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Illusionist银虫 (正式写手)
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[求助]
G09中的荧光计算部分的例子，第四部分
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前面几步都已经顺利完成。但是第四步的时候，Opt=RCFC加入之后就会出错。只用opt就好使。我想知道rcfc对这一步计算在结果上会不会有影响。 谢谢 %chk=step4.chk # B3LYP/6-31+G(d,p) TD=(Read,NStates=6,Root=1) SCRF=(Solvent=Ethanol) Geom=Modify Guess=Read Opt=RCFC Acetaldehyde: excited state opt Modify geometry to break Cs symmetry since first excited state is A" 0 1 4 1 2 3 10.0 5 1 2 7 -50.0 Initialization pass. Cartesian force constants read from checkpoint file: step4.chk Length of force constants on CHK file is 0 NAT3TT is 14196 Error termination via Lnk1e in /home/gaussian//g09/l103.exe at Wed Nov 30 08:32:45 2011. 官网有这段话： Note that Cartesian force constants are only available on the checkpoint file after a frequency calculation. You cannot use this option after an optimization dies because of a wrong number of negative eigenvalues in the approximate second derivative matrix. In the latter case, you may want to start from the most recent geometry and compute some derivatives numerically (see below). ReadCartesianFC is a synonym for RCFC. 但是第三步并没有进行频率计算啊··· [ Last edited by Illusionist on 2011-11-30 at 08:46 ] |
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Illusionist
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但是算例来看·第三步并没有计算什么震动啊······ Fluoresence example: Emission (Fluorescence) from First Excited State (n→π*) of Acetaldehyde Here we study the cycle: Acetaldehyde Excitation and Emission Cycle The primary process of interest is the emission, but this example shows how to study the complete cycle including the solvent effects. Step 1: Ground state geometry optimization and frequencies (equilibrium solvation). This is a standard Opt Freq calculation on the ground state including PCM equilibrium solvation. %chk=01-ac # B3LYP/6-31+G(d,p) Opt Freq SCRF=(Solvent=Ethanol) Acetaldehyde ground state 0 1 C C,1,RA X,2,1.,1,A O,2,RB,3,A,1,180.,0 X,1,1.,2,90.,3,0.,0 H,1,R1,2,A1,5,0.,0 H,1,R23,2,A23,5,B23,0 H,1,R23,2,A23,5,-B23,0 H,2,R4,1,A4,3,180.,0 RA=1.53643 RB=1.21718 R1=1.08516 R23=1.08688 R4=1.10433 A=62.1511 A1=110.51212 A23=109.88119 A4=114.26114 B23=120.56468 Step 2: Vertical excitation with linear response solvation. This is a TD-DFT calculation of the vertical excitation, therefore at the ground state equilibrium geometry, with the default solvation: linear response, non-equilibrium. We perform a single-point TD-DFT calculation, which defaults to non-equilibrium solvation. The results of this job will be used to identify which state or states are of interest and their ordering. These results give a reasonable description of the solvation of the excited state, but not quite as good as that from a state-specific solvation calculation. In this case, we see that the n->π* state is the first excited state. Next, we will use the state-specific method to produce a better description of the vertical excitation step. %chk=02-ac # B3LYP/6-31+G(d,p) TD=NStates=6 SCRF=(Solvent=Ethanol) Geom=Check Guess=Read Acetaldehyde: linear response vertical excited states 0 1 Step 3: State-specific solvation of the vertical excitation. This will require two job steps: first the ground state calculation is done, specifying NonEq=write in the PCM input section, in order to store the information about non-equilibrium solvation based on the ground state. Second, the actual state-specific calculation is done, reading in the necessary information for non-equilibrium solvation using NonEq=read. %chk=03-ac # B3LYP/6-31+G(d,p) SCRF=(Solvent=Ethanol,Read) Geom=Check Guess=Read Acetaldehyde: prepare for state-specific non-eq solvation by saving the solvent reaction field from the ground state 0 1 NonEq=write --link1-- %chk=03-ac # B3LYP/6-31+G(d,p) TD(NStates=6,Root=1) SCRF=(Solvent=Ethanol,StateSpecific,Read) Geom=Check Guess=Read Acetaldehyde: read non-eq solvation from ground state and compute energy of the first excited with the state-specific method 0 1 NonEq=read Step 4: Relaxation of the excited state geometry. Next, we perform a TD-DFT geometry optimization, with equilibrium, linear response solvation, in order to find the minimum energy point on the excited state potential energy surface. Since this is a TD-DFT optimization, the program defaults to equilibrium solvation. As is typical of such cases, the molecule has a plane of symmetry in the ground state but the symmetry is broken in the excited state, so the ground state geometry is perturbed slightly to break symmetry at the start of the optimization. %chk=04-ac # B3LYP/6-31+G(d,p) TD=(Read,NStates=6,Root=1) SCRF=(Solvent=Ethanol) Geom=Modify Guess=Read Opt=RCFC Acetaldehyde: excited state opt Modify geometry to break Cs symmetry since first excited state is A" 0 1 4 1 2 3 10.0 5 1 2 7 -50.0 |
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