This is a solution to the Schrödinger Equation for Hydrogen-like atoms using a power series. There are more elegant methods, but this one is fairly elementary. The tedious details are (mostly) included.

The Schrödinger Equation is:

.

The *Z* is included for generality; this way the description of other ions He^{1+}, Li^{2+}, ... (*etc*) automatically follows. Note that for a bound state, *E* is negative.

The angular dependence can be separated from the radial dependence, leaving only the factor proportional to *l*(*l*+1), with the requirement that *l* must be a non-negative integer. Find the separation of the Laplacian and the solutions of the angular eigenfunctions elsewhere if interested. The radial equation is:

The radial equation becomes simpler by means of a substitution,

.

It is convenient to define some constants to make the equations more tidy. To this end, the following uses

,

.

The radial equation becomes

,

or if we multiply everything by -*r*^{3},

.

Note that in the case where *l*=0 and *α*=0, this equation is harmonic. If there is no Coulomb potential (no nucleus), the remaining equation could describe a free particle or a constant potential. The corresponding solutions are plane waves or exponential decay. It would take an infinite number of terms in a series to reproduce these behaviors, which involve exponents. Hoping to find series solutions with a finite number of terms, we make one more substitution to remove this dependence:

The differential equation for *f*(*r*) is

Now we substitute a power series for *f*(*r*):

The first three terms in the equation for *f* become

,

, and

,

where the indices have been manipulated to express everything in terms of the coefficient that goes with *r ^{k}*. The full differential equation written in terms of the power series is

.

Now rewrite the series, using the substitution *k*=*m*+1 to obtain symmetry with *l* in the first term, which gives

.

For each *m*, *r*^{m+1} is linearly independent of all other powers, so that each coefficient in the power series must vanish to satisfy the differential equation.

The ground state should have *l*=0, since the energy increases proportional to *l*^{2}. To solve for the ground state, we substitute *l*=0 in the power series above and search for a solution. The equations corresponding to *m*=0, 1, and 2 are:

These equations are satisfied by setting all terms other than *c*_{1} to zero, provided that *α*=*λ* and *&lambda*^{2}=-2*β*. If the second condition looks strange, recall that for a bound state, we have *E*<0 and therefore *β*<0. Reinserting the constants that defined *α* and *β* to solve for the ground state energy, we find

Specifically for Hydrogen (*Z*=1), we have

This is the well-known ground-state energy of the hydrogen atom, which is reassuring. If we retrace the trail of substitutions, we also have the ground state radial wave function.

Now we seek more general solutions. Suppose we want to find a solution that terminates with a term *c*_{n} in the series, corresponding to *m*+1=*n*. Setting all *c _{m}*=0 for

,

which gives one of the conditions we found when solving for the ground state. The term from the differential equation proportional to *r* to the power *n*+1 becomes

The conditions from the *n*+1 and *n*+2 terms of the differential equation imply

The energy is

This is the classic expression for energy levels in a Hydrogen atom! Remembering the substitutions, for a series that terminates with *n*=*m*+1, the general form of the wave function is

.

The last remaining task is to determine the *c _{k}*, which can be obtained from the series form of the differential equation. They depend on the value of

.

Note that if *n* is not greater than or equal to *l*+1, the series is divergent. So long as *n* is greater than or equal to *l*+1, the series has *n*-*l* terms. This formula can be used to relate all the coefficients to the *n ^{th}* coefficient. Normalization will specify the final coefficient.