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Indeterminacy as the function of time

The position and momentum indeterminacy of a quantum state is defined as

$\begin{eqnarray} \Delta x & = & \sqrt { \left< {\bf {x}}^2 \right\gt - { \lef... ...< {\bf {k}}^2 \right\gt - { \left< {\bf {k}} \right\gt } ^ 2 } .\end{eqnarray}$

Indeterminacies for any given quantum state $\Psi (t)$ can be calculated directly and explicitly for any given value of the time t from the moments [5] of the wave function both in coordinate and wave number representation. There is, however, a general expression [2] for the time evolution of $\Delta x$ of a free particle which can be deduced without explicit use of the wave function. This formula makes it possible to calculate $(\Delta x) (t)$ from the $(\Delta x)_0$ and $(\Delta k)_0$ values of the initial state.

Time evolution of the x moments can be calculated [2,5] by expanding them in Taylor series in t :

$\begin{eqnarray} \left< {\bf {x}}^r \right\gt (t) &=& \sum_n \frac{1}{n!} \le... ... \frac{d^n \left< {\bf {k}}^r \right\gt}{d t^n} \right] _ 0 t^n\end{eqnarray}$

where the 0 indices denote evaluation at time t=0 .

The expectation value of an observable A is

$\begin{equation} <{\bf {A}}\gt = \int \psi^{\star} (x,t) {\bf {A}} \psi (x,t) dx\end{equation}$

By differentiating $<{\bf {A}}\gt$ directly we have

$\begin{eqnarray} \frac{d<{\bf {A}}\gt}{dt} & = & \int \frac{\partial \psi^{\sta... ...+ \int \psi^{\star} {\bf {A}} \frac{\partial \psi}{\partial t} dx\end{eqnarray}$

where we have assumed that $\bf {A}$ does not explicitly depend on t . Using the wave equation (13) this can be transformed into

$\begin{eqnarray} \frac{d<{\bf {A}}\gt}{dt} & = & \frac{1}{i} \left< [{\bf {A}},... ...\left< \left[ {\bf {A}}, \frac{{\bf {k}}^2}{2} \right] \right\gt\end{eqnarray}$

where the last equation is valid only in free space. First let us calculate the $<{\bf {k}} ^ r\gt$ moments. The time derivative of $<{\bf {k}}\gt$ is

$\begin{eqnarray} \frac{d<{\bf {k}}\gt}{dt} & = & \frac{1}{2 i} \left< \left[ {\bf {k}}, {\bf {k}}^2 \right] \right\gt = 0\end{eqnarray}$

because the momentum is a constant of the motion in free space. From (27) follows that any time derivative (for $n\gt$ ) of $<{\bf {k}} ^ r\gt$ is zero. This means that the $<{\bf {k}} ^ r\gt$ moments are constant in time

$\begin{eqnarray} \left< {\bf {k}}^r \right\gt (t) = \left< {\bf {k}}^r \right\gt _0\end{eqnarray}$

thus the wave number uncertainty is also constant in time

$\begin{eqnarray} (\Delta k) (t) = (\Delta k)_0 .\end{eqnarray}$

For the time derivative of $<{\bf {x}}\gt$ we have

$\begin{eqnarray} \frac{d<{\bf {x}}\gt}{dt} & = & \frac{1}{2 i} \left< \left[ {... ...{x}},{\bf {k}}]}_i \right\gt = <{\bf {k}}\gt = <{\bf {k}}\gt _0\end{eqnarray}$

where we have utilised the commutator rule $[{\bf {A}}, {\bf {B}} {\bf {C}}] = [{\bf {A}},{\bf {B}}]{\bf {C}} + {\bf {B}}[{\bf {A}},{\bf {C}}]$ ,the commutator relation (6) and our previous result 28 for the time dependence of $<{\bf {k}}\gt$ .Thus all higher order derivatives are zero in the series (22) $<{\bf {x}}\gt$ is a linear function of time

$\begin{equation} <{\bf {x}}\gt (t) = <{\bf {x}}\gt _0 + <{\bf {k}}\gt _0 t .\end{equation}$

The calculation of the time derivatives of $<{\bf {x}}^2\gt$ is also straightforward and leads to:

$\begin{eqnarray} \frac{d<{\bf {x}}^2\gt}{dt} & = & <{\bf {k}} {\bf {x}} + {\bf... ... \\ \frac{d^2<{\bf {x}}^2\gt}{dt^2} & = & 2 <{\bf {k}}^2\gt _0 .\end{eqnarray}$

Thus all higher order derivatives are zero in the series (23) and $<{\bf {x}}^2\gt$ is a quadratic function of time

$\begin{equation} <{\bf {x}}^2\gt (t) = <{\bf {x}}^2\gt _0 + <{\bf {k}} {\bf {x}} + {\bf {x}} {\bf {k}} \gt t + <{\bf {k}}^2\gt _0 t^2 .\end{equation}$

From the first (31) and second (34) moments the x indeterminacy given by (20):

$\begin{equation} ( \Delta x )^2 (t) = ( \Delta x )^2_0 + 2 \left[ \frac12 <{\b... ..._0 <{\bf {k}}\gt _0 \right] \cdot t + ( \Delta k )^2 \cdot t^2 .\end{equation}$

In the special case when $<{\bf {x}}\gt _0$ or $<{\bf {k}}\gt _0$ is zero the x - k cross terms (linear terms in t ) are zero and we get

$\begin{equation} ( \Delta x )^2 (t) = ( \Delta x )^2_0 + ( \Delta k )^2 \cdot t^2 .\end{equation}$

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