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Gaussian 09 中溶劑化的熒光發(fā)射
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Sample Text 在Gaussian 09 的計算熒光發(fā)射考慮了溶劑化效應的算例中(如下所示),在7步計算中,各自讀取的chk文件分別是哪個。窟有就是具體的“平衡溶解”和“非平衡溶解”是什么意思? 謝謝指導,苦惱了好幾天的一個問題了,拿出來大家討論一下 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 Step 5: Vibrational frequencies of the excited state structure. Now we run a frequency calculation to verify that the geometry located in step 4 is a minimum. The results could also be used as part of a Franck-Condon calculation if desired (see below). This is a numerical frequency calculation. %chk=05-ac # B3LYP/6-31+G(d,p) TD=(Read,NStates=6,Root=1) Freq SCRF=(Solvent=Ethanol) Geom=Check Guess=Read Acetaldehyde excited state freq 0 1 Step 6: Emission state-specific solvation (part 1). This step does state-specific equilibrium solvation of the excited state at its equilibrium geometry, writing out the solvation data for the next step via the PCM NonEq=write input. %chk=06-ac # B3LYP/6-31+G(d,p) TD=(Read,NStates=6,Root=1) SCRF=(Solvent=Ethanol,StateSpecific,Read) Geom=Check Guess=Read Acetaldehyde emission state-specific solvation at first excited state optimized geometry 0 1 NonEq=write Step 7: Emission to final ground state (part 2). Finally, we compute the ground state energy with non-equibrium solvation, at the excited state geometry and with the static solvation from the excited state. %chk=07-ac # B3LYP/6-31+G(d,p) SCRF=(Solvent=Ethanol,Read) Geom=Check Guess=Read Acetaldehyde: ground state non-equilibrium at excited state geometry. 0 1 NonEq=read |
計算 | Gaussian學習 | 經(jīng)驗 | 計算化學 |
銀蟲 (正式寫手)
送鮮花一朵 |
新手,有一個問題不明白,就是第四步進行結(jié)構優(yōu)化,里面說到要打破對稱性, 是不是就是最后 “4 1 2 3 10.0 5 1 2 7 -50.0” 但是這個怎么確定呢? 先在此謝過· %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 |
木蟲 (著名寫手)
瞌睡木蟲
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每一步計算分別使用前一步計算保存的chk文件,所以從第二個計算開始時,你需要手動將前一個計算chk文件復制一下,名字為新的計算的chk文件名。 關于平衡性和費平衡性計算主要是涉及使用溶劑模型的激發(fā)態(tài)計算,具體區(qū)分如下。 SCRF計算中,溶質(zhì)被激發(fā)時,溶劑會由于溶質(zhì)的狀態(tài)變化而被極化,其電子分布被改變,這是一個很快的過程,同時溶質(zhì)分子需要適應這種改變(比如進行轉(zhuǎn)動、平動等),但這是一個很慢的過程。此時計算就分為兩種: 平衡性計算:溶劑和溶質(zhì)有足夠的時間響應變化,調(diào)整到比較平衡的狀態(tài);激發(fā)態(tài)的幾何優(yōu)化及使用了ExternalIteration技術的計算默認使用平衡性計算。 非平衡性計算:近似描述因過程太快(比如電子的垂直激發(fā))使溶劑沒時間進行響應。激發(fā)態(tài)的單點計算默認使用非平衡性計算。 |

木蟲 (著名寫手)
瞌睡木蟲
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我分析了一下那幾步的計算,的確第6步需要第4步的chk 第3步使用第1或第2步的都行 第6步和第7步計算是和第3步的兩個計算相對應,只不過第6/7步是發(fā)射,第3步是激發(fā),第6/7兩步得到的能量之差就是經(jīng)過指定態(tài)的溶劑化效應校正之后的熒光發(fā)射能量,經(jīng)過轉(zhuǎn)換可以得到熒光的發(fā)射波長。 關于平衡和非平衡的問題是和statespecific計算有著不可分的關系的,你可以看一下下面的英文部分,這是我前一段時間從Gaussian官方技術顧問那里得到的詳細解釋,幾句話精練不出來,翻譯也要耗費一段時間,你耐心讀一讀。 另外,如果你是用的是Gaussian 09 A.01或者A.02,請使用ExternalIteration或者SelfConsistent代替Statespecific選項,因為后者有點問題,如果是G09 B.01,用這三個選項之一都行,意義相同。 =========================================== The "StateSpecific" approach in this context applies to the excited state, and yes it involves solving self-consistently the "fast" component of the solvent polarization for the target state and non-equilibrium solvation for the "slow" component of the solvent polarization (i.e. the "slow" component of the solvent polarization comes from the origin state). The total polarization is always partitioned into two components, "slow" and "fast". The "slow" part can be regarded as the reorganization of the solvent molecules as a response to a change in the electronic density of the solute. The "fast" part can be regarded as the response of the electrons in the solvent to a change in the electronic density of the solute. For a change in the electronic density of the solute such as a vertical electronic transition, the "slow" component of the polarization is much slower than the timescale of the electronic transition, so the solvent does not have time to respond in this way to the vertical electronic transition. The "fast" component of the polarization, on the other hand, is closer in timescale to the vertical electronic transition on the solute. In an equilibrium solvation calculation, both components are in equilibrium with the solute's density. By default, all ground state calculations assume equilibrium solvation, as well as geometry optimizations of excited states (and also any calculation that involves the computation of the relaxed density of the excited state). Again, in equilibrium solvation processes, both the "slow" and "fast" component of the solvent polarization are in equilibrium with the excited state density. In the case of a TD