\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} $$ V(x)=U\Theta(x) ,\quad E_{\mathrm F}=k_{\mathrm F}^2U $$ \hfil\includegraphics[height=5cm]{potential_square_thermal.pdf} \vfill \eject \title{Photonic emission} $$ V_t(x)=\Theta(x)(U-E_tx) ,\quad E_t=2\epsilon\omega\cos(\omega t) $$ \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: $$ \hat\psi_p(x)= \left\{\begin{array}{>\displaystyle l@{\ }l} c(p)e^{\sqrt{-ip}x}-\frac{ie^{ikx}}{-ip+k^2} &\mathrm{if\ }x<0\\[0.5cm] d(p)\varphi_p(x) &\mathrm{if\ }x> 0 \end{array}\right. $$ 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 $c$ and $d$ ensure that $\hat\psi_p(x)$ and $\partial\hat\psi_p(x)$ are continuous at $x=0$: $$ c(p)=\frac{i(ik\varphi_p(0)-\partial\varphi_p(0))}{(-ip+k^2)(\sqrt{-ip}\varphi_p(0)-\partial\varphi_p(0))} $$ $$ d(p)=-\frac{i}{(\sqrt{-ip}+ik)(\sqrt{-ip}\varphi_p(0)-\partial\varphi_p(0))}. $$ \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: $$ V_t(x)=\Theta(x)(U-2\epsilon\omega\cos(\omega t)x) $$ \item Magnetic gauge: $$ \Psi(x,t) :=\psi(x,t)e^{-ix\Theta(x)A(t)} ,\quad A(t):=\int_0^t ds\ 2\epsilon\omega\cos(\omega s) = 2\epsilon\sin(\omega t) $$ 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 $e^{ik^2t}$): $$ q_M=\sqrt{k^2+M\omega} ,\quad p_M=\pm\sqrt{k^2-U+M\omega-U_V} $$ and $$ U_V:=\frac\omega{2\pi}\int_0^{\frac{2\pi}\omega} d\tau\ A^2(\tau) =2\epsilon^2 . $$ \item The coefficients $R_M$ and $T_M$ are chosen such that $$ \Psi(x,t) ,\quad (-i\nabla+\Theta(x)A(t))\Psi(x,t) $$ are continuous at $x=0$. \end{itemize} \vfill \eject \title{Initial value problem} \begin{itemize} \item In Laplace space: $$ \hat\Psi_p(x):=\int_0^\infty dt\ e^{-pt}\Psi(x,t) $$ the equation is discrete: $$ \mathfrak f_n^{(\sigma)}(x):=\hat\Psi_{-ik^2-i\sigma-in\omega}(x) ,\quad \mathcal Re(\sigma)\in[{\textstyle-\frac\omega 2,\frac\omega 2}) $$ $$ \begin{array}{r} \left(-\Delta-k^2-\sigma-n\omega+\Theta(x)\left(U+2\epsilon^2\right)\right)\mathfrak f_n^{(\sigma)}(x) -\Theta(x)2\epsilon\nabla(\mathfrak f_{n+1}^{(\sigma)}(x)-\mathfrak f_{n-1}^{(\sigma)}(x)) \\[0.5cm] -\Theta(x)\epsilon^2(\mathfrak f_{n+2}^{(\sigma)}(x)+\mathfrak f_{n-2}^{(\sigma)}(x)) =-i\psi(x,0) \end{array} $$ \end{itemize} \vfill \eject \title{Initial value problem} \begin{itemize} \item This system of ODEs is {\it integrable} for $x<0$ and $x>0$, so we have closed form expressions for a family of solutions $\mathfrak f_n^{(\sigma)}(x)$, parametrized by two sequences $c_n^{(\sigma)}$ and $d_n^{(\sigma)}$: $$ \mathfrak f_n^{(\sigma)}(x)= \left\{\begin{array}{>\displaystyle ll} c_n^{(\sigma)}e^{-ix\sqrt{k^2+\sigma+n\omega}}+\frac{ie^{ikx}}{\sigma+n\omega} &,\ x<0 \\[0.5cm] \frac\omega{2\pi}\sum_{m\in\mathbb Z} d_m^{(\sigma)}e^{-\kappa_m^{(\sigma)}x}\int_0^{\frac{2\pi}\omega}dt\ e^{-i(n-m)\omega t}e^{\frac{i\epsilon^2}\omega\sin(2\omega t)+\kappa_m^{(\sigma)}\frac{4\epsilon}\omega\cos(\omega t)} &,\ x>0 \end{array}\right. $$ with $$ \kappa_m^{(\sigma)}:=\sqrt{U+2\epsilon^2-k^2-\sigma-m\omega} $$ \end{itemize} \vfill \eject \title{Initial value problem} \begin{itemize} \item The sequences $c_n$ and $d_n$ are determined by the continuity condition at $x=0$: $$ \sum_{m\in\mathbb Z}G_{n,m}^{(\sigma)}d_m^{(\sigma)}=v_n^{(\sigma)} ,\quad c_n^{(\sigma)}=\sum_{m\in\mathbb Z}H_{n,m}^{(\sigma)}d_m^{(\sigma)}+w_n^{(\sigma)} . $$ \item The long-time behavior of $\Psi$ depends on the singularities of $\hat\Psi_p$ with $p\in i\mathbb R$. \end{itemize} \vfill \eject \title{Work in progress} \begin{itemize} \item If $G^{(\sigma)}$ is invertible, then the imaginary poles of $\hat\Psi$ are at $-i(k^2+\omega\mathbb Z)$, and $\Psi(x,t)$ converges to the periodic solution as $t\to\infty$. \item We have made progress in proving this using a RAGE-like theorem and studying the spectrum of the Floquet operator. \item Numerical challenge: good approximation of the inverse of $G^{(\sigma)}$. \end{itemize} \end{document}