272 lines
14 KiB
TeX
272 lines
14 KiB
TeX
% !TeX root = ../script.tex
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\documentclass[../../script.tex]{subfiles}
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\begin{document}
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\section[Examples of the Stationary Schrödinger Equation]{Examples of the Stationary Schrödinger Equation}
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We now want to solve the Schrödinger equation
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\[
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\frac{-\hbar^2}{2m} \dv[2]{\psi}{x} + \epot\psi = E\psi
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\]
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for a few simple, one-dimensional problems. These examples will illustrate the description of classical particles as waves and the following physical consequences.
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\subsection{The Free Particle}
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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.
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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
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\begin{equation}\label{eq:onedschroedinger}
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\frac{-\hbar^2}{2m} \dv[2]{\psi}{x} = E\psi
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\end{equation}
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The total energy $E = \ekin + \epot$ is because of $\epot$ now
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\[
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E = \frac{p^2}{2m} = \frac{\hbar^2 k^2}{2m}
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\]
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Thus~\eqref{eq:onedschroedinger} gets reduced to
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\[
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\dv[2]{\psi}{x} = -k^2 \psi
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\]
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which has the general solution
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\begin{equation}\label{eq:onedsolution}
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\psi(x) = Ae^{ikx} + Be^{-ikx}
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\end{equation}
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The time-dependent wave function
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\begin{equation}\label{eq:onedsolutioncomplete}
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\psi(x, t) = \psi(x) \cdot e^{-i\omega t} = Ae^{i(kx - \omega t)} + Be^{-i(kx + \omega t)}
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\end{equation}
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represents the superposition of a planar wave travelling in the $+x$ and $-x$ direction.
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The coefficients $A$ and $B$ are the amplitudes of those waves, which are determined by the boundary conditions.
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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 $-x$.
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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.
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Using the normalization condition we get
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\begin{align*}
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&\int_0^L \abs{\psi(x)}^2 \dd{x} = 1 \\
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&\implies A^2 \cdot L = 1 \implies A = \frac{1}{\sqrt{L}}
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\end{align*}
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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}
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\begin{equation}
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\psi(x, t) = \int_{k_0 - \Delta k / 2}^{k_0 + \Delta k / 2} A(k) e^{i(kx - \omega t)} \dd{k}
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\end{equation}
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The location uncertainty of this packet at $t = 0$ is
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\[
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\Delta x \ge \frac{\hbar}{2 \Delta p_x} = \frac{1}{2\Delta k}
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\]
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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.
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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.
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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 $t$.
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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 $\Delta x$
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\[
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\dv{(\Delta x(t))}{t} = \Delta v(t = 0) = \frac{\hbar}{m} \Delta k(t = 0)
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\]
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changes proportionally to the initial impulse uncertainty.
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\subsection{Potential Step}
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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
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\[
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\phi(x) = \begin{cases}
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0 & x < 0 \\
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\phi_0 & x \ge 0
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\end{cases}
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\]
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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 $\phi(x \ge 0) = \phi_0 > 0$.
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The potential energy in this area is still constant $\epot = \energy{0}$. Thus, at $x = 0$ we have a potential jump $\Delta E = \energy{0}$.
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This problem has an equivalent in classic optics: a planar lightwave encountering a boundary between vacuum and material (e.g.\ a glass surface).
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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}
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for the location part of the wave function
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\[
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\wave[I](x) = Ae^{ikx} + Be^{-ikx}
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\]
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where $A$ is the amplitude of the incidental wave, and $B$ the amplitude of the wave reflected from the potential step.
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\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.
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In area II, the Schrödinger equation becomes
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\begin{equation}
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\dv[2]{\psi}{x} + \frac{2m}{\hbar^2}(E - \energy{0})\psi = 0
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\end{equation}
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If we use the shorthand $\alpha = \sqrt{2m(\energy{0} - E)} / \hbar$ we can reduce the equation to
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\begin{equation}\label{eq:potentialstep}
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\dv[2]{\psi}{x} - \alpha^2\psi = 0
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\end{equation}
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This equation has the solution
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\begin{equation}\label{eq:potentialstepsolution}
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\wave[II] = Ce^{+\alpha x} + De^{-\alpha x}
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\end{equation}
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If
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\[
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\psi(x) = \begin{cases}
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\wave[I] & x < 0 \\
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\wave[II] & x \ge 0
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\end{cases}
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\]
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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,
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or else the second derivative $\dd^2\psi / \dd x^2$ is not defined, and thus the Schrödinger equation is not applicable.