energy calculation (no excited state density or geometry optimization) for the computation of a vertical electronic excitation, the default is to do a non-equilibrium process. For this case then, the "fast" component of the polarization "responds" to the change in the solute density from ground to excited state, but the "slow" component did not have time to "respond" so it still comes from the one that was in equilibrium with the solute's ground state density. This is the case of both "Step 2" and "Step 3" in the example shown in the manual. In "Step 2", an energy calculation using TD is performed, thus it defaults to non-equilibrium solvation. The solvation effects on the excited states energies are computed by means of a linear response approach. The absorption energies via the linear response approach only are those reported directly in the output of this "Step 2" job. In "Step 3", a step further is taken and a correction of the linear response excitation energy is performed by solving the "fast" component of the solvent polarization self-consistently with the selected excited state density (the "State-Specific" approach). This is generally an improvement over the excitation energies obtained by linear response alone. Note that since this "State-Specific" approach involves the calculation of the excited state density, the program would default to doing an equilibrium solvation calculation on the excited state. However, the goal of "Step 3" is to compute the vertical excitation energy, so as mentioned above, we would like to use the "slow" component of the solvent polarization from the ground state calculation (in "Step 3", the first part does an equilibrium calculation on the ground state saving the solvent reaction field to the checkpoint file) and solving self-consistently the "fast" component with the excited state density (the second part of "Step 3" reads the reaction field from the checkpoint file, the one from the ground state calculation, keeps the "slow" component as is, and solves the "fast" component self-consistently with the excited state density). The absorption energy via the "State-Specific" approach is the energy difference between the excited state energy after all PCM corrections from the non-equilibrium calculation in "Step 3" and the ground state energy resulting from the equilibrium process (either first part of "Step 3" or final, optimized geometry, energy from "Step 1", the two ground state energies should be the same). In a TD geometry optimization of an excited state, since one is looking for the equilibrium geometry, the default is to do equilibrium solvation, so the two components, "slow" and "fast", of the polarization are in equilibrium with the solute's excited state density. All "Step 4", "Step 5" and "Step 6" use equilibrium solvation for the selected excited state. The emission energy (vertical energy of the excited to ground state transition) by means of a linear response approach can be found in the output of "Step 4". For the final (optimized) geometry in "Step 4", the "excitation energy" shown in this output would be equal to the emission energy since it is the result of an equilibrium calculation on the selected excited state. "Step 6" and "Step 7" are analogous to the two parts of "Step 3" but this time for the opposite transition (excited to ground states). Thus, "Step 6" is analogous to the first part of "Step 3", it is an equilibrium calculation on the origin state (now the excited state) in which both "slow" and "fast" components of solvent polarization are solved self-consistently with the excited state density (this calculation can be regarded as a correction of the excited state energy beyond the linear response approach, which was done in "Step 4" . Now, this "Step 6" saves the solvent reaction field to the checkpoint file. "Step 7" reads this information from the file (just like the second part of "Step 3" and performs a non-equilibrium calculation of the ground state energy, using the "slow" component of the solvent polarization from the excited state calculation ("Step 6" and only doing the "fast" component of the solvent polarization self-consistent with the ground state density. The emission energy via the "State-Specific" approach would be the energy difference between the excited state energy after all PCM corrections from "Step 6" and the ground state energy resulting from the non-equilibrium process in "Step 7". |

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