这是一份liverpool利物浦大学PHYS201的成功案例

The special case of a point charge at the origin, for which $\rho=q \delta(\mathbf{r})$ and $\mathbf{E}=q\left(\mathbf{r} / r^{3}\right)$, shows that $\nabla \cdot\left(\mathbf{r} / r^{3}\right)$ acts as if
$$
\nabla \cdot \frac{\mathbf{r}}{r^{3}}=4 \pi \delta(\mathbf{r})
$$
yields an equation for the electrostatic potential $\phi:$
$$
\nabla \cdot \mathbf{E}=-\nabla \cdot \Gamma \phi=4 \pi \rho \quad \text { or } \quad \nabla^{2} \phi=-4 \pi \rho .
$$
This is known as Poisson’s equation. In a portion of space where $\rho=0$, becomes
$$
\nabla^{2} \phi=0
$$

PHYS201 COURSE NOTES ：

$$
\psi_{0}(\mathbf{x}, 0)=e^{i \mathbf{k}{0}-\left(\mathbf{x}-\mathbf{x}{0}\right)} h\left(\mathbf{x}-\mathbf{x}{0}\right)+\text { c.c. } $$ where $$ h\left(\mathbf{x}-\mathbf{x}{0}\right)=\int d \mathbf{q} a(\mathbf{q}) e^{i \mathbf{q} \cdot\left(\mathbf{x}-\mathbf{x}_{0}\right)}
$$