Hi. As usual a great lecture. A question though: when basis are different in the white equations (mu different than nu) still the integral report the expectation value of the operator? Also, I got lost with the indexes: i: an original orbital of the Slater determinant (1..N) nu: a basis atomic orbital (1..K ) mu: dummy basis orbital (1..K) So, the green equation is for a given (mu), so what is the role of mu? or , in all cases mu=nu? Thanks
excellent explanation. However one thing i find confusing is that number of orbitals (psi) will be int(N/2) . Each of these orbitals can be expressed as a linear combination of the basis set as you mentioned. However there is no reason that these basis set will also have the total number K=int(N/2) . It should be any general number depending on the requirements like the STO-nG basis can have n to be anything.
Great job. Thank you! I have a question about the two-electron integrals. For example, I’m trying a calculation of oxygen atom using Hartree-Fock-Roothaan equation using the STO (1s, 2s, 2px, 2py, 2pz). There would be 120 two-electron integrals considering the 8 symmetry permutations. So too many integrals for calculating by hand even for a single atom. Is there any procedure to identify the vanishing integrals, and the integrals of the same value, without performing the integrations?
Yes, you can use the Cauchy-Schwartz inequality to get a bound and even approximation for some orbitals, I think there's a discussion in Szabo's Q. Chem. book.
Does the total molecular wavefunction equal to the product of spacial orbitals? What do spacial orbitals represent and what is the difference between the spacial orbitals and a molecular wavefunction in the Hartree-Fock formalism?
Hi Yevgeniy. The exact wavefunction of a molecule cannot be decomposed into one-particle functions. Orbitals are merely useful approximations in the pursuit of a molecular wavefunction. We start with a set of spin orbitals and multiply them all together for all electrons. This produces a Hartree product. We then add together all possible N! permutations of each electron appearing in each spin orbital (with appropriate sign flipping to satisfy the antisymmetry principle) to get a Slater determinant. The approximation of the molecular wavefunction in Hartree Fock theory is a Slater determinant whose spin orbitals are constructed from a linear combination of an atomic orbital basis set, whose coefficients are determined by minimizing the molecular energy with respect to those basis functions. So to summarize, we take a set of basis functions, figure out which coefficients in which to add them to get our spin orbitals, multiply those together to get a Hartree product, and then add / subtract all permutations of that to get the Slater determinant which is our molecular wavefunction.
TMP Chem Thanks for you reply. I have one more question. I know that choosing 100 basis functions for a Hydrogen molecule will result in 100 spin orbitals when we solve the matrix equations, namely 100 expansion coefficients. Out of these 100 spin orbitals, only the first one will be useful, but all other will represent the virtual spin orbitals which do not contribute to the wavefunction in the Slater Determinant expansion. Is my statement correct? I have read that these virtual orbitals could be used to perform Configuration Interaction procedure to add the correction to the wavefunction and the ground state energy? Thanks.
Correct. In a hydrogen molecule performing a restricted Hartree-Fock calculation with 100 basis functions there will be 1 doubly-occupied molecular orbital and 99 virtual molecular orbitals. Only the occupied molecular orbital contributes to the *ground state* determinant, but there are an exponentially large number of other possible Slater determinants in the Configuration Interaction wavefunction where all possible Slater determinants are included in a linear combination, and thus all molecular orbitals contribute to the energy of the molecule. In any post-Hartree-Fock method we will compute the correlation energy, and all virtual orbitals are used to calculate the correlation energy in the standard variants of all such methods.
HI Chnaller. A given spin orbital is a particular column of a Slater determinant. The columns of the Fock matrix describe what each spin orbital looks like, in terms of the coefficients in the linear combination of atomic orbital basis functions. It says "spin orbital X is composed of this much of basis function Y". That is all independent of the Slater determinant, which is a mathematical construct which tells us how to generate an anti-symmetric N-electron wavefunction. In practice, this wavefunction is extremely large, so we refer to the Slater determinant more as an abstraction rather than explicitly computing on it as we do in earlier quantum mechanics videos. The point of the HF Roothaan equations is to simplify the math down to a point where we can specify a manageable number of inputs (namely nuclear identities, Cartesian locations, and number of electrons) such that the rest of the math can be reduced to computer code and solved by black-box programs.
you don't sum all the elements, a single sum over two matrix elements is the element of another matrix, en.wikipedia.org/wiki/Matrix_multiplication#Definition
Thank you! For a clear explanation on how basis sets relate to the theory! Great Videos!
Thanks Jenni!
Diagonalization approach is very interesting
Hi. As usual a great lecture. A question though: when basis are different in the white equations (mu different than nu) still the integral report the expectation value of the operator?
