Schrödinger’s equation for the hydrogen atom
The most dramatic success in the history of the quantum mechanics was the understanding of the details of the spectra of some simple atoms and the understanding of the periodicities which are found in the table of chemical elements. In this chapter we will at last bring our quantum mechanics to the point of this important achievement, specifically to an understanding of the spectrum of the hydrogen atom. We will at the same time arrive at a qualitative explanation of the mysterious properties of the chemical elements. We will do this by studying in detail the behavior of the electron in a hydrogen atom—for the first time making a detailed calculation of a distribution-in-space according to the ideas we developed in Chapter 16.  For a complete description of the hydrogen atom we should describe the motions of both the proton and the electron. It is possible to do this in quantum mechanics in a way that is analogous to the classical idea of describing the motion of each particle relative to the center of gravity, but we will not do so. We will just discuss an approximation in which we consider the proton to be very heavy, so we can think of it as fixed at the center of the atom.  We will make another approximation by forgetting that the electron has a spin and should be described by relativistic laws of mechanics. Some small corrections to our treatment will be required since we will be using the nonrelativistic Schrödinger equation and will disregard magnetic effects. Small magnetic effects occur because from the electron’s point-of-view the proton is a circulating charge which produces a magnetic field. In this field the electron will have a different energy with its spin up than with it down. The energy of the atom will be shifted a little bit from what we will calculate. We will ignore this small energy shift. Also we will imagine that the electron is just like a gyroscope moving around in space always keeping the same direction of spin. Since we will be considering a free atom in space the total angular momentum will be conserved. In our approximation we will assume that the angular momentum of the electron spin stays constant, so all the rest of the angular momentum of the atom—what is usually called “orbital” angular momentum—will also be conserved. To an excellent approximation the electron moves in the hydrogen atom like a particle without spin—the angular momentum of the motion is a constant.  With these approximations the amplitude to find the electron at different places in space can be represented by a function of position in space and time. We let $\psi(x,y,z,t)$ be the amplitude to find the electron somewhere at the time $t$. According to the quantum mechanics the rate of change of this amplitude with time is given by the Hamiltonian operator working on the same function. From Chapter 16, \begin{equation} \label{Eq:III:19:1} i\hbar\,\ddp{\psi}{t}=\Hcalop\psi, \end{equation} with \begin{equation} \label{Eq:III:19:2} \Hcalop=-\frac{\hbar^2}{2m}\,\nabla^2+V(\FLPr). \end{equation} Here, $m$ is the electron mass, and $V(\FLPr)$ is the potential energy of the electron in the electrostatic field of the proton. Taking $V=0$ at large distances from the proton we can write1 \begin{equation*} V=-\frac{e^2}{r}. \end{equation*} The wave function $\psi$ must then satisfy the equation \begin{equation} \label{Eq:III:19:3} i\hbar\,\ddp{\psi}{t}=-\frac{\hbar^2}{2m}\,\nabla^2\psi -\frac{e^2}{r}\,\psi. \end{equation}  We want to look for definite energy states, so we try to find solutions which have the form \begin{equation} \label{Eq:III:19:4} \psi(\FLPr,t)=e^{-(i/\hbar)Et}\psi(\FLPr). \end{equation} The function $\psi(\FLPr)$ must then be a solution of \begin{equation} \label{Eq:III:19:5} -\frac{\hbar^2}{2m}\,\nabla^2\psi= \biggl(E+\frac{e^2}{r}\biggr)\psi, \end{equation} where $E$ is some constant—the energy of the atom.  Since the potential energy term depends only on the radius, it turns out to be much more convenient to solve this equation in polar coordinates rather than rectangular coordinates. The Laplacian is defined in rectangular coordinates by \begin{equation*} \nabla^2=\frac{\partial^2}{\partial x^2}+ \frac{\partial^2}{\partial y^2}+ \frac{\partial^2}{\partial z^2}. \end{equation*} We want to use instead the coordinates $r$, $\theta$, $\phi$ shown in Fig. 19–1. These coordinates are related to $x$, $y$, $z$ by \begin{equation*} x=r\sin\theta\cos\phi;\quad y=r\sin\theta\sin\phi;\quad z=r\cos\theta. \end{equation*} It’s a rather tedious mess to work through the algebra, but you can eventually show that for any function $f(\FLPr)=f(r,\theta,\phi)$,  \begin{equation} \label{Eq:III:19:6} \nabla^2f(r,\theta,\phi)=\frac{1}{r}\,\frac{\partial^2}{\partial r^2}\, (rf)+\frac{1}{r^2}\biggl\{ \frac{1}{\sin\theta}\,\ddp{}{\theta}\biggl( \sin\theta\,\ddp{f}{\theta}\biggr)+\frac{1}{\sin^2\theta}\, \frac{\partial^2f}{\partial\phi^2}\biggr\}. \end{equation} 
 \begin{align} \label{Eq:III:19:6} &\nabla^2f(r,\theta,\phi)=\\[1ex] &\frac{1}{r}\,\frac{\partial^2}{\partial r^2}\,(rf)\;+\notag\\[1ex] &\frac{1}{r^2}\biggl\{ \frac{1}{\sin\theta}\,\ddp{}{\theta}\biggl( \sin\theta\,\ddp{f}{\theta}\biggr)+\frac{1}{\sin^2\theta}\, \frac{\partial^2f}{\partial\phi^2}\biggr\}.\notag \end{align}  So in terms of the polar coordinates, the equation which is to be satisfied by $\psi(r,\theta,\phi)$ is  \begin{equation} \label{Eq:III:19:7} \frac{1}{r}\,\frac{\partial^2}{\partial r^2}\, (r\psi)+\frac{1}{r^2}\biggl\{ \frac{1}{\sin\theta}\,\ddp{}{\theta}\biggl( \sin\theta\,\ddp{\psi}{\theta}\biggr)+\frac{1}{\sin^2\theta}\, \frac{\partial^2\psi}{\partial \phi^2}\biggr\}= -\frac{2m}{\hbar^2}\biggl(E+\frac{e^2}{r}\biggr)\psi. \end{equation} 
 \begin{align} \label{Eq:III:19:7} &\frac{1}{r}\,\frac{\partial^2}{\partial r^2}\, (r\psi)\;+\\[1ex] &\frac{1}{r^2}\biggl\{ \frac{1}{\sin\theta}\,\ddp{}{\theta}\biggl( \sin\theta\,\ddp{\psi}{\theta}\biggr)+\frac{1}{\sin^2\theta}\, \frac{\partial^2\psi}{\partial \phi^2}\biggr\}\notag\\[1.5ex] &=-\frac{2m}{\hbar^2}\biggl(E+\frac{e^2}{r}\biggr)\psi.\notag \end{align} 