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Using~\eqref{eq:onedsolution} and~\eqref{eq:potentialstepsolution} this results in the boundary conditions
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\begin{subequations}\label{eq:boundaryconditions}
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\begin{equation}
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\begin{split}
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\wave[I](x = 0) &= \wave[II](x = 0) \\
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&\implies A + B = C + D
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\end{split}
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\end{equation}
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\begin{equation}
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\begin{split}
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\eval{\dv{\wave[I]}{x}}_0 &= \eval{\dv{\wave[II]}{x}}_0 \\
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&\implies ik(A - B) = \alpha (C - D)
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\end{split}
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\end{equation}
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\end{subequations}
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We can now investigate the two cases where the energy $\ekin = E$ of the incoming particle is smaller or larger than the potential step.
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\subsubsection{(a) $E < \energy{0}$}
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In this case, $\alpha$ is real valued and the coefficient $C$ in~\eqref{eq:potentialstepsolution} must be zero, because otherwise
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\[
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\wave[II] \xrightarrow{x \rightarrow +\infty} \pm\infty
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\]
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If this happens, the wave function is not normalizable. With the above boundary conditions this yields
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\begin{align}
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B = \frac{ik + \alpha}{ik - \alpha}A && \text{and} && D = \frac{2ik}{ik - \alpha} A
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\end{align}
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Thus the wave function for $x < 0$ becomes
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\begin{equation}
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\wave[I](x) = A \left[ e^{ikx} + \frac{ik + \alpha}{ik - \alpha}e^{-ikx} \right]
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\end{equation}
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The fraction of reflected particles is calculated as
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\begin{equation}\label{eq:reflectioncoefffirst}
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R = \frac{\abs{Be^{-ikx}}^2}{\abs{Ae^{ikx}}^2} = \frac{\abs{B}^2}{\abs{A}^2} = \abs{\frac{ik + \alpha}{ik - \alpha}}^2 = 1
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\end{equation}
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which means that \textit{all} particles are being reflected if $E < \energy{0}$. This corresponds to the expected classical behaviour or particles.
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However there is a notable difference to classic particle mechanics:
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\begin{tcolorbox}
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The particles are not being reflected at $x = 0$, but instead penetrate the domain $x > 0$ where $\epot = \energy{0} > \ekin$ before returning,
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even if their energy $\ekin < \energy{0}$ should not be enough to do so in a classical model.
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\end{tcolorbox}
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The probability $P(x)$ of finding a particle in $x > 0$ is
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\begin{equation}
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P(x) = \abs{\wave[II]}^2 = \abs{De^{-\alpha x}}^2 = \frac{4k^2}{\alpha^2 + k^2} \abs{A}^2 e^{-2\alpha x} = \frac{4k^2}{k_0^2} \abs{A}^2 e^{-2\alpha x}
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\end{equation}
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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 $x = 0$.
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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
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of the penetrating wave sinks to $1/e$ of its initial value after a distance $x = 1/(2k\kappa) = \lambda/(4\pi\kappa)$.
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\begin{tcolorbox}
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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.
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\end{tcolorbox}
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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.
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\subsubsection{(b) $E > \energy{0}$}
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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 $x > 0$,
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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
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\begin{equation}
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k' = \frac{\sqrt{2m(E - \energy{0})}}{\hbar} = i\alpha
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\end{equation}
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The solution~\eqref{eq:potentialstepsolution} then becomes
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\begin{equation}
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\wave[II] = Ce^{-ik'x} + De^{ik'x}
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\end{equation}
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Since there are no particles moving in the $-x$ direction, we know that $C = 0$, and we are left with $\wave[II] = De^{ik'x}$.
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From the boundary conditions~\eqref{eq:boundaryconditions} we can deduce
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\begin{align}
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B = \frac{k - k'}{k + k'}A && \text{and} && D = \frac{2k}{k + k'}A
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\end{align}
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thus leaving us with the wave functions
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\begin{equation}
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\begin{split}
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\wave[I](x) &= A \cdot \left(e^{ikx} + \frac{k - k'}{k + k'} \cdot e^{-ikx} \right) \\
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\wave[II](x) &= \frac{2k}{k + k'}Ae^{ik'x}
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\end{split}
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\end{equation}
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The \textbf{\textit{reflection coefficient}} $R$, i.e.\ the fraction of reflected particles is then analogous to the result from optics
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\begin{equation}\label{eq:reflectioncoeff}
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R = \frac{\abs{B}^2}{\abs{A}^2} = \abs{\frac{k - k'}{k + k'}}^2
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\end{equation}
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\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
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\[
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R = \abs{\frac{n_1 - n_2}{n_1 + n_2}}^2
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\]
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of a lightwave encountering a boundary surface between two mediums with refractive indexes $n_1$ and $n_2$.
