for a few simple, one-dimensional problems. These examples will illustrate the description of classical particles as waves and the following physical consequences.
\subsection{The Free Particle}
A particle is said to be free, if it is moving in a constant potention $\phi_0$, because then $\vec{F}=-\grad{\epot}$ means that no forces are acting on the particle.
Through a suitable choice of the zero point energy we can set $\phi_0=0$, i.e. $\epot=0$, and thus get the Schrödinger equation for a free particle
\begin{equation}\label{eq:onedschroedinger}
\frac{-\hbar^2}{2m}\dv[2]{\psi}{x} = E\psi
\end{equation}
The total energy $E =\ekin+\epot$ is because of $\epot$ now
represents the superposition of a planar wave travelling in the $+x$ and $-x$ direction.
The coefficients $A$ and $B$ are the amplitudes of those waves, which are determined by the boundary conditions.
For example, the wave function of electrons which are emitted from a cathode in $+x$ direction towards a detector, will have $B =0$, since there are no particles moving in $-x$.
From this experimental setup we know that the electrons are found along the length $L$ of the path between cathode and detector. This means their wave function can only be different from zero in this region of space.
Using the normalization condition we get
\begin{align*}
&\int_0^L \abs{\psi(x)}^2 \dd{x} = 1 \\
&\implies A^2 \cdot L = 1 \implies A = \frac{1}{\sqrt{L}}
\end{align*}
To determine the location of a particle at time $t$ more accurately, we will have to construct \textit{wave packets} in place of planar waves~\eqref{eq:onedsolution}
\begin{equation}
\psi(x, t) = \int_{k_0 - \Delta k / 2}^{k_0 + \Delta k / 2} A(k) e^{i(kx - \omega t)}\dd{k}
\end{equation}
The location uncertainty of this packet at $t =0$ is
\[
\Delta x \ge\frac{\hbar}{2 \Delta p_x} = \frac{1}{2\Delta k}
\]
and depends on the pulse width $\Delta p_x =\hbar\Delta k$. The larger $k$ is, the more certainly $\Delta x(t =0)$ can be determined, but the faster the wave packet spreads.
Experimentally, this can be illustrated as follows: If we apply a short voltage pulse to the cathode at time $t =0$, then electrons can start travelling towards the detector at this instance.
The emitted electrons have a velocity distribution $\Delta v$, such that electrons with differing velocities $v$ will not necessarily be in the same location $x$ at a later point in time $t$.
Instead they are spread over the interval $\Delta x(t)= t \cdot\Delta v$. The velocity distribution is described by $\Delta v \propto\Delta k$ of the wave packet, such that the location uncertainty $\Delta x$
We are still considering the particles from the previous example, however we introduce a potential step at $x =0$. This means we are considering the potential
\[
\phi(x) = \begin{cases}
0 & x < 0 \\
\phi_0 & x \ge 0
\end{cases}
\]
This means the particles are still moving in direction $+x$ and are free ($\epot=0$) if their position is $x < 0$. However at position $x =0$ they enter into an area of higher potential $\phi(x \ge0)=\phi_0 > 0$.
The potential energy in this area is still constant $\epot=\energy{0}$. Thus, at $x =0$ we have a potential jump $\Delta E =\energy{0}$.
This problem has an equivalent in classic optics: a planar lightwave encountering a boundary between vacuum and material (e.g.\ a glass surface).
We divide the domain $-\infty < x < +\infty$ into two areas I and II\@. For area I with $\epot=0$ we still have the equation~\eqref{eq:onedschroedinger} with the solution~\eqref{eq:onedsolution}
where $A$ is the amplitude of the incidental wave, and $B$ the amplitude of the wave reflected from the potential step.
\textbf{Note:} The complete solution is~\eqref{eq:onedsolutioncomplete}. The temporal part of the soltuion is often omitted, because it has no influence in the stationary problems considered here.
is a solution to the Schrödinger equation~\eqref{eq:potentialstep} on the entire domain $-\infty < x < +\infty$, then $\psi$ has to be continuously differentiable at every point,
or else the second derivative $\dd^2\psi/\dd x^2$ is not defined, and thus the Schrödinger equation is not applicable.
