# [molpro-user] calculating dipole moment using finite field

Tatiana Korona tania at tiger.chem.uw.edu.pl
Tue Apr 3 01:36:21 CEST 2018

```Hi,

"!dipole moment as first energy derivative"

got divided into two lines. The second line starts from the word "first", this
is why Molpro treats it as some unknown command.

Best wishes,

Tatiana

On Sat, 31 Mar 2018, Mahdi Yousefi Atashgah wrote:

> Hi,
> I'm trying to calculate AlCl dipole moment at equilibrium using finite
> field by below input:
>
>
> ***,AlCl finite field calculations
> ANGSTROM
> r=[2.130143]                !set geometry parameters
> geometry={Al;Cl,Al,r}
> basis=aug-cc-pv6z                        !define default basis
> field=[0,0.0003,-0.0003]             !define finite field strengths
> \$method=[hf,casscf,ccsd,ccsd(t)]
>
> k=0
> do i=1,#field                      !loop over fields
> dip,,,field(i)                   !add finite field to H
> do m=1,#method                   !loop over methods
>  k=k+1
>  \$method(m)                     !calculate energy
> !expec,dm   !calculate dipole moments for ACPF
>   e(k)=energy                    !save energy
> enddo
> enddo
> k=0
> n=#method
> do m=1,#method
> k=k+1
> energ(m)=e(k)
> dipmz(m)=(e(k+n)-e(k+2*n))/(field(2)-field(3))   !dipole moment as
> first energy derivative
>
> dpolz(m)=(e(k+n)+e(k+2*n)-2*e(k))/((field(2)-field(1))*(field(3)-field(1)))
> !polarizability as second der.
> enddo
>
> table,method,energ,dipmz,dpolz
> title,results for AlCl, r=\$R
>
>
> however, the output gives me below error:
> SETTING E(6)           =      -701.56161745  HARTREE
>
> SETTING K              =         0.00000000
>
> SETTING N              =         2.00000000
>
>
> DO M                   =         1.00000000
> SETTING K              =         1.00000000
>
> SETTING ENERG(1)       =      -701.51812000  HARTREE
>
> SETTING DIPMZ(1)       =        -0.57529364  HARTREE
>
> ? Error
> ? The problem occurs in get_commandset
>
>