That is an awesome lecture on basis sets. It's always been really hard for me to find this kind of information. Finally, being a physicist, I get the reasons for the existence of so many different basis set types. More importantly, many good guidelines were given, and now I think I'll be able to choose my basis sets much more soundly.
Thank you so much for your excellent lecture. I think in the bond lengths will be longer in the CI case. CI allows electrons to have some probabilty to enter the virtual orbitals, i.e., the excited states. And it is very common in photo-physics that once an electron gets excited by absorbing light, the bond length becomes longer as always being indicated in the Frank-Condon diagram. I guess this is because when an electron gets into the excited state (virtual orbital), energy of the molecular system gets higher and results in longer bond lengths.
Just a stupid answer Bond order is bonding - antibonding /2 so since I'm population antibonding my bond order will decrease so bond length will increase
Where can I find the answer to the thought question you ask at the very end. My attempted answer is - CI lowers the energy. So, lowered energy is extended bond. Does my simplistic argument hold?
Your answer is correct (the bonds are longer), but not your logic (if lengthening bonds automatically lowered the energy, why would there be any bonds at all?) To best answer the question, think about how the WAVE function changes going from HF to CI. Take a look at the H2 example earlier. The wave function is expressed in terms of orbitals, and we know how the occupation of certain kinds of ORBITALS ought to affect bond lengths. Hope that hint helps...
Hello Dr. Cramer, At 21:55 you mentioned that it would be better to relax HF orbitals. So, does CI do it automatically or one needs to add that in? I believe MCSCF does it!!
Yes, exactly. CI does NOT, but MCSCF does (within a limited active space). Full CI does not need to, because if all orbitals are active, the CI coefficients span the free variable space.
In CI, the electron correlation will lower the electron repulsion energy due to their correlated motion, hence I think CI bond length will be shorter for the molecule.
+Saleheen Noman Generally no, although it's certainly possible in any particular case. The simplest way to see this is to recognize that the CI wave function builds in correlation (reduced repulsion energy) by adjusting the wave function to LOWER the percentage of HF character (occupied bonding orbital(s)) and RAISING the percentage of excited configurations incorporating anti-bonding orbital(s) character. I.e., HF overbinds. This is a very well characterized phenomenon -- it also rationalizes why vibrational frequencies computed at the HF level are generally about 10% too large: too much bonding, too short bond lengths. [Exceptions can occur in more exotic cases, where the valence orbitals are themselves already antibonding, so that strongly correlating configurations could introduce additional bonding character at the expense of those antibonding reference configuration(s).]
Thanks for the explanation. I have a question: The additional correlation that is being considered here, is it due to Pauli's principle or the finite size of the particles. I think it is the latter as we have taken care of the former by the Slater determinant. Thanks again.
Well, think about it purely from the viewpoint of the variational theorem. If you make the energy go down, you MUST have done a better job accounting for electron correlation (because you can't go below the exact energy). Thus, the question of a more correlated wave function becomes a question of parameters -- with respect to how many parameters have you optimized the function? The more, the better (as long as the parameters are not redundant). That is why full CI in an infinite basis is exact -- the (hypothetical) optimization with respect to all possible parameters must give the exact energy and hence the resulting wave function is the exact (fully correlated) wave function.
Thank you for the wonderful lecture. The Configuration Interaction method considered here, is it use the same Hartree-Fock hamiltonian? Because we are only changing the basis set right?
I'm not 100% certain I understand the question, but one probably should not say "Hartree-Fock Hamiltonian". There is a Hartree-Fock OPERATOR, which is an approximation to the correct Hamiltonian (because it replaces the correct electron-electron interaction term with one involving an interaction between average distributions). The Hartree-Fock METHOD finds the optimum wave function for that operator, and it then evaluates the expectation value of the CORRECT Hamiltonian FOR that (approximate) wave function. The CI method involves building an IMPROVED (but still approximate, unless the CI considers all possible excitations (full CI) and uses an infinite basis set) wave function constructed from the Hartree-Fock reference, and evaluating THAT with the correct Hamiltonian. Hope that helps...
