12 - Electrostatic

Coulomb's Law

Let $e_1$ be a point charge at position $\mathbf{x}$
Let $e_2$ be a point charge at origin inducing conservative electrostatic force on $e_1$

$$\mathbf{F}=\frac{1}{4\pi {ϵ}_0}\frac{e_1{e}_2}{r^2}\frac{\mathbf{x}}{r}$$

where $r=|\mathbf{x}|$ and ${ϵ}_0$ is permittivity of free space

Coulomb Potential

Coulomb Potential from $\mathbf{F}=-\nabla V$ then

$$V(\mathbf{x})=V(r)=\frac{1}{4\pi {ϵ}_0}\frac{e_1{e}_2}{r}$$

Potential of Electron from Nucleus

Consider electron of atom with atomic number $Z$ then

$$V(r)=-\frac{Z{e}^2}{4\pi {ϵ}_{0}r}$$

Separation of Variables for a Central Potential

Consider Spherical Polar Coordinates $(r,\theta ,\varphi )$ so
Use Separation of Variables then

$$\psi =R(r)Y(\theta ,\varphi )$$

Then

$${\nabla }^2\psi =\frac{1}{r}\frac{d^2}{d{r}^2}[rR(r)]Y(\theta ,\varphi )+R(r)\frac{1}{r^2}\left(\frac{{\partial }^2}{\partial {\theta }^2}+\cot \theta \frac{\partial }{\partial \theta }+\frac{1}{{\sin }^2\theta }\frac{{\partial }^2}{\partial {\varphi }^2}\right)Y(\theta ,\varphi )$$

Substituting into Three-Dimensional Stationary State Schrödinger Equation

$$-\frac{{\hslash }^2}{2M}{\nabla }^2\psi +V(r)\psi =E\psi$$

where $M$ is the mass of electron and potential $V(r)$ then

$$\frac{1}{Y}\left(\frac{{\partial }^2}{\partial {\theta }^2}+\cot \theta \frac{\partial }{\partial \theta }+\frac{1}{{\sin }^2\theta }\frac{{\partial }^2}{\partial {\varphi }^2}\right)Y=\frac{2M{r}^2}{{\hslash }^2}(V(r)-E)-\frac{r(rR{)}^{\prime \prime }}{R}$$

As each side is independent of the other suppose it is equal to constant $-\lambda$ then

$$\left(\frac{{\partial }^2}{\partial {\theta }^2}+\cot \theta \frac{\partial }{\partial \theta }+\frac{1}{{\sin }^2\theta }\frac{{\partial }^2}{\partial {\varphi }^2}\right)Y(\theta ,\varphi )=-\lambda Y(\theta ,\varphi )$$

which has solutions of spherical harmonic

$$Y=Y_{l,m}(\theta ,\varphi )\text{ with eigenvalues }\lambda =l(l+1)$$

and for each $l$ then $m=-l,-l+1,\cdots ,l$

For right hand side suppose it equals $-\lambda =-l(l+1)$ so

$$\frac{1}{r}(rR{)}^{\prime \prime }-\frac{l(l+1)}{r^2}R-\frac{2M}{{\hslash }^2}V(r)R=-\frac{2ME}{{\hslash }^2}R$$

Bohr Radius

$$a\equiv \frac{4\pi {ϵ}_0{\hslash }^2}{M{e}^2}\simeq 5.2910^{-11}\mathrm{m}$$

where $M=m_{e_{-}}$ is mass of electron