Chapter 9 The Two-Body Problem

1.8. The Hydrogen Atom
We used the factorization method to find the eigenvalues of the Kepler problem, Eq. (8.154). Now let us apply the same method to a system composed of a positive charge interacting with an electron of charge . The Hamiltonian in this case is

Denoting the eigenvalues of the operators and by and . respectively we have

As we have seen in the case of one-dimensional motion we can write as

where

and and are real numbers to be determined. Similarly we can define the operators and by

and

Next we calculate

and in a similar way we calculate

By comparing obtained from (9.191) with the same operator found from (9.186) and we find

with

These equation have two sets of solutions:
We can either choose

or

The second set gives us a larger eigenvalue and therefore we choose (9.196) rather than (9.195). Also from (9.189) and (9.190) we have

Similarly for the general we choose

with the corresponding eigenvalue satisfying the relation

Again for a fixed we choose the solution which maximizes the eigenvalue;

Solving for and we find

For a hydrogen atom at rest and the energy eigenvalues are

where is a nonnegative integer and is a positive integer.
Hydrogen Atom Eigenstates - The wave function for the Coulomb potential can be found from Eq. (8.81) or from

The lowest state for a given is a solution of

where and are given by Eq. (9.201) and

is the radial momentum operator. Now by integrating this first order differential equation we find

For a given this wave function has no nodes, i.e. it is the wave function of the ground state which is found for in (9.203). If we write with or then

In Eq. (9.207) is the Bohr radius, represents the nodeless wave function for the principal quantum number .
Having obtained , we can construct

by the following method:
Let us consider the hydrogen atom for which . Then the normalized can be written as

Next we introduce the raising and lowering operators by

and

The lowering operator reduces the angular momentum quantum number by one unit [12]-[15]. Then we have

and

This lowering of the can be continued up to the point where becomes zero. In this way we can generate all of the wave functions of the hydrogen atom.
As an example consider the normalized which is given by

and the lowering operator which according to is

From these expressions we find

Likewise for , from we obtain

and from (9.211) we have

Thus

and from and we can calculate ,

Here the phase of the wave function is given by whereas this phase is 1 in the standard textbooks [7].
The complete wave function which includes the angular part can be written as

where is the spherical harmonics;