Subject: Re: CCL:Finite Fields in Gaussian

From: "T. Daniel Crawford" <crawdad..at..vt.edu>
Date: Sat, 17 Apr 2004 13:24:05 -0400 (EDT)

Dear Fernando:

I have carried out finite-difference field calculations myself (coded and
run), though admittedly not with Gaussian.  Thus, my answer and observations
may be only marginally helpful to you.

Most likely, the dipole perturbation is just a set of scaled dipole
integrals of the form, <p|z|q> * F, where p and q denote AO basis
functions, z is the z-component of the dipole operator, and F is the
(constant) field strength chosen in user input.  These integrals can be
added to the other one-electron integrals before the SCF calculation, thus
giving a homogeneous-field-perturbed energy.

The most difficult aspect of computing higher-order polarizabilities by
finite differences is the numerical error.  An example: the static 
polarizability is defined as the negative of the second derivative of the
energy with respect to the field, and a numerical second derivative formula
is:

alpha = - [ E(+d) - 2 * E(0) + E(-d) ]/ (2 d^2) + O(d^2)

where d denotes the field strength, E(+d) and E(-d) are the energies at
positive and negative field, and E(0) is the unperturbed energy.  The
numerical error is of the order d^2.

If I choose a typical finite-field strength of 0.0001 a.u., the numerical
error is about 10^-8.  However, this field strength produces a change in
the energy of only about 10^-7 a.u. for a typical small molecule like HF.
Thus, if you converge the energies to only 10 decimal places, you'll get
a polarizability that compares to the analytical result to only three
decimal places.  If you converge to 12-13, you can get 5-6 decimal places.
Beyond that, machine precision starts to interfere with the numerical
differentiation.

When you get to hyperpolarizabilities, however, the numerical errors get
larger, and it's harder to balance the wfn convergence criteria with the
inherent numerical errors in the finite-difference expressions.  (NB:
finite differences of analytically computed dipole moments can help.)

Finally, I note that the polarizability term in your polynomial expansions
differ by almost exactly a factor of two:

4.899299996216 * 2 = 9.798599992432, 

which is very close to the analytical result to about 5 decimal places.

I hope this helps.  If you already knew all these issues, I apologize for
wasting your time.

Sincerely,

-Daniel

On Thu, 15 Apr 2004, Fernando D. Vila wrote:

> For the energy:
>    -56.200911807600
>     -0.700079533332 * F
>     -4.899299996216 * F^2
>      3.333331240659 * F^3
>   -400.002401571933 * F^4
> 
> For the dipole moment:
>       -0.700079500000
>       -9.799183333333 * F
>        9.783333333291 * F^2
>     -466.666666569064 * F^3
>    66666.666631576154 * F^4
> 
> If you compare the coefficients of this polymomials with the coefficients
> in the formulas above you will see that there are many sign differences
> with respect to the analytic results I showed above. This problem is not
> new to me and the Gaussian manual very helpfully :-P informs us that "the
> coefficients are those of the Cartesian operator matrices; be careful of
> the choice of sign convention when interpreting the results".
> 
> Before I had always assumed that although Gaussian says that is adding an
> electric field, what is actually doing is adding a potential gradient (
> dV/di = V_i = -F_i ). This assumption always gave me correct results, but
> now it is totally irreconciliable with the sign I get for the
> hyperpolarizability.
> 
> So finally, my question: what is the CORRECT FORM of the perturbation 
> introduced by the Gaussian keyword Field.
> 
> Cheers and I will really appreciate any comments (Doug Fox, are you 
> there?? :-)
> 
> Fer.
> 

-- 
T. Daniel Crawford                           Department of Chemistry
crawdad..at..vt.edu                                    Virginia Tech
www.chem.vt.edu/faculty/crawford.php  Voice: 540-231-7760  FAX: 540-231-3255
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