Perfect localisation in momentum space

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Perfect localisation in momentum space

This is in fact the opposite case to the previous one (VE), $\rho (x)$ is constant, $\psi(x)$ is a plane wave, hence $\varphi(k)$ must be a Delta distribution.

$\begin{eqnarray} \psi_{kLoc} ( x; k_0 ) &=& \frac{1}{\sqrt{2 \pi}} e ^ { i k_0... ... \frac{1}{2 \pi} \\ \varphi_{kLoc}(k;k_0) &=& \delta ( k - k_0 )\end{eqnarray}$

The indeterminacies are:

$\begin{eqnarray} \Delta x = \infty \; \Delta k = 0 .\end{eqnarray}$

The properties of this plane wave state:

$\rho (x)$ is constant, hence this state is completely delocalised in x , the particle can be found everywhere with the same probability, it fills the whole space homogeneously.
It is completely localised in k space, because it has compact support in k and its support is the point. This state is the eigenstate of the wave number operator (5) because it consists of only one plane wave component.
$\Delta x = \infty$ which shows that it is also an idealised case because a real quantum particle can not spread into the whole space homogeneously in finite time.

Time evolution of the plane wave state (77) is simple because it has only one wave number component ().

$\begin{eqnarray} \psi_{kLoc} (x,t) &=& {\cal F}^{-1} \left[ \delta ( k - k_0... ... &=& { \vert \exp{ i \vartheta } \vert } ^ 2 = \frac{1}{2 \pi} .\end{eqnarray}$

The $\omega t$ phase factor in (81) describe the propagation of the plane wave. The phase speed of the propagation is $v = \omega / k_0 = k_0 / 2$ .

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