1 files changed, 7 insertions, 6 deletions
diff --git a/Costin_Costin_Jauslin_Lebowitz_2018.tex b/Costin_Costin_Jauslin_Lebowitz_2018.tex
index 84e9ece..e0ccaab 100644
--- a/Costin_Costin_Jauslin_Lebowitz_2018.tex
+++ b/Costin_Costin_Jauslin_Lebowitz_2018.tex
@@ -62,7 +62,7 @@ Solution of the time dependent Schr\"odinger equation\par
   \label{schrodinger}
   i\partial_t\psi(x,t)=(-{\textstyle\frac12}\partial_x^2+V(x))\psi(x,t)
 \end{equation}
-where
+(we write $\partial_x\equiv\frac\partial{\partial x}$) where
 \begin{equation}
   V(x)=\left\{ \begin{array}{ll} 0, & x<0\\
   U-E x, & x>0 \end{array}\right.  
@@ -91,7 +91,7 @@ in which $R_0$ and $T_0$ are the {\it reflection} and {\it transmission} coeffic
   .
   \label{R0T0}
 \end{equation}
-These constants ensure that $\psi_0(x)$ and $\partial\psi_0(x)$ are continuous at $x=0$. Note that, in this state, the current vanishes:
+These constants ensure that $\psi_0(x)$ and $\partial_x\psi_0(x)$ are continuous at $x=0$. Note that, in this state, the current vanishes:
 \begin{equation}
   j_0(x)=i(\psi_0\partial_x\psi_0^*-\psi_0^*\partial_x\psi_0)=0
   .
@@ -160,6 +160,7 @@ with
 \indent To obtain $\psi(x,t)$ we solve for $\hat\psi_p(x)$, the Laplace transform of $\psi(x,t)$,
 \begin{equation}
   \hat\psi_p(x):=\int_0^\infty dt\ e^{-pt}\psi(x,t)
+  \label{psip}
 \end{equation}
 which we obtain in closed form. We then compute, by inverting the Laplace transform, the long time asymptotics analytically, and the short time behavior numerically. This method provides an integral representation of the solution which can be evaluated numerically. It is thus better for our purposes than direct computations of the solution of\-~(\ref{schrodinger}). The latter requires cutoffs for the non-square integrable functions we are dealing with and cannot be used for long times. The Laplace transform of $\psi$ satisfies the equation
 \begin{equation}
@@ -196,7 +197,7 @@ and
   \eta_p(x)=e^{-\frac{i\pi}3}\mathrm{Ai}\left(-2^{\frac13}\left(E^{\frac13}x-E^{-\frac23}(U-ip)\right)\right)
   \label{varphi}
 \end{equation}
-are two independent solutions of $(-\frac12\partial_x^2+U-Ex-ip)f=0$. The phases $e^{-\frac{i\pi}3}$ and $-1$ are cube roots of $-1$. The constants $C_1(p)$ and $C_2(p)$ are set so that $\hat\psi_p$ and $\partial\hat\psi_p$ are continuous at $x=0$:
+are two independent solutions of $(-\frac12\partial_x^2+U-Ex-ip)f=0$. The phases $e^{-\frac{i\pi}3}$ and $-1$ are cube roots of $-1$. The constants $C_1(p)$ and $C_2(p)$ are set so that $\hat\psi_p$ and $\partial_x\hat\psi_p$ are continuous at $x=0$:
 \begin{equation}
   C_1(p)=-\frac{2iT_0}{\sqrt{-2ip}\varphi_p(0)-\partial\varphi_p(0)}\left(
     \frac{\sqrt{2U-k^2}\varphi_p(0)+\partial\varphi_p(0)}{-2ip+k^2}
@@ -256,7 +257,7 @@ where $\psi_E$ is the FN solution\-~(\ref{psiE}).
   ,\quad
   f(\alpha):=\hat\psi_{\alpha^2}(x)
 \end{equation}
-and write the integral along the branch cut as
+(recall the definition of $\hat\psi_p$ in~\-(\ref{psip})) and write the integral along the branch cut as
 \begin{equation}
   \psi^{(\mathrm{BC})}(x,t)
   :=
@@ -323,7 +324,7 @@ and integrated current (the current integrated over the supply function at 0 tem
 \begin{equation}
   J_{k_{\mathrm F}}(x,t):=\int_0^{k_{\mathrm F}}dk\ j_k(x,t)
 \end{equation}
-as a function of time at two different values of $x$: $x_0:=\frac{2U-k_{\mathrm F}^2}{2E}\approx 11\ \mathrm{nm}$ and $10x_0$ ($x_0$ is the point at which $V(x_0)=\frac{k_{\mathrm F}^2}2$), and at two different values of $E$: $4$ and $8\ \mathrm{V}\cdot\mathrm{nm}^{-1}$. We have normalized the current $j$ by $2k$, which is the current of the incoming wave $e^{ikx}$, and the integrated current $J$ by $k_{\mathrm F}^2$, which is the current of the incoming wave integrated over the supply function. We find that there is a transient regime that lasts a few femtoseconds before the system stabilizes to the FN value. There is a delay before the signal reaches $x_0$, and between $x_0$ and $10x_0$. As expected, the asymptotic value of the current is independent of $x$. Note that the current and density depend strongly on the field $E$.
+as a function of time at two different values of $x$: $x_0:=\frac{2U-k_{\mathrm F}^2}{2E}\approx 11\ \mathrm{nm}$ and $10x_0$ ($x_0$ is the point at which $V(x_0)=\frac{k_{\mathrm F}^2}2$), and at two different values of $E$: $4$ and $8\ \mathrm{V}\cdot\mathrm{nm}^{-1}$. We have normalized the current $j$ by $2k$, which is the current of the incoming wave $e^{ikx}$, and the integrated current $J$ by $k_{\mathrm F}^2$, which is the current of the incoming wave integrated over the supply function. We find that there is a transient regime that lasts a few femtoseconds before the system stabilizes to the FN value. Note that the approach to the FN regime has some ripples, which come from the imaginary parts of the poles in the $p$-plane (see Fig.\-~\ref{fig:contour}). There is a delay before the signal reaches $x_0$, and between $x_0$ and $10x_0$. As expected, the asymptotic value of the current is independent of $x$. Note that the current and density depend strongly on the field $E$.
 \bigskip
 
