meanings of the excess charge

How to solve

$$\nabla^2 V_H(r) = -4 \pi \rho(r )$$ (224)

when $V_H(x,y,z=z_0)$ and $V_H(x,y,z=z_{l+1})$ are given.

Define a parameter, which means the excess charge,

$$\rho_B = \rho(r)-\sum_{G\ne 0} \rho(G)$$ (225)

Then

$$\nabla^2 ( V(r) + V_B(r) ) = -4 \pi (\rho(r) + \rho_B )$$ (226)

$$\nabla^2 V(r) = -4 \pi \rho(r )$$ (227)

$$\nabla^2 V_B(r) = -4 \pi \rho_B$$ (228)

Eq.(228) is solved via FFT, while Eq.(229) is solved analytically,

$$V_B(z) = -4 \pi \rho_B (z - z_0 )^2 + a(z-z_0) + b$$ (229)

parameter $a$ and $b$ is determined via boundary condition. The sign of $\rho _B$ determines how $V_H(z)$ drops. (See figure.1.)

Note that the condition of no s-d bias voltage is satisfied, even if there exists some excess charge,

図 1: relationship between $\rho _B$ and $V(z)$