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【求助】高斯计算错误！！！
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请大家帮忙看看出错在哪里？怎么修改？谢谢！！ %chk=F:\PANH2O\PAN2H2O\out\TS2.chk %mem=1750mb %RWF=1,2999MB,2,2999MB,3,2999MB,4,2999MB,5,2999MB,6,2999MB,7,2999MB,8,2999MB Default route: MaxDisk=3000MB ------------------------------------------------------------- #ump2(Direct)/6-31++g(d,p) opt=(ts,noeigen,calcfc,maxcyc=200) ------------------------------------------------------------- 1/5=1,6=200,10=4,11=1,18=20,38=1/1,3; 2/9=110,17=6,18=5,40=1/2; 3/5=1,6=6,7=1111,11=2,16=1,25=1,30=1/1,2,3; 4/7=2/1; 5/5=2,38=5/2; 8/6=3,8=1,10=2,19=11,27=393216000,30=-1/1; 9/15=3,27=393216000/6; 11/6=1,8=1,15=11,17=12,24=-1,27=1,28=-2,29=300,32=6,42=3/1,2,10; 10/6=2,21=1/2; 8/6=4,8=1,10=2,19=11,27=393216000,30=-1/11,4; 10/5=1,20=4/2; 。。。。。。 **** Warning!!: The largest beta MO coefficient is 0.48247260D+02 Disk-based method using ON**2 memory for 8 occupieds at a time. Permanent disk used for amplitudes= 91082361 words. Estimated scratch disk usage= 220658040 words. Actual scratch disk usage= 167520120 words. JobTyp=2 Pass 1: I= 11 to 18 NPSUse= 1 ParTrn=F ParDer=F DoDerP=T. (rs|ai) integrals will be sorted in core. JobTyp=2 Pass 2: I= 19 to 26 NPSUse= 1 ParTrn=F ParDer=F DoDerP=T. (rs|ai) integrals will be sorted in core. JobTyp=2 Pass 3: I= 27 to 34 NPSUse= 1 ParTrn=F ParDer=F DoDerP=T. (rs|ai) integrals will be sorted in core. JobTyp=2 Pass 4: I= 35 to 41 NPSUse= 1 ParTrn=F ParDer=F DoDerP=T. (rs|ai) integrals will be sorted in core. Actual scratch disk usage= 167520120 words. JobTyp=3 Pass 1: I= 11 to 18 NPSUse= 1 ParTrn=F ParDer=F DoDerP=T. (rs|ai) integrals will be sorted in core. JobTyp=3 Pass 2: I= 19 to 26 NPSUse= 1 ParTrn=F ParDer=F DoDerP=T. (rs|ai) integrals will be sorted in core. JobTyp=3 Pass 3: I= 27 to 34 NPSUse= 1 ParTrn=F ParDer=F DoDerP=T. (rs|ai) integrals will be sorted in core. JobTyp=3 Pass 4: I= 35 to 41 NPSUse= 1 ParTrn=F ParDer=F DoDerP=T. (rs|ai) integrals will be sorted in core. Spin components of T(2) and E(2): alpha-alpha T2 = 0.8238513419D-01 E2= -0.2585933670D+00 alpha-beta T2 = 0.4174531006D+00 E2= -0.1342082110D+01 beta-beta T2 = 0.8238513419D-01 E2= -0.2585933670D+00 (S**2,0)= 0.00000D+00 (S**2,1)= 0.00000D+00 E(PUHF)= -0.65768126039D+03 E(PMP2)= -0.65954052923D+03 ANorm= 0.1257864607D+01 E2 = -0.1859268844D+01 EUMP2 = -0.65954052923020D+03 G2DrvN: will do 18 centers at a time, making 1 passes doing MaxLOS=2. FoFDir/FoFCou used for L=0 through L=2. Differentiating once with respect to electric field. with respect to dipole field. Differentiating once with respect to nuclear coordinates. Integrals replicated using symmetry in FoFDir. MinBra= 0 MaxBra= 2 Meth= 1. IRaf= 0 NMat= 54 IRICut= 54 DoRegI=T DoRafI=T ISym2E= 2 JSym2E=2. There are 54 degrees of freedom in the 1st order CPHF. 51 vectors were produced by pass 0. AX will form 51 AO Fock derivatives at one time. 51 vectors were produced by pass 1. 51 vectors were produced by pass 2. 51 vectors were produced by pass 3. 51 vectors were produced by pass 4. 51 vectors were produced by pass 5. 51 vectors were produced by pass 6. 24 vectors were produced by pass 7. 3 vectors were produced by pass 8. Inv2: IOpt= 1 Iter= 1 AM= 5.39D-15 Conv= 1.00D-12. Inverted reduced A of dimension 384 with in-core refinement. End of Minotr Frequency-dependent properties file 721 does not exist. MDV= 229376000. Form MO integral derivatives with frozen-active canonical formalism. Discarding MO integrals. Reordered first order wavefunction length = 366130574 Warning! MaxDisk= 393216000 but minimum for overlay 11= 1199537753.. WUsed= 135941079 WInt= 2382624 WEnd= 133938688 Dk804= 428524067. Dk1111= 0. Dk1112= 1061214050. MaxDsk= 393216000 LAFull= 366130574 DskLim= 393216000. NUsed= 64719605. 3035616501. 2191895760. 2071808345. 1758337531. 1549356989. In DefCFB: NBatch= 1, ICI= 41, ICA=191, LFMax= 28 Large arrays: LIAPS= 0, LIARS= 483266672 words. readwa-lseekm |
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3楼2008-10-19 16:10:03
abbott
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2楼2008-10-19 15:37:36
