hdh912

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[交流] 【求助】請教molpro計算中出現(xiàn)的一個問題

如題，我在對Na2分子的x1sigma+g, a3pi, A1sgma+u,B1PIU, C1PIU, D1PIU, E1PIU進行掃描的時候，我的輸入卡如下：
  memory,30,m
geometry={na1;na2,na1,r1(i)}
distance= [2.1,2.3,2.5,2.6,2.8,2.9,2.95,3.0,3.05,3.1,3.15,3.2,3.42,3.62,3.9,4.2,4.5,4.8,5.3,5.7,5.8,5.9,6.0,6.2,6.4,6.6,6.8,7.0,7.2,7.4,7.6,7.8,8.0,8.2,8.4,8.6,8.8,9.0] ang
basis=VTZ
r1(1)=distance(1)
hf,occ,4,1,1,0,3,1,1,0;core,3,1,1,0,3,1,1,0;
i=0
do ir1=1,#distance
i=i+1
r1(i)=distance(ir1)
multi;
occ,4,1,1,0,3,1,1,0;core,3,1,1,0,3,1,1,0;wf,22,1,0;state,2,refstate,1,2;wf,22,5,0;state,1;wf,22,2,0,state,4,refstate,1,2,3,4;lquant,1;
ecas(i)=energy(1)
ci;occ,4,1,1,0,3,1,1,0;core,3,1,1,0,3,1,1,0;wf,22,1,0;state,2;wf,22,5,0;state,1;wf,22,2,0,state,4
emrci(i)=energy(1)
emrciq(i)=energd(1)
enddo
但是計算的結(jié)果始終出現(xiàn)如下信息：
State symmetry 3

  State symmetry 3:    Projection for operator LZ    squared value = 1

Number of electrons:  2 Spin symmetry=Singlet    Space symmetry=2
Number of states:    1
Number of CSFs:    0 (    0 determinants,    1 intermediate states)

Valence orbitals read from record    2100.2  Type=RHF/CANONICAL (state 1.1)
Frozen core orbitals read from record 2100.2  Type=RHF/CANONICAL (state 1.1)

EXTRA SYMMETRY OF MOS IN SYMMETRY 1: 0 0 0 1 1 1 1 2 1 1 2 2 1 1 1
EXTRA SYMMETRY OF MOS IN SYMMETRY 2: 0 1 1 1 1 1 2 1
EXTRA SYMMETRY OF MOS IN SYMMETRY 3: 0 1 1 1 1 1 2 1
EXTRA SYMMETRY OF MOS IN SYMMETRY 5: 0 0 0 1 1 1 1 2 1 1 2 1 2 1 1
EXTRA SYMMETRY OF MOS IN SYMMETRY 6: 0 1 1 1 1 2 1 1
EXTRA SYMMETRY OF MOS IN SYMMETRY 7: 0 1 1 1 1 2 1 1

Wavefunction dump at record          2140.2

Convergence thresholds  0.10E-01 (gradient)  0.10E-05 (energy)  0.10E-02 (step length)

Weight factors for state symmetry  1: 0.25000 0.25000
Weight factors for state symmetry  2: 0.25000
Weight factors for state symmetry  3: 0.25000

Number of orbital rotations    11    ( 0 Core/Active 0 Core/Virtual 0 Active/Active 11 Active/Virtual)
Total number of variables =    13

ITER. MIC  NCI  NEG    ENERGY(VAR)    ENERGY(PROJ) ENERGY CHANGE    GRAD(0)  GRAD(ORB) GRAD(CI)    STEP    TIME

MORE STATES THAN CONFIGURATIONS SPECIFIED:  1  2
懇請高手幫我修改下輸入卡，看怎么樣才能消除這個錯誤，萬分感謝。

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1樓 2011-02-28 00:57:48

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fatpig8832

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hdh912(金幣+1): 謝謝，修改后還是一樣的錯誤。 2011-03-02 00:26:55

具體不清楚，不過下面這一行

wf,22,2,0,state,4,refstate,1,2,3,4;lquant,1;

