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
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).
Lab. of Organic Materials
ISSP Uni. of Latvia
On 25/04/18 19:06, Susi Lehtola susi.lehtola++alumni.helsinki.fi
Sent to CCL by: Susi Lehtola [susi.lehtola^^alumni.helsinki.fi]
On 04/25/2018 04:22 PM, Suresh Kumar Venkata Neelamraju
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