@@ChemProfCramerThank you for the detailed explanation but I don't quite understand the term "correct Hamiltonian." are you using perturbation theory to get expectation values?
@@Zephyrus006 For electronic structure theory the correct Hamiltonian is nuclear repulsion, electron-nuclear attraction, electron-electron repulsion, and electron kinetic energy. It's trivial to write it down, it's just impossible to solve the many-body problem that would deliver the correct wave function analytically. In HF theory, you modify the Hamiltonian operator to replace the electron-electron repulsion term with an electron-mean-field repulsion term. THAT can be solved for a wave function. The expectation value (of the correct Hamiltonian) FOR that wave function, which is NOT an eigenfunction of the correct Hamiltonian, will be an upper bound on the correct energy.
CI will lower energy of bonding MOs (and will increase anti-bonding), therefore making them more "stable", so as Chemist and going with chemical intuition :) I can guess that more stable bond is shorter bond, shorter bonds are stronger and needs more energy for dissociation, example is H2, very short and strong bond. Also, you said pick a molecule, so the answer is suppose to be same for every molecule, so I went with very easy H2 approach for my answer :D.
Note that CI doesn't change the one-electron orbitals -- rather, it changes the MANY-electron wave function by reducing the weight of the HF Slater determinant and adding in some weight for determinants defined by excitations from HF occupied orbitals to HF virtual orbitals. And, since the HF occupied orbitals are usually BONDING (lower in energy) and the virtual orbitals are usually ANTIbonding (by needing to be orthogonal to the occupied orbitals), adding weight for determinants with less bonding and more antibonding actually LENGTHENS bonds in the structure that is optimized at the CI level. This generally DOES improve agreement with experiment, i.e., HF overbinds.
@@ChemProfCramer Yes, I understand know. I was looking only from HF determinant perspective, but HF determinant is HF determinant and was mistake by me thinking that CI will alter it, also not considering that in CI there is weight on (doubly) excited determinant in case of H2 (antibonding) and weight less than 1 for HF determinant is another mistake. I have clearer picture now. Thank you for clarifications. Usually I use only DFT in my work and I don't like when I need to check which functional is good for that particular problem, but what I like in WFT methods that there is well defined path on how to improve the results (WF).
CAS implies a full CI (all possible excitations) within a given active space, so the number of configuration state functions (CSFs) gets enormous very quickly. RAS "restricts" how many electrons can leave the occupied window and how many can enter the virtual window. This permits many more orbitals to be included but at the cost of possibly missing some CSFs that need more excitations to describe. Something with a large, conjugated pi system, for example, or multiple metals with active d orbitals, would present too many active space orbitals for a practical CAS but might be accessibly with a well-designed RAS.
Thank you, Professor, for delivering such a wonderful lecture. I thoroughly enjoyed your presentation and found it to be exceptionally insightful.
Very kind of you to offer your appreciative comments, thank you.
That is an awesome lecture on basis sets. It's always been really hard for me to find this kind of information. Finally, being a physicist, I get the reasons for the existence of so many different basis set types. More importantly, many good guidelines were given, and now I think I'll be able to choose my basis sets much more soundly.
I appreciate hearing the feedback, thanks!
Thank you so much for your excellent lecture. I think in the bond lengths will be longer in the CI case. CI allows electrons to have some probabilty to enter the virtual orbitals, i.e., the excited states. And it is very common in photo-physics that once an electron gets excited by absorbing light, the bond length becomes longer as always being indicated in the Frank-Condon diagram. I guess this is because when an electron gets into the excited state (virtual orbital), energy of the molecular system gets higher and results in longer bond lengths.
Your answer (and more importantly, your reasoning about the CI wave function partially populating virtual (antibonding) orbitals) is correct.
@@ChemProfCramer Thank you so much for your reply!
