一般来说,我们都是用Landau-Zener跃迁几率公式进行计算,可是文献中M.S.Gordon等人却直接从图11中读出了几率的大小,很是疑惑,请大家帮忙解答下!谢谢!
J. Chem. Phys., Vol. 112, No. 23, 15 June 2000.pdf
(2008-06-18 16:17:28, Size: 313 KB, Downloads: 16)
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跃迁几率计算的问题
ccidnetcom 发表于: 2008-6-18 16:17 来源: 化学吧 - 化学论坛 - 学术论坛
最近看了篇文献,里面涉及到自旋-轨道耦合(SOC)和不同反应势能面跃迁几率的计算.
一般来说,我们都是用Landau-Zener跃迁几率公式进行计算,可是文献中M.S.Gordon等人却直接从图11中读出了几率的大小,很是疑惑,请大家帮忙解答下!谢谢!
一般来说,我们都是用Landau-Zener跃迁几率公式进行计算,可是文献中M.S.Gordon等人却直接从图11中读出了几率的大小,很是疑惑,请大家帮忙解答下!谢谢!
最新回复
况且直接可以算出跃迁几率公式中的所有未知参数,如: 瞬时速率v, 不同势能面在交叉点附近的梯度差等.
但是对于过渡金属,尤其是重金属,相关很详细的报道还没见过
Molpro好象也能算SOC,但是大多好象是GAMESS,毕竟GAMESS-US是免费的,呵呵~
Dice,好长时间不见啊
Based on Landau-Zener formula,
P(LZ) = exp[(-2pi*A2)/(hv*delta F)].
The case Gordon studied is a crossing between doublet and quartet. The spin are different! Calculations may give a 0 value for derivative coupling between two different spins?
Sometimes, I saw some people used SOC to estimate magnitude of the crossing. I guess he used SOC to replace derivative coupling for Landau-Zener formula??? What's F paper in 2000? How they get velocity of the motion? Amazing!?!?
Do you have experience to calculate Landau-Zener formula?? I have to consider that in my study.
SOC can be calculated by Gaussian, Molpro too, but I did not use before.
Calculations may give a 0 value for derivative coupling between two different spins?
不同自旋态之间的耦合系数应该有等于0的情况,但是我们主要计算关键电子和轨道之间的作用,所以耦合系数一般不为0,具体你可以参考这篇文献:
http://g.zhubajie.com/urllink.php?id=2580504uybdea2e5drxrohr
I guess he used SOC to replace derivative coupling for Landau-Zener formula???
应该不是吧,文献中已经明确提到了室温下的跃迁几率为13.4%.况且直接以SOC的大小衡量Landau-Zener的几率大小也是不准确的,见参考文献:
http://g.zhubajie.com/urllink.php?id=258050709l1ag5qz1cb12op
http://g.zhubajie.com/urllink.php?id=2580534wyz21qux5pz1f5fe
What's F paper in 2000?
F, fluoride atom.
How they get velocity of the motion? Amazing!?!?
CASSCF的CONICAL关键词好象可以计算得到,我没算过
Do you have experience to calculate Landau-Zener formula?? I have to consider that in my study.
和你一样,第一次接触,欢迎多交流
[quote]原帖由 ccidnetcom 于 2008-6-19 16:24 发表
个人见解,仅供参考.
Calculations may give a 0 value for derivative coupling between two different spins?
不同自旋态之间的耦合系数应该有等于0的情况,但是我们主要计算关键电子和轨道之间的作用,所以耦合系数一般不为0,具体你可以参考这篇文献:
http://g.zhubajie.com/urllink.php?id=2580504uybdea2e5drxrohr
I guess he used SOC to replace derivative coupling for Landau-Zener formula???
应该不是吧,文献中已经明确提到了室温下的跃迁几率为13.4%.况且直接以SOC的大小衡量Landau-Zener的几率大小也是不准确的,见参考文献:
http://g.zhubajie.com/urllink.php?id=258050709l1ag5qz1cb12op
http://g.zhubajie.com/urllink.php?id=2580534wyz21qux5pz1f5fe
From the molpro 2006 manual, when we locate MECP between differentspin, we don't need to calculate derivative coupling (DC). "skip the DCevaluation if the conical intersection involves states with differentspin (e.g., a Singlet/Triplet crossing) because the coupling is thenzero." Interestingly, Shaik exactly used SOC to appromixate DC in LZ equation.
What's F paper in 2000?
F, fluoride atom.
D. G. Fedorov and M. S. Gordon, J. Chem. Phys. 112, 5611 ~2000!.
How they get velocity of the motion? Amazing!?!?
