We begin with the time-independent Schrödinger equation:
\begin{equation}
\left( -\frac{\hbar^2}{2m_e}\Delta - \frac{e^2}{4\pi\varepsilon r} \right)\psi = E\psi
\label{eq:a1_1}
\end{equation}
Where $m_e$ is the mass of the electron, which can be found mostly around the nucleus in a cloud of probability. Since the hydrogen atom problem has a central potential $1/r$, it means it has a rotational invariance symmetry $SO(3)$, which means we're going to adapt a Laplacian $\Delta$ adequately:
\begin{equation}
\Delta f = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial f}{\partial r} \right)
+ \frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \left( \sin \theta \frac{\partial f}{\partial \theta} \right) + \frac{1}{r^2 \sin^2 \theta} \frac{\partial^2 f}{\partial \phi^2}
\label{eq:a1_2}
\end{equation}
We define the orbital angular momentum operator $\hat{L}^2$ (Casimir invariant in $\mathfrak{so}(3)$) as follows:
\begin{equation}
\hat{L}^2 f = \frac{1}{\sin \theta} \frac{\partial}{\partial \theta} \left( \sin \theta \frac{\partial f}{\partial \theta} \right) + \frac{1}{\sin^2 \theta} \frac{\partial^2 f}{\partial \phi^2}
\label{eq:a1_3}
\end{equation}
Which makes our Laplacian take the following form:
\begin{equation}
\Delta f = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial f}{\partial r} \right) + \frac{\hat{L}^2 f}{r^2}
\label{eq:a1_4}
\end{equation}
Upon applying the orbital angular momentum operator on a random state we get:
\begin{equation}
\hat{L}^2 f(\theta,\phi) = \ell(\ell+1) f(\theta,\phi)
\label{eq:a1_5}
\end{equation}
01 Radial Separation
Isolating the radial part $\psi_r$ and including the centrifugal term:
\begin{equation}
\frac{\partial^2\psi_r}{\partial r^2} + \frac{2}{r}\frac{\partial\psi_r}{\partial r} + \left[ \frac{2m_eE}{\hbar^2} - \frac{\ell(\ell+1)}{r^2} + \frac{2m_ee^2}{4\pi\varepsilon\hbar^2 r} \right]\psi_r = 0
\label{eq:a1_6}
\end{equation}
Next we try to compactify the form of the differential equation by defining some parameters, Bohr radius $a$ and the wavevector $k$ as follows:
\begin{equation}
a=\frac{m_ee^2}{4\pi\varepsilon\hbar^2} \ \ \ \ \ \ k^2=\frac{2m_eE}{\hbar^2}
\label{eq:a1_7}
\end{equation}
Which makes our differential equation look much cleaner:
\begin{equation}
\frac{\partial^2\psi_r}{\partial r^2} + \frac{2}{r}\frac{\partial\psi_r}{\partial r} + \left[ k^2 - \frac{\ell(\ell+1)}{r^2} + \frac{2}{a r} \right]\psi_r = 0
\label{eq:a1_8}
\end{equation}
02 Transformation to Whittaker Form
In order to find the Whittaker differential equation form, we need to perform a small change of variable:
\begin{equation}
\phi(r)= \frac{\psi_r(r)}{r}
\label{eq:a1_9}
\end{equation}
This will end up removing the term $\frac{2}{r}\frac{\partial\psi_r}{\partial r}$, then we multiply our resultant differential equation with $1/\beta^2$:
\begin{equation}
\frac{\partial^2\phi}{\partial (\beta r)^2} + \left[ \frac{k^2}{\beta^2} - \frac{\ell(\ell+1)}{(\beta r)^2} + \frac{2}{a \beta (\beta r)} \right]\phi = 0
\label{eq:a1_10}
\end{equation}
We define a new variable $\rho$:
\begin{equation}
\rho = \beta r
\label{eq:a1_11}
\end{equation}
Which allows us to write the differential equation as follows:
\begin{equation}
\frac{\partial^2\phi}{\partial \rho^2} + \left[ \frac{k^2}{\beta^2} + \frac{2/a\beta}{\rho} - \frac{\ell(\ell+1)}{\rho^2} \right]\phi = 0
\label{eq:a1_12}
\end{equation}
By setting the constants to be equal to the following:
\begin{equation}
\frac{k^2}{\beta^2}=-\frac{1}{4} \ \ \ \ \ \ \frac{2}{a\beta}=n
\label{eq:a1_13}
\end{equation}
The equation transforms into the Whittaker form:
\begin{equation}
\frac{\partial^2\phi}{\partial \rho^2} + \left[ -\frac{1}{4} + \frac{n}{\rho} - \frac{\ell(\ell+1)}{\rho^2} \right]\phi = 0
\label{eq:a1_14}
\end{equation}
This means that we uncovered what $\beta$ equals, meaning that the energy is pretty much found:
\begin{equation}
\beta=\frac{2}{na} \implies k^2=-\frac{1}{4}\beta^2 \implies \frac{2m_eE}{\hbar^2}=-\frac{1}{n^2a^2} \implies E_n=-\frac{\hbar^2}{2m_ea^2n^2}
\label{eq:a1_15}
\end{equation}
Which is the correct answer for the energy eigenvalues of the hydrogen atom.
03 The Total Normalized Wavefunction
Using the Whittaker function $M_{n, \ell+1/2}$, we define the radial normalization constant $A$:
\begin{equation}
A = \sqrt{ \frac{2}{na \int_0^{+\infty} M_{n, \ell+1/2}^2(\rho) d\rho} }
\label{eq:a1_16}
\end{equation}
The total wavefunction $\psi_{n,l,m,\alpha}$ including the spherical harmonics and the Pauli spinor $\alpha$ is then:
\begin{equation}
\psi_{n,\ell,m,\alpha}(\mathbf{r}) = \frac{1}{r} A \cdot M_{n, \ell+1/2}\left( \frac{2r}{na} \right) Y_{\ell m}(\theta, \phi) \otimes \alpha
\label{eq:a1_17}
\end{equation}
Where the unmixed states projected to the spherical coordinates of $S^2$ are defined as follows:
\begin{equation}
\langle \theta, \phi | [|\ell m_\ell \rangle \otimes | s m_s\rangle] = Y_{\ell m}(\theta, \phi) \otimes \alpha
\label{eq:a1_18}
\end{equation}
So the goal is to find the entangled states, which are the result of this tensor product $|\ell m_\ell \rangle \otimes | s m_s\rangle$, they can be found, with the help of some tables, as follows:
\begin{equation}
|\ell m_\ell \rangle \otimes | s m_s\rangle = \sum_{\tau=|\ell-s|}^{\ell+s} C_{\ell m_\ell s m_s}^{\tau m_\tau} |\tau m_\ell+m_s\rangle
\label{eq:a1_19}
\end{equation}
Where $C_{\ell m_\ell s m_s}^{\tau m_\tau}$ are the Clebsch-Gordan coefficients.
References
- Whittaker, E. T., & Watson, G. N. (1927). A Course of Modern Analysis. Cambridge University Press.
- Griffiths, D. J. (2018). Introduction to Quantum Mechanics. Cambridge University Press.
