Dear collegues, excuse me in advance for asking, probably, stupid questions, because I'm new in Dalton.
I tried to obtain energy of oxygen atom O(3P) and O(1D), but failed. For example, my singleinput file is:

BASIS
ccpVDZ
Oatom symmetries in D2h are: 3Pg: 4,6,7 1Dg: 1,1,4,6,7, 1Sg:1
1
8. 1
O .0 .0 .0
**DALTON
.RUN RESPONSE
**WAVE FUNCTIONS
.HF
.MP2
*SCF INPUT
.DOUBLY OCCUPIED
2 0 0 1 0 0 0 0
.SINGLY OCCUPIED
0 0 0 0 0 1 1 0
**END OF

gives in output file:
@ Spin multiplicity: 3
@ Spatial symmetry: 4 ( irrep B1g in D2h )
@ Final HF energy: 63.256994617873
this HF energy is very far from the energy of the ground state of the oxygen atom (near 74.7 a.u.). What is wrong?
best wishes, Alexey Chichinin
input for O(3P) and O(1D) atom

 Posts: 2
 Joined: 07 Dec 2015, 22:46
 First name(s): Alexey
 Last name(s): Chichinin
 Affiliation: Institute fur Physicalische und Theoretische Chemie, Braunschweig
 Country: Germany
Re: input for O(3P) and O(1D) atom
Did you that check the orbitals in the output?You will most likely find that your openshell orbitals are not porbitals, but dorbitals.
The porbitals show up in the ungerade symmetries (by default usually 2, 3, 5)
Regads,
Olav
The porbitals show up in the ungerade symmetries (by default usually 2, 3, 5)
Regads,
Olav

 Posts: 589
 Joined: 15 Oct 2013, 05:37
 First name(s): Peter
 Middle name(s): Robert
 Last name(s): Taylor
 Affiliation: Tianjin University
 Country: China
Re: input for O(3P) and O(1D) atom
Olav has already answered and is certainly correct: can I suggest also that you look at the detailed manual section on symmetry in Dalton (disclaimer: I wrote it...)? The comment in your input about the irreps for the O states is correct only for the manyelectron states, 3Pg and 1Dg. But in the SCF input you specify the symmetry of the individual orbitals that comprise your open shells, not the overall symmetry of the state.
Best regards
Pete
Best regards
Pete

