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标题: Gaussian频率分析中出现的问题，计算意外停止 [打印本页]

作者
Author: -ZQ- 时间: 2023-12-24 14:42
标题: Gaussian频率分析中出现的问题，计算意外停止
本帖最后由 -ZQ- 于 2025-3-20 09:15 编辑

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在gaussian频率分析的时候意外停止，不知道是什么原因，还望各位老师帮忙解决，以下是log输出文件的最后一部分

**** Warning!!: The largest alpha MO coefficient is  0.87166728D+02

Semi-Direct transformation.
ModeAB=          4 MOrb=          20 LenV= 1578415135
LASXX=    94671445 LTotXX= 94671445 LenRXX= 190638925
LTotAB= 95967480 MaxLAS= 124208640 LenRXY=          0
NonZer= 285310370 LenScr= 429644800 LnRSAI= 124208640
LnScr1= 187062272 LExtra=          0 Total= 931554637
MaxDsk=       -1 SrtSym=          T ITran=          4
JobTyp=0 Pass  1:  I= 1 to  20.
(rs|ai) integrals will be sorted in core.
SymMOI:  orbitals are not symmetric.
Spin components of T(2) and E(2):
   alpha-alpha T2 =    0.5697191625D-01 E2=    -0.1406423048D+00
   alpha-beta  T2 =    0.3010022824D+00 E2=    -0.7969753183D+00
   beta-beta T2 =    0.5697191625D-01 E2=    -0.1406423048D+00
ANorm= 0.1189515076D+01
E2 = -0.1078259928D+01 EUMP2 = -0.30869612100846D+03
Keep R2 and R3 ints in memory in canonical form, NReq=642076266.
CIDS: In Core Option IDoMem=          1.
Iterations=  50 Convergence= 0.100D-06
Produce multiple copies of IABC intergrals
Iteration Nr. 1
**********************
DD1Dir will call FoFMem 1 times, MxPair=    420
NAB= 210 NAA=    0 NBB=    0.
MP4(R+Q)=  0.32377206D-01
E3=    -0.45258058D-01       EUMP3=    -0.30874137907D+03
E4(DQ)= 0.34944377D-02       UMP4(DQ)= -0.30873788463D+03
E4(SDQ)=  -0.40085684D-02       UMP4(SDQ)=  -0.30874538763D+03
DE(Corr)=  -1.0911017    E(Corr)=    -308.70896278
NORM(A)= 0.11927555D+01
Iteration Nr. 2
**********************
DD1Dir will call FoFMem 1 times, MxPair=    420
NAB= 210 NAA=    0 NBB=    0.
DE(Corr)=  -1.1230011    E(CORR)=    -308.74086218    Delta=-3.19D-02
NORM(A)= 0.12094299D+01
Iteration Nr. 3
**********************
DD1Dir will call FoFMem 1 times, MxPair=    420
NAB= 210 NAA=    0 NBB=    0.
DE(Corr)=  -1.1256519    E(CORR)=    -308.74351300    Delta=-2.65D-03
NORM(A)= 0.12123093D+01
Iteration Nr. 4
**********************
DD1Dir will call FoFMem 1 times, MxPair=    420
NAB= 210 NAA=    0 NBB=    0.
DE(Corr)=  -1.1284213    E(CORR)=    -308.74628242    Delta=-2.77D-03
NORM(A)= 0.12135283D+01
Iteration Nr. 5
**********************
DD1Dir will call FoFMem 1 times, MxPair=    420
NAB= 210 NAA=    0 NBB=    0.
DE(Corr)=  -1.1287546    E(CORR)=    -308.74661565    Delta=-3.33D-04
NORM(A)= 0.12138415D+01
Iteration Nr. 6
**********************
DD1Dir will call FoFMem 1 times, MxPair=    420
NAB= 210 NAA=    0 NBB=    0.
DE(Corr)=  -1.1288002    E(CORR)=    -308.74666128    Delta=-4.56D-05
NORM(A)= 0.12138681D+01
Iteration Nr. 7
**********************
DD1Dir will call FoFMem 1 times, MxPair=    420
NAB= 210 NAA=    0 NBB=    0.
DE(Corr)=  -1.1288052    E(CORR)=    -308.74666624    Delta=-4.96D-06
NORM(A)= 0.12138702D+01
Iteration Nr. 8
**********************
DD1Dir will call FoFMem 1 times, MxPair=    420
NAB= 210 NAA=    0 NBB=    0.
DE(Corr)=  -1.1288029    E(CORR)=    -308.74666401    Delta= 2.22D-06
NORM(A)= 0.12138753D+01
Iteration Nr. 9
**********************
DD1Dir will call FoFMem 1 times, MxPair=    420
NAB= 210 NAA=    0 NBB=    0.
DE(Corr)=  -1.1288031    E(CORR)=    -308.74666422    Delta=-2.03D-07
NORM(A)= 0.12138756D+01
Iteration Nr.  10
**********************
DD1Dir will call FoFMem 1 times, MxPair=    420
NAB= 210 NAA=    0 NBB=    0.
DE(Corr)=  -1.1288038    E(CORR)=    -308.74666488    Delta=-6.61D-07
NORM(A)= 0.12138767D+01
Iteration Nr.  11
**********************
DD1Dir will call FoFMem 1 times, MxPair=    420
NAB= 210 NAA=    0 NBB=    0.
DE(Corr)=  -1.1288042    E(CORR)=    -308.74666524    Delta=-3.59D-07
NORM(A)= 0.12138772D+01
Iteration Nr.  12
**********************
DD1Dir will call FoFMem 1 times, MxPair=    420
NAB= 210 NAA=    0 NBB=    0.
DE(Corr)=  -1.1288044    E(CORR)=    -308.74666548    Delta=-2.44D-07
NORM(A)= 0.12138775D+01
Iteration Nr.  13
**********************
DD1Dir will call FoFMem 1 times, MxPair=    420
NAB= 210 NAA=    0 NBB=    0.
DE(Corr)=  -1.1288045    E(CORR)=    -308.74666559    Delta=-1.09D-07
NORM(A)= 0.12138777D+01
Iteration Nr.  14
**********************
DD1Dir will call FoFMem 1 times, MxPair=    420
NAB= 210 NAA=    0 NBB=    0.
DE(Corr)=  -1.1288046    E(CORR)=    -308.74666564    Delta=-4.66D-08
NORM(A)= 0.12138778D+01
Wavefunction amplitudes converged. E(Corr)=    -308.74666564
Largest amplitude= 5.07D-02

作者
Author: mfdsrax2 时间: 2023-12-24 15:59
用的方法是CISD(T)?
算(T)的时候非常耗资源，确保内存和硬盘足够，且服务器没有进行其他任务

作者
Author: -ZQ- 时间: 2023-12-24 16:05

mfdsrax2 发表于 2023-12-24 15:59
用的方法是CISD(T)?
算(T)的时候非常耗资源，确保内存和硬盘足够，且服务器没有进行其他任务

用的W1U算的一个小结构，没有提示内存和硬盘不足（No space left on device），就是不知道为啥意外停止了

作者
Author: ionexchangeC 时间: 2023-12-24 16:53
超过十个原子且体系没有对称性就不要考虑W1了

作者
Author: -ZQ- 时间: 2023-12-24 18:10

ionexchangeC 发表于 2023-12-24 16:53
超过十个原子且体系没有对称性就不要考虑W1了

明白，谢谢

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