----- GAMESS execution script 'rungms' ----- This job is running on host PaulDirac under operating system Linux at Thu Jun 20 13:28:15 EDT 2013 Available scratch disk space (Kbyte units) at beginning of the job is Filesystem 1K-blocks Used Available Use% Mounted on /dev/sdb1 709627400 192661664 480337136 29% /mnt/disk2 GAMESS temporary binary files will be written to /mnt/disk2/nikita/scr GAMESS supplementary output files will be written to /mnt/disk2/nikita/scr Copying input file exam20.inp to your run's scratch directory... cp tests/standard/exam20.inp /mnt/disk2/nikita/scr/exam20.F05 unset echo /mnt/disk2/nikita/gamess/ddikick.x /mnt/disk2/nikita/gamess/gamess.00.x exam20 -ddi 1 1 PaulDirac -scr /mnt/disk2/nikita/scr Distributed Data Interface kickoff program. Initiating 1 compute processes on 1 nodes to run the following command: /mnt/disk2/nikita/gamess/gamess.00.x exam20 ****************************************************** * GAMESS VERSION = 1 MAY 2013 (R1) * * FROM IOWA STATE UNIVERSITY * * M.W.SCHMIDT, K.K.BALDRIDGE, J.A.BOATZ, S.T.ELBERT, * * M.S.GORDON, J.H.JENSEN, S.KOSEKI, N.MATSUNAGA, * * K.A.NGUYEN, S.J.SU, T.L.WINDUS, * * TOGETHER WITH M.DUPUIS, J.A.MONTGOMERY * * J.COMPUT.CHEM. 14, 1347-1363(1993) * **************** 64 BIT INTEL VERSION **************** SINCE 1993, STUDENTS AND POSTDOCS WORKING AT IOWA STATE UNIVERSITY AND ALSO IN THEIR VARIOUS JOBS AFTER LEAVING ISU HAVE MADE IMPORTANT CONTRIBUTIONS TO THE CODE: IVANA ADAMOVIC, CHRISTINE AIKENS, YURI ALEXEEV, POOJA ARORA, ANDREY ASADCHEV, ROB BELL, PRADIPTA BANDYOPADHYAY, JONATHAN BENTZ, BRETT BODE, GALINA CHABAN, WEI CHEN, CHEOL HO CHOI, PAUL DAY, ALBERT DEFUSCO, TIM DUDLEY, DMITRI FEDOROV, GRAHAM FLETCHER, MARK FREITAG, KURT GLAESEMANN, DAN KEMP, GRANT MERRILL, NORIYUKI MINEZAWA, JONATHAN MULLIN, TAKESHI NAGATA, SEAN NEDD, HEATHER NETZLOFF, BOSILJKA NJEGIC, RYAN OLSON, MIKE PAK, JIM SHOEMAKER, LYUDMILA SLIPCHENKO, SAROM SOK, JIE SONG, TETSUYA TAKETSUGU, SIMON WEBB, SOOHAENG YOO, FEDERICO ZAHARIEV ADDITIONAL CODE HAS BEEN PROVIDED BY COLLABORATORS IN OTHER GROUPS: IOWA STATE UNIVERSITY: JOE IVANIC, LAIMUTIS BYTAUTAS, KLAUS RUEDENBERG UNIVERSITY OF TOKYO: KIMIHIKO HIRAO, TAKAHITO NAKAJIMA, TAKAO TSUNEDA, MUNEAKI KAMIYA, SUSUMU YANAGISAWA, KIYOSHI YAGI, MAHITO CHIBA, SEIKEN TOKURA, NAOAKI KAWAKAMI UNIVERSITY OF AARHUS: FRANK JENSEN UNIVERSITY OF IOWA: VISVALDAS KAIRYS, HUI LI NATIONAL INST. OF STANDARDS AND TECHNOLOGY: WALT STEVENS, DAVID GARMER UNIVERSITY OF PISA: BENEDETTA MENNUCCI, JACOPO TOMASI UNIVERSITY OF MEMPHIS: HENRY KURTZ, PRAKASHAN KORAMBATH UNIVERSITY OF ALBERTA: TOBY ZENG, MARIUSZ KLOBUKOWSKI UNIVERSITY OF NEW ENGLAND: MARK SPACKMAN MIE UNIVERSITY: HIROAKI UMEDA MICHIGAN STATE UNIVERSITY: KAROL KOWALSKI, MARTA WLOCH, JEFFREY GOUR, JESSE LUTZ, WEI LI, PIOTR PIECUCH UNIVERSITY OF SILESIA: MONIKA MUSIAL, STANISLAW KUCHARSKI FACULTES UNIVERSITAIRES NOTRE-DAME DE LA PAIX: OLIVIER QUINET, BENOIT CHAMPAGNE UNIVERSITY OF CALIFORNIA - SANTA BARBARA: BERNARD KIRTMAN INSTITUTE FOR MOLECULAR SCIENCE: KAZUYA ISHIMURA, MICHIO KATOUDA, AND SHIGERU NAGASE UNIVERSITY OF NOTRE DAME: DAN CHIPMAN KYUSHU UNIVERSITY: HARUYUKI NAKANO, FENG LONG GU, JACEK KORCHOWIEC, MARCIN MAKOWSKI, AND YURIKO AOKI, HIROTOSHI MORI AND EISAKU MIYOSHI PENNSYLVANIA STATE UNIVERSITY: TZVETELIN IORDANOV, CHET SWALINA, JONATHAN SKONE, SHARON HAMMES-SCHIFFER WASEDA UNIVERSITY: MASATO KOBAYASHI, TOMOKO AKAMA, TSUGUKI TOUMA, TAKESHI YOSHIKAWA, YASUHIRO IKABATA, HIROMI NAKAI NANJING UNIVERSITY: SHUHUA LI UNIVERSITY OF NEBRASKA: PEIFENG SU, DEJUN SI, NANDUN THELLAMUREGE, YALI WANG, HUI LI UNIVERSITY OF ZURICH: ROBERTO PEVERATI, KIM BALDRIDGE N. COPERNICUS UNIVERSITY AND JACKSON STATE UNIVERSITY: MARIA BARYSZ EXECUTION OF GAMESS BEGUN Thu Jun 20 13:28:15 2013 ECHO OF THE FIRST FEW INPUT CARDS - INPUT CARD>! EXAM 20. INPUT CARD>! Optimize an orbital exponent. INPUT CARD>! The SBKJC basis for I consists of 5 gaussians, in a -41 INPUT CARD>! type split. The exponent of a diffuse L shell for INPUT CARD>! iodide ion is optimized (6th exponent overall). The INPUT CARD>! optimal exponent turns out to be 0.036713, with a INPUT CARD>! corresponding FINAL energy of -11.3010023066 INPUT CARD>! INPUT CARD> $CONTRL SCFTYP=RHF RUNTYP=TRUDGE ICHARG=-1 PP=SBKJC $END INPUT CARD> $SYSTEM TIMLIM=1 $END INPUT CARD> $TRUDGE OPTMIZ=BASIS NPAR=1 IEX(1)=6 $end INPUT CARD> $GUESS GUESS=HUCKEL $END INPUT CARD> $DATA INPUT CARD>I- ion INPUT CARD>Dnh 2 INPUT CARD> INPUT CARD>Iodine 53.0 INPUT CARD> SBKJC INPUT CARD> L 1 INPUT CARD> 1 0.02 1.0 INPUT CARD> INPUT CARD> $END 1000000 WORDS OF MEMORY AVAILABLE RUN TITLE --------- I- ion THE POINT GROUP OF THE MOLECULE IS DNH THE ORDER OF THE PRINCIPAL AXIS IS 2 ATOM ATOMIC COORDINATES (BOHR) CHARGE X Y Z IODINE 53.0 0.0000000000 0.0000000000 0.0000000000 INTERNUCLEAR DISTANCES (ANGS.) ------------------------------ 1 IODI 1 IODI 0.0000000 * ... LESS THAN 3.000 ATOMIC BASIS SET ---------------- THE CONTRACTED PRIMITIVE FUNCTIONS HAVE BEEN UNNORMALIZED THE CONTRACTED BASIS FUNCTIONS ARE NOW NORMALIZED TO UNITY SHELL TYPE PRIMITIVE EXPONENT CONTRACTION COEFFICIENT(S) IODINE 1 L 1 2.6250000 0.073659961641 -0.008880001403 1 L 2 1.0140000 -0.836869564198 -0.257351040669 1 L 3 0.5009000 0.656246658258 0.455368071961 1 L 4 0.2023000 0.900743530935 0.760107120118 2 L 5 0.0780000 1.000000000000 1.000000000000 3 L 6 0.0200000 1.000000000000 1.000000000000 TOTAL NUMBER OF BASIS SET SHELLS = 3 NUMBER OF CARTESIAN GAUSSIAN BASIS FUNCTIONS = 12 NUMBER OF ELECTRONS = 54 CHARGE OF MOLECULE = -1 SPIN MULTIPLICITY = 1 NUMBER OF OCCUPIED ORBITALS (ALPHA) = 27 NUMBER OF OCCUPIED ORBITALS (BETA ) = 27 TOTAL NUMBER OF ATOMS = 1 THE NUCLEAR REPULSION ENERGY IS 0.0000000000 NOTE THIS RUN IS USING CORE POTENTIALS, AND THE NUMBER OF ELECTRONS, OCCUPIED ORBITALS, AND NUCLEAR REPULSION ENERGY WILL BE ADJUSTED BELOW AFTER REMOVAL OF THE CORE CHARGES. THIS MOLECULE IS RECOGNIZED AS BEING LINEAR, ORBITAL LZ DEGENERACY TOLERANCE ETOLLZ= 1.00E-06 $CONTRL OPTIONS --------------- SCFTYP=RHF RUNTYP=TRUDGE EXETYP=RUN MPLEVL= 0 CITYP =NONE CCTYP =NONE VBTYP =NONE DFTTYP=NONE TDDFT =NONE MULT = 1 ICHARG= -1 NZVAR = 0 COORD =UNIQUE PP =SBKJC RELWFN=NONE LOCAL =NONE NUMGRD= F ISPHER= -1 NOSYM = 0 MAXIT = 30 UNITS =ANGS PLTORB= F MOLPLT= F AIMPAC= F FRIEND= NPRINT= 7 IREST = 0 GEOM =INPUT NORMF = 0 NORMP = 0 ITOL = 20 ICUT = 9 INTTYP=BEST GRDTYP=BEST QMTTOL= 1.0E-06 $SYSTEM OPTIONS --------------- REPLICATED MEMORY= 1000000 WORDS (ON EVERY NODE). DISTRIBUTED MEMDDI= 0 MILLION WORDS IN AGGREGATE, MEMDDI DISTRIBUTED OVER 1 PROCESSORS IS 0 WORDS/PROCESSOR. TOTAL MEMORY REQUESTED ON EACH PROCESSOR= 1000000 WORDS. TIMLIM= 1.00 MINUTES, OR 0.0 DAYS. PARALL= F BALTYP= DLB KDIAG= 0 COREFL= F MXSEQ2= 300 MXSEQ3= 150 -------------- ECP POTENTIALS -------------- PARAMETERS FOR "SBKJC-53" ON ATOM 1 WITH ZCORE 46 AND LMAX 3 ARE FOR L= 3 COEFF N ZETA 1 -3.69639 1 0.97628 2 -14.00305 1 4.33343 FOR L= 0 COEFF N ZETA 1 12.11123 0 4.13071 2 -41.09206 2 1.33375 3 70.73761 2 1.49121 FOR L= 1 COEFF N ZETA 1 10.59271 0 3.04692 2 -46.02273 2 1.06339 3 65.05047 2 1.14405 FOR L= 2 COEFF N ZETA 1 9.73089 0 3.93063 2 13.98880 2 1.06920 THE ECP RUN REMOVES 46 CORE ELECTRONS, AND THE SAME NUMBER OF PROTONS. NUMBER OF ELECTRONS KEPT IN THE CALCULATION IS = 8 NUMBER OF OCCUPIED ORBITALS (ALPHA) KEPT IS = 4 NUMBER OF OCCUPIED ORBITALS (BETA ) KEPT IS = 4 THE ADJUSTED NUCLEAR REPULSION ENERGY= 0.0000000000 ECP ANGULAR