CASSCF的CONICAL关键词好象可以计算得到,我没算过
It should be wrong. CONICAL关键词 calculates DC and 2 Gradients (to get gradient difference) only. People need velocity of trajectory when it crosses. Shaik guessed velocity by "the sum of translational energies of the two reactants at 300 K". Interestingly.
Shaik guessed velocity by "the sum of translational energies of the two reactants at 300 K".
S.Shaik 很巧妙的回避了这样定性猜测速率的不合理性,但倘若我们要仔细地讨论两不同势能面交叉点附近的自旋翻转行为的话则不能这样说明.
回归到这个帖子的原始问题上,Gordon等人是用什么方法计算做出图11并得到跃迁几率的?用GAMESS计算速率和交叉点附近势能面梯度差?还是用G03的CASSCF方法?
CASSCF我也是刚入手,好多问题还不懂,请大家指教!
谢谢
势能面梯度差 can be done by many QM softwares. You can also use other methods such as DFT for 势能面梯度差, when you get correct SCF solutions for these two states. CASSCF sometimes is expensive.
However, velocity can't be obtained easily. For my case, I have to run MD to get velocity.
No other idea!
QUOTE:
DFT方法计算势能面梯度差我没试过,麻烦你能再深入介绍下吗?
十分感谢你的帮助!
QUOTE:
http://g.zhubajie.com/urllink.php?id=2588815rz88ok3sxzyluvhj
文献很不错,我看的不多,以后还希望你指点!
顺便还有问题,我用J.N.Harvey的Crossing2004程序包计算了两势能面的交叉点所对应的最低能量交叉点(MECP),如果我要计算MECP点的Landau-Zener跃迁几率,这时MECP点处的梯度差是否和G03下CP点附近的梯度差相同呢?
这是Crossing2004找到MECP后的部分输入文件:
Geometry at Step 18
6 -1.8285865 -0.0366090 0.1594269
1 -2.3166355 -0.5805176 -0.6551690
1 -2.1388912 1.0102687 0.1009388
1 -2.1913298 -0.4454553 1.1077786
1 1.6058601 0.8704737 -0.4442845
75 0.3074879 -0.1753686 0.0126932
Energy of First State: -118.7147209450
Energy of Second State: -118.7147207560
Convergence Check (Actual Value, then Threshold, then Status):
Max Gradient El.: 0.000320 (0.000700) YES
RMS Gradient El.: 0.000135 (0.000500) YES
Max Change of X: 0.000704 (0.004000) YES
RMS Change of X: 0.000291 (0.002500) YES
Difference in E: 0.000000 (0.000050) YES
Overall Effective Gradient:
1 -0.00022589 -0.00000281 0.00009367
2 0.00011155 0.00009810 0.00000019
3 0.00009820 -0.00032041 0.00010019
4 0.00008381 0.00027594 -0.00012887
5 -0.00003501 0.00005792 0.00000540
6 -0.00003265 -0.00010872 -0.00007057
Difference Gradient: (RMS * DE: 0.020900)
1 0.02385443 0.04572778 -0.02059291
2 0.01149468 -0.00117398 -0.00601337
3 -0.00622857 -0.00707550 0.00145379
4 0.01892810 -0.00331754 0.00854815
5 -0.05129095 0.00119207 0.00356336
6 0.00324231 -0.03535284 0.01304098
Parallel Gradient: (RMS: 0.000135)
1 -0.00022526 -0.00000160 0.00009312
2 0.00011185 0.00009807 0.00000003
3 0.00009803 -0.00032060 0.00010022
4 0.00008431 0.00027585 -0.00012864
5 -0.00003637 0.00005795 0.00000549
6 -0.00003256 -0.00010966 -0.00007023
The MECP Optimization has CONVERGED at that geometry !!!
Goodbye and fly with us again...
http://g.zhubajie.com/urllink.php?id=25914152sxwmzwwmahn22s9
http://g.zhubajie.com/urllink.php?id=2591468tm3ia4mftw9fyoks
If I remember correctly.Harvey's MECP code calls G03 to calculate E and gradient at the different spin states. Then, gradient difference can be obtained in the MECP code. G03 does not do gradient difference, nor the MECP code calculate gradient.
I am not a physical chemist. I also don't know equations. I heard there is another way to get transition probability, Zhu-Nakamura formula.
QUOTE:
We use the kinetic theory of gases to evaluate relative velocities (this is a gas phase reaction).But of course getting probability values from CASSCF is perfect.I hope this answers your question.
等于没有说嘛