I am trying to calculate the spin-orbit coupling value for cyclobutadine. Here is my input.
# casscf=(2,2,spinorbit)/3-21g
But it is stopping at
Enter MCSCF program.
NO. OF ORBITALS = 44 NO. OF CORE-ORBITALS = 13
NO. OF VALENCE-ORBITALS = 2 NO. OF VIRTUAL-ORBITALS = 29
USED ACCURACY IN CHECKING CONVERGENCE = 1.00D-08
I am not sure what exactly the issue is? I think i am missing something in the input. Kindly help me out.
Mr. Shekar,
You should write CASSCF(x,x,spin)/... in your input file in order to receive a table:
*****************************
Spin-Orbit coupling program.
*****************************
.................
in the output.
But in your question, there are not shown the route sections of the "links". SO couplings are obtained by multi-step computations.
Please pay attention to an example of H2C=O at RHF/SDD level of theory:
........................
EIGENVALUES AND EIGENVECTORS OF CI MATRIX
( 1) EIGENVALUE -113.8230793853
( 1) 0.9995226 ( 4)-0.0308950 ( 3) 0.0000000 ( 2) 0.0000000 (
( 2) EIGENVALUE -113.7208016287
( 2) 0.7071068 ( 3)-0.7071068 ( 1) 0.0000000 ( 4) 0.0000000 (
Final one electron symbolic density matrix:
1 2
1 0.100000D+01
2 0.148940D-23 0.100000D+01
*****************************
Spin-Orbit coupling program.
*****************************
Number of configs= 4
1st state is 1
2nd state is 2
Transition Spin Density Matrix
1 2
1 -0.514078D-12 0.141354D+01
2 0.436922D-01 0.514078D-12
Magnitude in x-direction= 0.0 cm-1
Magnitude in y-direction= 0.0 cm-1
Magnitude in z-direction= 70.6 cm-1
Total magnitude= 70.6 cm-1
...............................
Thanks Prof. Ivanova for your reply. Here is the input file that i was using. I am unable to figure out what exactly the problem in my input?
-------------------------------------------------------------------------------
%nprocshared=4
%chk=cyclobutadiene_guess_casscf_ex1.chk
# casscf(2,4,spinorbit)/6-31+G(d) guess=(alter) pop=naturalorbital test
Title Card Required
0 1
C 0.80110400 0.66169600 0.00004000
C -0.80110700 0.66169400 0.00015600
C -0.80110600 -0.66169600 -0.00004200
C 0.80110800 -0.66169300 -0.00015600
H 1.55022800 1.41963300 0.00010300
H -1.55020800 1.41965200 0.00034300
H -1.55023000 -1.41963200 -0.00010200
H 1.55021000 -1.41965100 -0.00032900
12,13
16,28
--Link1--
%chk=cyclobutadiene_guess_casscf_ex1.chk
# casscf(2,4,nroot=2,spinorbit)/6-31+G(d) pop=naturalorbital test
Title card
0 1
C 0.80110400 0.66169600 0.00004000
C -0.80110700 0.66169400 0.00015600
C -0.80110600 -0.66169600 -0.00004200
C 0.80110800 -0.66169300 -0.00015600
H 1.55022800 1.41963300 0.00010300
H -1.55020800 1.41965200 0.00034300
H -1.55023000 -1.41963200 -0.00010200
H 1.55021000 -1.41965100 -0.00032900
------------------------------------------------------------------------
Thanks in advance.
Mr. Shekar,
To compute spin orbit coupling between states, you should identify these states before.
For identifying, you should enter your MCSCF by specifying (NO. OF ORBITALS, NO. OF CORE-ORBITALS,
NO. OF VALENCE-ORBITALS and NO. OF VIRTUAL-ORBITALS.
Mr. Shekar,
You, in fact, have provided a Gaussian job with link of two almost identical operations, only involving "Guess=Alter", which is used to modify the orbitals, towards occupancy. So, why you have used it...?!
In the second link, you have given pop=nbo analysis, but you do not need it?! The NBO analysis you can do as a third link, using optimized geometry of your molecule.
If your idea is account for perturbation towards the occupancy and % distribution as Mr. Sabor pointed out,you should determine those perturbation preliminary. In your inputs, however are shown fixed number of orbitals 4 and electrons 2.
