the case of $G_\parallel \ne 0$

Let's solve $G$ with the boundary conditon, and . First the solution in the free condition is.

$$G(G_\parallel ,k) = 1/( G_\parallel ^2 + k^2 )$$ (197)

$$G(G_\parallel ,z) = \int dk e^{ikz} \frac{1}{G_\parallel ^2+ k^2}$$ (198)

$$= \left\{ \begin{array}{l} \frac{2\pi i}{2\pi} \frac{ e^{-\vert G_\... ...rallel \vert z}}{2 \vert G_\parallel \vert}, \mbox{($z<0$)} \end{array} \right.$$ (199)

Next, $G$ with the boundary conditon, and , $G$ can be solved by applying a method of images

$$G(G_\parallel ,z,z') = \frac{ e^{-G \vert z-z'\vert} + e^{-... ...- e^{-G(z+z'-2 z_0)} } { 2 G (1- e^{-2 G(z_{l+1} -z_0) } ) }$$ (200)

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

Green's theorem

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

can be read in this case, when $V_H(z_{0})$ and $V_H(z_{l+1})$ are given,

$$\left[ G(z,z') \partial_{z'} V_H(z') - V_{H}(z') \partial_{z'} G(z,z') \right]^{z_{l+1}}_{z_0}$$ (202)

$$= \int dz' \left( G(z,z') \partial_{z'}^2 V_H(z') - V_{H}(z') \partial_{z'}^2 G(z,z') \right)$$ (203)

$$= \int dz' \left( G(z,z') \left( G_\parallel ^2 V_H(z') - 4\pi \rho... ...ight) - V_{H}(z') \left( G_\parallel ^2 G(z,z') - \delta (z-z') \right) \right)$$ (204)

$$= -4 \pi \int_{z_0}^{z_{l+1}} dz' G(z,z') \rho(z') + V_H(z)$$ (205)

Thus

$$V_H(z) = 4 \pi \int_{z_0}^{z_{l+1}} dz' G(z,z') \rho(z') + \left[ G(z,z') \partial_{z'} V_H(z') - V_{H}(z') \partial_{z'} G(z,z') \right]^{z_{l+1}}_{z_0}$$ (206)

$$= 4 \pi \int_{z_0}^{z_{l+1}} dz' G(z,z') \rho(z') - \left[ V_{H}(z') \partial_{z'} G(z,z') \right]^{z'=z_{l+1}}_{z'=z_0}$$ (207)

$$V_H(G_\parallel ,z) = 4 \pi \int_{z_0}^{z_{l+1}} dz' G(G_\parallel ,z,z') \rho(G_\parallel ,z')$$

$$+ \left\{ V_H(G_\parallel ,z_{l+1}) \left( e^{-G(z_{l+1} - z) } - e^{-G( z_{l+1} -2 z_0 +z)} \right) \right.$$

$$+ \left. V_H(G_\parallel ,z_{0}) \left( e^{-G(z - z_0) } - e^{-G(2 z_{l+1} - z_0 -z)} \right) \right\} / \left( 1- e^{-2G( z_{l+1} -z_0) } \right)$$ (208)

$$= 4 \pi \int \! dk\; \frac{ \rho(G_\parallel ,k) }{G^2+k^2} e^{i k z}$$

$$+ \left\{ V_H(G_\parallel ,z_{l+1}) \left( e^{-G(z_{l+1} - z) } - e^{-G( z_{l+1} -2 z_0 +z)} \right) \right.$$

$$+ \left. V_H(G_\parallel ,z_{0}) \left( e^{-G(z - z_0) } - e^{-G(2 z_{l+1} - z_0 -z)} \right) \right\} / \left( 1- e^{-2G( z_{l+1} -z_0) } \right)$$

$$V_H(r) = 4 \pi \int \! dk\; \frac{ \rho(G_\parallel ,k) }{G^2+k^2} e^{i (G_\parallel r_\parallel + k z) }$$

$$+ \int \! dG_\parallel \; e^{i G_\parallel r_\parallel } \left\{ ... ...{l+1}) \left( e^{-G(z_{l+1} - z) } - e^{-G( z_{l+1} -2 z_0 +z)} \right) \right.$$

$$+ \left. V_H(G_\parallel ,z_{0}) \left( e^{-G(z - z_0) } - e^{-G(2 z_{l+1} - z_0 -z)} \right) \right\} / \left( 1- e^{-2G( z_{l+1} -z_0) } \right)$$ (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.