INTS......... 0.00 SECONDS ---------------- PROPERTIES INPUT ---------------- MOMENTS FIELD POTENTIAL DENSITY IEMOM = 1 IEFLD = 0 IEPOT = 0 IEDEN = 0 WHERE =COMASS WHERE =NUCLEI WHERE =NUCLEI WHERE =NUCLEI OUTPUT=BOTH OUTPUT=BOTH OUTPUT=BOTH OUTPUT=BOTH IEMINT= 0 IEFINT= 0 IEDINT= 0 MORB = 0 EXTRAPOLATION IN EFFECT SOSCF IN EFFECT ORBITAL PRINTING OPTION: NPREO= 1 12 2 1 ------------------------------- INTEGRAL TRANSFORMATION OPTIONS ------------------------------- NWORD = 0 CUTOFF = 1.0E-09 MPTRAN = 0 DIRTRF = F AOINTS =DUP ---------------------- INTEGRAL INPUT OPTIONS ---------------------- NOPK = 1 NORDER= 0 SCHWRZ= F ------------------------------------------ THE POINT GROUP IS DNH, NAXIS= 2, ORDER= 8 ------------------------------------------ DIMENSIONS OF THE SYMMETRY SUBSPACES ARE AG = 3 AU = 0 B3U = 3 B3G = 0 B1G = 0 B1U = 3 B2U = 3 B2G = 0 ..... DONE SETTING UP THE RUN ..... STEP CPU TIME = 0.01 TOTAL CPU TIME = 0.0 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.0 SECONDS, CPU UTILIZATION IS 100.00% -------------------------------- NON-GRADIENT ENERGY MINIMIZATION -------------------------------- $TRURST KSTART=-1 JSTART= 0 TOLF= 0.001000 TOLR= 0.050000 FNOISE= 0.000500 FNOT= 0.000000000 $END ----- TRUDGE RESTART DATA AT NSTEP 0 -------- $TRUDGE OPTMIZ=BASIS NPAR= 1 IEX(1)= 6, P(1)= 0.020000, $END ******************** 1 ELECTRON INTEGRALS ******************** TIME TO DO ORDINARY INTEGRALS= 0.00 TIME TO DO ECP INTEGRALS= 0.00 ...... END OF ONE-ELECTRON INTEGRALS ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.0 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.0 SECONDS, CPU UTILIZATION IS 50.00% ------------- GUESS OPTIONS ------------- GUESS =HUCKEL NORB = 0 NORDER= 0 MIX = F PRTMO = F PUNMO = F TOLZ = 1.0E-08 TOLE = 1.0E-05 SYMDEN= F PURIFY= F INITIAL GUESS ORBITALS GENERATED BY HUCKEL ROUTINE. HUCKEL GUESS REQUIRES 1936 WORDS. SYMMETRIES FOR INITIAL GUESS ORBITALS FOLLOW. BOTH SET(S). 4 ORBITALS ARE OCCUPIED ( 0 CORE ORBITALS). 1=AG 2=B3U 3=B2U 4=B1U 5=AG 6=AG 7=B3U 8=B3U 9=B1U 10=B1U 11=B2U 12=B2U ...... END OF INITIAL ORBITAL SELECTION ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.0 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.0 SECONDS, CPU UTILIZATION IS 33.33% ---------------------- AO INTEGRAL TECHNOLOGY ---------------------- S,P,L SHELL ROTATED AXIS INTEGRALS, REPROGRAMMED BY KAZUYA ISHIMURA (IMS) AND JOSE SIERRA (SYNSTAR). S,P,D,L SHELL ROTATED AXIS INTEGRALS PROGRAMMED BY KAZUYA ISHIMURA (INSTITUTE FOR MOLECULAR SCIENCE). S,P,D,F,G SHELL TO TOTAL QUARTET ANGULAR MOMENTUM SUM 5, ERIC PROGRAM BY GRAHAM FLETCHER (ELORET AND NASA ADVANCED SUPERCOMPUTING DIVISION, AMES RESEARCH CENTER). S,P,D,F,G,L SHELL GENERAL RYS QUADRATURE PROGRAMMED BY MICHEL DUPUIS (PACIFIC NORTHWEST NATIONAL LABORATORY). -------------------- 2 ELECTRON INTEGRALS -------------------- THE -PK- OPTION IS OFF, THE INTEGRALS ARE NOT IN SUPERMATRIX FORM. STORING 15000 INTEGRALS/RECORD ON DISK, USING 12 BYTES/INTEGRAL. TWO ELECTRON INTEGRAL EVALUATION REQUIRES 89362 WORDS OF MEMORY. II,JST,KST,LST = 1 1 1 1 NREC = 1 INTLOC = 1 II,JST,KST,LST = 2 1 1 1 NREC = 1 INTLOC = 17 II,JST,KST,LST = 3 1 1 1 NREC = 1 INTLOC = 139 TOTAL NUMBER OF NONZERO TWO-ELECTRON INTEGRALS = 570 1 INTEGRAL RECORDS WERE STORED ON DISK FILE 8. ...... END OF TWO-ELECTRON INTEGRALS ..... STEP CPU TIME = 0.01 TOTAL CPU TIME = 0.0 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.0 SECONDS, CPU UTILIZATION IS 50.00% -------------------------- RHF SCF CALCULATION -------------------------- NUCLEAR ENERGY = 0.0000000000 MAXIT = 30 NPUNCH= 2 EXTRAP=T DAMP=F SHIFT=F RSTRCT=F DIIS=F DEM=F SOSCF=T DENSITY MATRIX CONV= 1.00E-05 SOSCF WILL OPTIMIZE 32 ORBITAL ROTATIONS, SOGTOL= 0.250 MEMORY REQUIRED FOR RHF ITERS= 31304 WORDS. ITER EX DEM TOTAL ENERGY E CHANGE DENSITY CHANGE ORB. GRAD 1 0 0 -11.2631582817 -11.2631582817 0.293171916 0.000000000 ---------------START SECOND ORDER SCF--------------- 2 1 0 -11.2996006463 -0.0364423646 0.034663215 0.014636326 3 2 0 -11.3003787752 -0.0007781289 0.009057044 0.001082140 4 3 0 -11.3004116289 -0.0000328538 0.000438033 0.000191454 5 4 0 -11.3004118582 -0.0000002292 0.000030657 0.000017350 6 5 0 -11.3004118586 -0.0000000004 0.000003659 0.000001332 7 6 0 -11.3004118586 0.0000000000 0.000000041 0.000000030 ----------------- DENSITY CONVERGED ----------------- TIME TO FORM FOCK OPERATORS= 0.0 SECONDS ( 0.0 SEC/ITER) TIME TO SOLVE SCF EQUATIONS= 0.0 SECONDS ( 0.0 SEC/ITER) FINAL RHF ENERGY IS -11.3004118586 AFTER 7 ITERATIONS LZ VALUE ANALYSIS FOR THE MOS ---------------------------------------- MO 1 ( 1) HAS LZ(WEIGHT)= 0.00(100.0%) MO 2 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 3 ( 2) HAS LZ(WEIGHT)= 0.00(100.0%) MO 4 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 5 ( 3) HAS LZ(WEIGHT)= 0.00(100.0%) MO 6 ( 3) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 7 ( 3) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 8 ( 4) HAS LZ(WEIGHT)= 0.00(100.0%) MO 9 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 10 ( 5) HAS LZ(WEIGHT)= 0.00(100.0%) MO 11 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 12 ( 6) HAS LZ(WEIGHT)= 0.00(100.0%) ------------ EIGENVECTORS ------------ 1 2 3 4 5 -0.6065 -0.1233 -0.1233 -0.1233 0.1825 AG B3U B1U B2U B1U 1 I 1 S 0.838144 0.000000 0.000000 0.000000 0.000000 2 I 1 X 0.000000 0.689927 0.000000 0.000000 0.000000 3 I 1 Y 0.000000 0.000000 0.000000 0.689927 0.000000 4 I 1 Z 0.000000 0.000000 0.689927 0.000000 -0.064944 5 I 1 S 0.171601 0.000000 0.000000 0.000000 0.000000 6 I 1 X 0.000000 0.355035 0.000000 0.000000 0.000000 7 I 1 Y 0.000000 0.000000 0.000000 0.355035 0.000000 8 I 1 Z 0.000000 0.000000 0.355035 0.000000 -0.509927 9 I 1 S 0.026648 0.000000 0.000000 0.000000 0.000000 10 I 1 X 0.000000 0.067600 0.000000 0.000000 0.000000 11 I 1 Y 0.000000 0.000000 0.000000 0.067600 0.000000 12 I 1 Z 0.000000 0.000000 0.067600 0.000000 1.198456 6 7 8 9 10 0.1825 0.1825 0.1880 0.6258 0.6258 B3U B2U AG B3U B1U 1 I 1 S 0.000000 0.000000 -0.187833 0.000000 0.000000 2 I 1 X -0.064944 0.000000 0.000000 -1.321612 0.000000 3 I 1 Y 0.000000 -0.064944 0.000000 0.000000 0.000000 4 I 1 Z 0.000000 0.000000 0.000000 0.000000 -1.321612 5 I 1 S 0.000000 0.000000 -0.550119 0.000000 0.000000 6 I 1 X -0.509927 0.000000 0.000000 1.681232 0.000000 7 I 1 Y 0.000000 -0.509927 0.000000 0.000000 0.000000 8 I 1 Z 0.000000 0.000000 0.000000 0.000000 1.681232 9 I 1 S 0.000000 0.000000 1.326357 0.000000 0.000000 10 I 1 X 1.198456 0.000000 0.000000 -0.511491 0.000000 11 I 1 Y 0.000000 1.198456 0.000000 0.000000 0.000000 12 I 1 Z 0.000000 0.000000 0.000000 0.000000 -0.511491 11 12 0.6258 4.2042 B2U AG 1 I 1 S 0.000000 -1.903127 2 I 1 X 0.000000 0.000000 3 I 1 Y -1.321612 0.000000 4 I 1 Z 0.000000 0.000000 5 I 1 S 0.000000 2.669106 6 I 1 X 0.000000 0.000000 7 I 1 Y 1.681232 0.000000 8 I 1 Z 0.000000 0.000000 9 I 1 S 0.000000 -0.970330 10 I 1 X 0.000000 0.000000 11 I 1 Y -0.511491 0.000000 12 I 1 Z 0.000000 0.000000 ...... END OF RHF CALCULATION ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.0 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.1 SECONDS, CPU UTILIZATION IS 28.57% ---------------------------------------------------------------- PROPERTY VALUES FOR THE RHF SELF-CONSISTENT FIELD WAVEFUNCTION ---------------------------------------------------------------- ----------------- ENERGY COMPONENTS ----------------- WAVEFUNCTION NORMALIZATION = 1.0000000000 ONE ELECTRON ENERGY = -20.6481505684 TWO ELECTRON ENERGY = 9.3477387099 NUCLEAR REPULSION ENERGY = 0.0000000000 ------------------ TOTAL ENERGY = -11.3004118586 ELECTRON-ELECTRON POTENTIAL ENERGY = 9.3477387099 NUCLEUS-ELECTRON POTENTIAL ENERGY = -23.9362143695 NUCLEUS-NUCLEUS POTENTIAL ENERGY = 0.0000000000 ------------------ TOTAL POTENTIAL ENERGY = -14.5884756596 TOTAL KINETIC ENERGY = 3.2880638010 VIRIAL RATIO (V/T) = 4.4367982321 --------------------------------------- MULLIKEN AND LOWDIN POPULATION ANALYSES --------------------------------------- ATOMIC MULLIKEN POPULATION IN EACH MOLECULAR ORBITAL 1 2 3 4 2.000000 2.000000 2.000000 2.000000 1 2.000000 2.000000 2.000000 2.000000 ----- POPULATIONS IN EACH AO ----- MULLIKEN LOWDIN 1 I 1 S 1.66571 1.32562 2 I 1 X 1.31738 1.20402 3 I 1 Y 1.31738 1.20402 4 I 1 Z 1.31738 1.20402 5 I 1 S 0.30702 0.58848 6 I 1 X 0.62499 0.70676 7 I 1 Y 0.62499 0.70676 8 I 1 Z 0.62499 0.70676 9 I 1 S 0.02727 0.08590 10 I 1 X 0.05763 0.08922 11 I 1 Y 0.05763 0.08922 12 I 1 Z 0.05763 0.08922 ----- MULLIKEN ATOMIC OVERLAP POPULATIONS ----- (OFF-DIAGONAL ELEMENTS NEED TO BE MULTIPLIED BY 2) 1 1 8.0000000 TOTAL MULLIKEN AND LOWDIN ATOMIC POPULATIONS ATOM MULL.POP. CHARGE LOW.POP. CHARGE 1 IODINE 8.000000 -1.000000 8.000000 -1.000000 ------------------------------- BOND ORDER AND VALENCE ANALYSIS BOND ORDER THRESHOLD=0.050 ------------------------------- BOND BOND BOND ATOM PAIR DIST ORDER ATOM PAIR DIST ORDER ATOM PAIR DIST ORDER TOTAL BONDED FREE ATOM VALENCE VALENCE VALENCE 1 IODINE 0.000 0.000 0.000 --------------------- ELECTROSTATIC MOMENTS --------------------- POINT 1 X Y Z (BOHR) CHARGE 0.000000 0.000000 0.000000 -1.00 (A.U.) DX DY DZ /D/ (DEBYE) 0.000000 0.000000 0.000000 0.000000 ...... END OF PROPERTY EVALUATION ...... STEP CPU TIME = 0.01 TOTAL CPU TIME = 0.0 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.1 SECONDS, CPU UTILIZATION IS 37.50% TRUDGE ENERGY VALUE AT NSTEP= 0 IS -11.3004118586 ----- TRUDGE RESTART DATA AT NSTEP 1 -------- $TRUDGE OPTMIZ=BASIS NPAR= 1 IEX(1)= 6, P(1)= 0.022103, $END ...... END OF ONE-ELECTRON INTEGRALS ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.0 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.1 SECONDS, CPU UTILIZATION IS 33.33% TOTAL NUMBER OF NONZERO TWO-ELECTRON INTEGRALS = 570 1 INTEGRAL RECORDS WERE STORED ON DISK FILE 8. ...... END OF TWO-ELECTRON INTEGRALS ..... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.0 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.1 SECONDS, CPU UTILIZATION IS 33.33% -------------------------- RHF SCF CALCULATION -------------------------- DENSITY MATRIX CONV= 1.00E-05 ITER EX DEM TOTAL ENERGY E CHANGE DENSITY CHANGE ORB. GRAD ---------------START SECOND ORDER SCF--------------- 1 0 0 -11.3004828415 -11.3004828415 0.011765613 0.002078553 2 1 0 -11.3005831101 -0.0001002686 0.005091350 0.000567223 3 2 0 -11.3005920528 -0.0000089427 0.000251418 0.000079165 4 3 0 -11.3005920735 -0.0000000207 0.000046579 0.000012771 5 4 0 -11.3005920741 -0.0000000005 0.000002368 0.000001412 6 5 0 -11.3005920741 0.0000000000 0.000000148 0.000000104 ----------------- DENSITY CONVERGED ----------------- TIME TO FORM FOCK OPERATORS= 0.0 SECONDS ( 0.0 SEC/ITER) TIME TO SOLVE SCF EQUATIONS= 0.0 SECONDS ( 0.0 SEC/ITER) FINAL RHF ENERGY IS -11.3005920741 AFTER 6 ITERATIONS LZ VALUE ANALYSIS FOR THE MOS ---------------------------------------- MO 1 ( 1) HAS LZ(WEIGHT)= 0.00(100.0%) MO 2 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 3 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 4 ( 2) HAS LZ(WEIGHT)= 0.00(100.0%) MO 5 ( 3) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 6 ( 3) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 7 ( 3) HAS LZ(WEIGHT)= 0.00(100.0%) MO 8 ( 4) HAS LZ(WEIGHT)= 0.00(100.0%) MO 9 ( 5) HAS LZ(WEIGHT)= 0.00(100.0%) MO 10 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 11 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 12 ( 6) HAS LZ(WEIGHT)= 0.00(100.0%) ...... END OF RHF CALCULATION ...... STEP CPU TIME = 0.01 TOTAL CPU TIME = 0.0 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.1 SECONDS, CPU UTILIZATION IS 33.33% TRUDGE ENERGY VALUE AT NSTEP= 1 IS -11.3005920741 LOOKS = 1 ALF = 0.1000 FUNC = -11.3005920741 AMIN = 0.1000 FMIN = -11.3005920741 ANEXT = 0.0000 FNEXT = -11.3004118586 ----- TRUDGE RESTART DATA AT NSTEP 2 -------- $TRUDGE OPTMIZ=BASIS NPAR= 1 IEX(1)= 6, P(1)= 0.021025, $END ...... END OF ONE-ELECTRON INTEGRALS ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.0 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.1 SECONDS, CPU UTILIZATION IS 33.33% TOTAL NUMBER OF NONZERO TWO-ELECTRON INTEGRALS = 570 1 INTEGRAL RECORDS WERE STORED ON DISK FILE 8. ...... END OF TWO-ELECTRON INTEGRALS ..... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.0 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.1 SECONDS, CPU UTILIZATION IS 33.33% -------------------------- RHF SCF CALCULATION -------------------------- DENSITY MATRIX CONV= 1.00E-05 ITER EX DEM TOTAL ENERGY E CHANGE DENSITY CHANGE ORB. GRAD ---------------START SECOND ORDER SCF--------------- 1 0 0 -11.3004703980 -11.3004703980 0.005912292 0.001173294 2 1 0 -11.3005024236 -0.0000320256 0.002717401 0.000347152 3 2 0 -11.3005056686 -0.0000032450 0.000158884 0.000046475 4 3 0 -11.3005056879 -0.0000000193 0.000034854 0.000010503 5 4 0 -11.3005056883 -0.0000000004 0.000003350 0.000000982 6 5 0 -11.3005056883 0.0000000000 0.000000192 0.000000129 ----------------- DENSITY CONVERGED ----------------- TIME TO FORM FOCK OPERATORS= 0.0 SECONDS ( 0.0 SEC/ITER) TIME TO SOLVE SCF EQUATIONS= 0.0 SECONDS ( 0.0 SEC/ITER) FINAL RHF ENERGY IS -11.3005056883 AFTER 6 ITERATIONS LZ VALUE ANALYSIS FOR THE MOS ---------------------------------------- MO 1 ( 1) HAS LZ(WEIGHT)= 0.00(100.0%) MO 2 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 3 ( 2) HAS LZ(WEIGHT)= 0.00(100.0%) MO 4 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 5 ( 3) HAS LZ(WEIGHT)= 0.00(100.0%) MO 6 ( 3) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 7 ( 3) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 8 ( 4) HAS LZ(WEIGHT)= 0.00(100.0%) MO 9 ( 5) HAS LZ(WEIGHT)= 0.00(100.0%) MO 10 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 11 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 12 ( 6) HAS LZ(WEIGHT)= 0.00(100.0%) ...... END OF RHF CALCULATION ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.0 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.2 SECONDS, CPU UTILIZATION IS 26.67% TRUDGE ENERGY VALUE AT NSTEP= 2 IS -11.3005056883 LOOKS = 2 ALF = 0.0500 FUNC = -11.3005056883 AMIN = 0.1000 FMIN = -11.3005920741 ANEXT = 0.0500 FNEXT = -11.3005056883 ----- TRUDGE RESTART DATA AT NSTEP 3 -------- $TRUDGE OPTMIZ=BASIS NPAR= 1 IEX(1)= 6, P(1)= 0.023237, $END ...... END OF ONE-ELECTRON INTEGRALS ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.0 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.2 SECONDS, CPU UTILIZATION IS 26.67% TOTAL NUMBER OF NONZERO TWO-ELECTRON INTEGRALS = 570 1 INTEGRAL RECORDS WERE STORED ON DISK FILE 8. ...... END OF TWO-ELECTRON INTEGRALS ..... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.0 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.2 SECONDS, CPU UTILIZATION IS 26.67% -------------------------- RHF SCF CALCULATION -------------------------- DENSITY MATRIX CONV= 1.00E-05 ITER EX DEM TOTAL ENERGY E CHANGE DENSITY CHANGE ORB. GRAD ---------------START SECOND ORDER SCF--------------- 1 0 0 -11.3005563636 -11.3005563636 0.012936473 0.002179299 2 1 0 -11.3006615625 -0.0001051989 0.005487423 0.000593288 3 2 0 -11.3006705922 -0.0000090297 0.000267103 0.000079373 4 3 0 -11.3006706161 -0.0000000239 0.000050046 0.000013121 5 4 0 -11.3006706167 -0.0000000006 0.000002683 0.000001502 6 5 0 -11.3006706167 0.0000000000 0.000000158 0.000000111 ----------------- DENSITY CONVERGED ----------------- TIME TO FORM FOCK OPERATORS= 0.0 SECONDS ( 0.0 SEC/ITER) TIME TO SOLVE SCF EQUATIONS= 0.0 SECONDS ( 0.0 SEC/ITER) FINAL RHF ENERGY IS -11.3006706167 AFTER 6 ITERATIONS LZ VALUE ANALYSIS FOR THE MOS ---------------------------------------- MO 1 ( 1) HAS LZ(WEIGHT)= 0.00(100.0%) MO 2 ( 2) HAS LZ(WEIGHT)= 0.00(100.0%) MO 3 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 4 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 5 ( 3) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 6 ( 3) HAS LZ(WEIGHT)= 0.00(100.0%) MO 7 ( 3) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 8 ( 4) HAS LZ(WEIGHT)= 0.00(100.0%) MO 9 ( 5) HAS LZ(WEIGHT)= 0.00(100.0%) MO 10 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 11 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 12 ( 6) HAS LZ(WEIGHT)= 0.00(100.0%) ...... END OF RHF CALCULATION ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.0 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.2 SECONDS, CPU UTILIZATION IS 22.22% TRUDGE ENERGY VALUE AT NSTEP= 3 IS -11.3006706167 LOOKS = 3 ALF = 0.1500 FUNC = -11.3006706167 AMIN = 0.1500 FMIN = -11.3006706167 ANEXT = 0.1000 FNEXT = -11.3005920741 ----- TRUDGE RESTART DATA AT NSTEP 4 -------- $TRUDGE OPTMIZ=BASIS NPAR= 1 IEX(1)= 6, P(1)= 0.024428, $END ...... END OF ONE-ELECTRON INTEGRALS ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.0 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.2 SECONDS, CPU UTILIZATION IS 22.22% TOTAL NUMBER OF NONZERO TWO-ELECTRON INTEGRALS = 570 1 INTEGRAL RECORDS WERE STORED ON DISK FILE 8. ...... END OF TWO-ELECTRON INTEGRALS ..... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.0 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.2 SECONDS, CPU UTILIZATION IS 22.22% -------------------------- RHF SCF CALCULATION -------------------------- DENSITY MATRIX CONV= 1.00E-05 ITER EX DEM TOTAL ENERGY E CHANGE DENSITY CHANGE ORB. GRAD ---------------START SECOND ORDER SCF--------------- 1 0 0 -11.3007076553 -11.3007076553 0.007376299 0.001212433 2 1 0 -11.3007383636 -0.0000307083 0.003116248 0.000338768 3 2 0 -11.3007409882 -0.0000026245 0.000143916 0.000040490 4 3 0 -11.3007410004 -0.0000000122 0.000031991 0.000008148 5 4 0 -11.3007410006 -0.0000000002 0.000002787 0.000000823 6 5 0 -11.3007410006 0.0000000000 0.000000134 0.000000079 ----------------- DENSITY CONVERGED ----------------- TIME TO FORM FOCK OPERATORS= 0.0 SECONDS ( 0.0 SEC/ITER) TIME TO SOLVE SCF EQUATIONS= 0.0 SECONDS ( 0.0 SEC/ITER) FINAL RHF ENERGY IS -11.3007410006 AFTER 6 ITERATIONS LZ VALUE ANALYSIS FOR THE MOS ---------------------------------------- MO 1 ( 1) HAS LZ(WEIGHT)= 0.00(100.0%) MO 2 ( 2) HAS LZ(WEIGHT)= 0.00(100.0%) MO 3 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 4 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 5 ( 3) HAS LZ(WEIGHT)= 0.00(100.0%) MO 6 ( 3) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 7 ( 3) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 8 ( 4) HAS LZ(WEIGHT)= 0.00(100.0%) MO 9 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 10 ( 5) HAS LZ(WEIGHT)= 0.00(100.0%) MO 11 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 12 ( 6) HAS LZ(WEIGHT)= 0.00(100.0%) ...... END OF RHF CALCULATION ...... STEP CPU TIME = 0.01 TOTAL CPU TIME = 0.0 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.2 SECONDS, CPU UTILIZATION IS 23.81% TRUDGE ENERGY VALUE AT NSTEP= 4 IS -11.3007410006 LOOKS = 4 ALF = 0.2000 FUNC = -11.3007410006 AMIN = 0.2000 FMIN = -11.3007410006 ANEXT = 0.1500 FNEXT = -11.3006706167 ----- TRUDGE RESTART DATA AT NSTEP 5 -------- $TRUDGE OPTMIZ=BASIS NPAR= 1 IEX(1)= 6, P(1)= 0.025681, $END ...... END OF ONE-ELECTRON INTEGRALS ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.0 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.2 SECONDS, CPU UTILIZATION IS 23.81% TOTAL NUMBER OF NONZERO TWO-ELECTRON INTEGRALS = 570 1 INTEGRAL RECORDS WERE STORED ON DISK FILE 8. ...... END OF TWO-ELECTRON INTEGRALS ..... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.0 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.2 SECONDS, CPU UTILIZATION IS 22.73% -------------------------- RHF SCF CALCULATION -------------------------- DENSITY MATRIX CONV= 1.00E-05 ITER EX DEM TOTAL ENERGY E CHANGE DENSITY CHANGE ORB. GRAD ---------------START SECOND ORDER SCF--------------- 1 0 0 -11.3007677028 -11.3007677028 0.008170798 0.001276918 2 1 0 -11.3008003139 -0.0000326112 0.003374793 0.000355608 3 2 0 -11.3008029834 -0.0000026695 0.000175863 0.000040479 4 3 0 -11.3008029985 -0.0000000151 0.000033950 0.000008303 5 4 0 -11.3008029987 -0.0000000002 0.000003728 0.000000880 6 5 0 -11.3008029987 0.0000000000 0.000000140 0.000000075 ----------------- DENSITY CONVERGED ----------------- TIME TO FORM FOCK OPERATORS= 0.0 SECONDS ( 0.0 SEC/ITER) TIME TO SOLVE SCF EQUATIONS= 0.0 SECONDS ( 0.0 SEC/ITER) FINAL RHF ENERGY IS -11.3008029987 AFTER 6 ITERATIONS LZ VALUE ANALYSIS FOR THE MOS ---------------------------------------- MO 1 ( 1) HAS LZ(WEIGHT)= 0.00(100.0%) MO 2 ( 2) HAS LZ(WEIGHT)= 0.00(100.0%) MO 3 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 4 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 5 ( 3) HAS LZ(WEIGHT)= 0.00(100.0%) MO 6 ( 3) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 7 ( 3) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 8 ( 4) HAS LZ(WEIGHT)= 0.00(100.0%) MO 9 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 10 ( 5) HAS LZ(WEIGHT)= 0.00(100.0%) MO 11 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 12 ( 6) HAS LZ(WEIGHT)= 0.00(100.0%) ...... END OF RHF CALCULATION ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.0 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.2 SECONDS, CPU UTILIZATION IS 20.00% TRUDGE ENERGY VALUE AT NSTEP= 5 IS -11.3008029987 LOOKS = 5 ALF = 0.2500 FUNC = -11.3008029987 AMIN = 0.2500 FMIN = -11.3008029987 ANEXT = 0.2000 FNEXT = -11.3007410006 ----- TRUDGE RESTART DATA AT NSTEP 6 -------- $TRUDGE OPTMIZ=BASIS NPAR= 1 IEX(1)= 6, P(1)= 0.026997, $END ...... END OF ONE-ELECTRON INTEGRALS ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.0 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.3 SECONDS, CPU UTILIZATION IS 19.23% TOTAL NUMBER OF NONZERO TWO-ELECTRON INTEGRALS = 570 1 INTEGRAL RECORDS WERE STORED ON DISK FILE 8. ...... END OF TWO-ELECTRON INTEGRALS ..... