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A solution of this problem was already given by Lindhard [50]. The energy loss $ \frac{dE}{d\vec{r}}$ of the projectile is determined by the force resulting from the electric field $ \vec{E}(\vec{r},t)$ which is generated in the plasma by the penetrating charged particle with a charge $ Z_1$ moving with the velocity $ \vec{v}$. For his solution Lindhard applied the linear response theory.

$\displaystyle \frac{dE}{dR} = \frac{\vec{v}}{\vert\vec{v}\vert}\cdot \vec{E}(\vec{r},t)\cdot \rho_0(\vec{r},t)$ (3.100)

$\displaystyle \rho_0(\vec{r},t) = Z_1\cdot e\cdot \delta(\vec{r}-\vec{v}\cdot t)$ (3.101)

$\displaystyle \mathrm{div} \left(\epsilon(\vec{r},t)\cdot \vec{E}(\vec{r},t)\right) = 4\cdot \pi\cdot \rho_0(\vec{r},t)$ (3.102)

$ \epsilon(\vec{r},t)$ is the linear dielectric response function of the medium on the charge.

This set of differential equations can be solved by Fourier transformation [50] ( $ v = \vert\vec{v}\vert$).

$\displaystyle \frac{dE_l}{dR} = \frac{Z_1^2\cdot e^2}{\pi\cdot v}\cdot Im\left\...\limits_{kv}^{-kv} d\omega\cdot \frac{\omega}{\epsilon_l(k,\omega)} \right\}$ (3.103)

$\displaystyle \frac{dE_{tra}}{dR} = \frac{Z_1^2\cdot e^2}{\pi\cdot v}\cdot Im\l...
...k^2\cdot v^2}}{k^2-\frac{\omega^2}{c^2}\cdot \epsilon_{tra}(k,\omega)} \right\}$ (3.104)

The contributions to the energy loss by the transversal electric field $ \vec{E}_{tra}$ and the longitudinal electric field $ \vec{E}_l$ are treated separately, which requires the definition of a transversal $ \epsilon_{tra}(k,\omega)$ (Fourier transform of $ \epsilon_{tra}(\vec{r},t)$) and a longitudinal $ \epsilon_l(k,\omega)$ (Fourier transform of $ \epsilon_l(\vec{r},t)$) dielectric response function. $ \epsilon_{tra}(k,\omega)$ and $ \epsilon_l(k,\omega)$ are defined in Fourier space via the scalar potential $ \Phi(k,\omega)$ and vector potential $ \vec{A}(k,\omega)$, describing the electromagnetic field as shown in (3.105) and (3.106), and the Maxwell equations.

$\displaystyle \vec{E} = - \mathrm{grad}\Phi(\vec{r},t) - \frac{1}{c}\cdot \frac{\partial \vec{A}(\vec{r},t)}{\partial t}$ (3.105)

$\displaystyle \vec{B} = \mathrm{rot}\vec{A}(\vec{r},t)$ (3.106)

$ \epsilon_l(k,\omega)$ is defined by

$\displaystyle \epsilon_l(k,\omega)\cdot k^2 \cdot \Phi(k,\omega) = 4\cdot \pi\cdot \rho_0(k,\omega)$ (3.107)

while $ \epsilon_{tra}(k,\omega)$ is defined by

$\displaystyle (k^2-\frac{\omega^2}{c^2}\cdot \epsilon_{tra}(k,\omega))\cdot \vec{A}_{tra}(k,\omega) = \frac{4\cdot \pi}{c}\cdot j_{0tra}(k,\omega)$ (3.108)

The transversal field and the transversal current density are characterized by

$\displaystyle \vec{k}\cdot \vec{A}_{tra}(k,\omega) = 0$ (3.109)

$\displaystyle \vec{k}\cdot \vec{j}_{0tra}(k,\omega) = 0$ (3.110)

The total energy loss is a sum of the longitudinal and the transversal contribution, respectively. The transversal contribution is negligible since the particle velocity is non-relativistic [50]. This requirement holds for typical ion implantation conditions.

Lindhard derived an expression for the longitudinal dielectric response function $ \epsilon_l(k,\omega)$ by applying a self-consistent treatment of the electron gas. He calculates the charge density $ \rho_0$ and the current density $ j_0$ induced in the plasma by the forces resulting from the scalar potential $ \Phi$ and the vector potential $ \vec{A}$. These induced sources produce an electro-magnetic field which acts on the electron gas. Therefore the electro-magnetic field must be self-consistent with the induced sources. A quantum-mechanical calculation of this problem results in

\begin{displaymath}\begin{split}\epsilon_l &= 1+\frac{2\cdot m^2\cdot \omega_0^2...
...\omega+i\cdot \frac{\gamma}{\hbar}\right)} \right\} \end{split}\end{displaymath} (3.111)


$\displaystyle \omega_0^2 = \frac{4\cdot \pi\cdot e^2\cdot \varrho}{m}$ (3.112)

being the classical resonance frequency of the gas. $ f(E_n)$ is the electron density distribution function, $ m$ is the electron mass, $ E_n$ is the energy, $ k_n$ the wave vector of the electron in the nth state and $ \gamma$ is a small damping factor. By applying this dielectric response function to (3.103) Lindhard derived the energy loss in an electron gas described by the Fermi distribution in the ground state with a maximum energy $ E_0 = \frac{m\cdot v_0^2}{2}$.

$\displaystyle \frac{dE}{dR} = \frac{4\cdot \pi\cdot z_1^2\cdot e^4}{m\cdot v^2}\cdot\varrho\cdot L$ (3.113)

$\displaystyle L = \frac{6}{\pi}\int\limits_0^{\frac{v}{v_0}}u\cdot \;du\cdot \i...
...z\cdot \frac{f_2(u,z)}{(z^2+\chi^2\cdot f_1(u,z))^2 + (\chi^2\cdot f_2(u,z))^2}$ (3.114)

\begin{displaymath}\begin{split}f(u',z) &= \frac{1}{2}+ \frac{1}{8\cdot z}\cdot ...
... \right\}\cdot \ln\left(\frac{z+u'+1}{z+u'-1}\right)\end{split}\end{displaymath} (3.115)

$\displaystyle u' = \frac{\omega+i\cdot \frac{\gamma}{\hbar}}{k\cdot v_0}$ (3.116)

$\displaystyle u = \frac{\vert\omega\vert}{k\cdot v_0}$ (3.117)

$\displaystyle \chi^2 = \frac{e^2}{\pi\cdot \hbar\cdot v_0}$ (3.118)

Assuming a small damping factor $ \gamma$ and a small particle velocity $ v$, (3.113) can be simplified to an electronic stopping equation which is proportional to velocity.

$\displaystyle \frac{dE}{dR} = \frac{4\cdot z_1^2\cdot e^4\cdot m^2}{3\cdot \pi\...
...cdot \int\limits_0^1\frac{z^3\cdot dz}{\left(z^2+\chi^2\cdot f_1(0,z)\right)^2}$ (3.119)

$ f_1(u,z)$ is the real part of (3.115), while $ f_2(u,z)$ is the imaginary part.

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A. Hoessiger: Simulation of Ion Implantation for ULSI Technology