User:Batamtig/Sandbox/Potential

Potential

We're working with two materials, with dielectric constants $$\epsilon_1$$, $$\epsilon_2$$. We will always use the convention that $$\epsilon_1 > \epsilon_2$$.

We break the potential into several parts:


 * $$V=V_{step} + V_C + V_i,

$$ where $$V_{step}$$ confines the electron to material 1,


 * $$V_{step}(r)=\left\{ \begin{array}{ll}0 & z \ge 0\\

V_0 & z < 0,\\ \end{array}\right.$$ $$V_C$$ is the Coulombic potential from the nucleus (see Nakayama & Kayanuma),
 * $$V_C(r)=-\frac{2}{\epsilon_1 + \epsilon_2}\frac{e^2}{r}

=-\frac{1}{\epsilon_{ave}}\frac{e^2}{r} ,$$ and $$V_i$$ is the electron image potential,
 * $$V_i(r)=\left\{\begin{array}{ll}

\frac{1}{\epsilon_1} \frac{\epsilon_1-\epsilon_2}{\epsilon_1+\epsilon_2} \frac{e^2}{2z} =\frac{1}{\epsilon_1} \frac{\Delta\epsilon}{\epsilon_{ave}} \frac{e^2}{4z} & z \ge 0\\ \ & \ \\                    \frac{1}{\epsilon_2} \frac{\epsilon_1-\epsilon_2}{\epsilon_1+\epsilon_2} \frac{e^2}{2z} =\frac{1}{\epsilon_2} \frac{\Delta\epsilon}{\epsilon_{ave}} \frac{e^2}{4z} & z < 0,\\ \end{array}\right.$$

Cut-off

We set a cut-off to fix the asymptotic value of the image potential as $$z\rightarrow 0$$, since the image potential behavior is macroscopically derived and not expected to continue at small values of $$z$$. Thus we assume that $$V_i$$ is bounded by constant values,
 * $$V_i(r)=\left\{\begin{array}{ll}

=\frac{1}{\epsilon_1} \frac{\Delta\epsilon}{\epsilon_{ave}} \frac{e^2}{4\delta} & \delta > z \ge 0\\ \ & \ \\                   =\frac{1}{\epsilon_2} \frac{\Delta\epsilon}{\epsilon_{ave}} \frac{e^2}{4\delta} & -\delta < z < 0,\\ \end{array}\right.$$