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.0 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.3 SECONDS, CPU UTILIZATION IS 19.23% -------------------------- RHF SCF CALCULATION -------------------------- DENSITY MATRIX CONV= 1.00E-05 ITER EX DEM TOTAL ENERGY E CHANGE DENSITY CHANGE ORB. GRAD ---------------START SECOND ORDER SCF--------------- 1 0 0 -11.3008189260 -11.3008189260 0.009085187 0.001347130 2 1 0 -11.3008537336 -0.0000348076 0.003665503 0.000373731 3 2 0 -11.3008564564 -0.0000027228 0.000225246 0.000040178 4 3 0 -11.3008564750 -0.0000000186 0.000035918 0.000008412 5 4 0 -11.3008564752 -0.0000000002 0.000004737 0.000001034 6 5 0 -11.3008564752 0.0000000000 0.000000137 0.000000069 ----------------- DENSITY CONVERGED ----------------- TIME TO FORM FOCK OPERATORS= 0.0 SECONDS ( 0.0 SEC/ITER) TIME TO SOLVE SCF EQUATIONS= 0.0 SECONDS ( 0.0 SEC/ITER) FINAL RHF ENERGY IS -11.3008564752 AFTER 6 ITERATIONS LZ VALUE ANALYSIS FOR THE MOS ---------------------------------------- MO 1 ( 1) HAS LZ(WEIGHT)= 0.00(100.0%) MO 2 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 3 ( 2) HAS LZ(WEIGHT)= 0.00(100.0%) MO 4 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 5 ( 3) HAS LZ(WEIGHT)= 0.00(100.0%) MO 6 ( 4) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 7 ( 4) HAS LZ(WEIGHT)= 0.00(100.0%) MO 8 ( 4) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 9 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 10 ( 5) HAS LZ(WEIGHT)= 0.00(100.0%) MO 11 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 12 ( 6) HAS LZ(WEIGHT)= 0.00(100.0%) ...... END OF RHF CALCULATION ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.0 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.3 SECONDS, CPU UTILIZATION IS 17.24% TRUDGE ENERGY VALUE AT NSTEP= 6 IS -11.3008564752 LOOKS = 6 ALF = 0.3000 FUNC = -11.3008564752 AMIN = 0.3000 FMIN = -11.3008564752 ANEXT = 0.2500 FNEXT = -11.3008029987 ----- TRUDGE RESTART DATA AT NSTEP 7 -------- $TRUDGE OPTMIZ=BASIS NPAR= 1 IEX(1)= 6, P(1)= 0.028381, $END ...... END OF ONE-ELECTRON INTEGRALS ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.0 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.3 SECONDS, CPU UTILIZATION IS 17.24% TOTAL NUMBER OF NONZERO TWO-ELECTRON INTEGRALS = 570 1 INTEGRAL RECORDS WERE STORED ON DISK FILE 8. ...... END OF TWO-ELECTRON INTEGRALS ..... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.0 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.3 SECONDS, CPU UTILIZATION IS 16.67% -------------------------- RHF SCF CALCULATION -------------------------- DENSITY MATRIX CONV= 1.00E-05 ITER EX DEM TOTAL ENERGY E CHANGE DENSITY CHANGE ORB. GRAD ---------------START SECOND ORDER SCF--------------- 1 0 0 -11.3008612164 -11.3008612164 0.010144663 0.001423838 2 1 0 -11.3008985774 -0.0000373610 0.003995207 0.000393319 3 2 0 -11.3009013639 -0.0000027865 0.000286084 0.000043363 4 3 0 -11.3009013867 -0.0000000228 0.000037895 0.000008470 5 4 0 -11.3009013869 -0.0000000002 0.000005787 0.000001170 6 5 0 -11.3009013869 0.0000000000 0.000000127 0.000000062 ----------------- DENSITY CONVERGED ----------------- TIME TO FORM FOCK OPERATORS= 0.0 SECONDS ( 0.0 SEC/ITER) TIME TO SOLVE SCF EQUATIONS= 0.0 SECONDS ( 0.0 SEC/ITER) FINAL RHF ENERGY IS -11.3009013869 AFTER 6 ITERATIONS LZ VALUE ANALYSIS FOR THE MOS ---------------------------------------- MO 1 ( 1) HAS LZ(WEIGHT)= 0.00(100.0%) MO 2 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 3 ( 2) HAS LZ(WEIGHT)= 0.00(100.0%) MO 4 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 5 ( 3) HAS LZ(WEIGHT)= 0.00(100.0%) MO 6 ( 4) HAS LZ(WEIGHT)= 0.00(100.0%) MO 7 ( 4) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 8 ( 4) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 9 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 10 ( 5) HAS LZ(WEIGHT)= 0.00(100.0%) MO 11 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 12 ( 6) HAS LZ(WEIGHT)= 0.00(100.0%) ...... END OF RHF CALCULATION ...... STEP CPU TIME = 0.01 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.3 SECONDS, CPU UTILIZATION IS 18.18% TRUDGE ENERGY VALUE AT NSTEP= 7 IS -11.3009013869 LOOKS = 7 ALF = 0.3500 FUNC = -11.3009013869 AMIN = 0.3500 FMIN = -11.3009013869 ANEXT = 0.3000 FNEXT = -11.3008564752 ----- TRUDGE RESTART DATA AT NSTEP 8 -------- $TRUDGE OPTMIZ=BASIS NPAR= 1 IEX(1)= 6, P(1)= 0.029836, $END ...... END OF ONE-ELECTRON INTEGRALS ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.3 SECONDS, CPU UTILIZATION IS 18.18% TOTAL NUMBER OF NONZERO TWO-ELECTRON INTEGRALS = 570 1 INTEGRAL RECORDS WERE STORED ON DISK FILE 8. ...... END OF TWO-ELECTRON INTEGRALS ..... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.3 SECONDS, CPU UTILIZATION IS 18.18% -------------------------- RHF SCF CALCULATION -------------------------- DENSITY MATRIX CONV= 1.00E-05 ITER EX DEM TOTAL ENERGY E CHANGE DENSITY CHANGE ORB. GRAD ---------------START SECOND ORDER SCF--------------- 1 0 0 -11.3008945417 -11.3008945417 0.011381161 0.001507954 2 1 0 -11.3009348915 -0.0000403499 0.004372670 0.000414579 3 2 0 -11.3009377547 -0.0000028632 0.000360789 0.000049419 4 3 0 -11.3009377825 -0.0000000278 0.000039878 0.000008474 5 4 0 -11.3009377828 -0.0000000002 0.000006856 0.000001287 6 5 0 -11.3009377828 0.0000000000 0.000000111 0.000000053 ----------------- DENSITY CONVERGED ----------------- TIME TO FORM FOCK OPERATORS= 0.0 SECONDS ( 0.0 SEC/ITER) TIME TO SOLVE SCF EQUATIONS= 0.0 SECONDS ( 0.0 SEC/ITER) FINAL RHF ENERGY IS -11.3009377828 AFTER 6 ITERATIONS LZ VALUE ANALYSIS FOR THE MOS ---------------------------------------- MO 1 ( 1) HAS LZ(WEIGHT)= 0.00(100.0%) MO 2 ( 2) HAS LZ(WEIGHT)= 0.00(100.0%) MO 3 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 4 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 5 ( 3) HAS LZ(WEIGHT)= 0.00(100.0%) MO 6 ( 4) HAS LZ(WEIGHT)= 0.00(100.0%) MO 7 ( 4) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 8 ( 4) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 9 ( 5) HAS LZ(WEIGHT)= 0.00(100.0%) MO 10 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 11 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 12 ( 6) HAS LZ(WEIGHT)= 0.00(100.0%) ...... END OF RHF CALCULATION ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.4 SECONDS, CPU UTILIZATION IS 16.22% TRUDGE ENERGY VALUE AT NSTEP= 8 IS -11.3009377828 LOOKS = 8 ALF = 0.4000 FUNC = -11.3009377828 AMIN = 0.4000 FMIN = -11.3009377828 ANEXT = 0.3500 FNEXT = -11.3009013869 ----- TRUDGE RESTART DATA AT NSTEP 9 -------- $TRUDGE OPTMIZ=BASIS NPAR= 1 IEX(1)= 6, P(1)= 0.031366, $END ...... END OF ONE-ELECTRON INTEGRALS ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.4 SECONDS, CPU UTILIZATION IS 16.22% TOTAL NUMBER OF NONZERO TWO-ELECTRON INTEGRALS = 570 1 INTEGRAL RECORDS WERE STORED ON DISK FILE 8. ...... END OF TWO-ELECTRON INTEGRALS ..... STEP CPU TIME = 0.01 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.4 SECONDS, CPU UTILIZATION IS 18.42% -------------------------- RHF SCF CALCULATION -------------------------- DENSITY MATRIX CONV= 1.00E-05 ITER EX DEM TOTAL ENERGY E CHANGE DENSITY CHANGE ORB. GRAD ---------------START SECOND ORDER SCF--------------- 1 0 0 -11.3009189388 -11.3009189388 0.012835584 0.001600574 2 1 0 -11.3009628120 -0.0000438733 0.004809233 0.000437754 3 2 0 -11.3009657678 -0.0000029557 0.000452353 0.000055695 4 3 0 -11.3009658014 -0.0000000337 0.000041860 0.000008424 5 4 0 -11.3009658017 -0.0000000002 0.000007933 0.000001381 6 5 0 -11.3009658017 0.0000000000 0.000000092 0.000000043 ----------------- DENSITY CONVERGED ----------------- TIME TO FORM FOCK OPERATORS= 0.0 SECONDS ( 0.0 SEC/ITER) TIME TO SOLVE SCF EQUATIONS= 0.0 SECONDS ( 0.0 SEC/ITER) FINAL RHF ENERGY IS -11.3009658017 AFTER 6 ITERATIONS LZ VALUE ANALYSIS FOR THE MOS ---------------------------------------- MO 1 ( 1) HAS LZ(WEIGHT)= 0.00(100.0%) MO 2 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 3 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 4 ( 2) HAS LZ(WEIGHT)= 0.00(100.0%) MO 5 ( 3) HAS LZ(WEIGHT)= 0.00(100.0%) MO 6 ( 4) HAS LZ(WEIGHT)= 0.00(100.0%) MO 7 ( 4) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 8 ( 4) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 9 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 10 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 11 ( 5) HAS LZ(WEIGHT)= 0.00(100.0%) MO 12 ( 6) HAS LZ(WEIGHT)= 0.00(100.0%) ...... END OF RHF CALCULATION ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.4 SECONDS, CPU UTILIZATION IS 17.07% TRUDGE ENERGY VALUE AT NSTEP= 9 IS -11.3009658017 LOOKS = 9 ALF = 0.4500 FUNC = -11.3009658017 AMIN = 0.4500 FMIN = -11.3009658017 ANEXT = 0.4000 FNEXT = -11.3009377828 ----- TRUDGE RESTART DATA AT NSTEP 10 -------- $TRUDGE OPTMIZ=BASIS NPAR= 1 IEX(1)= 6, P(1)= 0.032974, $END ...... END OF ONE-ELECTRON INTEGRALS ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.4 SECONDS, CPU UTILIZATION IS 17.07% TOTAL NUMBER OF NONZERO TWO-ELECTRON INTEGRALS = 570 1 INTEGRAL RECORDS WERE STORED ON DISK FILE 8. ...... END OF TWO-ELECTRON INTEGRALS ..... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.4 SECONDS, CPU UTILIZATION IS 17.07% -------------------------- RHF SCF CALCULATION -------------------------- DENSITY MATRIX CONV= 1.00E-05 ITER EX DEM TOTAL ENERGY E CHANGE DENSITY CHANGE ORB. GRAD ---------------START SECOND ORDER SCF--------------- 1 0 0 -11.3009345030 -11.3009345030 0.014560941 0.001703023 2 1 0 -11.3009825601 -0.0000480571 0.005319722 0.000463127 3 2 0 -11.3009856278 -0.0000030677 0.000564548 0.000062170 4 3 0 -11.3009856683 -0.0000000405 0.000043847 0.000008326 5 4 0 -11.3009856685 -0.0000000002 0.000009019 0.000001455 6 5 0 -11.3009856685 0.0000000000 0.000000078 0.000000033 ----------------- DENSITY CONVERGED ----------------- TIME TO FORM FOCK OPERATORS= 0.0 SECONDS ( 0.0 SEC/ITER) TIME TO SOLVE SCF EQUATIONS= 0.0 SECONDS ( 0.0 SEC/ITER) FINAL RHF ENERGY IS -11.3009856685 AFTER 6 ITERATIONS LZ VALUE ANALYSIS FOR THE MOS ---------------------------------------- MO 1 ( 1) HAS LZ(WEIGHT)= 0.00(100.0%) MO 2 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 3 ( 2) HAS LZ(WEIGHT)= 0.00(100.0%) MO 4 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 5 ( 3) HAS LZ(WEIGHT)= 0.00(100.0%) MO 6 ( 4) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 7 ( 4) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 8 ( 4) HAS LZ(WEIGHT)= 0.00(100.0%) MO 9 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 10 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 11 ( 5) HAS LZ(WEIGHT)= 0.00(100.0%) MO 12 ( 6) HAS LZ(WEIGHT)= 0.00(100.0%) ...... END OF RHF CALCULATION ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.4 SECONDS, CPU UTILIZATION IS 15.91% TRUDGE ENERGY VALUE AT NSTEP= 10 IS -11.3009856685 LOOKS = 10 ALF = 0.5000 FUNC = -11.3009856685 AMIN = 0.5000 FMIN = -11.3009856685 ANEXT = 0.4500 FNEXT = -11.3009658017 ----- TRUDGE RESTART DATA AT NSTEP 11 -------- $TRUDGE OPTMIZ=BASIS NPAR= 1 IEX(1)= 6, P(1)= 0.034665, $END ...... END OF ONE-ELECTRON INTEGRALS ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.5 SECONDS, CPU UTILIZATION IS 15.56% TOTAL NUMBER OF NONZERO TWO-ELECTRON INTEGRALS = 570 1 INTEGRAL RECORDS WERE STORED ON DISK FILE 8. ...... END OF TWO-ELECTRON INTEGRALS ..... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.5 SECONDS, CPU UTILIZATION IS 15.56% -------------------------- RHF SCF CALCULATION -------------------------- DENSITY MATRIX CONV= 1.00E-05 ITER EX DEM TOTAL ENERGY E CHANGE DENSITY CHANGE ORB. GRAD ---------------START SECOND ORDER SCF--------------- 1 0 0 -11.3009413726 -11.3009413726 0.016626850 0.001816926 2 1 0 -11.3009944367 -0.0000530641 0.005923754 0.000491029 3 2 0 -11.3009976401 -0.0000032034 0.000702228 0.000068832 4 3 0 -11.3009976884 -0.0000000483 0.000045864 0.000008188 5 4 0 -11.3009976886 -0.0000000002 0.000010121 0.000001509 6 5 0 -11.3009976886 0.0000000000 0.000000065 0.000000024 ----------------- DENSITY CONVERGED ----------------- TIME TO FORM FOCK OPERATORS= 0.0 SECONDS ( 0.0 SEC/ITER) TIME TO SOLVE SCF EQUATIONS= 0.0 SECONDS ( 0.0 SEC/ITER) FINAL RHF ENERGY IS -11.3009976886 AFTER 6 ITERATIONS LZ VALUE ANALYSIS FOR THE MOS ---------------------------------------- MO 1 ( 1) HAS LZ(WEIGHT)= 0.00(100.0%) MO 2 ( 2) HAS LZ(WEIGHT)= 0.00(100.0%) MO 3 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 4 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 5 ( 3) HAS LZ(WEIGHT)= 0.00(100.0%) MO 6 ( 4) HAS LZ(WEIGHT)= 0.00(100.0%) MO 7 ( 4) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 8 ( 4) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 9 ( 5) HAS LZ(WEIGHT)= 0.00(100.0%) MO 10 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 11 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 12 ( 6) HAS LZ(WEIGHT)= 0.00(100.0%) ...... END OF RHF CALCULATION ...... STEP CPU TIME = 0.01 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.5 SECONDS, CPU UTILIZATION IS 16.67% TRUDGE ENERGY VALUE AT NSTEP= 11 IS -11.3009976886 LOOKS = 11 ALF = 0.5500 FUNC = -11.3009976886 AMIN = 0.5500 FMIN = -11.3009976886 ANEXT = 0.5000 FNEXT = -11.3009856685 ----- TRUDGE RESTART DATA AT NSTEP 12 -------- $TRUDGE OPTMIZ=BASIS NPAR= 1 IEX(1)= 6, P(1)= 0.036442, $END ...... END OF ONE-ELECTRON INTEGRALS ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.5 SECONDS, CPU UTILIZATION IS 16.33% TOTAL NUMBER OF NONZERO TWO-ELECTRON INTEGRALS = 570 1 INTEGRAL RECORDS WERE STORED ON DISK FILE 8. ...... END OF TWO-ELECTRON INTEGRALS ..... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.5 SECONDS, CPU UTILIZATION IS 16.33% -------------------------- RHF SCF CALCULATION -------------------------- DENSITY MATRIX CONV= 1.00E-05 ITER EX DEM TOTAL ENERGY E CHANGE DENSITY CHANGE ORB. GRAD ---------------START SECOND ORDER SCF--------------- 1 0 0 -11.3009397074 -11.3009397074 0.019126102 0.001944301 2 1 0 -11.3009988152 -0.0000591078 0.006647644 0.000521858 3 2 0 -11.3010021837 -0.0000033685 0.000871746 0.000075678 4 3 0 -11.3010022408 -0.0000000571 0.000047964 0.000008022 5 4 0 -11.3010022410 -0.0000000002 0.000011258 0.000001547 6 5 0 -11.3010022410 0.0000000000 0.000000017 0.000000015 ----------------- DENSITY CONVERGED ----------------- TIME TO FORM FOCK OPERATORS= 0.0 SECONDS ( 0.0 SEC/ITER) TIME TO SOLVE SCF EQUATIONS= 0.0 SECONDS ( 0.0 SEC/ITER) FINAL RHF ENERGY IS -11.3010022410 AFTER 6 ITERATIONS LZ VALUE ANALYSIS FOR THE MOS ---------------------------------------- MO 1 ( 1) HAS LZ(WEIGHT)= 0.00(100.0%) MO 2 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 3 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 4 ( 2) HAS LZ(WEIGHT)= 0.00(100.0%) MO 5 ( 3) HAS LZ(WEIGHT)= 0.00(100.0%) MO 6 ( 4) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 7 ( 4) HAS LZ(WEIGHT)= 0.00(100.0%) MO 8 ( 4) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 9 ( 5) HAS LZ(WEIGHT)= 0.00(100.0%) MO 10 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 11 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 12 ( 6) HAS LZ(WEIGHT)= 0.00(100.0%) ...... END OF RHF CALCULATION ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.5 SECONDS, CPU UTILIZATION IS 15.38% TRUDGE ENERGY VALUE AT NSTEP= 12 IS -11.3010022410 LOOKS = 12 ALF = 0.6000 FUNC = -11.3010022410 AMIN = 0.6000 FMIN = -11.3010022410 ANEXT = 0.5500 FNEXT = -11.3009976886 ----- TRUDGE RESTART DATA AT NSTEP 13 -------- $TRUDGE OPTMIZ=BASIS NPAR= 1 IEX(1)= 6, P(1)= 0.038311, $END ...... END OF ONE-ELECTRON INTEGRALS ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.5 SECONDS, CPU UTILIZATION IS 15.09% TOTAL NUMBER OF NONZERO TWO-ELECTRON INTEGRALS = 570 1 INTEGRAL RECORDS WERE STORED ON DISK FILE 8. ...... END OF TWO-ELECTRON INTEGRALS ..... