Subject: Re: bsse and second derivatives in G98

From: AM <[email-masked]>
Date: Thu, 27 Jan 2000 20:10:10 -0500

Tanja, Dominik, and all,
There is no phisical law forbidding frequency calculations on Ar atom.
And Hessian (as opposed to frequencies) makes sence at any point on
potencial surface, not only at the crytical points. In fact, this is how Opt=CalcFC
works in Gaussian.
If you are interested in Hessian, there is a workarownd:
use keyword FORCE instead of FREQ. Try HF first.
We found that in some cases FORCE on single atom won't work with MP2.
If you try GAMESS-UK, you may get Hessian in more readable format :)

Artem

> What you are trying to do here is to calulate the frequency of the Ar
> atom (albeit with some basis functions on the formic acid ghost atoms).
> This is not possible.
>
> Similarly, you cannot calculate the frequencies of formic acid in the
> presence of ghost functions like this (i.e. by replacing the last 5 lines
> by "6 Nuc 0.0"). To calculate a hessian, your molecule has to be in the minimum
> (for the given method/basis), and this is not the case (I assume the geometry
> above comes from an optimization on the cluster).
>
> I am afraid that there is not an easy way to calculate counterpoise-corrected
> frequencies automatically. You would need a potential energy surface and
> calculate the frequencies on it numerically, but this is not practical on any
> higher level of theory.
>
> Best regards,
>
> Tanja

>
>  > to do the calculation with ghost atoms for the organic molecule and
> > electrons only at the argon atom. Unfortunately, with G98 this type of
> > calculation fails: 'Molecule is non-linear but NTrRo=3'. Who knows how
> > to fix this problem, or is there another convenient way to obtain the
> > second derivatives without this problem. I used the following input:
> >

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From: Herbert Fruchtl <[email-masked]>
Date: Fri, 28 Jan 2000 11:04:22 +0000

> There is no phisical law forbidding frequency calculations on Ar atom.

Am I missing something completely? All methods mentioned are based on
the Born-Oppenheimer approximation. The Argon can't feel a force,
because in this approximation forces only exist between nuclei. The
Hessian is the derivative of the force, which is also zero. Everything
else is spurious.

To the original poster: as I understand Counterpoise, it can only be
used to calculate wavefunction properties (e.g. the energy, but not the
Hessian). Then you can calculate other properties (gradients, the
Hessian, etc.) from the potential energy surface. I don't think you can
apply the counterpoise method directly to the Hessian. Even if your
program happens to give you non-zero numbers. 

You will of course get different energies when you move ghost atoms
around (and therefore a numerical gradient/Hessian will be non-zero),
but that has nothing to do with a physical frequency.

  Herbert

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From: "Jochen K�pper" <[email-masked]>
Date: Fri, 28 Jan 2000 13:19:56 +0100 (CET)

Herbert Fruchtl writes:
 > the Born-Oppenheimer approximation.

But then BOA isn't a physical law :-O

 > The Argon can't feel a force,
 > Hessian is the derivative of the force, which is also zero.

Well, the derivative of nothing isn't actually zero - but nothing.
In the BOA it simply doesn't make sense to talk about _that_ Hessian.

 > but that has nothing to do with a physical frequency.

It actually means leaving BOA behind. You should then probably try to
do that from the beginning - if you really want :-)

Jochen

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From: Richard Wood <[email-masked]>
Date: Fri, 28 Jan 2000 12:23:48 +36000

> In the BOA it simply doesn't make sense to talk about _that_ Hessian.

Isn't zero a constant, and so the derivative of a constant is zero?

Richard

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From: Herbert Fruchtl <[email-masked]>
Date: Fri, 28 Jan 2000 16:45:07 +0000

> Isn't zero a constant, and so the derivative of a constant is zero?
Of course you're right. Just the quote that you generously attribute to me
isn't mine. 

  Herbert

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From: "Jochen Küpper" <[email-masked]>
Date: Fri, 28 Jan 2000 17:32:03 +0100 (CET)

> Isn't zero a constant, and so the derivative of a constant is zero?

Your statement as such ios correct.
But if you read the earlier messages (e.g the one you cited), you will
see that he was calcualting the derivative outside the definition
range of the function.

Consider
   t: ]-1,+1[ ---> R
      x       |--> tan( x )

What's the derivative of dt/dx at x = 5 ?

Jochen

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From: Richard Wood <[email-masked]>
Date: Fri, 28 Jan 2000 16:29:01 +36000

But i WAS responding to what was written, namely the derivative of 0 not 
being zero.

What does 0 have to do with dt/dx anyway?

Look at it this way, if dt/dx=0, doesn't d2t/dx2=0 also?

Richard