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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)
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one has to consider that the velocities in both areas are different.
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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}}
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\begin{equation}\label{eq:transmissioncoeff}
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T = \frac{v' \abs{D}^2}{v \abs{A}^2} = \frac{4k \cdot k'}{(k + k')^2}
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\end{equation}
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From comparing~\eqref{eq:reflectioncoeff} and~\eqref{eq:transmissioncoeff} we can see that
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\[
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T + R = 1
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\]
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which makes sense because the amount of particles overall has to be conserved.
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\textbf{\textit{Remarks.}}
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\begin{itemize}
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\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 $R = 1$
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\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.
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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.
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\end{itemize}
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\subsection{Quantum Tunnelling}
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Now we will consider the case where the area in which the potential energy $\epot(x) = \energy{0}$ has only a finite width $\Delta x = a$,
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such that for $x < 0$ and $x > a$ we have $\epot(x) = 0$, and for $0 \le x \le a$ we have $\epot(x) = \energy{0}$.
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The domain is then split into three parts I, II and III, for which we repeat our previous approaches for the calculation of the wave functions
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\begin{equation}
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\begin{split}
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\wave[I] &= Ae^{ikx} + Be^{-ikx} \\
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\wave[II] &= Ce^{\alpha x} + De^{-\alpha x} \\
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\wave[III] &= A'e^{ikx}
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\end{split}
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\end{equation}
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We also find the boundary conditions
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\begin{align}
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\wave[I](0) &= \wave[II](0), & \wave[II](a) &= \wave[III](a) \nonumber \\
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\wave[I]'(0) &= \wave[II]'(0), & \wave[II]'(a) &= \wave[III]'(a)
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\end{align}
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Similar to the previous example we then find the following conditions
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\begin{align*}
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A + B &= C + D \\
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Ce^{\alpha a} + De^{-\alpha a} &= A'e^{ika} \\
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ik(A - B) &= \alpha (C - D) \\
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\alpha Ce^{\alpha a} - \alpha D e^{-\alpha a} &= ikA' e^{ika}
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\end{align*}
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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
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\begin{equation}
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^T = \frac{v \cdot \abs{A'}^2}{v \cdot \abs{A}^2}
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\end{equation}
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\begin{figure}
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\begin{center}
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\begin{tikzpicture}
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\fill[red!10] (3, 0) rectangle (5, 2.5);
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\draw[black] (0, 0) -- node[above] {I} (3, 0) -- node[above] {II} (5, 0) -- node[above] {III} (8, 0);
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\draw[red] (0, 0) -- (3, 0) -- (3, 2.5) -- (5, 2.5) -- (5, 0) -- (8, 0);
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\draw[black] (5.05, 2.5) -- (5.35, 2.5);
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\draw[black, <->] (5.2, 0) -- node[below right] {$\energy{0}$} (5.2, 2.5);
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\draw[black, <->] (3, 0.5) -- node[above] {$a$} (5, 0.5);
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\draw[black] (0, 1.5) -- (8, 1.5);
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\draw[red, domain=0:3, smooth, variable=\x] plot ({\x}, {sin(5*\x r) * 0.7 + 1.5}) ;
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\draw[red, domain=3:5, smooth, variable=\x] plot ({\x}, {exp(-(\x - 3)) + 0.96});
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\draw[red, domain=5:8, smooth, variable=\x] plot ({\x}, {sin(5*(\x + 0.95) r) * 0.4 + 1.5});
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\node at (1.5, 2.8) {$\lambda = \frac{2\pi}{k}$};
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\node at (6.5, 2.5) {$\lambda = \frac{2\pi}{k}$};
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\end{tikzpicture}
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\end{center}
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\caption{Quantum tunnelling through a rectangular potential barrier}
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\label{fig:quantumtunnelling}
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\end{figure}
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Inserting $k = p/h$ and $E = p^2/2m$ gives
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\begin{equation}
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T = \frac{1 - E/\energy{0}}{(1 - E/\energy{0}) + (\energy{0} / 4E) \cdot \sinh^2(\alpha \cdot a)}
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\end{equation}
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with $\alpha = \sqrt{2m(\energy{0} - E)}/\hbar$.
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\end{document} |