Perfect localisation in coordinate space

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Perfect localisation in coordinate space

For a completely localised wave packet $\Delta x = 0$ , i.e. all position measurements on this state give exactly the same result - say

- value. Hence this $\psi_{xLoc} ( x; x_0 )$ state must be an eigenstate of the position operator (4) with eigenvalue

$\begin{equation} x \cdot \psi_{xLoc} ( x; x_0 ) = x_0 \cdot \psi_{xLoc} ( x; x_0 ) .\end{equation}$

This eigenvalue equation has only nontrivial solution in the space of the generalised functions or distributions [4]:

$\begin{equation} \psi_{xLoc} ( x; x_0 ) = \delta ( x - x_0 ) .\end{equation}$

The $\psi_{xLoc} ( x; x_0 )$ functions are normalised in the following sense: $\left\{ \delta ( x - x_0 ) : x \in ( -\infty, +\infty ) \right\}$ form an orthonormal base:

$\begin{equation} \int_{-\infty}^{\infty} \delta ( x - x_1 ) \delta ( x - x_2 ) dx = \delta ( x_1 - x_2 ) .\end{equation}$

The wave number space wave function of the position eigenstate is the Fourier transform of the Delta distribution:

$\begin{eqnarray} \varphi_{xLoc} ( k; x_0 ) &=& \frac{1}{\sqrt{2 \pi}} e ^ { - i x_0 k } \\ \varrho_{xLoc} ( k; x_0 ) &=& \frac{1}{2 \pi} .\end{eqnarray}$

The indeterminacies are:

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

The important properties of the x localised state are:

It has compact support in x , in fact its support is the point. All position measurements find the particle at with probability one.
It has infinite support in k and it is completely delocalised in k because $\varrho (k)$ does not decrease at all with increasing k because it is constant.
$\Delta k = \infty$ , hence $E = \infty$ . This means that this state is also an idealisation and can not be prepared exactly because one would have to perform an infinite amount of work to squeeze a quantum particle into an interval of zero length. Hence this state can only be approximately prepared.

Time evolution of the Dirac delta initial state (70) can be calculated easily with the convolution formula (18) :

$\begin{eqnarray} \psi_{xLoc} (x,t) &=& \delta ( x - x_0 ) * P_{Free}(x,t) = P_... ... { \vert P_{Free} ( x - x_0, t ) \vert } ^ 2 = \frac 1 {2 \pi t}\end{eqnarray}$

because the Dirac delta is the unit element of the convolution operation: $f * \delta = f$ .

This means that a completely localised initial state not only fills the whole space after an infinitesimal dt as all compact supported initial states do but fills the whole space homogeneously.

Next: Perfect localisation in momentum Up: Examples (with figures) Previous: A ''good'' initial state