@@azizrouichi3844 Hi . Unfortunately I can't speak French (I used Google translate) . Moreover , I am sorry, I am yet to read about it. Maybe in a few months I would be able to do that 😬
@azizrouichi38444 жыл бұрын
@@zainabm809 ok thank you, good luck
@daniellinzel19949 жыл бұрын
THANK YOU! This was so clear and I needed this for my exams next week. Huge thanks!
@ChemProfCramer9 жыл бұрын
+Daniël Linzel you're very welcome! Good luck on the exams.
@studistudi6 жыл бұрын
Thank you, this was very clear. And also your presentation style was very agreeable and well paced.
@ChemProfCramer6 жыл бұрын
It's very kind of you to let me know. I'm glad you found the presentation useful, and wish you will for the future!
@dumitru-claudiusergentu67208 жыл бұрын
Thank you Chris, I took some of your basic ideas for my PhD thesis.
@ChemProfCramer8 жыл бұрын
+Sergentu Claudiu Delighted you found it useful, and congratulations on completing your degree!
@hurolinci59866 жыл бұрын
Thank you. Great video!
@ChemProfCramer6 жыл бұрын
You're very welcome.
@maryamdaneen71902 жыл бұрын
Thank you so much, Sir
@lalitasharma6687 Жыл бұрын
Thanks for the lecture professor I have two questions. 1) is it only applicable for ground state only . 2) in the term of when you said it's an exact theory but it's exchange correlation energy functional if that's the issue here so can we just “ignore” it since it's very small term and the remaining term can be treated as by average value of them as Born Oppenheimer approximation. I hope it make sense
@ramniwasprajapat33827 жыл бұрын
Thank you chris for help me
@quantumdonut71792 жыл бұрын
Thank you for this series. This is the first time I feel like I get a proper overview. When reading about DFT in books, I got lost everytime. I have a question though. In the LSDA approximation, where is the correction to the kinetic energy? Delta T had suddenly disappeared, and I am not sure where to. Is it in the exchange term? Am I assuming correctly that it is just a correction to alpha?
@ChemProfCramer2 жыл бұрын
Yes, you're exactly right. Very few approaches to a separate kinetic energy correction have been pursued (successfully, anyway...), so in essence any parameterization that goes into the exchange-correlation functional rolls in the kinetic energy correction in some average way based on the parameterization procedure.
@quantumdonut71792 жыл бұрын
@@ChemProfCramer Oh, what a quick response. I see, thank you very much!
@aminouxphy3 жыл бұрын
very interesting
@dariomiric29582 жыл бұрын
What is the name of the textbook you are reffering to at the beggining (section 4.5.1)?
Hello Dr. Cramer, By the same reason that HF ended up with four indices integrals if one plugs equ 15 in eq 20, DFT ends up with the same integrals. Honestly, I didn't understand how four indices integrals showed up (I already watched the video on HF). Do you consider a four indices integral? Thank you so much for your fast responses
@trollsofalabama8 жыл бұрын
Hello Chris, I wanted to ask you really quick. The treatments I've seen involves doing Lagrange multiplier optimization to show that the interacting and non-interacting cases have the same form. Is that all that portion of the treatment shows? Just an inspiration to solve the interacting case in this basically perturbation expansion way with the non-interacting case as the origin of expansion?
@ChemProfCramer8 жыл бұрын
I'm probably not the best person to ask about really fundamental aspects of the theory. However, I would say that the invocation of the non-interacting reference was not motivated by an original desire to generate a perturbation theory machinery. Rather, it was a creative realization that the kinetic energy in the non-interacting form had a terrifically simple means of computation (the kinetic energy operator applied to the KS orbital Slater determinant) and the hope that the remaining (correlated) portion of the kinetic energy would either be small or not too hard to approximate. Prior to that, most DFT theory dealt with much more generalized density formulations, for which kinetic energy was (and continues to be) much more difficult to estimate.
@Tyomas14 жыл бұрын
Real and fictitious systems do not have the same nuclei positions! This is a factual error already in the first slide!
@ganjaboy57175 жыл бұрын
You can't say that DFT is an exact method.
@vittoriobariosco80064 жыл бұрын
No he can say, because DFT is correct while calculation methods developed instead are approximated.