[molpro-user] Davidson and Pople correction in MRCI

Lorenzo Lodi l.lodi at ucl.ac.uk
Sun Mar 16 22:01:34 GMT 2008

I am trying to understand how the Davidson and Pople corrections are 
computed in MRCI. In the CISD signle-reference case I have no problems 
and I am able to reproduce the values given by molpro from the given 
C(0). I am also aware that what molpro calls "Davidson correction" isn't 
(1-C(0))^2 * E_corr as it was originally defined but (1-C(0))^2/C(0)^2 * 

For example, consider this small input for water:
r1=1.7, r2=1.9, theta=104.5
geometry={O;  H1,O,r1;  H2,O,r2,H1,theta}
{MCSCF; OCC, 7, 2 ; CLOSED, 1,0 ; FROZEN, 0,0}
{MRCI ; OCC, 7, 2 ; CORE, 1,0}

I get:
!MCSCF STATE 1.1 ENERGY              -76.141139431398
!MRCI STATE 1.1 ENERGY               -76.233689766943
Coefficient of reference function:   C(0) = 0.98999087
Cluster corrected energies           -76.23555251 (Davidson) 
-76.23555251 (externals)
                                       -76.23510092 (Pople) 
-76.23510092 (externals)

On the other hand, if I compute the correction from the formulae and the 
given C(0) I get

Davidson = -76.23557065536305
Pople (using n=8)=-76.2351148021805

Can anyone explain to me the reason of these discrepancies? 
Incidentally, if I try to derive C(0) working backwards from the MCSCF, 
MRCI and Davidson/Pople energies I get two contrasting numbers for the 
Davidson and Pople formula, namely C(0)=0.9899872295746449 and 
C(0)=0.9900859782702703 (all of which are different from the one given 
by molpro).

Any suggestion would be appreciated.

Lorenzo Lodi

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