 \begin{figure}
@@ -371,7 +372,7 @@ in which $\varphi_p$ is the solution of
 When $p=-ik^2/2$, (\ref{eqvarphi}) coincides with the equation\-~(\ref{eqpsiE}) for $\psi_E$. Assuming that $R$ and $T$ do not introduce any new poles on the imaginary axis (as we showed is the case), this implies that $\psi(x,t)$ converges to $e^{-i\frac12k^2t}\psi_E(x)$ as $t\to\infty$.
 \bigskip
 
-\point{\bf Potentials.} The exact form of the potential $V(x)=U-Ex$ was not really used in much of the computation above, so it can be carried out in very much the same way for many other $V(x)$. For instance, one could round off the triangular barrier as occurs in the Schottky effect\-~\cite{Fo08}. We could also consider a square barrier. The only real constraint on the potential is that it not introduce bound states. To make this into a precise statement, one would also have to put constraints on the regularity and asymptotic properties of $V$, which we will not do here.
+\point{\bf Potentials.} The exact form of the potential $V(x)=U-Ex$ was not really used in much of the computation above, so it can be carried out in very much the same way for many other $V(x)$. For instance, one could round off the triangular barrier as occurs in the Schottky effect\-~\cite{Fo08}. We could also consider a square barrier. The only real constraint on the potential is that it does not introduce bound states. To make this into a precise statement, one would also have to put constraints on the regularity and asymptotic properties of $V$, which we will not do here.
 \bigskip
 
 This leaves open the possibility of studying trains of pulses, in which the field is turned on and off repeatedly. The regime in which the field is off corresponds to a potential $V(x)=U\Theta(x)$, which can be studied using the method described above. Provided the time between the pulses is long enough, the system would stabilize to the stationary state in the time between each field switching.