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Well, if you can organize the necessary funding for visiting my university, we could probably figure out a project you could work on. But it very much depends also on what your background and skills are.

Yes, I am a contributor to Dalton, but this is not really the appropriate channel for asking about problems with the Dalton program. Please either put your question at the user support page of Dalton (gitlab.com/dalton/user-support) or send me personally an e-mail.

Hi, I am not sure, whether I understand your question. Do you ask, how I made the video? Well, I made my slides with the beamer package in LaTeX and then I used ExplainEverything on an iPad.

I used "Explain everything" on my Ipad. It works very nicely

Thanks a lot for your comment. Well, the other course: "Computational Chemistry", which is our Masters course is actually a practical course, where the students get experience with carrying out different types of calculations employing different quantum chemical codes like but not exclusively "our own" Dalton program.

I don't think so, because the zeroth order term is time-independent.

@@SPASauer ok, I see. The |Psi^(0)_{0}> here is the eigenvector of the time-independent Hamiltonian (H^{0})

Yes, here and everywhere in the book.

Hi Jonah, no I will only discuss MP and CC methods here. I am sorry.

I am glad to hear that you liked it. Yes, the two-electron operator is included in the Fock operator but in the form of the Coulomb and exchange operators, which are effective one-electron operators. This means, that the Coulomb repulsion between the electrons is of course included in the HF-method, otherwise you would not get any meaningful results, but it is included in an averaged way. When we talk about "electron correlation" we do not talk about the Coulomb interaction (repulsion) between electrons but about the deviation from the averaged Coulomb interaction as it is included in the HF method. Thus HF does not include electron correlation by definition, as electron correlation is defined as the difference between the exact behavior and the behavior as predicted by the HF-method. Therefore (spin-)orbital energies as calculated by the HF method are also by definition without electron correlation. Koopmans' theorem assigns some meaning to the orbital energies, i.e. an approximation to the energy, which is necessary to remove a particular electron from the system. Otherwise, there is not much which relates orbital energies to something you can measure and the sum of the orbital energies is definitely not a good approximation to the total energy of the system, because in each orbital energy you have the averaged interaction with all the other electrons, which on summing adds up to twice the averaged electron repulsion between all the electrons, which of course is wrong.

I am glad to hear, that you liked the lecture. The point is, as you write yourself, that electron 2 is just a dummy variable. It is the interaction between electron 1, because we look at \psi(1) and another electron. So you could look at "1" and "2" as actually just saying the first electron I look at and the second electron or the electron I look at and another electron, which then has to have a different number. But why does that lead to an average? Well, your integral in (7) is over the whole space meaning all possible positions of electron 2, where at each position you sample |\psi_j(2)|^2 meaning the probability density of the second electron occupying orbital \psi_j, so you weigh the interaction at each position x_2 with the probability for the electron 2 to be at this position. This means that you are not looking at the interaction between electron 1 at the position x_1 and electron 2 at a particular position x_2, but you take the "sum" of the interactions of electron 1 at x_1 with another electron in all possible points in space and as the sum of the probability density of electron 2 |\psi_j(2)|^2 over all points in space is equal to 1, this means, that this sum of the interactions is also divided by the number of points and thus you calculate the average of this interaction.

@@SPASauer thanks for the reply! I think understand it better now. So the “averaging” is being done through the multiplication by the probability of finding electron 2 at all possible points? We then do this for each spin orbital? Basically (in loop form), For each spin orbital, compute integral of (rho(r2)/r1-r2) dr2 Here I guess is where I am still a little confused on the averaging. In the integral above, if we did not have the 1/r1-r2 term, we would just have the integral of the density, which would be 1. So when we do this integration of only rho for all N spin orbitals, won’t we obtain the number N? I feel like this is not an average. Thank you again!

I guess that my confusion comes from what the sum over J is really supposed to be. If J is the average interaction of electron 1 with “electron two” in a given spin-orbital, then the sum over J for all spin orbitals would mean that the potential electron 1 feels is the sum of the average interactions with electron 2 over each spin orbital. In a nutshell, It seems that we are computing the sum of the average interactions. And most of the time I hear that “electron 1 feels the average interaction with all other electron”. The problem is that the sum of averages is not always equal to the average of the sum

You should think about it like, when you take in statistics an average value of some event, e.g. throwing a dice. You would take each possible outcome of throwing a dice (1, 2, 3, 4, 5, 6), multiply it with the probability for this outcome (1/6, if you're using a good dice :-)) and sum over them. Here we do the same: our outcome of the throwing the dice is 1/(r1-r2), the probability is calculated as |\psi_j(2)|^2 and the summation has to be an integral here, because our outcome is a continuous quantity. So, we just take the average of the interaction!

@@StephanPASauer thanks again for the reply! This was really helpful. I can see this in a whole new way now. My final question regards the summation over J. Shouldn’t the summation be averaged? If we do not average the summation, then this means that the potential is larger for higher numbers of other electrons? This is how I am confused about the averaging. If we do not average the sum, then we have the sum of average interactions with the other electrons

I am sorry to hear that the sound is not very good. Well with MCSCF you get a similar problem: choosing configurations/determinants for different systems so that you include a consistent amount of electron correlation in your calculation. The size-extensivity problem does not affect orbital coefficients. It is a problem affecting correlated wavefunction calculations and the energies obtained with these. If you consider a process, where the molecule you consider does not change its number of electrons, then your CI wavefunction will still not be size-extensive. But the real problem comes in, when you compare the energies of systems with different numbers of electrons, like in the H2 and 2 times H2 problem. And in your example, I guess this is not the case.

States higher in energy than the ground state are the excited states. HF doesn't represent the natural system because it doesn't include any electron correlation. CASSCF does include the static part of it, so, it's an improvement over the HF wavefunction.

Sure, I can remember that :-) It is Frank Jensen's: Introduction to Computational Chemistry: www.wiley.com/en-us/Introduction+to+Computational+Chemistry%2C+3rd+Edition-p-9781118825990 In the 3rd. edition the mentioned chapter is 11.5 starting in page 352

Ja, det er en fejl. På den øvre tavle er “1/RT” der stadigvæk. Så jeg må have glemt at skrive den på den nedre tavle og ingen af de tilstedeværende studere havde bemærket det.

I am glad to hear, that you like the lectures.

In a photochemical reaction you have to shine light on your reaction vessel for the reaction to proceed, which means that one or more of the reactions will be excited, while in a thermal reaction you just heat the reaction vessel.

I know this .... but when we say it is thermal or light reaction possible or not possible Or what is the difference in terms of mechanism.... I mean..thermal and photochemical cycloaddition reaction..

The mechanism is supposed to be the same, i.e. a concerted reaction.

Hej Emil, det glæder mig at høre. Bose-Einstein kondensation er desværre ikke et emne i vores fysisk kemi kursus. Så den kommer ikke lige på denne kanal.