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.5 SECONDS, CPU UTILIZATION IS 15.09% -------------------------- RHF SCF CALCULATION -------------------------- DENSITY MATRIX CONV= 1.00E-05 ITER EX DEM TOTAL ENERGY E CHANGE DENSITY CHANGE ORB. GRAD ---------------START SECOND ORDER SCF--------------- 1 0 0 -11.3009296596 -11.3009296596 0.022184483 0.002087688 2 1 0 -11.3009961327 -0.0000664730 0.007527268 0.000556088 3 2 0 -11.3009997024 -0.0000035698 0.001081577 0.000082717 4 3 0 -11.3009997695 -0.0000000671 0.000050242 0.000007842 5 4 0 -11.3009997697 -0.0000000002 0.000012458 0.000001572 6 5 0 -11.3009997697 0.0000000000 0.000000000 0.000000000 ----------------- DENSITY CONVERGED ----------------- TIME TO FORM FOCK OPERATORS= 0.0 SECONDS ( 0.0 SEC/ITER) TIME TO SOLVE SCF EQUATIONS= 0.0 SECONDS ( 0.0 SEC/ITER) FINAL RHF ENERGY IS -11.3009997697 AFTER 6 ITERATIONS LZ VALUE ANALYSIS FOR THE MOS ---------------------------------------- MO 1 ( 1) HAS LZ(WEIGHT)= 0.00(100.0%) MO 2 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 3 ( 2) HAS LZ(WEIGHT)= 0.00(100.0%) MO 4 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 5 ( 3) HAS LZ(WEIGHT)= 0.00(100.0%) MO 6 ( 4) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 7 ( 4) HAS LZ(WEIGHT)= 0.00(100.0%) MO 8 ( 4) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 9 ( 5) HAS LZ(WEIGHT)= 0.00(100.0%) MO 10 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 11 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 12 ( 6) HAS LZ(WEIGHT)= 0.00(100.0%) ...... END OF RHF CALCULATION ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.6 SECONDS, CPU UTILIZATION IS 14.29% TRUDGE ENERGY VALUE AT NSTEP= 13 IS -11.3009997697 LOOKS = 13 ALF = 0.6500 FUNC = -11.3009997697 AMIN = 0.6000 FMIN = -11.3010022410 ANEXT = 0.6500 FNEXT = -11.3009997697 ----- TRUDGE RESTART DATA AT NSTEP 14 -------- $TRUDGE OPTMIZ=BASIS NPAR= 1 IEX(1)= 6, P(1)= 0.036713, $END ...... END OF ONE-ELECTRON INTEGRALS ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.6 SECONDS, CPU UTILIZATION IS 14.29% TOTAL NUMBER OF NONZERO TWO-ELECTRON INTEGRALS = 570 1 INTEGRAL RECORDS WERE STORED ON DISK FILE 8. ...... END OF TWO-ELECTRON INTEGRALS ..... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.6 SECONDS, CPU UTILIZATION IS 14.29% -------------------------- RHF SCF CALCULATION -------------------------- DENSITY MATRIX CONV= 1.00E-05 ITER EX DEM TOTAL ENERGY E CHANGE DENSITY CHANGE ORB. GRAD ---------------START SECOND ORDER SCF--------------- 1 0 0 -11.3009391710 -11.3009391710 0.018569293 0.001958876 2 1 0 -11.3009986736 -0.0000595026 0.006392713 0.000563327 3 2 0 -11.3010021940 -0.0000035204 0.001319338 0.000103459 4 3 0 -11.3010023062 -0.0000001123 0.000063615 0.000010104 5 4 0 -11.3010023066 -0.0000000003 0.000016046 0.000002111 6 5 0 -11.3010023066 0.0000000000 0.000000124 0.000000116 ----------------- DENSITY CONVERGED ----------------- TIME TO FORM FOCK OPERATORS= 0.0 SECONDS ( 0.0 SEC/ITER) TIME TO SOLVE SCF EQUATIONS= 0.0 SECONDS ( 0.0 SEC/ITER) FINAL RHF ENERGY IS -11.3010023066 AFTER 6 ITERATIONS LZ VALUE ANALYSIS FOR THE MOS ---------------------------------------- MO 1 ( 1) HAS LZ(WEIGHT)= 0.00(100.0%) MO 2 ( 2) HAS LZ(WEIGHT)= 0.00(100.0%) MO 3 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 4 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 5 ( 3) HAS LZ(WEIGHT)= 0.00(100.0%) MO 6 ( 4) HAS LZ(WEIGHT)= 0.00(100.0%) MO 7 ( 4) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 8 ( 4) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 9 ( 5) HAS LZ(WEIGHT)= 0.00(100.0%) MO 10 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 11 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 12 ( 6) HAS LZ(WEIGHT)= 0.00(100.0%) ...... END OF RHF CALCULATION ...... STEP CPU TIME = 0.01 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.6 SECONDS, CPU UTILIZATION IS 15.00% TRUDGE ENERGY VALUE AT NSTEP= 14 IS -11.3010023066 LOOKS = 14 ALF = 0.6074 FUNC = -11.3010023066 AMIN = 0.6074 FMIN = -11.3010023066 ANEXT = 0.6000 FNEXT = -11.3010022410 NEW CURVATURE = 0.14047E-02 ALPHA = 0.6074074 OLD CURVATURE = 0.00000E+00 ALFSET= 0.2000000 ----- TRUDGE RESTART DATA AT NSTEP 15 -------- $TRUDGE OPTMIZ=BASIS NPAR= 1 IEX(1)= 6, P(1)= 0.044842, $END ...... END OF ONE-ELECTRON INTEGRALS ...... STEP CPU TIME = 0.01 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.6 SECONDS, CPU UTILIZATION IS 16.67% TOTAL NUMBER OF NONZERO TWO-ELECTRON INTEGRALS = 570 1 INTEGRAL RECORDS WERE STORED ON DISK FILE 8. ...... END OF TWO-ELECTRON INTEGRALS ..... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.6 SECONDS, CPU UTILIZATION IS 16.39% -------------------------- RHF SCF CALCULATION -------------------------- DENSITY MATRIX CONV= 1.00E-05 ITER EX DEM TOTAL ENERGY E CHANGE DENSITY CHANGE ORB. GRAD ---------------START SECOND ORDER SCF--------------- 1 0 0 -11.2999634408 -11.2999634408 0.125663725 0.008189303 2 1 0 -11.3009186631 -0.0009552223 0.038420165 0.001652779 3 2 0 -11.3009518469 -0.0000331838 0.001869635 0.000241258 4 3 0 -11.3009519893 -0.0000001424 0.000060052 0.000045951 5 4 0 -11.3009519901 -0.0000000008 0.000022316 0.000002614 6 5 0 -11.3009519901 0.0000000000 0.000003270 0.000000402 7 6 0 -11.3009519901 0.0000000000 0.000000000 0.000000000 ----------------- DENSITY CONVERGED ----------------- TIME TO FORM FOCK OPERATORS= 0.0 SECONDS ( 0.0 SEC/ITER) TIME TO SOLVE SCF EQUATIONS= 0.0 SECONDS ( 0.0 SEC/ITER) FINAL RHF ENERGY IS -11.3009519901 AFTER 7 ITERATIONS LZ VALUE ANALYSIS FOR THE MOS ---------------------------------------- MO 1 ( 1) HAS LZ(WEIGHT)= 0.00(100.0%) MO 2 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 3 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 4 ( 2) HAS LZ(WEIGHT)= 0.00(100.0%) MO 5 ( 3) HAS LZ(WEIGHT)= 0.00(100.0%) MO 6 ( 4) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 7 ( 4) HAS LZ(WEIGHT)= 0.00(100.0%) MO 8 ( 4) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 9 ( 5) HAS LZ(WEIGHT)= 0.00(100.0%) MO 10 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 11 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 12 ( 6) HAS LZ(WEIGHT)= 0.00(100.0%) ...... END OF RHF CALCULATION ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.6 SECONDS, CPU UTILIZATION IS 15.63% TRUDGE ENERGY VALUE AT NSTEP= 15 IS -11.3009519901 LOOKS = 1 ALF = 0.2000 FUNC = -11.3009519901 AMIN = 0.0000 FMIN = -11.3010023066 ANEXT = 0.2000 FNEXT = -11.3009519901 ----- TRUDGE RESTART DATA AT NSTEP 16 -------- $TRUDGE OPTMIZ=BASIS NPAR= 1 IEX(1)= 6, P(1)= 0.030058, $END ...... END OF ONE-ELECTRON INTEGRALS ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.6 SECONDS, CPU UTILIZATION IS 15.63% TOTAL NUMBER OF NONZERO TWO-ELECTRON INTEGRALS = 570 1 INTEGRAL RECORDS WERE STORED ON DISK FILE 8. ...... END OF TWO-ELECTRON INTEGRALS ..... