Xavier Andrade edited photoemission2.tex  over 9 years ago

Commit id: fced61b9a8d20a96beb15ac3ed9f7c4bcd2b6dee

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\end{align}  and the additional set of equations,  \begin{align}\label{eq:FMM_prop_aux}  \begin{array}{l}  \varphi^A_i(\vec{r},t+\Delta t) t)&  = M \hat{U}(\Delta t) \phi^A_i(\vec{r},t)\ ,\\ \varphi^B_i(\vec{r},t+\Delta t) t)&  = \frac{M}{(2\pi)^{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) t)&  = \frac1{(2\pi)^{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) t)&  = \hat{U}_{\rm v}(\Delta t) \phi^B_i(\vec{k},t) - \\  &-  \frac1{(2\pi)^{3/2}} \int {\rm d}\vec{r} e^{-\mathrm{i}\vec{k}\cdot\vec{r}} \varphi^B_i(\vec{r},t+\Delta t)\ .  \end{array}  \end{align}  The momentum-resolved photoelectron probability is then obtained directly from   the momentum components as~\cite{DeGiovannini_2012}