Coulomb Potential

Energy Eigenvalue Equation

For an atom with one electron and one proton, the potential energy of the electron is $$V(r) = - \frac{e^2}{r}$$ Let the wavefunction be $$\psi_{Elm} = \frac{1}{r}R(r)Y_{lm}(\theta,\phi)$$ By separation of variable, the radial part of the energy eigenvalue equation is $$-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial r^2}u(r) + \left(\frac{\hbar}{2m}\frac{l(l+1)}{r^2} -\frac{e^2}{r}\right)u = E u$$

Dimension Analysis

From the potential energy, we know that the dimension of \(e^2\) is energy times length $$[e^2] = E \cdot L$$ From kinetic energy, we know that the dimension of energy is $$[E] = \frac{p^2}{2m}$$ So, $$[e^2] = \frac{p^2L}{2m}$$ From \(p = \hbar k\), we know that the dimension of \(\hbar\) is $$[\hbar] = p\cdot L$$ So, we have $$r_0 = \frac{\hbar^2}{2me^2}$$ The energy ground state is $$E_0 = \frac{e^2}{r_0} = \frac{2me^4}{\hbar^2}$$

Solve for Wavefunction

Define variables that are dimensionless \begin{align} r &= r_0\rho \\ E &= - E_0\epsilon \end{align} The boundary conditions are \begin{align} \rho \to \infty, &\space u \to e^{-\sqrt{\epsilon}\rho} \\ \rho \to 0, &\space u ~ \rho^{l +1}\\ \end{align} So, $$u = \rho^{l+1}e^{-\sqrt{\epsilon}\rho}v(\rho)$$ The energy eigenvalue equation becomes $$\rho v'' + 2(1+l-\sqrt{\epsilon}\rho)v' + [1-2\sqrt{\epsilon}(l+1)]v = 0$$ The solution to this equation is $$v = \sum_j a_j \rho^j$$ where $$a_{j+1} = \frac{2\sqrt{\epsilon}(j+l+1)-1}{(j+1)(j+2l+2)}a_j$$ Since there exists a max \(j = j_{max}\) $$a_{j_{max}+1} = 0$$ So, $$\epsilon = \frac{1}{4n^2}$$ where $$n = j_{max} + l + 1$$ Thus, $$ \bbox[5px,border:2px solid #666] { E_{nlm} = -\frac{E_0}{4n^2} } $$