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Xavier Andrade edited photoemission2.tex
over 9 years ago
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The result can be written in a closed form for $\phi^A_i(\vec{r},t)$, represented in real space, and
$\phi^B_i(\vec{k},t)$, in momentum space, with the following structure:
\begin{align}\label{eq:FMM_prop}
\left\{
\begin{array}{l}
\phi^A_i(\vec{r},t+\Delta t) =
\varphi^A_i(\vec{r},t+\Delta t) +\varphi^B_i(\vec{r},t+\Delta
t)\\ t)\ ,\\
\phi^B_i(\vec{k},t+\Delta t) =
\vartheta^A_i(\vec{k},t+\Delta t)+\vartheta^B_i(\vec{k},t+\Delta
t) t)\ ,
\end{array}
\right.\,,
\end{align}
and
with the additional set of equations,
\begin{align}\label{eq:FMM_prop_aux}
\left\{
\begin{array}{l}
\varphi^A_i(\vec{r},t+\Delta t) = M \hat{U}(\Delta t)
\phi^A_i(\vec{r},t)\\ \phi^A_i(\vec{r},t)\ ,\\
\varphi^B_i(\vec{r},t+\Delta t) =
\frac{M}{(2\pi)^{\frac{3}{2}}}\int {\rm d}\vec{k} e^{\mathrm{i}\vec{k}\cdot\vec{r}} \hat{U}_{\rm v}(\Delta t)
\phi^B_i(\vec{k},t)
\\ \ ,\\
\vartheta^A_i(\vec{k},t+\Delta t) =
\frac1{(2\pi)^{\frac{3}{2}}} \int {\rm d}\vec{r} e^{-\mathrm{i}\vec{k}\cdot\vec{r}} (1-M) \hat{U}(\Delta t)
\phi^A_i(\vec{r},t)
\\ \ ,\\
\vartheta^B_i(\vec{k},t+\Delta t) =
\hat{U}_{\rm v}(\Delta t) \phi^B_i(\vec{k},t) -
\frac1{(2\pi)^{\frac{3}{2}}} \int {\rm d}\vec{r} e^{-\mathrm{i}\vec{k}\cdot\vec{r}}
\varphi^B_i(\vec{r},t+\Delta
t) t)\ .
\end{array}
\right. .
\end{align}
The momentum-resolved photoelectron probability is then obtained directly from
the momentum components as~\cite{DeGiovannini_2012}