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Thomas

thomaspijper@hotmail.com

Dear Sanya,

Again, thank you for your reply, I’m learning a lot here.

I’ve attached an MCQDPT2 output file this time. The input file looks like this:

$CONTRL SCFTYP=MCSCF RUNTYP=ENERGY MPLEVL=2 UNITS=ANGS $END

$SYSTEM TIMLIM=300 MEMORY=3000000 MKLNP=2 $END

$P2P P2P=.True. DLB=.True. $end

$BASIS GBASIS=N21 NGAUSS=3 $END

$MCSCF CISTEP=ALDET NTRACK=2 $END

$DET NCORE=19 NACT=6 NELS=6 NSTATE=10 WSTATE(1)=1,1 $END

$MCQDPT KSTATE(1)=1,1,1,1,1,1,1,1,1,1 EDSHFT=0.02 $END

$GUESS GUESS=MOREAD NORB=70 $END

> After setting NSTATE=10, look carefully at the CI

> states (either before or after CASSCF iterations)

> and make sure that the state you need is among the

> averaged states (i.e., if you are averaging over 2

> states, make sure that the state you are interested

> in is #1 or #2; otherwise, you should average over

> more states).

Is it really necessary for them to be #1 or #2? Here’s how I would interpret the output file:

So, I’m interested in the energies of the two lowest roots (the ground state and the first excited state). These seem to be states #1 and #4. During the CASSCF part of the calculation, Firefly reports:

THE DENSITIES ARE STATE AVERAGED OVER 2 ROOT(S)

STATE= 1 ENERGY= -230.6178180831 WEIGHT= 0.50000 S= 0.00

STATE= 4 ENERGY= -230.3357925186 WEIGHT= 0.50000 S= 0.00

So, the correct states are being averaged. I also notice that, between the unconverged and converged CASSCF, states #5 and #6 switch places, but the roots themselves stay in the same order, and besides I’m only interested in states 1 and 4, so this should not affect the results of the calculation.

After the MCQDPT part, I get the following lines:

1 E(MCSCF)= -230.6137982782 E(MP2)= -231.1030362807

2 E(MCSCF)= -230.3641630425 E(MP2)= -230.8788353319

...etc.

The “E(MCSCF)” energies correspond to those of states #1 and #4, the first and second roots in the CASSCF part. And the corresponding “E(MP2)” energies are the two lowest energies in the list. So, those are the energies of the two lowest roots.

Does that make sense?

> I may be wrong, but I see no difference between

> the CI roots and states.

The chances that I’m wrong are much, much higher, but still I find it hard to see states and roots as the same if Firefly seemingly treats them differently. :-/

Thank you for bearing with me.

Kind regards,

Thom

On Thu Oct 21 '10 1:05pm, sanya wrote

-------------------------------------

>>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).

>Not exactly. After setting NSTATE=10, look carefully at the CI states (either before or after CASSCF iterations) and make sure that the state you need is among the averaged states (i.e., if you are averaging over 2 states, make sure that the state you are interested in is #1 or #2; otherwise, you should average over more states). In further runs (for example, if CASSCF has not converged in your first run), you may reduce NSTATE to a desired minimum (but leave a couple of extra states to be on the safe side). In general, the first run is done to get the idea of how many states to average (hence, you do not need to wait for CASSCF to converge), and in further runs you will set correct WSTATE and, possibly, smaller NSTATE. If your active space and/or basis set is large, excessive NSTATE may require extra disk space.

>>Then, during the MCQDPT step, the effective Hamiltonian should include enough (at least 10) states in order to obtain good results.

>Yes, and the size of the effective Hamiltonian does not depend on NSTATE in $DET/$DRT

>>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,â€¦ could be also written as NSTATE=10)

>>

>Not quite so, again. WSTATE(1)=0,1 may cause convergence problems. It is better to set WSTATE(1)=1.0,0.1 for the first state and WSTATE(1)=0.1,1.0 for the second state. Or WSTATE(1)=1.,1. for both states.

>>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)?

>I may be wrong, but I see no difference between the CI roots and states. And, I'm afraid, the lines you cited do not adequately show how QDPT procedure corrected (and reordered) the states. I mean that the energies are just listed in ascending order, while the character of, say, state #2 may be different in MCSCF and MCQDPT.

>>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.

>CASSCF wavefunction is a zero-order approximation to MCQDPT. I'm not sure but I guess that this is a state-averaged CASSCF energy. What WSTATE was in the corresponding MCSCF run?

This message contains the 104 kb attachment [ OUTPUT_1.rar ] SA-CASSCF/MCQDPT2 output file |

Thu Oct 21 '10 4:25pm

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