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背景:计算一个有机分子的荧光发射。
1. opt td的方法优化S1,得到如下的输出:
Excited State 1: Singlet-A 0.7064 eV 1755.06 nm f=0.0001 <S**2>=0.000
149 ->150 0.69974
This state for optimization and/or second-order correction.
Total Energy, E(TD-HF/TD-DFT) = -1800.49595170
Copying the excited state density for this state as the 1-particle RhoCI density.
Excited State 2: Singlet-A 2.2855 eV 542.47 nm f=0.7530 <S**2>=0.000
148 ->150 0.70412
Excited State 3: Singlet-A 2.8615 eV 433.28 nm f=0.0122 <S**2>=0.000
147 ->150 -0.15126
149 ->151 0.58907
149 ->152 -0.16620
149 ->154 0.14120
149 ->161 -0.20913
149 ->163 -0.13502
这样感觉是S2发光,所有优化了S2
2. opt td (root=2),输出结果如下:
Excited State 1: Singlet-A 2.2855 eV 542.48 nm f=0.7532 <S**2>=0.000
148 ->150 -0.70412
Excited State 2: Singlet-A 0.7059 eV 1756.35 nm f=0.0001 <S**2>=0.000
149 ->150 0.69974
This state for optimization and/or second-order correction.
Total Energy, E(TD-HF/TD-DFT) = -1800.49594700
Copying the excited state density for this state as the 1-particle RhoCI density.
Excited State 3: Singlet-A 2.8611 eV 433.35 nm f=0.0120 <S**2>=0.000
147 ->150 -0.14937
149 ->151 0.58957
149 ->152 -0.16583
149 ->154 0.14135
149 ->161 -0.20925
149 ->163 -0.13509
但是,这样S1的振子强度又是最大的了。到底是哪个发光?如何做呢?势能面交叉了?什么样子呢?
我基于优化S2的结构,重新优化S1,结果如下:
Excited State 1: Singlet-A 0.7071 eV 1753.49 nm f=0.0001 <S**2>=0.000
149 ->150 0.69974
This state for optimization and/or second-order correction.
Total Energy, E(TD-HF/TD-DFT) = -1800.49535344
Copying the excited state density for this state as the 1-particle RhoCI density.
Excited State 2: Singlet-A 2.2852 eV 542.55 nm f=0.7532 <S**2>=0.000
148 ->150 0.70413
又是S2大。这应该如何处理呢?
问题总结:优化S1得到的结果S2的f大。优化S2,得到的结果S1的f大。
是一个什么样的势能面交叉图像呢?
这样的问题应该如何处理呢?
谢谢各位老师指点。
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