Subject: Summary: Localised Kohn-Sham orbitals

From: Brian Salter-Duke <[email-masked]>
Date: Fri, 6 Sep 2002 11:34:21 +0930

My original question was:-

Is the Kohn-Sham DFT energy and electron density (as generated for example 
by Gaussian98 using BLYP or other functionals) invariant to a unitary 
transformation of the Kohn-Sham orbitals to localised orbitals by any or
all of the standard methods, in the same way that the Hartree-Fock
wavefunction is invariant to unitary transformations of the Hartree-Fock
orbitals? I will summarise if this turns out to be an interesting
question.

I had two replies:-

-------------------------------------------------------------------------
Huub van Dam                               E-mail: [email-masked]
CCLRC Daresbury Laboratory                  phone: +44-1925-603362
Daresbury, Warrington                         fax: +44-1925-603634
Cheshire, UK
WA4 4AD

replied:-

Hi Brian,

I think the answer must be yes, DFT is invariant for unitary transformations
among orbitals with the same occupation. The reason is quite simple:

Call the orbitals Y and their occupation numbers n. Because the Kohn-Sham
density matrix is diagonal (single determinant approximation) the density P is
given by

P = sum_i n_i Y_i^* Y_i

Using the fact that mixing two generate eigenfunctions results in yet again an
eigenfunction with the same eigenvalue, I can mix all occupied orbitals among
each other by an unitary transformation and get an equally valid density
matrix. The occupation numbers remain the same, and the new orbitals will be
eigenfunctions of this density matrix. Therefore I can apply the same
expression to get the density which will be the same as before. Finally as the
energy is a functional of the density it follows that if the density is the
same the energy has to be the same as well.

Perhaps there is a more compact explaination but I think this covers most
aspects.

Good luck,

    Huub

-------------------------------------------------------------------------
  Douglas J. Fox
  Technical Support
  Gaussian, Inc.
  [email-masked]

replied:-
   
  Brian,

   My gut answer was yes, the same machinery is used in both cases and
all the MO integrals are retained so the most common semi-empirical
problems are avoided.

   My experimental side ran the following,

%Chk=lw2tBL
#p B3LYP/6-311G**

Cyclic Water Dimer RBLYP/6-311G** Optimisation

0 1
O      -0.441888678   -3.645331531   -0.949088929
H      -0.784006389   -3.840454337   -1.851910976
H       0.537637876   -3.616431553   -1.071306833
O       1.570274523   -1.575840084   -0.777018993
H       1.907288823   -1.418001616    0.134591502
H       0.594354979   -1.678408943   -0.656798044

--link1--
%Chk=lw2tBL
# B3LYP/6-311G** Guess=(Read,Local,Save,Only) Geom=AllCheck

--link1--
%Chk=lw2tBL
#p B3LYP/6-311G** Guess=(Read) Geom=AllCheck

which takes two cycles in the last run.  The energy matches to 8 significant
figures which is all I expect for single point accuracy.  It takes two cycles
because the coefficients change substantially even if the density and energy
are invariant.

-------------------------------------------------------------------------

Doug's reply did not quite convince me, but after reruning his example
with changes I was convinced.

Adding "scf=tight" gives the energy matching exactly to the digits
printed. It also means that the 3rd step converges in one cycle not two.

Adding pop=full to steps 1 and 3 shows that the density is identical at
all three steps while the orbitals, being localised, are different at
step 2 from steps 1 and 3.

Replacing B3LYP with RHF gives the exact same pattern as expected.

So I am convinced that the answer to my question is "Yes".

Thanks to both Huub and Doug for replying.

Brian.

-- 
            Brian Salter-Duke (Brian Duke) [email-masked]  
  Honorary Fellow in Chemistry, NT University, Darwin, NT 0909, Australia.
             Post: Box 1028, Humpty Doo, NT 0836, Australia.
Phone 08-89881600.  Fax 08-89881302.  http://www.octa4.net.au/linden/brian/