Please pay attention to Gaussian manual, about details associated with CASSCF, and input parameters:
http://www.gaussian.com/g_tech/g_ur/k_casscf.htm
Despite, this, you should obtain, computing your system as it is given in yout posting, the following data (the optimization has been carried out by RHF/SDD method; CASSCF(2,4)):
......................
Charge = 0 Multiplicity = 1
C,0,0.801104,0.661696,0.00004
C,0,-0.801107,0.661694,0.000156
C,0,-0.801106,-0.661696,-0.000042
C,0,0.801108,-0.661693,-0.000156
H,0,1.550228,1.419633,0.000103
H,0,-1.550208,1.419652,0.000343
H,0,-1.55023,-1.419632,-0.000102
H,0,1.55021,-1.419651,-0.000329
Recover connectivity data from disk.
NAtoms= 8 NQM= 8 NQMF= 0 NMic= 0 NMicF= 0 NTot= 8.
Isotopes and Nuclear Properties:
(Nuclear quadrupole moments (NQMom) in fm**2, nuclear magnetic moments (NMagM)
in nuclear magnetons)
Atom 1 2 3 4 5 6 7 8
IAtWgt= 12 12 12 12 1 1 1 1
AtmWgt= 12.0000000 12.0000000 12.0000000 12.0000000 1.0078250 1.0078250 1.0078250 1.0078250
NucSpn= 0 0 0 0 1 1 1 1
AtZEff= -3.6000000 -3.6000000 -3.6000000 -3.6000000 -1.0000000 -1.0000000 -1.0000000 -1.0000000
NQMom= 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000
NMagM= 0.0000000 0.0000000 0.0000000 0.0000000 2.7928460 2.7928460 2.7928460 2.7928460
Leave Link 101 at Thu Apr 21 07:51:19 2016, MaxMem= 33554432 cpu: 0.0
(Enter C:\g09w\l202.exe)
Input orientation:
---------------------------------------------------------------------
Center Atomic Atomic Coordinates (Angstroms)
Number Number Type X Y Z
---------------------------------------------------------------------
1 6 0 0.801104 0.661696 0.000040
2 6 0 -0.801107 0.661694 0.000156
3 6 0 -0.801106 -0.661696 -0.000042
4 6 0 0.801108 -0.661693 -0.000156
5 1 0 1.550228 1.419633 0.000103
6 1 0 -1.550208 1.419652 0.000343
7 1 0 -1.550230 -1.419632 -0.000102
8 1 0 1.550210 -1.419651 -0.000329
---------------------------------------------------------------------
................................
Truncation Level= 99999
a= 1 b= 0 c= 3
a=N/2 - s b=2s c=n- (a+b)
no. active orbitals (n) 4
no. active ELECTRONS (N)= 2
SLATER DETERMINANT BASIS
IRREPS TO BE RETAINED = 1
GROUP IRREP. MULT. TABLE
1
IRREP. LABELS FOR ORBITALS
1 1 1 1
BOTTOM WEIGHT= 2 TOP WEIGHT= 8
Configuration 1 Symmetry 1 1000
Configuration 2 Symmetry 1 ba00
Configuration 3 Symmetry 1 ab00
Configuration 4 Symmetry 1 b0a0
Configuration 5 Symmetry 1 0100
Configuration 6 Symmetry 1 a0b0
Configuration 7 Symmetry 1 b00a
Configuration 8 Symmetry 1 0ba0
Configuration 9 Symmetry 1 0ab0
Configuration 10 Symmetry 1 a00b
Configuration 11 Symmetry 1 0b0a
Configuration 12 Symmetry 1 0010
Configuration 13 Symmetry 1 0a0b
Configuration 14 Symmetry 1 00ba
Configuration 15 Symmetry 1 00ab
Configuration 16 Symmetry 1 0001
NO OF BASIS FUNCTIONS = 16 NO TO BE DELETED = 0
CI Matrix Elements calculated here
NO. OF CONFIGURATIONS IN REFERENCE SPACE = 1
SECONDARY SPACE = 16
TERTIARY SPACE = 16
NO. OF ORBITALS = 4
NO. OF ELECTRONS = 2
NO. OF WEIGHTS = 7
REFERENCE STATE CONFIGURATIONS ARE: 0
NO. OF CORE ORBITALS = 0
OPTION: NON-DIAGONAL HOLE LINE INTERACTIONS INCLUDED
...........................................
