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Minimising wave packet

(Click to enlarge image)

  This is the wave packet for which the Heisenberg inequality (1) becomes an equality (2) which leads to a differential equation for the minimising wave packet. From the solution of this differential equation it follows [1] that this packet is the following Gaussian:

  \begin{eqnarray}
 \psi_{Gauss} ( x ; a, x_0, k_0 ) &=&
 \sqrt[4] { \frac {2} { \...
 ...\exp{ \left( - 2 \cdot \frac { ( x - x_0 )^2 } { a^2 } \right) } .\end{eqnarray}

The Fourier transform of the coordinate wave function $ \psi_{Gauss} ( x ) $gives the wave number wave function $ \varphi_{Gauss} ( k ) $which is also Gaussian:

  \begin{eqnarray}
 \varphi_{Gauss} ( k ; a, x_0, k_0 ) &=&
 \sqrt[4] { \frac { a^...
 ... }
 \cdot
 \exp{ \left( - \frac{ a^2 } {2} ( k - k_0 )^2 \right) }\end{eqnarray}

The Gaussian wave packet is a product of three factors:

The position and wave number indeterminacies $\Delta x$ and $\Delta k$ can be calculated from the x (20) and k (21) moments of the wave function. For the Gaussian wave packet these integrals can be calculated easily with Gaussian integrals and we get

% latex2html id marker 1958
$\textstyle\parbox{5.0cm}{
\begin{eqnarray}
 \left< ...
 ...right\gt &=& \frac{a^2}{4} + x_0^2
 \\  \Delta x &=& \frac{a}{2}\end{eqnarray}}$

% latex2html id marker 1959
$\textstyle\parbox{5.0cm}{
\begin{eqnarray}
 \left< ...
 ...ght\gt &=& \frac{1}{a^2} + k_0^2
 \\  \Delta k &=& \frac{1}{a} .\end{eqnarray}}$

Multiplying these $\Delta x$ and $\Delta k$ values we can verify that their product gives 1/2 hence this Gaussian really satisfies the Heisenberg equality (2).

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

$\psi(x,t)$ is given by a straightforward evaluation of (16) or (18) using Gaussian integrals [6]:

    \begin{eqnarray}
 \psi_{Gauss} ( x, t ; a, x_0, k_0 ) &=&
 \sqrt[4] { \frac {2} ...
 ... -i \frac{k_0^2}{2} t \right) }
 \\  a(t) &=&
 a + 2 i \frac{t}{a}\end{eqnarray}

where ${\rm arg} \, z$ is the phase of the complex number z , i.e. ${\rm arg} R e^{i \varphi} = \varphi$.Our $\psi(x,t)$ has three main factors (47, 48 and 49 ). The first factor (47) is a product of two pure real coefficients and a plane wave. This plane wave part of Factor 1. and the entire second (48) and third (49) factors are pure phase factors, i.e. their magnitude is one. Hence it is very easy to calculate the probability density $\rho (x,t) = \vert \psi (x,t) \vert ^ 2$:one has only to calculate the square of the two pure real coefficients of Factor 1. which gives:

 \begin{equation}
 \rho_{Gauss} ( x, t ; a, x_0, k_0 ) =
 \sqrt { \frac {2} { \pi...
 ...x_0 + k_0 t ) \right] } ^ 2 }
 { \vert a(t)\vert^2 }
 \right)
 } .\end{equation}

The three terms of $\psi_{Gauss} (x,t)$ are as follows:

Factor 1.
(Cf. 47.) A Gaussian of the form (37). This is an expression having the same form as the initial $\psi_0 (x)$ but the centre of gravity of the Gaussian is moving with speed $u_g = k_0$ and its width is increased from a to $\vert a(t)\vert = \sqrt { a^2 + 4 t^2 / a^2 }$. The maximum value of $\rho (x,t)$ is decreasing as its width increases making the area under $\rho (x,t)$ constant.

Factor 2.
(Cf. 48.) An x and t dependent phase factor which is quadratic in x . This factor oscillates faster for larger |x| values which accounts for the fact that the higher wave number components of the initial Gaussian $\psi_0 (x)$ move with higher velocities.

Factor 3.
(Cf. 49.) An x independent (but still t dependent) phase factor. This phase factor is a product of two terms. The first term is a monotonic function of time while the second one is oscillating. The phase of the first term is zero for t=0 (a(t=0) is pure real) and $- \pi /4$ for $t=\infty$ ($a(t=\infty)$ is pure imaginary). The second term is $\exp{ \left( - i \omega_0 t \right) }$ and it accounts for the time development of the plane wave component $\exp{ \left( i k_0 x \right) }$ in Factor 1.


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