next up previous
次へ: the case of 上へ: 戻る:

the case of $G_\parallel \ne 0$

Let's solve $G$ with the boundary conditon, \( G(z=z_0, z')=0 \) and \( G(z=z_{l+1}, z')=0 \). First the solution in the free condition is.
$\displaystyle G(G_\parallel ,k)$ $\textstyle =$ $\displaystyle 1/( G_\parallel ^2 + k^2 )$ (197)
$\displaystyle G(G_\parallel ,z)$ $\textstyle =$ $\displaystyle \int dk e^{ikz} \frac{1}{G_\parallel ^2+ k^2}$ (198)
  $\textstyle =$ $\displaystyle \left\{ \begin{array}{l}
\frac{2\pi i}{2\pi} \frac{ e^{-\vert G_\...
...rallel \vert z}}{2 \vert G_\parallel \vert}, \mbox{($z<0$)}
\end{array} \right.$ (199)

Next, $G$ with the boundary conditon, \( G(z=z_0, z')=0 \) and \( G(z=z_{l+1}, z')=0 \), $G$ can be solved by applying a method of images
\begin{displaymath}
G(G_\parallel ,z,z') = \frac{
e^{-G \vert z-z'\vert} + e^{-...
...- e^{-G(z+z'-2 z_0)} }
{ 2 G (1- e^{-2 G(z_{l+1} -z_0) } ) }
\end{displaymath} (200)

where $G=\vert G_\parallel \vert$.

Green's theorem

$\displaystyle \int_S dS' n \cdot (u \nabla v - v\nabla u ) =
\int_V dV' (u \triangle v - v \triangle u )$     (201)

can be read in this case, when $V_H(z_{0})$ and $V_H(z_{l+1})$ are given,
    $\displaystyle \left[ G(z,z') \partial_{z'} V_H(z')
- V_{H}(z') \partial_{z'} G(z,z') \right]^{z_{l+1}}_{z_0}$ (202)
  $\textstyle =$ $\displaystyle \int dz' \left( G(z,z') \partial_{z'}^2 V_H(z')
- V_{H}(z') \partial_{z'}^2 G(z,z')
\right)$ (203)
  $\textstyle =$ $\displaystyle \int dz' \left( G(z,z') \left( G_\parallel ^2 V_H(z') - 4\pi \rho...
...ight)
- V_{H}(z') \left( G_\parallel ^2 G(z,z') - \delta (z-z') \right)
\right)$ (204)
  $\textstyle =$ $\displaystyle -4 \pi \int_{z_0}^{z_{l+1}} dz' G(z,z') \rho(z') + V_H(z)$ (205)

Thus

$\displaystyle V_H(z)$ $\textstyle =$ $\displaystyle 4 \pi \int_{z_0}^{z_{l+1}} dz' G(z,z') \rho(z') +
\left[ G(z,z') \partial_{z'} V_H(z')
- V_{H}(z') \partial_{z'} G(z,z') \right]^{z_{l+1}}_{z_0}$ (206)
  $\textstyle =$ $\displaystyle 4 \pi \int_{z_0}^{z_{l+1}} dz' G(z,z') \rho(z') -
\left[
V_{H}(z') \partial_{z'} G(z,z') \right]^{z'=z_{l+1}}_{z'=z_0}$ (207)
$\displaystyle V_H(G_\parallel ,z)$ $\textstyle =$ $\displaystyle 4 \pi \int_{z_0}^{z_{l+1}} dz' G(G_\parallel ,z,z') \rho(G_\parallel ,z')$  
    $\displaystyle +
\left\{
V_H(G_\parallel ,z_{l+1}) \left(
e^{-G(z_{l+1} - z) } - e^{-G( z_{l+1} -2 z_0 +z)} \right) \right.$  
    $\displaystyle +
\left.
V_H(G_\parallel ,z_{0}) \left(
e^{-G(z - z_0) } - e^{-G(2 z_{l+1} - z_0 -z)} \right)
\right\} /
\left(
1- e^{-2G( z_{l+1} -z_0) } \right)$ (208)
  $\textstyle =$ $\displaystyle 4 \pi \int \! dk\; \frac{ \rho(G_\parallel ,k) }{G^2+k^2} e^{i k z}$  
    $\displaystyle +
\left\{
V_H(G_\parallel ,z_{l+1}) \left(
e^{-G(z_{l+1} - z) } - e^{-G( z_{l+1} -2 z_0 +z)} \right) \right.$  
    $\displaystyle +
\left.
V_H(G_\parallel ,z_{0}) \left(
e^{-G(z - z_0) } - e^{-G(2 z_{l+1} - z_0 -z)} \right)
\right\} /
\left(
1- e^{-2G( z_{l+1} -z_0) } \right)$  
$\displaystyle V_H(r)$ $\textstyle =$ $\displaystyle 4 \pi \int \! dk\; \frac{ \rho(G_\parallel ,k) }{G^2+k^2} e^{i (G_\parallel r_\parallel + k z) }$  
    $\displaystyle + \int \! dG_\parallel \; e^{i G_\parallel r_\parallel }
\left\{
...
...{l+1}) \left(
e^{-G(z_{l+1} - z) } - e^{-G( z_{l+1} -2 z_0 +z)} \right) \right.$  
    $\displaystyle +
\left.
V_H(G_\parallel ,z_{0}) \left(
e^{-G(z - z_0) } - e^{-G(2 z_{l+1} - z_0 -z)} \right)
\right\} /
\left(
1- e^{-2G( z_{l+1} -z_0) } \right)$ (209)

Note that $\rho(G,z_0)=0$ and $\rho(G,z_{l+1})=0$. Thus the Fourier transform of $\rho(G,z)$ is executed in the region from $z_1$ to $z_l$.

$V_H(G_\parallel \ne 0,z_{0})=0$ and $V_H(G_\parallel \ne 0,z_{l+1})=0$ when the left and right electrodes are jellium.


next up previous
次へ: the case of 上へ: 戻る:
kino 平成18年4月17日