The Schrödinger equation can be used to describe an electron in a Hydrogen like atom. A Hydrogen like atom is a nucleus with a single electron. The nucleus can be Hydrogen or a heavier element such as Helium or Lithium that has been ionised to have just one electron. The electron can also be replaced by the heavier Muon. The electron has a charge \(-e\), the nucleus has a charge \(+Ze\), where \(Z\) is the number of protons.
The atom needs to be represented in spherical polar coodinates centred of the centre of mass of the electon and the nucleus. Given that the mass of the electron is \(m_e\) and the mass of the nucleus is \(m_n\), then we need to use the reduced mass of the elecron \(\mu_e\) and the reduced mass of the nucleus \(\mu_n\).
\[\mu_e = \frac{m_e}{m_n + m_e} \quad \mu_n = \frac{m_n}{m_n + m_e}\]
The laplacian operator needs to be in spherical polar coordinates.
\[\nabla^2 = \frac{\partial^2}{\partial r^2} + \frac{2}{r}\frac{\partial}{\partial r} + \frac{1}{r^2}\left[\frac{\partial^2}{\partial \theta^2} + \cot\theta\frac{\partial}{\partial\theta} + cosec^2\theta\frac{\partial^2}{\partial \phi^2}\right]\]
The potential is only a function of the radius \(r\) from the centre of mass.
\[V(r) = -\frac{Ze^2}{4\pi\epsilon_0 r}\]
The time independent Schrödinger equation for the electron is:
\[-\frac{\hbar^2}{2\mu_e}\nabla^2\psi(r, \theta, \phi) + V(r)\psi(r, \theta, \phi) = E\psi(r, \theta, \phi)\]
Separate the radial and rotation components of the wave function \(\psi(r, \theta, \phi) = R(r)Y(\theta, \phi)\).
\[-\frac{\hbar^2}{2\mu_e}\left[\frac{d^2 R}{dr^2}Y + \frac{2}{r}\frac{dR}{dr}Y + \frac{R}{r^2}\left[\frac{\partial^2Y}{\partial \theta^2} + \cot\theta\frac{\partial Y}{\partial\theta} + cosec^2\theta\frac{\partial^2 Y}{\partial \phi^2}\right]\right] - \frac{Ze^2}{4\pi\epsilon_0 r}RY = ERY\]
Multiply by \(-\frac{2\mu_e r^2}{\hbar^2 RY}\) and rearrange. As the left hand side is a function of \(r\) and the right hand side is a function of \(\theta\) and \(\phi\), then the two sides must be equal to a separation constant that we will call \(l(l + 1)\).
\[\frac{r^2}{R}\left[\frac{d^2 R}{dr^2} + \frac{2}{r}\frac{dR}{dr}\right] + \frac{2\mu_e r^2}{\hbar^2}\left[E + \frac{Ze^2}{4\pi\epsilon_0 r}\right] = -\frac{1}{Y}\left[\frac{\partial^2Y}{\partial \theta^2} + \cot\theta\frac{\partial Y}{\partial\theta} + cosec^2\theta\frac{\partial^2 Y}{\partial \phi^2}\right] = l(l + 1)\]
Take the left hand side and multiply by \(R/r^2\).
\[\frac{d^2 R}{dr^2} + \frac{2}{r}\frac{dR}{dr} + \frac{2\mu_e R}{\hbar^2}\left[E + \frac{Ze^2}{4\pi\epsilon_0 r}\right] - \frac{l(l + 1)R}{r^2} = 0\]
Take the right hand side and multiple by \(-Y\).
\[\frac{\partial^2Y}{\partial \theta^2} + \cot\theta\frac{\partial Y}{\partial\theta} + cosec^2\theta\frac{\partial^2 Y}{\partial \phi^2} + Yl(l + 1) = 0\]
Use separate of variables to solve for the rotational components \(Y(\theta, phi) = \Theta(\theta)\Phi(\phi)\).
\[\Phi\frac{d^2\Theta}{d\theta^2} + \Phi\cot\theta\frac{d\Theta}{d\theta} + \Theta cosec^2\theta\frac{d^2\Phi}{d\phi^2} + \Theta\Phi l(l + 1) = 0\]
Multiply by \(\sin^2\theta/\Theta\Phi\) and separate. As the two sides are independent, they are equal to a separation constant \(m^2\).
\[\frac{\sin\theta}{\Theta}\left(\sin\theta\frac{d^2\Theta}{d\theta^2} + \cos\theta\frac{d\Theta}{d\theta} - l( l + 1)\sin\theta\right) = -\frac{1}{\Phi}\frac{d^2\Phi}{d\phi^2} = m^2\]
Multiple the right hand side by \(-\Phi\).
\[\frac{d^2\Phi}{d\phi^2} + m^2\Phi = 0\]
The boundary condition is that the \(\Phi\) function must be single valued \(\Phi(phi + 2\pi) = \Phi(\phi)\).
The differential equation can be solved by substituting \(\Phi = e^{k\phi}\).
\[\frac{d\Phi}{d\phi} = ke^{k\phi} \quad \frac{d^2\Phi}{d\phi^2} = k^2e^{k\phi}\]
Substitute.
\[k^2 + m^2 = 0 \quad k = \pm im\]