x^3  \times \delta p^3 \simeq \frac{\hbar}{2}.\end{aligned}$$ The numerical result for $\psi_{wp}^3$ is shown in Fig. \[well\], where we set $\delta x^1=8.0\times 10^{-10}[\text{m}]$, and the width of the potential is taken from 0 to $1.5\times \delta x^1$ and the potential depth is of the order of the average energy of the wave packet. In this region,  $\delta x^3 \times \delta p^3$ is approximately adiabatic invariant and the minimum wave packet remains nearly minimum after potential scattering. On the other hand, $\delta x^2 \times \delta p^2$ becomes large. The large change of $\delta x^2 \times \delta p^2$ from $\delta x^1
\times \delta p^1$ is generated, because there are two boundaries of the potential barrier/well.

\[well\]

 Potential scatterings in three dimensions
------------------------------------------

The Schrödinger equation with a one-dimensional potential $U( z)$ $$\begin{aligned}
\left[-{{\hbar}^2 \over 2m} {\vec \nabla}^2 +U(z)\right]\psi({\vec x})=E \psi({\vec x})\end{aligned}$$ has solutions of the form, $$\begin{aligned}
\psi({\vec x})=e^{i{\vec p}_T\cdot{\vec x}_T}u(z).\end{aligned}$$ The first part on the right-hand side is the plane wave in the transverse direction ${\vec p}_T=(p_x,p_y),{\vec x}_T=(x,y)$ and the second part is the function of $z$.

The wave packet is formed from the above wave function as $$\begin{aligned}
\psi_{w p}=\int d{\vec k}_T N({\vec k-k_0}_T)e^{i{\vec p}_T\cdot{\vec x}_T}  \int d k_z  N(k_z-k_z^0)u(z),\end{aligned}$$ where 