# sign of spin orbit matrix elements?

Peter Knowles peterk at tc.bham.ac.uk
Thu Jan 4 11:30:41 GMT 2001

Dear Millard,

I think I disagree with your hand calculation. For the first matrix
element you mention,

\hat S_x \beta = (\hbar/2) \alpha
\hat L_x p_z = -i\hbar (y d/dz-z d/dy) p_z = -i \hbar p_y

and so

\hat S_x \hat L_x p_z = - i \hbar^2 /2  p_y

and the matrix element of the spin-orbit operator should then be -i*a/2

The result produced by molpro is indeed a negative imaginary number:

geometry={al}
basis={spd,al,vdz}
rhf
mcscf;wf,13,2,1;wf,13,3,1;wf,13,5,1
mrci;wf,13,2,1;save,5002.1
mrci;wf,13,3,1;save,5003.1
mrci;wf,13,5,1;save,5005.1
lsint
mrci;tranls,5003.1,5005.1,-1,1,lsx

Spin-orbit matrix elements for mean field operator:

!MRCI MATR. ELEMENT <1.3|LSX|1.5>     -0.00013562i A.U.            -29.76594258i CM-1

Spin-orbit matrix elements using full Breit-Pauli operator for internal part:

!MRCI MATR. ELEMENT <1.3|LSX|1.5>     -0.00013546i A.U.            -29.73009831i CM-1

However, exactly what you get of course depends on the phases of the
wavefunctions.  I think (but am not absolutely sure) that the MCSCF
code does indeed try to make the largest coefficient in each orbital
positive, and the CI code makes its largest coefficient positive. This
should be enough to ensure that your P_x, P_y, P_z states transform
like a right-handed set of cartesian axes!

Did I make a mistake?

Best wishes
Peter

millard alexander wrote at 11:55 on 3 January 2001:
> dear molpro users and molpro experts:
>
> i am determining the spin-orbit matrix elements for an atom in a 2P
> state (Aluminum).  when i calculate the matrix elements by hand,
> i find that the <py | lx sx | pz-bar > matrix element is i*a/2, where a
>  is the spin-orbit constant, and, similarly, the
>  <pz |ly sy |px-bar> matrix element is i*a/2
>
> when i run the following molpro input file
>
> ***,almrci
> memory,1,m
> file,1,al.i,scratch
> file,3,al.wf,new
>
> basis={spd,al,vdz}
>
> ncalc=0
> geometry={x,y;a1,al}
> saveorb=3514.3
>
> int;-;pri,-1
> lsint
> rhf;occ,5,1,1;wf,13,1,1;start,2410.3;save,2411.3;
> !orbprint
> multi;occ,6,2,2;closed,3,1,1;
> wf,13,1,1;wf,13,2,1;wf,13,3,1
> orbprint
> natorb,,ci
>
> e1A1mc(ncalc)=energy(1)
> e2A1mc(ncalc)=energy(2)
> e1B2mc(ncalc)=energy(3)
> ci;occ,6,2,2;closed,3,1,1
> wf,13,1,1;state,1
> ref,2
> ref,3
> save,3164.1
> e1A1ci(ncalc)=energy(1)
> e1A1d(ncalc)=energd(1)
> ci;occ,6,2,2;closed,3,1,1
> wf,13,2,1;state,1
> save,3174.1
> ref,1
> ref,3
> e1B2ci(ncalc)=energy(1)
> e1B2d(ncalc)=energd(1)
> ci;occ,6,2,2;closed,3,1,1
> wf,13,3,1;state,1
> save,3184.1
> ref,1
> ref,2
> e1B1ci(ncalc)=energy(1)
> e1B1d(ncalc)=energd(1)
> ci;tranls,3174.1,3184.1,1,1,lsz
> set,lsz_xy,trlsz
> ci;tranls,3184.1,3174.1,1,1,lsz
> set,lsz_yx,trlsz
> ci;tranls,3164.1,3174.1,1,-1,lsy
> set,lsy_zxb,trlsy
> ci;tranls,3174.1,3164.1,1,-1,lsy
> set,lsy_xzb,trlsy
> ci;tranls,3164.1,3184.1,1,-1,lsx
> set,lsx_zyb,trlsx
> ci;tranls,3184.1,3164.1,1,-1,lsx
> set,lsx_yzb,trlsx
> show,ls*
> ---
>
> i get
>
>  LSZ_XY                 =         -.00013559
>  LSZ_YX                 =          .00013559
>  LSY_ZXB                =         -.00013559
>  LSY_XZB                =          .00013559
>  LSX_ZYB                =          .00013559
>  LSX_YZB                =         -.00013559
>
> which corresponds to the opposite signs than i would predict
> for the matrix elements of lxsx and lysy, but the correct
> signs for the matrix elements of lzsz.
>
> perhaps i am making an error?
>
> any insight would be appreciated.
>
> millard alexander
>
> --
> Millard Alexander, Distinguished University Professor
> Department of Chemistry and Biochemistry
> University of Maryland, College Park, MD 20742-2021
> tel:  1.301.405.1823     fax:  1.301.314.9121
> email:  mha at mha-ibm4.umd.edu
> www:  http://www-mha.umd.edu/~mha
>
>

--
Prof. Peter J. Knowles
Email P.J.Knowles at bham.ac.uk  Phone +44-121-414-7472  Fax +44-121-414-7471
School of Chemistry, Univ. of Birmingham, Edgbaston, Birmingham, B15 2TT, UK.
WWW http://www.tc.bham.ac.uk/~peterk/