Relating the Energy Minimization to the Differential Equation

We begin by minimizing the energy given in Rydbergs (h/2p =1, 2m=1, e²=2, E_Hyd=-1) by

(1.1)

the functional derivative of the energy is

(1.2)

_______________________________________________

Note that the part coming from the variation of the denominator includes the energy to yield

(1.3)

Integration by parts utilizing dy=0 on the boundaries yields

(1.4)
With A = H, this is not quite, but almost

../integration/MonteCarlo/MonteCarlo3.doc#MCVariance

Start with

(1.5)

Figure 1 Nuclei and electrons

The selection method is described in ..\integration\MonteCarlo\FMonte\Fmonte.doc .htm

The codes are in moltest.for and moltest2\moltest.for. I have not looked at these for some time. Look at analytical4.doc .htm and reproduce integrals of the appropriate Gaussians before starting.

Born Oppenheimer approximation

(1.7)

Where

(1.8)

And

(1.9)

L²

The operator[1]

(1.10)

So that

(1.11)

The second term is a bit of a surprise, making the Schroeinger equation in the limit that rà0

(1.12)

Examining the equation at small r

a) Angular momentum > 0

For l > 0, the dominant term as r à 0 is the l(l+1)/r² term which must be exactly cancelled.

Try the form so that the first two terms of (1.12) becomes

(1.13)

This implies so that for small r

(1.14)

Note that this makes for a hole in the probability density near the nucleus.

b) Angular momentum = 0

In this case the dominant term in (1.12) is the Ze²/r singularity in V(r). Try the form so that

This implies that a=Z and that too a very good approximation E = Z²

Examining the equation at large r

If we assume that for sufficiently large r, the potential and the angular momentum terms (L(L+1)/r²) are negligible, then equation (1.12) reduces to

(1.15)

For a solution of the form u=exp(-ar), so that this becomes

(1.16)

(1.17)

The - solution is the one that we want, but the + solution is also an exact solution to the differential equation.

Product form

This form is almost ready to use in

../integration/MonteCarlo/MonteCarlo3.doc#MCVariance

Lennard-Jones Lennard Jones Potential.htm .doc

The Fermi repulsion generated by the fact that more than two electrons cannot occupy the same location results in an E(R) rather well approximated by a/R¹² as the two nuclei come together.

For two nuclei interacting with a Lennard-Jones potential as Rà 0, (1.12) becomes

(2.1)

Consider a trial wave function of the form

(2.2)

Again in spherical coordinates the del² is given by

(2.3)

So that

(2.4)

Substituting (2.4) into (2.1) yields

(2.5)

For n=-5, (2.5) becomes

(2.6)

Then solve

(2.7)

To reduce (2.6) to

(2.8)

The remaining singularity is unfortunately not exactly canceled by the attractive part of the Lennard Jones potential, but is a lot easier to deal with than the much higher one just removed.