Also, I got lost with the indexes:
i: an original orbital of the Slater determinant (1..N)
nu: a basis atomic orbital (1..K )
mu: dummy basis orbital (1..K)
So, the green equation is for a given (mu), so what is the role of mu? or , in all cases mu=nu?
Thanks
I got it. The diagonal matrix of energies selects the right mu (mu= nu). right?
@@mikel5264 i have the same question, so mu = nu?
excellent explanation. However one thing i find confusing is that number of orbitals (psi) will be int(N/2) . Each of these orbitals can be expressed as a linear combination of the basis set as you mentioned. However there is no reason that these basis set will also have the total number K=int(N/2) . It should be any general number depending on the requirements like the STO-nG basis can have n to be anything.
Same here, already found a satisfactory explanation for it?
Great job. Thank you! I have a question about the two-electron integrals. For example, I’m trying a calculation of oxygen atom using Hartree-Fock-Roothaan equation using the STO (1s, 2s, 2px, 2py, 2pz). There would be 120 two-electron integrals considering the 8 symmetry permutations. So too many integrals for calculating by hand even for a single atom. Is there any procedure to identify the vanishing integrals, and the integrals of the same value, without performing the integrations?
Yes, you can use the Cauchy-Schwartz inequality to get a bound and even approximation for some orbitals, I think there's a discussion in Szabo's Q. Chem. book.
Does the total molecular wavefunction equal to the product of spacial orbitals? What do spacial orbitals represent and what is the difference between the spacial orbitals and a molecular wavefunction in the Hartree-Fock formalism?
Hi Yevgeniy. The exact wavefunction of a molecule cannot be decomposed into one-particle functions. Orbitals are merely useful approximations in the pursuit of a molecular wavefunction. We start with a set of spin orbitals and multiply them all together for all electrons. This produces a Hartree product. We then add together all possible N! permutations of each electron appearing in each spin orbital (with appropriate sign flipping to satisfy the antisymmetry principle) to get a Slater determinant. The approximation of the molecular wavefunction in Hartree Fock theory is a Slater determinant whose spin orbitals are constructed from a linear combination of an atomic orbital basis set, whose coefficients are determined by minimizing the molecular energy with respect to those basis functions. So to summarize, we take a set of basis functions, figure out which coefficients in which to add them to get our spin orbitals, multiply those together to get a Hartree product, and then add / subtract all permutations of that to get the Slater determinant which is our molecular wavefunction.
TMP Chem Thanks for you reply. I have one more question. I know that choosing 100 basis functions for a Hydrogen molecule will result in 100 spin orbitals when we solve the matrix equations, namely 100 expansion coefficients. Out of these 100 spin orbitals, only the first one will be useful, but all other will represent the virtual spin orbitals which do not contribute to the wavefunction in the Slater Determinant expansion. Is my statement correct? I have read that these virtual orbitals could be used to perform Configuration Interaction procedure to add the correction to the wavefunction and the ground state energy? Thanks.
Correct. In a hydrogen molecule performing a restricted Hartree-Fock calculation with 100 basis functions there will be 1 doubly-occupied molecular orbital and 99 virtual molecular orbitals. Only the occupied molecular orbital contributes to the *ground state* determinant, but there are an exponentially large number of other possible Slater determinants in the Configuration Interaction wavefunction where all possible Slater determinants are included in a linear combination, and thus all molecular orbitals contribute to the energy of the molecule. In any post-Hartree-Fock method we will compute the correlation energy, and all virtual orbitals are used to calculate the correlation energy in the standard variants of all such methods.
Great explanation! But there’s something which isn‘t that clear to me: Are the spin-orbitals psi_i not slater determinants anymore?
HI Chnaller. A given spin orbital is a particular column of a Slater determinant. The columns of the Fock matrix describe what each spin orbital looks like, in terms of the coefficients in the linear combination of atomic orbital basis functions. It says "spin orbital X is composed of this much of basis function Y". That is all independent of the Slater determinant, which is a mathematical construct which tells us how to generate an anti-symmetric N-electron wavefunction. In practice, this wavefunction is extremely large, so we refer to the Slater determinant more as an abstraction rather than explicitly computing on it as we do in earlier quantum mechanics videos. The point of the HF Roothaan equations is to simplify the math down to a point where we can specify a manageable number of inputs (namely nuclear identities, Cartesian locations, and number of electrons) such that the rest of the math can be reduced to computer code and solved by black-box programs.
Now I got it. Thank you!
How the sum of matrix elements give another matrix?
you don't sum all the elements, a single sum over two matrix elements is the element of another matrix, en.wikipedia.org/wiki/Matrix_multiplication#Definition
First.