# [molpro-user] hexadecapole moment of CH4 using finite-field

Omololu prayerz at yahoo.com
Wed Jun 9 11:52:42 BST 2010

```Yuri,

Your finite field _method_ is OK but:

1. Your applied field is too small. Try f=[0.0001,-0.0001]
2. Be sure your geometry (& geometry section) is correct. E.g.,  CH bondlength in methane is about 1.099 Angstroms but your CH bondlengths are 1.18912*sqrt(3) Angstroms.

When you are off by less than orders of magnitude, check:
3. Methane has 2 independent non-zero hexadecapole moments (Q40, Q43, say). If you are trying to calculate Q40, Check the prefactor. I am _guessing_ there would be a factor of (35/8) to multiply your <zzzz>

My \$0.02.

o.

--- On Wed, 6/9/10, Yurii Vlasov <vlasov.yura at gmail.com> wrote:

From: Yurii Vlasov <vlasov.yura at gmail.com>
Subject: [molpro-user] hexadecapole moment of CH4 using finite-field
To: molpro-user at molpro.net
Date: Wednesday, June 9, 2010, 6:09 AM

Dear Molpro Community,

I have a problem with the calculation of the hexadecapole moment of CH4 molecule using finite-field method. But I've got the value of Hexadec mom =  -2.01067478D+07 ! (It should be ~7 a.u. according to Maroulis, or if to calculate the expectation value). May be I didn't understand correctly the idea of finite field method...

Could anyone help me. I would be very appreciated

memory,20,m

GTHRESH, oneint=1.d-14, energy=1.d-9, step=1.d-5,
twoint=1.d-14, prefac=1.d-14

GTHRESH, coeff=1.0d-5, zero=1.0d-14
symmetry,nosym

geometry={
1 C1,,   0., 0., 0.,
2 H2,, 1.18912    1.18912    1.18912,
3 H3,, -1.18912   -1.18912    1.18912,
4 H4,, -1.18912    1.18912   -1.18912

5 H5,, 1.18912   -1.18912   -1.18912,
}
basis=avtz

f=[0.000001,-0.000001]
method=[hf]
k=0
do i=1,#f
field,zzzz,f(i)
hf
e(k)=energy
end do

k=0
n=1
do m=1,1
k=k+1

hexz(m)=(e(k)-e(k+n))/(f(1)-f(2))
end do

table,method,hexz

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