From 4ba162a2c06335bcbb73d713802a9f5e76048425 Mon Sep 17 00:00:00 2001 From: Lauchmelder Date: Sun, 16 Oct 2022 00:19:06 +0200 Subject: [PATCH] started quantum tunnelling --- chapters/fundamentals/examples.tex | 85 ++++++++++++++++++++++++----- script.pdf | Bin 325043 -> 332415 bytes script.tex | 9 +++ 3 files changed, 81 insertions(+), 13 deletions(-) diff --git a/chapters/fundamentals/examples.tex b/chapters/fundamentals/examples.tex index 0c54a4c..ae6a0da 100644 --- a/chapters/fundamentals/examples.tex +++ b/chapters/fundamentals/examples.tex @@ -77,7 +77,7 @@ This problem has an equivalent in classic optics: a planar lightwave encounterin 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} for the location part of the wave function \[ - \psi_{\text{I}}(x) = Ae^{ikx} + Be^{-ikx} + \wave[I](x) = Ae^{ikx} + Be^{-ikx} \] where $A$ is the amplitude of the incidental wave, and $B$ the amplitude of the wave reflected from the potential step. @@ -93,13 +93,13 @@ If we use the shorthand $\alpha = \sqrt{2m(\energy{0} - E)} / \hbar$ we can redu \end{equation} This equation has the solution \begin{equation}\label{eq:potentialstepsolution} - \psi_{\text{II}} = Ce^{+\alpha x} + De^{-\alpha x} + \wave[II] = Ce^{+\alpha x} + De^{-\alpha x} \end{equation} If \[ \psi(x) = \begin{cases} - \psi_{\text{I}} & x < 0 \\ - \psi_{\text{II}} & x \ge 0 + \wave[I] & x < 0 \\ + \wave[II] & x \ge 0 \end{cases} \] 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, @@ -108,14 +108,14 @@ Using~\eqref{eq:onedsolution} and~\eqref{eq:potentialstepsolution} this results \begin{subequations}\label{eq:boundaryconditions} \begin{equation} \begin{split} - \psi_{\text{I}}(x = 0) &= \psi_{\text{II}}(x = 0) \\ + \wave[I](x = 0) &= \wave[II](x = 0) \\ &\implies A + B = C + D \end{split} \end{equation} \begin{equation} \begin{split} - \eval{\dv{\psi_{\text{I}}}{x}}_0 &= \eval{\dv{\psi_{\text{II}}}{x}}_0 \\ + \eval{\dv{\wave[I]}{x}}_0 &= \eval{\dv{\wave[II]}{x}}_0 \\ &\implies ik(A - B) = \alpha (C - D) \end{split} \end{equation} @@ -125,7 +125,7 @@ We can now investigate the two cases where the energy $\ekin = E$ of the incomin \subsubsection{(a) $E < \energy{0}$} In this case, $\alpha$ is real valued and the coefficient $C$ in~\eqref{eq:potentialstepsolution} must be zero, because otherwise \[ - \psi_{\text{II}} \xrightarrow{x \rightarrow +\infty} \pm\infty + \wave[II] \xrightarrow{x \rightarrow +\infty} \pm\infty \] If this happens, the wave function is not normalizable. With the above boundary conditions this yields \begin{align} @@ -133,7 +133,7 @@ If this happens, the wave function is not normalizable. With the above boundary \end{align} Thus the wave function for $x < 0$ becomes \begin{equation} - \psi_{\text{I}}(x) = A \left[ e^{ikx} + \frac{ik + \alpha}{ik - \alpha}e^{-ikx} \right] + \wave[I](x) = A \left[ e^{ikx} + \frac{ik + \alpha}{ik - \alpha}e^{-ikx} \right] \end{equation} The fraction of reflected particles is calculated as \begin{equation}\label{eq:reflectioncoefffirst} @@ -147,7 +147,7 @@ However there is a notable difference to classic particle mechanics: \end{tcolorbox} The probability $P(x)$ of finding a particle in $x > 0$ is \begin{equation} - P(x) = \abs{\psi_{\text{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} + 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} \end{equation} 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 $x = 0$. @@ -166,9 +166,9 @@ losing speed in the process, because their kinetic energy decreases to $\ekin = \end{equation} The solution~\eqref{eq:potentialstepsolution} then becomes \begin{equation} - \psi_{\text{II}} = Ce^{-ik'x} + De^{ik'x} + \wave[II] = Ce^{-ik'x} + De^{ik'x} \end{equation} -Since there are no particles moving in