In depth The energy levels of the hydrogen atom

In the chapter we set up the eigenvalue problem for the energy and limited ourselves to reporting the levels of the hydrogen atom. Here we solve it.

The problem. The nucleus of the hydrogen atom has charge $+e$ and produces the potential $V(r)=e/(4\pi\varepsilon_0 r)$; the electron orbiting it has charge $q=-e$, so its potential energy is $qV(r)=-e^2/(4\pi\varepsilon_0 r)$, negative as it must be for an attraction. The eigenvalue problem of the chapter becomes

Let us write it for wave functions. The operator $\overline{P}$ maps $\psi$ to the function $-i\hbar\overline{\nabla}\psi$, so $\overline{P}^2$ maps it to $-\hbar^2\nabla^2\psi$, and the equation becomes

First restriction. Let us narrow the search to functions that depend only on the distance $r$ from the nucleus, that is, to spherically symmetric states. As in the fourth chapter, narrowing the search is not an assumption about the result: in the end we shall test what we find against the original equation.

For a function of $r$ alone the Laplacian reads

The last equality is checked by differentiating the product twice

This suggests the substitution $u(r)=r\psi(r)$: multiplying the equation by $r$, in place of the Laplacian we get exactly $d^2u/dr^2$, and we obtain an equation in a single variable

We look for the bound states, those in which the electron stays near the nucleus: for these the energy is negative, so $-2mE/\hbar^2$ is a positive quantity. Let us then set $\kappa^2=-2mE/\hbar^2$ and $A=me^2/(2\pi\varepsilon_0\hbar^2)$, and multiplying by $-2m/\hbar^2$ the equation becomes

call it 1a. We make two demands on the solution: $u(0)=0$, since otherwise $\psi=u/r$ would diverge at the nucleus, and $u\to0$ as $r\to\infty$, since otherwise the electron would not be bound.

Second restriction. For large $r$ the term $A/r$ is negligible and 1a reduces to $d^2u/dr^2=\kappa^2u$, whose solutions are $e^{\kappa r}$ and $e^{-\kappa r}$: only the second goes to zero. Let us then look for $u$ in the form

with $w$ to be determined. The derivatives are

and on substituting into 1a the term in $\kappa^2w$ cancels against $-\kappa^2u$; cancelling then the factor $e^{-\kappa r}$, which never vanishes, we are left with

call it 2a.

Third restriction. Let us look for $w$ among the functions expandable in a power series, and with no constant term, since $u(0)=0$:

The three terms of 2a become

where in the first line we shifted the summation index by one — the term with $k=1$ vanishes — so as to have the same power $r^{k-1}$ in all three terms. Equation 2a then reads

This must hold for every $r$, and a power series vanishes identically only if all its coefficients vanish; therefore

call it 3a. Once $c_1$ is fixed, 3a determines all the other coefficients: the solution is unique up to a constant factor.

The series must terminate. Suppose it does not. For large $k$ the term $A$ becomes negligible compared with $2\kappa k$, and 3a gives

but this is exactly the ratio between two successive coefficients of the expansion $e^{2\kappa r}=\sum_{k=0}^{\infty}(2\kappa)^k r^k/k!$. Then $w$ would behave like $e^{2\kappa r}$ and $u=we^{-\kappa r}$ like $e^{\kappa r}$, which does not go to zero: the solution would not be a bound state.

There must then exist an integer $n\geq1$ with $c_n\neq0$ and $c_{n+1}=0$. By 3a this happens if and only if

This is where the integer comes from: we did not impose it, it is imposed by the requirement that the electron stay bound to the nucleus.

The levels. Recalling that $\kappa^2=-2mE/\hbar^2$ and that $A=me^2/2\pi\varepsilon_0\hbar^2$, we have

and finally, substituting $\hbar=h/2\pi$,

Which is what we set out to prove.

Check. Take the first level, $n=1$. Equation 3a immediately gives $c_2=0$, so $w=c_1r$, $u=c_1re^{-\kappa r}$ and the wave function is $\psi=c_1e^{-\kappa r}$ with $\kappa=A/2=me^2/(4\pi\varepsilon_0\hbar^2)$. The Laplacian is

and substituting into the original equation

The $1/r$ term vanishes because $\hbar^2\kappa/m=e^2/(4\pi\varepsilon_0)$, which is precisely the definition of $\kappa$ for $n=1$; there remains $E=-\hbar^2\kappa^2/2m$, that is, the formula we found.

What we left out. We searched only among the spherically symmetric functions, those with zero angular momentum. There are also solutions depending on the angles, with non-zero angular momentum, and they are needed to describe the states of the atom; but they give no new levels: every $E_n$ already appears among the solutions found here. For the energy levels, which is what we needed, the restriction took nothing away.

The number $n$ is the one that appears in the ninth chapter in the jumps between levels: an atom passing from a level of energy $E_i$ to one of energy $E_f$ emits a photon of energy $E_i-E_f$.