State Average Calculation. The weights are:
St.: 1 w.=0.500000 # St.: 2 w.=0.500000 # St.:
2ND ORD PT ENERGY CV -0.000010 CU -0.000595 UV -0.005057
TOTAL -153.528833
ITN= 1 MaxIt= 64 E= -153.5231709921 DE=-1.54D+02 Acc= 1.00D-08 Lan= 0
ITN= 2 MaxIt= 64 E= -153.5328475836 DE=-9.68D-03 Acc= 1.00D-08 Lan= 0
ITN= 3 MaxIt= 64 E= -153.5334652118 DE=-6.18D-04 Acc= 1.00D-08 Lan= 0
ITN= 4 MaxIt= 64 E= -153.5336082275 DE=-1.43D-04 Acc= 1.00D-08 Lan= 0
ITN= 5 MaxIt= 64 E= -153.5335740001 DE= 3.42D-05 Acc= 1.00D-08 Lan= 0
ITN= 6 MaxIt= 64 E= -153.5335704618 DE= 3.54D-06 Acc= 1.00D-08 Lan= 0
ITN= 7 MaxIt= 64 E= -153.5335628309 DE= 7.63D-06 Acc= 1.00D-08 Lan= 0
ITN= 8 MaxIt= 64 E= -153.5335614321 DE= 1.40D-06 Acc= 1.00D-08 Lan= 0
ITN= 9 MaxIt= 64 E= -153.5335602326 DE= 1.20D-06 Acc= 1.00D-08 Lan= 0
ITN= 10 MaxIt= 64 E= -153.5335599537 DE= 2.79D-07 Acc= 1.00D-08 Lan= 0
ITN= 11 MaxIt= 64 E= -153.5335597677 DE= 1.86D-07 Acc= 1.00D-08 Lan= 0
ITN= 12 MaxIt= 64 E= -153.5335597163 DE= 5.14D-08 Acc= 1.00D-08 Lan= 0
ITN= 13 MaxIt= 64 E= -153.5335596867 DE= 2.96D-08 Acc= 1.00D-08 Lan= 0
ITN= 14 MaxIt= 64 E= -153.5335596774 DE= 9.27D-09 Acc= 1.00D-08 Lan= 0
... Do an extra-iteration for final printing.
EIGENVALUES AND EIGENVECTORS OF CI MATRIX
( 1) EIGENVALUE -153.6053427533
( 1) 0.9824534 ( 5)-0.1563767 ( 16)-0.1010426 ( 12)-0.0110488 ( 13)-0.0000176 ( 11)-0.0000176 ( 7)-0.0000003
( 10)-0.0000003 ( 8)-0.0000001 ( 9)-0.0000001 ( 6) 0.0000000 ( 4) 0.0000000 ( 3) 0.0000000 ( 2) 0.0000000
( 14) 0.0000000 ( 15) 0.0000000 (
( 2) EIGENVALUE -153.5335596726
( 3)-0.7071068 ( 2) 0.7071068 ( 7) 0.0000037 ( 10)-0.0000037 ( 11)-0.0000003 ( 13) 0.0000003 ( 14) 0.0000002
( 15)-0.0000002 ( 9) 0.0000000 ( 8) 0.0000000 ( 6) 0.0000000 ( 4) 0.0000000 ( 5) 0.0000000 ( 1) 0.0000000
( 12) 0.0000000 ( 16) 0.0000000 (
Final one electron symbolic density matrix:
1 2 3 4
1 0.100000D+01
2 -0.487174D-11 0.100000D+01
3 -0.845368D-07 -0.678305D-07 0.630368D-13
4 0.938980D-06 0.103706D-04 -0.391409D-12 0.271676D-10
*****************************
Spin-Orbit coupling program.
*****************************
Number of configs= 16
1st state is 1
2nd state is 2
Transition Spin Density Matrix
1 2 3 4
1 -0.380396D-10 0.138940D+01 -0.471216D-07 0.000000D+00
2 0.221150D+00 0.283380D-10 -0.934185D-08 0.103828D-06
3 0.142871D-06 0.314398D-07 -0.711049D-14 -0.383036D-08
4 0.000000D+00 -0.443672D-06 0.350302D-07 0.970863D-11
Magnitude in x-direction= 0.0 cm-1
Magnitude in y-direction= 0.0 cm-1
Magnitude in z-direction= 0.8 cm-1
Total magnitude= 0.8 cm-1
MCSCF converged.
...................................
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