Ian Jauslin
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+\documentclass{ian-presentation}
+
+\usepackage[hidelinks]{hyperref}
+\usepackage{graphicx}
+\usepackage{array}
+
+\begin{document}
+\pagestyle{empty}
+\hbox{}\vfil
+\bf\Large
+\hfil Field electron emission\par
+\smallskip
+\hfil and the Fowler-Nordheim equation\par
+\vfil
+\large
+\hfil Ian Jauslin
+\normalsize
+\vfil
+\hfil\rm joint with {\bf Ovidiu Costin}, {\bf Rodica Costin}, and {\bf Joel L. Lebowitz}\par
+\vfil
+arXiv:{\tt \href{http://arxiv.org/abs/1808.00936}{1808.00936}}\hfill{\tt \href{http://ian.jauslin.org}{http://ian.jauslin.org}}
+\eject
+
+\setcounter{page}1
+\pagestyle{plain}
+
+\title{Field emission}
+\vfill
+\hfil\includegraphics[height=5cm]{emitter.jpg}
+\vfill
+\eject
+
+\title{Field emission}
+$$
+ V(x)=U\Theta(x)
+ ,\quad
+ E_{\mathrm F}=k_{\mathrm F}^2<U
+$$
+\hfil\includegraphics[height=5cm]{potential_square.pdf}
+\vfill
+\eject
+
+\title{Thermal emission}
+$$
+ V(x)=U\Theta(x)
+ ,\quad
+ E_{\mathrm F}=k_{\mathrm F}^2>U
+$$
+\hfil\includegraphics[height=5cm]{potential_square_thermal.pdf}
+\vfill
+\eject
+
+\title{Photonic emission}
+$$
+ V_t(x)=\Theta(x)(U-E_tx)
+ ,\quad
+ E_t=\epsilon\omega\cos(\omega t)
+$$
+\hfil\includegraphics[height=5cm]{potential_square_photonic.pdf}
+\vfill
+\eject
+
+\title{Field emission}
+$$
+ V(x)=\Theta(x)(U-Ex)
+$$
+\hfil\includegraphics[height=5cm]{potential.pdf}
+\vfill
+\eject
+
+\title{Field emission}
+\begin{itemize}
+ \item \href{https://doi.org/10.1073\%2Fpnas.14.1.45}{[Millikan, Lauritsen, 1928]}: experimental plot of the logarithm of the current against $1/E$
+\end{itemize}
+\hfil\includegraphics[height=4.5cm]{Millikan-Lauritsen_current.png}
+\vfill
+\eject
+
+\title{Field emission through a triangular barrier}
+\vfill
+\begin{itemize}
+ \item \href{https://doi.org/10.1098/rspa.1928.0091}{[Fowler, Nordheim, 1928]}: predicted that the current is, for small $E$,
+ $$
+ J\approx CE^2e^{-\frac aE}
+ $$
+ \item (\href{https://doi.org/10.1088/1751-8113/44/5/05530@}{[Rokhlenko, 2011]}: studied the range of applicability of the approximation, and found more accurate approximations for larger fields.)
+\end{itemize}
+\vfill
+\eject
+
+\title{Fowler-Nordheim equation}
+\begin{itemize}
+ \item Schr\"odinger equation
+ $$
+ i\partial_t\psi=-\Delta\psi+\Theta(x)(U-Ex)\psi
+ $$
+ \item Fowler-Nordheim: stationary solution: $\psi_{\mathrm{FN}}(x,t)=e^{-ik^2t}\varphi_{\mathrm{FN}}(x)$
+ $$
+ \varphi_{\mathrm{FN}}(x)=
+ \left\{ \begin{array}{l@{\ }l}
+ e^{ikx}+R_Ee^{-ikx} & x<0\\
+ T_E\mathrm{Ai}(e^{-\frac{i\pi}3}(E^{\frac13}x-E^{-\frac23}(U-k^2)) & x>0
+ \end{array}\right.
+ $$
+ $R_E$ and $T_E$ are chosen so that $\varphi_{\mathrm{FN}}$ and $\partial\varphi_{\mathrm{FN}}$ are continuous at $x=0$.
+\end{itemize}
+\vfill
+\eject
+
+\title{Fowler-Nordheim equation}
+\vfill
+\hfil\includegraphics[height=5.5cm]{asymptotic.pdf}
+\vfill
+\eject
+
+\title{Initial value problem}
+\begin{itemize}
+ \item Initial condition:
+ $$
+ \psi(x,0)=
+ \left\{ \begin{array}{l@{\ }l}
+ e^{ikx}+R_0e^{-ikx} & x<0\\
+ T_0 e^{-\sqrt{U-k^2}x} & x>0
+ \end{array}\right.
+ $$
+ $R_0$ and $T_0$ ensure that $\psi$ and $\partial\psi$ are continuous.
