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Thomas
thomaspijper@hotmail.com

Sayna, let me see if I get it now. MCQDPT uses the optimized CASSCF orbitals, which are determined by the values of WSTATE and NSTATE. Because states might get reordered during the MCQDPT step, it is best to include more states than needed by setting NSTATE to a high enough value (e.g. NSTATE=10). Then, during the MCQDPT step, the effective Hamiltonian should include enough (at least 10) states in order to obtain good results.

So, when trying to calculating the energy of the first excited state, the CASSCF/MCQDPT part of the input file could look like this (state-specific example):


$DET GROUP=C1 NCORE=19 NACT=6 NELS=6 NSTATE=10 WSTATE(1)=0,1 $END
$MCQDPT KSTATE(1)=1,1,1,1,1,1,1,1,1,1 EDSHFT=0.02 $END

(in which KSTATE(1)=1,1,… could be also written as NSTATE=10)


That leaves me with two questions about the output of a MCQDPT calculation. First, the above method gave me a list of energies:


*** MC-QDPT2 ENERGIES ***
-----------------------------------------------------------------------
1 E(MCSCF)= -1365.5486114818 E(MP2)= -1367.1030309626
2 E(MCSCF)= -1365.3205961740 E(MP2)= -1367.0468868024
3 E(MCSCF)= -1365.2696589473 E(MP2)= -1366.9541974175
4 E(MCSCF)= -1365.2608241655 E(MP2)= -1366.8545831671
5 E(MCSCF)= -1365.2519965112 E(MP2)= -1366.8430834090
6 E(MCSCF)= -1365.2308933371 E(MP2)= -1366.8176982023
7 E(MCSCF)= -1365.2266705813 E(MP2)= -1366.7604573556
8 E(MCSCF)= -1365.2085431094 E(MP2)= -1366.7489911227
9 E(MCSCF)= -1365.2074885849 E(MP2)= -1366.7273932680
10 E(MCSCF)= -1365.2018793382 E(MP2)= -1366.2020115102


Are these energies of roots or of CI states (I'm guessing roots)?

Second question: the output also contained 10 sections called "Properties for the zero-order QDPT density". Here's one of them:

WAVEFUNCTION NORMALIZATION = 1.0000000000

ONE ELECTRON ENERGY = -4703.5993126161
TWO ELECTRON ENERGY = 1914.3537631053
NUCLEAR REPULSION ENERGY = 1423.7688450077
------------------
TOTAL ENERGY = -1365.4767045030

ELECTRON-ELECTRON POTENTIAL ENERGY = 1914.3537631053
NUCLEUS-ELECTRON POTENTIAL ENERGY = -6062.3804478862
NUCLEUS-NUCLEUS POTENTIAL ENERGY = 1423.7688450077
------------------
TOTAL POTENTIAL ENERGY = -2724.2578397731
TOTAL KINETIC ENERGY = 1358.7811352701
VIRIAL RATIO (V/T) = 2.0049276289


What exactly are these energies? They seem to be CASSCF energies, but they don't match those in the "*** MC-QDPT2 ENERGIES ***" section above.

Again, thank you very much for your help.


Kind regards,
Thom



On Tue Oct 19 '10 2:02pm, sanya wrote
-------------------------------------
>>This part I find a bit confusing, perhaps because we're talking about both states and roots. From what I understand, both WSTATE (in $DET) and KSTATE (in $MCQDPT) refer to roots, whereas NSTATE (in $DET) refers to CI states. I assume that when you speak of "no less than 10 states", you are talking about NSTATE in $DET (NSTATE=10 or higher).

>No, no! I was talking about KSTATE. In fact, I see no difference between roots (solutions of CI equations) and states (linear combinations of Slater determinants on optimized orbitals with the corresponding CI coefficients) at CASSCF stage. In MCQDPT, the CI roots included in the effective Hamiltonian are corrected in the second order of perturbation theory (which includes their mixing, reordering, etc.). Therefore, PT-corrected states may go far from the initial CASSCF states (CI roots).

>NSTATE in $DET ($DRT) defines how many states (CI roots) we want to calculate. WSTATE defines the states whose densities are averaged to obtain the optimized orbitals (state-averaged CASSCF). I.e., if you need the ground and lowest excited state, WSTATE(1)=1.,1. NSTATE is always greater or equal to WSTATE. The number of states in the active space is just the number of all possible excitations from occupied active orbitals to unoccupied ones; this number grows dramatically with the size of the active space. The program prints the number of CSFs (for GUGA) or determinants (for ALDET), this equals the number of states in the active space and may range from 4 (for CAS(2,2)) to several millions and even more.

>NSTATE from $DET or $DRT as well as WSTATE is not used by MCQDPT routine. MCQDPT uses only optimized orbitals (which depend on both NSTATE and WSTATE). KSTATE defines the effective Hamiltonian (which appears at the MCQDPT stage). It is the size of the effective Hamiltonian that should be large enough to obtain sensible results. For two states (ground and first excited state), it should include no less than 10 states (KSTATE=1,1,1,1,1,1,1,1,1,1).

>>Having that said, I would like to provide the following hypothetical case to see if I got the point: the cyclization of hexatriene to cyclohexadiene. This is a 6-electron pericyclic reaction, so CASSCF(6,6) would be a good active space for this. May I ask you to comment on the following input sections which I would use for such a case? (BTW, I do understand that for bigger systems it is better to run the CASSCF part parallel and the MCQDPT part multithreaded in two separate runs).

>CASSCF(6,6) is a minimum active space for this system. If possible, it is advisable to add some extra orbitals (usually, pi) for better description of the active orbitals. Anyway, if you face problems (poor convergence, different starting MOs leading to different CASSCF solutions, some orbitals initially included in the active space being missing after iterations, etc.), first try to expand the active space.

>Some more points about WSTATE. WSTATE(1)=1.,0.,... defines a pure state-specific CASSCF for the ground state. But if you want to perform a state-specific CASSCF for the first excited state, WSTATE(1)=0.,1. may cause problems. In such cases, it is advisable to set WSTATE(1)=0.1,1.0 to perform an "almost state-specific" CASSCF for the excited state (and, correspondingly, WSTATE(1)=1.0,0.1 for the ground state for consistency). But state-averaged CASSCF (WSTATE(1)=1.,1.) is better.

>How many states should be included in WSTATE, is also not so evident. In some cases, CASSCF is not a good approximation. The order of roots may be not quite correct. If all the necessary states are included in the effective Hamiltonian, they'll be corrected by MCQDPT and go into their places. But if some of them are not included in WSTATE (say, the necessary state is #7 and WSTATE included only 4 lowest states), their density is not taken into account in orbital optimization, and these states just fall out of the active space. No state -- nothing to correct. So, for the first run, set NSTATE certainly larger than you'll need in all further calculations (say, 10 instead of 5) and look at all the CI roots. Probably, the state you need is higher than you expect. In further runs, you may set NSTATE as necessary.