Subject: OVERLAP INTEGRAL AND SLATER FUNNC

From: "E. Lewars" <[email-masked]>
Date: Sun, 2 Nov 1997 16:54:11 -0500 (EST)

1997 Nov 2

                          SLATER OVERLAP INTEGRALS
>From   E. Lewars
To     CCL

  Hello,

 There is some info' about overlap integrals and Slater functions that I am
finding it hard to dig out of the literature.  I should state at the outset
that _I know_:

that there are programs for calculating overlap integrals

that in most molecular quantum chem Gaussian functions, not Slater functions,
are used

that Mulliken et al published a long paper (J Chem Phys, 17 (1949) 1248-1267))
with recipes for calculating overlap integrals

that all kinds of tricks exist for "simplifying" the calculation of overlap
integrals
---------
OK, but what I want is simply this:

(1)  In the molecule H-He+ (yes, protonated helium) if the H nucleus has
     Cartesians H1(x_1, y_1, z_1) and the He nucleus He2(x_2, y_2, z_2),
     where the coordinates are in Angstroms (_not_ atomic units or Bohrs)

     and if we write

     for H:  phi_1 = a_1 exp{b_1[(x-x_1)^2 + (y-y_1)^2 + (z-z_1)^2]^1/2}
 and for He: phi_2 = a_2 exp{b_2[(x-x_2)^2 + (y-y_2)^2 + (z-z_2)^2]^1/2}

      (usually a is expressed in terms of zeta and pi, b = -zeta, and
      the variable is a radius vector r)
   QUESTION:  what are a_1 and b_1, as numbers, not Greek letters?
              what are a_2,and b_2 as  numbers, not Greek letters?
              (i.e a_1 = 0.5647 or whatever, etc etc (4 decimals).

    S_12 = the triple integral from 0--> infinity of phi_1.phi_2 dxdydz
    must = the correct overlap, if the six Cartesian coordinates are in
    Angstroms.  And of course integral of S_11 = S_22 = 1.

(2)  If (x_1, y_1, z_1) = (0, 0, 0) and (x_2, y_2, z_2) = (0, 0, 0.800),
     (0.800 Angstroms) what is S_12 (as a number, e.g 0.4273 or whatever)?

(3)  For (x_1, y_1, z_1) = (0, 0, 0) and (x_2, y_2, z_2) = (0, 0, d),
     is there a simple function S_12 = f(d) ? (for H, He?  for any two
     1s orbitals)?  Of course f will depend parametrically on the a's and b's.
-------------
  Thanks     E. Lewars
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