Using~\eqref{eq:onedsolution} and~\eqref{eq:potentialstepsolution} this results in the boundary conditions
where $k_0^2=2m\energy{0}/\hbar^2$. After a distance $x =1/(2\alpha)$, the penetration probability is reduced to $1/e$ of its value at $x =0$.
We already know this result from wave optics. Even if waves are reflected totally at a boundary with refraction index $n = n' - i\kappa$, the wave penetrates the surface of the medium before returning, and the intensity
of the penetrating wave sinks to $1/e$ of its initial value after a distance $x =1/(2k\kappa)=\lambda/(4\pi\kappa)$.
\begin{tcolorbox}
Particles with energy $E$ can penetrate into potential areas $\energy{0} > E$ with a certain probability, even if they shouldn't be able to according to classic particle physics.
\end{tcolorbox}
Once we accept that particles are described by waves, we can come to the conclusion that particles are allowed to exist in \textit{classically forbidden} locations.
In this case, the kinetic energy $ekin = E$ of the incidental particle is greater than the potential jump $\energy{0}$, and in the classical model all particles would enter the area $x > 0$,
losing speed in the process, because their kinetic energy decreases to $\ekin= E -\energy{0}$. What does this look like in the wave model? The quantity $\alpha$ is now purely imaginary, so we introduce the real quantity
\begin{equation}
k' = \frac{\sqrt{2m(E - \energy{0})}}{\hbar} = i\alpha
\end{equation}
The solution~\eqref{eq:potentialstepsolution} then becomes
The \textbf{\textit{reflection coefficient}}$R$, i.e.\ the fraction of reflected particles is then analogous to the result from optics
\begin{equation}\label{eq:reflectioncoeff}
R = \frac{\abs{B}^2}{\abs{A}^2} = \abs{\frac{k - k'}{k + k'}}^2
\end{equation}
\textbf{Remark.} Since the wave number $k$ is proportional to the refractive index in optics ($k = n \cdot k_0$), the result~\eqref{eq:reflectioncoeff} can easily be transformed into the reflection coefficient
\[
R = \abs{\frac{n_1 - n_2}{n_1 + n_2}}^2
\]
of a lightwave encountering a boundary surface between two mediums with refractive indexes $n_1$ and $n_2$.
To compute the amount of particles being transmitted per unit of time (i.e.\ the number of particles passing through the area $x = x_0 > 0$ divided by the amount of particles passing through an area $x < 0$ per unit of time)
one has to consider that the velocities in both areas are different.
The ratio $v' / v = k' / k =\lambda/\lambda'$ is given by the ratio of the wave numbers. That is why the \textbf{\textit{transmission coefficient}}
From comparing~\eqref{eq:reflectioncoeff} and~\eqref{eq:transmissioncoeff} we can see that
\[
T + R = 1
\]
which makes sense because the amount of particles overall has to be conserved.
\textbf{\textit{Remarks.}}
\begin{itemize}
\item Total reflection also occurs for $E =\energy{0}$. In this case $\alpha=0$ and $k' =0$, and thus according to~\eqref{eq:reflectioncoefffirst} or~\eqref{eq:reflectioncoeff} we see that $R =1$
\item Instead of a positive potential barrier one can also consider a negative one with $\energy{0} < 0$, where reflection and transmission occurs as well.
One has to let the wave incide from the right in order to yield the same equations. The optical equivalent would be the transition from a medium with higher optical density into one with a lower optical density.
Similar to the previous example we then find the following conditions
\begin{align*}
A + B &= C + D \\
Ce^{\alpha a} + De^{-\alpha a}&= A'e^{ika}\\
ik(A - B) &= \alpha (C - D) \\
\alpha Ce^{\alpha a} - \alpha D e^{-\alpha a}&= ikA' e^{ika}
\end{align*}
Solving the last equation for $A'$ and inserting the relationships between $C$ and $A$ and $D$ and $A$ from the first three equations then gives us the transmission coefficient of the barrier
