<![CDATA[Multiwfn forum / Ratio of CT and LE using the ADCH model]]> http://sobereva.com/wfnbbs/viewtopic.php?id=437 Fri, 22 Jan 2021 01:45:35 +0000 FluxBB <![CDATA[Re: Ratio of CT and LE using the ADCH model]]> http://sobereva.com/wfnbbs/viewtopic.php?pid=1563#p1563 MLCT is just a special case of CT. You can use exactly the same way to obtain CT percentage, then LE % = 1 - CT%

Currently there is no general way of deriving %CT based on hole-electron analysis.

]]> Fri, 22 Jan 2021 01:45:35 +0000 http://sobereva.com/wfnbbs/viewtopic.php?pid=1563#p1563 <![CDATA[Ratio of CT and LE using the ADCH model]]> http://sobereva.com/wfnbbs/viewtopic.php?pid=1561#p1561 I wonder that how obtain the CT and LE contributions to excited state.

I saw the answer related to my question in this page and blog article, but I don't understand it.
Your blog article shows the MLCT percentages(based on the atomic charge) and it is the percentage of charge transfer between the two fragment in the electronic transition.

But I want to obtain the ratio of CT and LE (CT means the spatial seperation of hole and electron, LE means the hole and electron occupy similar spatial region).

I think what I want to obtain and your previous answer is different.

Could you please explain to me in this regard??

]]> Thu, 21 Jan 2021 08:03:18 +0000 http://sobereva.com/wfnbbs/viewtopic.php?pid=1561#p1561