前面既然寫了"state,4"，后面的 lquant 是不是應(yīng)該相應(yīng)的跟著寫上四個數(shù)字，而不是只寫一個呢？

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3樓2011-03-01 11:12:51

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hdh912

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怎么沒有兄弟會嗎？

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2樓2011-03-01 09:46:07

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beefly

新蟲

hdh912(金幣+3): 謝謝老大的回答，但是我將refstate刪掉后出錯信息還是一樣。 2011-03-02 00:26:19

輸入輸出明顯不符。refstate不僅在casscf計算中沒有意義，而且把你的casscf輸入結(jié)構(gòu)破壞了。

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4樓2011-03-01 13:01:36

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hdh912

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引用回帖:

Originally posted by beefly at 2011-03-01 13:01:36:
輸入輸出明顯不符。refstate不僅在casscf計算中沒有意義，而且把你的casscf輸入結(jié)構(gòu)破壞了。

老哥，你好我的全部輸入卡修改為
**,Na2
memory,30,m
geometry={na1;na2,na1,r1(i)}
distance= [2.1,2.3,2.5,2.6,2.8,2.9,2.95,3.0,3.05,3.1,3.15,3.2,3.42,3.62,3.9,4.2,4.5,4.8,5.3,5.7,5.8,5.9,6.0,6.2,6.4,6.6,6.8,7.0,7.2,7.4,7.6,7.8,8.0,8.2,8.4,8.6,8.8,9.0] ang
basis=VTZ
r1(1)=distance(1)
hf,occ,4,1,1,0,3,1,1,0;core,3,1,1,0,3,1,1,0;
i=0
do ir1=1,#distance
i=i+1
r1(i)=distance(ir1)
multi;
occ,4,1,1,0,3,1,1,0;core,3,1,1,0,3,1,1,0;wf,22,1,0;state,2;wf,22,5,0;state,1;wf,22,2,0,state,4;lquant,1,1,1,1;
ecas(i)=energy(1)
ci;occ,4,1,1,0,3,1,1,0;core,3,1,1,0,3,1,1,0;wf,22,1,0;state,2;wf,22,5,0;state,1;wf,22,2,0,state,4
emrci(i)=energy(1)
emrciq(i)=energd(1)
enddo
table,r1,ecas,emrci,emrciq
head,r1,ecas,emrci,emrciq
save,na2.tab
輸出信息還是
State symmetry 3

  State symmetry 3:    Projection for operator LZ    squared value = 1

Number of electrons:  2 Spin symmetry=Singlet    Space symmetry=2
Number of states:    1
Number of CSFs:    0 (    0 determinants,    1 intermediate states)

Valence orbitals read from record    2100.2  Type=RHF/CANONICAL (state 1.1)
Frozen core orbitals read from record 2100.2  Type=RHF/CANONICAL (state 1.1)

EXTRA SYMMETRY OF MOS IN SYMMETRY 1: 0 0 0 1 1 1 1 2 1 1 2 2 1 1 1
EXTRA SYMMETRY OF MOS IN SYMMETRY 2: 0 1 1 1 1 1 2 1
EXTRA SYMMETRY OF MOS IN SYMMETRY 3: 0 1 1 1 1 1 2 1
EXTRA SYMMETRY OF MOS IN SYMMETRY 5: 0 0 0 1 1 1 1 2 1 1 2 1 2 1 1
EXTRA SYMMETRY OF MOS IN SYMMETRY 6: 0 1 1 1 1 2 1 1
EXTRA SYMMETRY OF MOS IN SYMMETRY 7: 0 1 1 1 1 2 1 1

Wavefunction dump at record          2140.2

Convergence thresholds  0.10E-01 (gradient)  0.10E-05 (energy)  0.10E-02 (step length)

Weight factors for state symmetry  1: 0.25000 0.25000
Weight factors for state symmetry  2: 0.25000
Weight factors for state symmetry  3: 0.25000

Number of orbital rotations    11    ( 0 Core/Active 0 Core/Virtual 0 Active/Active 11 Active/Virtual)
Total number of variables =    13

ITER. MIC  NCI  NEG    ENERGY(VAR)    ENERGY(PROJ) ENERGY CHANGE    GRAD(0)  GRAD(ORB) GRAD(CI)    STEP    TIME

MORE STATES THAN CONFIGURATIONS SPECIFIED:  1  2
懇請老哥幫忙。謝謝

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5樓2011-03-02 00:29:23

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