Do the terms [aa|bb] & [ab|ba] represent the exact coulomb and exchange integrals (with double integration over x1 and x2), or do they represent the mean field approximation (with a single integral over solely x2?) I can’t tell if the Fock operator results from the substitution of the approximate mean-field integrals for J and K, or if the Fock operator is a natural result (without any need for substitution in this derivation) of assuming the total electronic wave-function takes the form of a slater determinant of one-electron spin orbitals
Hi, in the pink term on the left, shouldn't there be a minus sign in front of all the exchange terms? Furtheremore, I did not understand exactly which two terms you combined in order to factor out the two and cancel the 1/2 in front of the sum sign.
You are correct, there should be negative signs around each of the final 4 exchange terms. For the second part, I exploited the permutational symmetry in purple to get the remaining green pairs on the right. If I list the terms from 1 through 8, I paired 1 and 3, 2 and 4, 5 and 7, and 6 and 8 to get the remaining 4 terms in green, effectively doubling each one, thus cancelling the 1/2.
TMP Chem I know this is a long shot to try and get an answer, but how can you combine those terms? For instance, you said that you are combining terms 1 and 3, or the terms [daa|bb] and [aa|dbb], where d is the delta. Since you have the variation on a and then b, how do you apply the symmetry principles since they are different spin orbitals? Also I’m really enjoying your videos and their very straightforward explanations!
Could you explain a little bit more the last line in the left, how is possible the exchange of electron 1 in orbital a to the orbital b in the following expression, [delta aa|bb]=[bb|delta aa], it is not completely clear what do you mean by dummy indexes?
Hi Said. By dummy indices, I mean that the labels electron 1 and electron 2 are arbitrary. We could have just as well called the first one electron 2 and the second one electron 1. Or A and B, r and s, etc. So if we exchange the labels of electrons 1 and 2, it becomes apparent that it must be that [aa|bb] = [bb|aa], because the second expression is equal to the first expression after that index exchange.
Hi, am I correct in saying that the pairwise sum has b = a + 1 and no half in front or alternatively there is only one sum sign with no half in front and a < b underneath the sum sign.
Hi. I'm confused. Are 'a' and 'b' orbitals or electrons? Also, I note that now the basis orbitals are being differentiated directly (what is the meaning of that?); while in the last slide were the coefficients of the linear combination.
Hi, am I right in saying that there are only two rules regarding how you can exchange the terms. 1) exchange [𝛿aa/b𝛿b] into [b𝛿b/𝛿aa] because they are just dummy indices 2) exchange there complex conjugates [𝛿aa/b𝛿b] into [a𝛿a/b𝛿b]*, in this case the whole term [a𝛿a/b𝛿b]* is the complex conjugate of [𝛿a/b𝛿b] and since they are Hermitian, they are equal to each other.
There are three distinct, fundamental permutational symmetry operations for two-electron integrals: [ab|cd] = [ba|cd], [ab|cd] = [cd|ab], and [ab|cd] = [ba|dc]*. These operations can be applied in succession and are transitive, so there are acutally 2**3 = 8 distinct two-electron integrals which [ab|cd] is equal to. Additionally there is an implicit time / phase factor within the orbitals which we can always arbitrarily choose such that all orbitals will be real, and the complex conjugate business goes away.
Do the terms [aa|bb] & [ab|ba] represent the exact coulomb and exchange integrals (with double integration over x1 and x2), or do they represent the mean field approximation (with a single integral over solely x2?)
I can’t tell if the Fock operator results from the substitution of the approximate mean-field integrals for J and K, or if the Fock operator is a natural result (without any need for substitution in this derivation) of assuming the total electronic wave-function takes the form of a slater determinant of one-electron spin orbitals
Hi. I don't get why the Lagrangian multipliers epsilon_ab are energies.
Hi, in the pink term on the left, shouldn't there be a minus sign in front of all the exchange terms? Furtheremore, I did not understand exactly which two terms you combined in order to factor out the two and cancel the 1/2 in front of the sum sign.
You are correct, there should be negative signs around each of the final 4 exchange terms. For the second part, I exploited the permutational symmetry in purple to get the remaining green pairs on the right. If I list the terms from 1 through 8, I paired 1 and 3, 2 and 4, 5 and 7, and 6 and 8 to get the remaining 4 terms in green, effectively doubling each one, thus cancelling the 1/2.
TMP Chem I know this is a long shot to try and get an answer, but how can you combine those terms? For instance, you said that you are combining terms 1 and 3, or the terms [daa|bb] and [aa|dbb], where d is the delta. Since you have the variation on a and then b, how do you apply the symmetry principles since they are different spin orbitals? Also I’m really enjoying your videos and their very straightforward explanations!
Could you explain a little bit more the last line in the left, how is possible the exchange of electron 1 in orbital a to the orbital b in the following expression, [delta aa|bb]=[bb|delta aa], it is not completely clear what do you mean by dummy indexes?
Hi Said. By dummy indices, I mean that the labels electron 1 and electron 2 are arbitrary. We could have just as well called the first one electron 2 and the second one electron 1. Or A and B, r and s, etc. So if we exchange the labels of electrons 1 and 2, it becomes apparent that it must be that [aa|bb] = [bb|aa], because the second expression is equal to the first expression after that index exchange.
Hi, am I correct in saying that the pairwise sum has b = a + 1 and no half in front or alternatively there is only one sum sign with no half in front and a < b underneath the sum sign.
yes, a double sum of a = 1, ..., N; b = a + 1, ..., N is equivalent to a single sum with index a < b.
Hi. I'm confused. Are 'a' and 'b' orbitals or electrons?
Also, I note that now the basis orbitals are being differentiated directly (what is the meaning of that?); while in the last slide were the coefficients of the linear combination.
@Thomas Asa definitely, have been using flixzone} for since december myself =)
@Thomas Asa definitely, I've been watching on Flixzone} for months myself :D
Hi, am I right in saying that there are only two rules regarding how you can exchange the terms. 1) exchange [𝛿aa/b𝛿b] into [b𝛿b/𝛿aa] because they are just dummy indices 2) exchange there complex conjugates [𝛿aa/b𝛿b] into [a𝛿a/b𝛿b]*, in this case the whole term [a𝛿a/b𝛿b]* is the complex conjugate of [𝛿a/b𝛿b] and since they are Hermitian, they are equal to each other.
There are three distinct, fundamental permutational symmetry operations for two-electron integrals: [ab|cd] = [ba|cd], [ab|cd] = [cd|ab], and [ab|cd] = [ba|dc]*. These operations can be applied in succession and are transitive, so there are acutally 2**3 = 8 distinct two-electron integrals which [ab|cd] is equal to. Additionally there is an implicit time / phase factor within the orbitals which we can always arbitrarily choose such that all orbitals will be real, and the complex conjugate business goes away.
First.