There is nothing to wonder at this however because we used a non-relativistic description of our quantum particle. In a relativistic formalism the wave front moves with speed c .

One could think that this infinite fast spreading phenomenon follows from the fact that the ideal square wave packet has infinite kinetic energy. It can be proven [9] that this phenomenon does exist for all wave packets with bounded support.

Looking at Fig. 3. one can verify that the square wave packet spreads quite fast, much faster than the other packets. This is because of its wide spectral distribution.

Next: Compact supported triangular shape Up: Examples (with figures) Previous: Minimising wave packet