+ \item Behaves asymptotically like $\psi_{\mathrm{FN}}$:
+ $$
+ \psi(x,t)e^{ik^2t}\mathop{\longrightarrow}_{t\to\infty}\varphi_{\mathrm{FN}}(x)
+ $$
+\end{itemize}
+\vfill
+\eject
+
+\title{Initial value problem}
+\begin{itemize}
+ \item Laplace transform:
+ $$
+ \hat\psi_p(x):=\int_0^\infty dt\ e^{-pt}\psi(x,t)
+ $$
+ \item Schr\"odinger equation:
+ $$
+ (-\Delta+\Theta(x)V(x)-ip)\psi_p(x)=-i\psi(x,0)
+ ,\quad
+ V(x):=U-Ex
+ $$
+\end{itemize}
+\vfill
+\eject
+
+\title{Solution in Laplace space}
+\begin{itemize}
+ \item For simplicity, $R_0\equiv T_0\equiv0$.
+ \item Solution:
+ $$
+ \hat\psi_p(x)=
+ \left\{\begin{array}{>\displaystyle l@{\ }l}
+ C_1(p)e^{\sqrt{-ip}x}-\frac{ie^{ikx}}{-ip+k^2}
+ &\mathrm{if\ }x<0\\[0.5cm]
+ C_2(p)\varphi_p(x)
+ &\mathrm{if\ }x> 0
+ \end{array}\right.
+ $$
+ with
+ $$
+ (-\Delta+V(x)-ip)\varphi_p(x)=0
+ $$
+ $$
+ \varphi_p(x)=\mathrm{Ai}\left(e^{-\frac{i\pi}3}\left(E^{\frac13}x-E^{-\frac23}(U-ip)\right)\right)
+ $$
+\end{itemize}
+\vfill
+\eject
+
+\title{Solution in Laplace space}
+\begin{itemize}
+ \item $C_1$ and $C_2$ ensure that $\hat\psi_p(x)$ and $\partial\hat\psi_p(x)$ are continuous at $x=0$:
+ $$
+ C_1(p)=\frac{i(ik\varphi_p(0)-\partial\varphi_p(0))}{(-ip+k^2)(\sqrt{-ip}\varphi_p(0)-\partial\varphi_p(0))}
+ $$
+ $$
+ C_2(p)=-\frac{i}{(\sqrt{-ip}+ik)(\sqrt{-ip}\varphi_p(0)-\partial\varphi_p(0))}.
+ $$
+\end{itemize}
+\vfill
+\eject
+
+\title{Poles in Laplace plane}
+\vfill
+\hfil\includegraphics[height=5.5cm]{contour.pdf}
+\vfill
+\eject
+
+\title{Asymptotic behavior}
+\begin{itemize}
+ \item As $t\to\infty$:
+ $$
+ \psi(x,t)
+ =\psi_{\mathrm{FN}}(x,t)+\left(\frac{t}{\tau_E(x)}\right)^{-\frac32}+O(t^{-\frac52})
+ .
+ $$
+
+ \item If $k<0$ (reflected wave), then there is no pole on the imaginary axis, so there is no contribution as $t\to\infty$.
+ \item Similarly, the transmitted wave in the initial condition does not contribute.
+\end{itemize}
+\vfill
+\eject
+
+\title{Laser field}
+\begin{itemize}
+ \item Time dependent potential:
+ $$
+ V_t(x)=\Theta(x)(U-\epsilon\omega\cos(\omega t)x)
+ $$
+ \item Magnetic gauge:
+ $$
+ \Psi(x,t)
+ :=\psi(x,t)e^{-ix\Theta(x)A(t)}
+ ,\quad
+ A(t):=\int_0^t ds\ \epsilon\omega\cos(\omega s)
+ =
+ \epsilon\sin(\omega t)
+ $$
+ satisfies
+ $$
+ i\partial_t\Psi(x,t)=\left((-i\nabla+\Theta(x)A(t))^2+\Theta(x)U\right)\Psi(x,t)
+ $$
+\end{itemize}
+\vfill
+\eject
+
+\title{Periodic solution}
+\begin{itemize}
+ \item A solution:
+ $$
+ \Psi(x,t)=\left\{\begin{array}{ll}
+ \Psi_I(x,t)+\Psi_R(x,t)&\mathrm{\ if\ }x<0\\
+ \Psi_T(x,t)&\mathrm{\ if\ }x>0
+ \end{array}\right.
+ $$
+ $$
+ \Psi_I(x,t)=e^{ikx}\exp\left(-ik^2t\right)
+ $$
+ $$
+ \Psi_R(x,t)=\sum_{M\in\mathbb Z}R_Me^{iq_Mx}\exp\left(-iq_M^2t\right)
+ $$
+ $$
+ \Psi_T(x,t)=\sum_{M\in\mathbb Z}T_Me^{ip_Mx}\exp\left(-iUt-i\int_0^td\tau\ (p_M+A(\tau))^2\right)
+ $$
+\end{itemize}
+\vfill
+\eject
+
+\title{Periodic solution}
+\begin{itemize}
+ \item Choose $q_M$ and $p_M$ to make the solution periodic (up to the phase $e^{ik^2t}$):
+ $$
+ q_M=\pm\sqrt{k^2+M\omega}
+ ,\quad
+ p_M=\pm\sqrt{k^2-U+M\omega-U_V}
+ $$
+ and
+ $$
+ U_V:=\frac\omega{2\pi}\int_0^{\frac{2\pi}\omega} d\tau\ A^2(\tau)
+ =\frac{\epsilon^2}2
+ .
+ $$
+\end{itemize}
+
+
+\end{document}