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hdh912

金虫 (正式写手)


[交流] 【求助】请教molpro计算中出现的一个问题

如题,我在对Na2分子的x1sigma+g, a3pi, A1sgma+u,B1PIU, C1PIU, D1PIU, E1PIU进行扫描的时候,我的输入卡如下:
  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
但是计算的结果始终出现如下信息:
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
恳请高手帮我修改下输入卡,看怎么样才能消除这个错误,万分感谢。
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hdh912(金币+3): 谢谢老大的回答,但是我将refstate删掉后出错信息还是一样。 2011-03-02 00:26:19
输入输出明显不符。refstate不仅在casscf计算中没有意义,而且把你的casscf输入结构破坏了。
4楼2011-03-01 13:01:36
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hdh912

金虫 (正式写手)


怎么没有兄弟会吗?
2楼2011-03-01 09:46:07
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fatpig8832

铁杆木虫 (著名写手)


hdh912(金币+1): 谢谢,修改后还是一样的错误。 2011-03-02 00:26:55
具体不清楚,不过下面这一行

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

前面既然写了"state,4",后面的 lquant 是不是应该相应的跟着写上四个数字,而不是只写一个呢?
3楼2011-03-01 11:12:51
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hdh912

金虫 (正式写手)


引用回帖:
Originally posted by beefly at 2011-03-01 13:01:36:
输入输出明显不符。refstate不仅在casscf计算中没有意义,而且把你的casscf输入结构破坏了。

老哥,你好我的全部输入卡修改为
**,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
输出信息 还是
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
恳请老哥帮忙。谢谢
5楼2011-03-02 00:29:23
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