the $-x$ direction, we know that $C = 0$, and we are left with $\psi_{\text{II}} = De^{ik'x}$. +Since there are no particles moving in the $-x$ direction, we know that $C = 0$, and we are left with $\wave[II] = De^{ik'x}$. From the boundary conditions~\eqref{eq:boundaryconditions} we can deduce \begin{align} B = \frac{k - k'}{k + k'}A && \text{and} && D = \frac{2k}{k + k'}A @@ -176,8 +176,8 @@ From the boundary conditions~\eqref{eq:boundaryconditions} we can deduce thus leaving us with the wave functions \begin{equation} \begin{split} - \psi_{\text{I}}(x) &= A \cdot \left(e^{ikx} + \frac{k - k'}{k + k'} \cdot e^{-ikx} \right) \\ - \psi_{\text{II}}(x) &= \frac{2k}{k + k'}Ae^{ik'x} + \wave[I](x) &= A \cdot \left(e^{ikx} + \frac{k - k'}{k + k'} \cdot e^{-ikx} \right) \\ + \wave[II](x) &= \frac{2k}{k + k'}Ae^{ik'x} \end{split} \end{equation} The \textbf{\textit{reflection coefficient}} $R$, i.e.\ the fraction of reflected particles is then analogous to the result from optics @@ -210,4 +210,63 @@ which makes sense because the amount of particles overall has to be conserved. \end{itemize} \subsection{Quantum Tunnelling} +Now we will consider the case where the area in which the potential energy $\epot(x) = \energy{0}$ has only a finite width $\Delta x = a$, +such that for $x < 0$ and $x > a$ we have $\epot(x) = 0$, and for $0 \le x \le a$ we have $\epot(x) = \energy{0}$. + +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 +\begin{equation} + \begin{split} + \wave[I] &= Ae^{ikx} + Be^{-ikx} \\ + \wave[II] &= Ce^{\alpha x} + De^{-\alpha x} \\ + \wave[III] &= A'e^{ikx} + \end{split} +\end{equation} +We also find the boundary conditions +\begin{align} + \wave[I](0) &= \wave[II](0), & \wave[II](a) &= \wave[III](a) \nonumber \\ + \wave[I]'(0) &= \wave[II]'(0), & \wave[II]'(a) &= \wave[III]'(a) +\end{align} +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 +\begin{equation} + ^T = \frac{v \cdot \abs{A'}^2}{v \cdot \abs{A}^2} +\end{equation} + +\begin{figure} + \begin{center} + \begin{tikzpicture} + \fill[red!10] (3, 0) rectangle (5, 2.5); + \draw[black] (0, 0) -- node[above] {I} (3, 0) -- node[above] {II} (5, 0) -- node[above] {III} (8, 0); + \draw[red] (0, 0) -- (3, 0) -- (3, 2.5) -- (5, 2.5) -- (5, 0) -- (8, 0); + + \draw[black] (5.05, 2.5) -- (5.35, 2.5); + \draw[black, <->] (5.2, 0) -- node[below right] {$\energy{0}$} (5.2, 2.5); + \draw[black, <->] (3, 0.5) -- node[above] {$a$} (5, 0.5); + + \draw[black] (0, 1.5) -- (8, 1.5); + + \draw[red, domain=0:3, smooth, variable=\x] plot ({\x}, {sin(5*\x r) * 0.7 + 1.5}) ; + \draw[red, domain=3:5, smooth, variable=\x] plot ({\x}, {exp(-(\x - 3)) + 0.96}); + \draw[red, domain=5:8, smooth, variable=\x] plot ({\x}, {sin(5*(\x + 0.95) r) * 0.4 + 1.5}); + + \node at (1.5, 2.8) {$\lambda = \frac{2\pi}{k}$}; + \node at (6.5, 2.5) {$\lambda = \frac{2\pi}{k}$}; + \end{tikzpicture} + \end{center} + + \caption{Quantum tunnelling through a rectangular potential barrier} + \label{fig:quantumtunnelling} +\end{figure} + +Inserting $k = p/h$ and $E = p^2/2m$ gives +\begin{equation} + T = \frac{1 - E/\energy{0}}{(1 - E/\energy{0}) + (\energy{0} / 4E) \cdot \sinh^2(\alpha \cdot a)} +\end{equation} +with $\alpha = \sqrt{2m(\energy{0} - E)}/\hbar$. \end{document} \ No newline at end of file diff --git a/script.pdf b/script.pdf index 6b644657f952521b3fb10e60443bf87b49d42d39..6c7eb9c390afad5dca7f6204e0ac07c497eed2f7 100644 GIT binary patch delta 31847 zcmZs?V{~QRv+$c8CmlQK*tTukwmW8rJGQlBc5HQQ+v$#NJL%~5^FQx-?{n@wUskPI zvuYN`SRdxE#%er6|NMbbXGR0mHXI7zo`c$cxra?P6;FK%z}Rbu(Z1Lud@CE_g|=OI z!*M*ayV6);)V3K8hC74~-E-JDKl@|)b-i{hYsK2C2Sylu&7s63eHi@^8{^iSqkqby zgPSk*n;U}sP5LU7;aAC5_DiH8exJrNXdzDFD%b#$+ 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