\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}
