CCL:G: Magnitude of static field used for calculation of polarizability



Dear Suresh,

By default, Gaussian uses CPHF for polarizabilities (quite different from the response formalism, as far as I can judge – can anyone comment on that?), and finite-field (which utilizes field strengths) is needed only for post-SCF methods and the third differentiation in calculation of Gamma (the second hyperpolarizability).
Numerical differentiation field step is by default 0.0003 a.u. (about 0.01542 V/Å). For double-numerical differentiation it is 0.001 a. u. (so 0.051 V/Å). For BOTH cases, this corresponds to keyword argument Step=10 (see this post from my rusty blog).

Best regards,
Igors Mihailovs
Lab. of Organic Materials
ISSP Uni. of Latvia

On 25/04/18 19:06, Susi Lehtola susi.lehtola++alumni.helsinki.fi wrote:

Sent to CCL by: Susi Lehtola [susi.lehtola^^alumni.helsinki.fi]
On 04/25/2018 04:22 PM, Suresh Kumar Venkata Neelamraju
nvs.kumar^^iiit.ac.in wrote:

Sent to CCL by: "Suresh Kumar Venkata Neelamraju"
[nvs.kumar%x%iiit.ac.in] Dear members of CCL,

Can anyone please inform the magnitude of static field used by
Gaussian09 for calculation of static polarizability.

One does not need a field to calculate static polarizability. It can be solved at zero field as the second derivative of the molecular energy using a response formalism. This appears to be the approach used in Gaussian.

If you do not have a program that supports this approach, you can always estimate the static polarizability by finite differences. (A static field is very easy to implement, as it only requires dipole integrals.) You can then run calculations at several values of the field (as well as all field directions), and estimate the polarizability using e.g. a three- or five-point stencil.