 Posts: 2
 Joined: 07 Dec 2015, 22:46
 First name(s): Alexey
 Last name(s): Chichinin
 Affiliation: Institute fur Physicalische und Theoretische Chemie, Braunschweig
 Country: Germany
Re: input for O(3P) and O(1D) atom
Dear Olav, dear Pete, thank you very much for your very informative reply. Now I can make a singleinput files to obtain Oatom properties. Here is one of them:
BASIS
ccpVDZ
Oatom symmetries in D2h: 3Pg:4,6,7; 1Dg:1,1,
4,6,7; 1Sg:1. Single Pu orbital is 2,3,5.
Atomtypes=1
8. 1
O .0 .0 .0
**DALTON
.RUN RESPONSE
**WAVE FUNCTIONS
.HF
.MP2
*SCF INPUT
.DOUBLY OCCUPIED
2 1 0 0 0 0 0 0
.SINGLY OCCUPIED
0 0 1 0 1 0 0 0 0
**END OF
I made all possible combinations of doubly/singly occupied orbitals and obtained the results:
Doubly: 21000000 Singly: 00101000 > 3Xg, Sym=7, E=74.7861742664
Doubly: 20001000 Singly: 01100000 > 3Xg, Sym=4, E=74.7875130746
Doubly: 20100000 Singly: 01001000 > 3Xg, Sym=6, E=74.7861742664
Doubly: 21100000 Singly: 00000000 > 1Xg, Sym=1, E=74.665278726428
Doubly: 20101000 Singly: 00000000 > 1Xg, Sym=1, E=74.665278726428
Doubly: 21001000 Singly: 00000000 > 1Xg, Sym=1, E=74.665278726428
First three solutions are ground state triplet, they have correct symmetry and are evidently O(3P).
The next three solutions have the same energy, spin, symmetry, and they are higher 3P by
0.1208955*27.2116= 3.29 eV. I can not understand what they mean.
(I also don't know, where in *.out file to read orbital moment of the atom (S, P, or D?) )
Note that in oxygen \Delta E ( 1D3P)= 2 eV, and \Delta E(1S1D)= 2 eV, and hence the 3 states of symmetry=1
can not have the same energy.
Then I proposed to Dalton to find correct states by CAS SCF:
BASIS
ccpVDZ
Oatom symmetries in D2h: 3Pg:4,6,7; 1Dg:1,1,
4,6,7; 1Sg:1. Single Pu orbital is 2,3,5.
Atomtypes=1
8. 1
O .0 .0 .0
**DALTON
.RUN WAVE FUNCTIONS
**WAVE FUNCTIONS
.MCSCF
*CONFIGURATION INPUT
.SYMMETRY
1
.SPIN MULTIPLICITY
1
.INACTIVE ORBITALS
2 0 0 0 0 0 0 0
.ELECTRONS
4
.CAS SPACE READ
0 1 1 0 1 0 0 0
*POPULATION ANALYSIS
.ALL
**END OF DALTON INPUT
and obtained 1Xg symmetry=1, Final MCSCF energy: 74.705522205724, which is 2.2 eV above the ground state O(3P). I like to think that I have obtained here O(1D), because of correct energy, but I don't know if I am right.
But why it have not found ground state O(3P)?
Help me please, how to obtain O(1D) energy and wave functions?
Also, what should I add in last scropt to see properties of all excited states ?
Best wishes, Alexey
BASIS
ccpVDZ
Oatom symmetries in D2h: 3Pg:4,6,7; 1Dg:1,1,
4,6,7; 1Sg:1. Single Pu orbital is 2,3,5.
Atomtypes=1
8. 1
O .0 .0 .0
**DALTON
.RUN RESPONSE
**WAVE FUNCTIONS
.HF
.MP2
*SCF INPUT
.DOUBLY OCCUPIED
2 1 0 0 0 0 0 0
.SINGLY OCCUPIED
0 0 1 0 1 0 0 0 0
**END OF
I made all possible combinations of doubly/singly occupied orbitals and obtained the results:
Doubly: 21000000 Singly: 00101000 > 3Xg, Sym=7, E=74.7861742664
Doubly: 20001000 Singly: 01100000 > 3Xg, Sym=4, E=74.7875130746
Doubly: 20100000 Singly: 01001000 > 3Xg, Sym=6, E=74.7861742664
Doubly: 21100000 Singly: 00000000 > 1Xg, Sym=1, E=74.665278726428
Doubly: 20101000 Singly: 00000000 > 1Xg, Sym=1, E=74.665278726428
Doubly: 21001000 Singly: 00000000 > 1Xg, Sym=1, E=74.665278726428
First three solutions are ground state triplet, they have correct symmetry and are evidently O(3P).
The next three solutions have the same energy, spin, symmetry, and they are higher 3P by
0.1208955*27.2116= 3.29 eV. I can not understand what they mean.
(I also don't know, where in *.out file to read orbital moment of the atom (S, P, or D?) )
Note that in oxygen \Delta E ( 1D3P)= 2 eV, and \Delta E(1S1D)= 2 eV, and hence the 3 states of symmetry=1
can not have the same energy.
Then I proposed to Dalton to find correct states by CAS SCF:
BASIS
ccpVDZ
Oatom symmetries in D2h: 3Pg:4,6,7; 1Dg:1,1,
4,6,7; 1Sg:1. Single Pu orbital is 2,3,5.
Atomtypes=1
8. 1
O .0 .0 .0
**DALTON
.RUN WAVE FUNCTIONS
**WAVE FUNCTIONS
.MCSCF
*CONFIGURATION INPUT
.SYMMETRY
1
.SPIN MULTIPLICITY
1
.INACTIVE ORBITALS
2 0 0 0 0 0 0 0
.ELECTRONS
4
.CAS SPACE READ
0 1 1 0 1 0 0 0
*POPULATION ANALYSIS
.ALL
**END OF DALTON INPUT
and obtained 1Xg symmetry=1, Final MCSCF energy: 74.705522205724, which is 2.2 eV above the ground state O(3P). I like to think that I have obtained here O(1D), because of correct energy, but I don't know if I am right.
But why it have not found ground state O(3P)?
Help me please, how to obtain O(1D) energy and wave functions?
Also, what should I add in last scropt to see properties of all excited states ?
Best wishes, Alexey

 Posts: 589
 Joined: 15 Oct 2013, 05:37
 First name(s): Peter
 Middle name(s): Robert
 Last name(s): Taylor
 Affiliation: Tianjin University
 Country: China
Re: input for O(3P) and O(1D) atom
With Dalton you specify the spin multiplicity for the state of interest in your CASSCF calculation, as you have done by explicitly choosing the multiplicity 1. This is a singlet in you (that is, what you specify is 2S+1, where S is the total spin, not just the M_S value). If you want to get the components of the 3P state, you must specify the spin multiplicity as 3. Thus the 3P and 1D will need to be done in two separate calculations.
By the way, these "CAS" calculations are single configurations, and since no correlation at all is included one would expect, as is typical with HartreeFock, that higher multiplicity states will be favoured. Hence you should expect the 1D state to come out somewhat further above 3P than is observed in nature.
Best regards
Pete
By the way, these "CAS" calculations are single configurations, and since no correlation at all is included one would expect, as is typical with HartreeFock, that higher multiplicity states will be favoured. Hence you should expect the 1D state to come out somewhat further above 3P than is observed in nature.
Best regards
Pete
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