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.7 SECONDS, CPU UTILIZATION IS 15.38% -------------------------- RHF SCF CALCULATION -------------------------- DENSITY MATRIX CONV= 1.00E-05 ITER EX DEM TOTAL ENERGY E CHANGE DENSITY CHANGE ORB. GRAD ---------------START SECOND ORDER SCF--------------- 1 0 0 -11.2880509569 -11.2880509569 0.166085198 0.027219313 2 1 0 -11.3000996160 -0.0120486591 0.048918019 0.009099891 3 2 0 -11.3008710326 -0.0007714166 0.022456037 0.002344050 4 3 0 -11.3009421271 -0.0000710945 0.000852546 0.000210608 5 4 0 -11.3009424389 -0.0000003118 0.000329558 0.000060856 6 5 0 -11.3009424578 -0.0000000189 0.000009196 0.000003449 7 6 0 -11.3009424578 0.0000000000 0.000001218 0.000000946 8 7 0 -11.3009424578 0.0000000000 0.000000143 0.000000145 ----------------- DENSITY CONVERGED ----------------- TIME TO FORM FOCK OPERATORS= 0.0 SECONDS ( 0.0 SEC/ITER) TIME TO SOLVE SCF EQUATIONS= 0.0 SECONDS ( 0.0 SEC/ITER) FINAL RHF ENERGY IS -11.3009424578 AFTER 8 ITERATIONS LZ VALUE ANALYSIS FOR THE MOS ---------------------------------------- MO 1 ( 1) HAS LZ(WEIGHT)= 0.00(100.0%) MO 2 ( 2) HAS LZ(WEIGHT)= 0.00(100.0%) MO 3 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 4 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 5 ( 3) HAS LZ(WEIGHT)= 0.00(100.0%) MO 6 ( 4) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 7 ( 4) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 8 ( 4) HAS LZ(WEIGHT)= 0.00(100.0%) MO 9 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 10 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 11 ( 5) HAS LZ(WEIGHT)= 0.00(100.0%) MO 12 ( 6) HAS LZ(WEIGHT)= 0.00(100.0%) ...... END OF RHF CALCULATION ...... STEP CPU TIME = 0.01 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.7 SECONDS, CPU UTILIZATION IS 16.18% TRUDGE ENERGY VALUE AT NSTEP= 16 IS -11.3009424578 LOOKS = 2 ALF = -0.2000 FUNC = -11.3009424578 AMIN = 0.0000 FMIN = -11.3010023066 ANEXT = 0.2000 FNEXT = -11.3009519901 ----- TRUDGE RESTART DATA AT NSTEP 17 -------- $TRUDGE OPTMIZ=BASIS NPAR= 1 IEX(1)= 6, P(1)= 0.037032, $END ...... END OF ONE-ELECTRON INTEGRALS ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.7 SECONDS, CPU UTILIZATION IS 15.94% TOTAL NUMBER OF NONZERO TWO-ELECTRON INTEGRALS = 570 1 INTEGRAL RECORDS WERE STORED ON DISK FILE 8. ...... END OF TWO-ELECTRON INTEGRALS ..... STEP CPU TIME = 0.01 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.7 SECONDS, CPU UTILIZATION IS 17.39% -------------------------- RHF SCF CALCULATION -------------------------- DENSITY MATRIX CONV= 1.00E-05 ITER EX DEM TOTAL ENERGY E CHANGE DENSITY CHANGE ORB. GRAD ---------------START SECOND ORDER SCF--------------- 1 0 0 -11.3002986031 -11.3002986031 0.071979372 0.006559306 2 1 0 -11.3009704507 -0.0006718476 0.023451693 0.001495642 3 2 0 -11.3010020779 -0.0000316272 0.000803721 0.000241875 4 3 0 -11.3010021890 -0.0000001111 0.000054416 0.000033819 5 4 0 -11.3010021899 -0.0000000009 0.000018316 0.000002321 6 5 0 -11.3010021899 0.0000000000 0.000000250 0.000000061 ----------------- DENSITY CONVERGED ----------------- TIME TO FORM FOCK OPERATORS= 0.0 SECONDS ( 0.0 SEC/ITER) TIME TO SOLVE SCF EQUATIONS= 0.0 SECONDS ( 0.0 SEC/ITER) FINAL RHF ENERGY IS -11.3010021899 AFTER 6 ITERATIONS LZ VALUE ANALYSIS FOR THE MOS ---------------------------------------- MO 1 ( 1) HAS LZ(WEIGHT)= 0.00(100.0%) MO 2 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 3 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 4 ( 2) HAS LZ(WEIGHT)= 0.00(100.0%) MO 5 ( 3) HAS LZ(WEIGHT)= 0.00(100.0%) MO 6 ( 4) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 7 ( 4) HAS LZ(WEIGHT)= 0.00(100.0%) MO 8 ( 4) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 9 ( 5) HAS LZ(WEIGHT)= 0.00(100.0%) MO 10 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 11 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 12 ( 6) HAS LZ(WEIGHT)= 0.00(100.0%) ...... END OF RHF CALCULATION ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.7 SECONDS, CPU UTILIZATION IS 16.44% TRUDGE ENERGY VALUE AT NSTEP= 17 IS -11.3010021899 LOOKS = 3 ALF = 0.0087 FUNC = -11.3010021899 AMIN = 0.0000 FMIN = -11.3010023066 ANEXT = 0.0087 FNEXT = -11.3010021899 NEW CURVATURE = 0.13771E-02 ALPHA = 0.0000000 OLD CURVATURE = 0.14047E-02 ALFSET= 0.0500000 F= -11.3010023066 DELTA F=0.00000E+00 DELTA R=0.00000E+00 ----- TRUDGE RESTART DATA AT NSTEP 18 -------- $TRUDGE OPTMIZ=BASIS NPAR= 1 IEX(1)= 6, P(1)= 0.036713, $END ...... END OF ONE-ELECTRON INTEGRALS ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.7 SECONDS, CPU UTILIZATION IS 16.44% TOTAL NUMBER OF NONZERO TWO-ELECTRON INTEGRALS = 570 1 INTEGRAL RECORDS WERE STORED ON DISK FILE 8. ...... END OF TWO-ELECTRON INTEGRALS ..... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.7 SECONDS, CPU UTILIZATION IS 16.44% -------------------------- RHF SCF CALCULATION -------------------------- DENSITY MATRIX CONV= 1.00E-05 ITER EX DEM TOTAL ENERGY E CHANGE DENSITY CHANGE ORB. GRAD ---------------START SECOND ORDER SCF--------------- 1 0 0 -11.3009999936 -11.3009999936 0.003615798 0.000374812 2 1 0 -11.3010021758 -0.0000021821 0.001251947 0.000105210 3 2 0 -11.3010023032 -0.0000001275 0.000222981 0.000017955 4 3 0 -11.3010023066 -0.0000000033 0.000011182 0.000001790 5 4 0 -11.3010023066 0.0000000000 0.000002726 0.000000363 6 5 0 -11.3010023066 0.0000000000 0.000000015 0.000000014 ----------------- DENSITY CONVERGED ----------------- TIME TO FORM FOCK OPERATORS= 0.0 SECONDS ( 0.0 SEC/ITER) TIME TO SOLVE SCF EQUATIONS= 0.0 SECONDS ( 0.0 SEC/ITER) FINAL RHF ENERGY IS -11.3010023066 AFTER 6 ITERATIONS LZ VALUE ANALYSIS FOR THE MOS ---------------------------------------- MO 1 ( 1) HAS LZ(WEIGHT)= 0.00(100.0%) MO 2 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 3 ( 2) HAS LZ(WEIGHT)= 0.00(100.0%) MO 4 ( 2) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 5 ( 3) HAS LZ(WEIGHT)= 0.00(100.0%) MO 6 ( 4) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 7 ( 4) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 8 ( 4) HAS LZ(WEIGHT)= 0.00(100.0%) MO 9 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 10 ( 5) HAS LZ(WEIGHT)=-1.00( 50.0%) 1.00( 50.0%) MO 11 ( 5) HAS LZ(WEIGHT)= 0.00(100.0%) MO 12 ( 6) HAS LZ(WEIGHT)= 0.00(100.0%) ...... END OF RHF CALCULATION ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 0.8 SECONDS, CPU UTILIZATION IS 15.79% TRUDGE ENERGY VALUE AT NSTEP= 18 IS -11.3010023066 NON-GRADIENT ENERGY MINIMIZATION ... CONVERGED CONJUGATE DIRECTION = 0 PARAMETER = 1 VALUE OF THE FUNCTION = -11.3010023066 FNOT = -11.3010023066 $TRURST KSTART= 0 JSTART= 1 TOLF= 0.001000 TOLR= 0.050000 FNOISE= 0.000500 FNOT= -11.301002307 RNOT(1)= 0.607407 CURV(1)= 0.001377 ALPH(1)= 0.050000 V(1)= 1.000000 $END ...... END OF NON-GRADIENT ENERGY MINIMIZATION ...... STEP CPU TIME = 0.00 TOTAL CPU TIME = 0.1 ( 0.0 MIN) TOTAL WALL CLOCK TIME= 1.5 SECONDS, CPU UTILIZATION IS 8.28% 580000 WORDS OF DYNAMIC MEMORY USED EXECUTION OF GAMESS TERMINATED NORMALLY Thu Jun 20 13:28:17 2013 DDI: 263624 bytes (0.3 MB / 0 MWords) used by master data server. ---------------------------------------- CPU timing information for all processes ======================================== 0: 0.64 + 0.64 = 0.129 ---------------------------------------- ddikick.x: exited gracefully. unset echo ----- accounting info ----- Files used on the master node PaulDirac were: -rw-r--r-- 1 nikita 50003 Jun 20 13:28 /mnt/disk2/nikita/scr/exam20.dat -rw-r--r-- 1 nikita 540 Jun 20 13:28 /mnt/disk2/nikita/scr/exam20.F05 -rw-r--r-- 1 nikita 180016 Jun 20 13:28 /mnt/disk2/nikita/scr/exam20.F08 -rw-r--r-- 1 nikita 1701440 Jun 20 13:28 /mnt/disk2/nikita/scr/exam20.F10 ls: No match. ls: No match. ls: No match. Thu Jun 20 13:28:20 EDT 2013 0.249u 0.156s 0:04.87 8.0% 0+0k 0+0io 0pf+0w