The notion that the region form –infinity to infinity could be considered by a change of variables is explored in cofv.htm.
An unexpected result of taking the region from –infinity to infinity with a capped potential is the fact that the energy ground state is not zero at r=0. Then when I attempted to do the assignment at the end of this document, I found it impossible to converge on excited states. This led to the development of a variance minimization technique for solving the one-dimensional equation Schroedinger equation. Variance minimization files describe this technique. An annotated listing of the files leading up to this is in Welcome.htm.
A consideration of the method described below in curve fit notation so that a matrix is used to find the next solution is explored in the Neqns.htm, while the precise nature of the iterative solution below is considered in ItSol.htm.
When I attempted to do the assignment at the end of this document, I found it impossible to converge on excited states.
in Rydbergs is
First consider the small r limit and assume that
. Then
for
. Next consider the
large r limit
Thus the starting assumption is ![]()
The Fourier transforms of u and U are defined by
Fourier sums are considered in DFT.htm. The simplest Fourier sum starts at zero in r. This works but introduces a discontinuity in at least the first derivative at the origin making the sum more slowly convergent -- requiring a larger N -- than it needs with a better choice of range DFT- one_sided.
For r=-R/N, the first
negative point this yields
![]()
As r®¥,
u(r) and all of its derivatives go to zero, but this is not the case as r ® 0. For this
reason u has been extended into the negative r region as
so that
the integral is from
-R/2 to R/2 in finding U(f). The
function of r in equation 6 is
So that

Note – Bob Coldwell was unable to make the following converge on excited states. The method that works is variance minimization.
Equation
1
This implies
where a, b, c, and d can be determined to make the states
have desired properties.

Figure 1 Zeff(r) ZEFF.FORwith a = b = c = d = 0.
The resulting effective potential is
![]()
Solve the 1-d Schroedinger equation for the 3s and 3p states with the above effective potential using the method outlined below.
Assume u is given by equation . Use a coarse grid e.g. 16 points. Calculate ![]()
u1(ri) on this grid.
Use the FFT to find U1(fk) on this grid.
Calculate 
As ![]()
Note that for real u

Note that ![]()
Later on note that 

Note that
for real u(r)
Find A from the condition
![]()

Note that this
relation is merely a proof that u(r) is real, it makes no sense to truncate the
fft.
Note that this
needs to be on a symmetric range.
Test for convergence, use Aitkin’s extrapolation very third step. If not converged go to step 2.
If the number of points is >= Nmax then stop. Otherwise use the FFT as described in DFT- Symmetric_range_in_r_and_f_with_FFT to double the number of points and then go to step 2