Just a stupid answer
Bond order is bonding - antibonding /2 so since I'm population antibonding my bond order will decrease so bond length will increase
Well, if by "stupid" you mean "logically followed to the correct conclusion in a concise and brilliant manner", then yes! Exactly what you wrote 🙂
Just one critique; I think you were referring to Brillouin's theorem, not Koopman's theorem which is about IPs.
Your videos are awesome!
Absolutely correct! Slip-o-the-tongue... Nice catch.
Where can I find the answer to the thought question you ask at the very end.
My attempted answer is - CI lowers the energy. So, lowered energy is extended bond. Does my simplistic argument hold?
Your answer is correct (the bonds are longer), but not your logic (if lengthening bonds automatically lowered the energy, why would there be any bonds at all?) To best answer the question, think about how the WAVE function changes going from HF to CI. Take a look at the H2 example earlier. The wave function is expressed in terms of orbitals, and we know how the occupation of certain kinds of ORBITALS ought to affect bond lengths. Hope that hint helps...
@@ChemProfCramer thank you so much!
Hello Dr. Cramer, you mentioned Koopmans theorem at 9 : 27, what's the relation beetween the theorem and the single excited SD?
Nice catch. Should've invoked Brillouin's theorem, not Koopman's theorem. I'll add an erratum to the description, and the credit is yours!
Hello Dr. Cramer,
At 21:55 you mentioned that it would be better to relax HF orbitals. So, does CI do it automatically or one needs to add that in? I believe MCSCF does it!!
Yes, exactly. CI does NOT, but MCSCF does (within a limited active space). Full CI does not need to, because if all orbitals are active, the CI coefficients span the free variable space.
In CI, the electron correlation will lower the electron repulsion energy due to their correlated motion, hence I think CI bond length will be shorter for the molecule.
+Saleheen Noman Generally no, although it's certainly possible in any particular case. The simplest way to see this is to recognize that the CI wave function builds in correlation (reduced repulsion energy) by adjusting the wave function to LOWER the percentage of HF character (occupied bonding orbital(s)) and RAISING the percentage of excited configurations incorporating anti-bonding orbital(s) character. I.e., HF overbinds. This is a very well characterized phenomenon -- it also rationalizes why vibrational frequencies computed at the HF level are generally about 10% too large: too much bonding, too short bond lengths. [Exceptions can occur in more exotic cases, where the valence orbitals are themselves already antibonding, so that strongly correlating configurations could introduce additional bonding character at the expense of those antibonding reference configuration(s).]
+Chris Cramer That makes a lot of sense now, Thanks Dr. Cramer!
Thanks for the explanation. I have a question:
The additional correlation that is being considered here, is it due to Pauli's principle or the finite size of the particles. I think it is the latter as we have taken care of the former by the Slater determinant. Thanks again.
Hi Dr. Cramer. I still don't see how by including empty orbitals (and permutations) the issue about the instantaneous correlation is addressed.
Well, think about it purely from the viewpoint of the variational theorem. If you make the energy go down, you MUST have done a better job accounting for electron correlation (because you can't go below the exact energy). Thus, the question of a more correlated wave function becomes a question of parameters -- with respect to how many parameters have you optimized the function? The more, the better (as long as the parameters are not redundant). That is why full CI in an infinite basis is exact -- the (hypothetical) optimization with respect to all possible parameters must give the exact energy and hence the resulting wave function is the exact (fully correlated) wave function.
Chris Cramer Makes perfect sense. Thanks
Hello sir !
While calculating Fukui function, what should v take Hirshfeld charges or population??
Kindly, make me this clear...
Thank you for the wonderful lecture. The Configuration Interaction method considered here, is it use the same Hartree-Fock hamiltonian? Because we are only changing the basis set right?
I'm not 100% certain I understand the question, but one probably should not say "Hartree-Fock Hamiltonian". There is a Hartree-Fock OPERATOR, which is an approximation to the correct Hamiltonian (because it replaces the correct electron-electron interaction term with one involving an interaction between average distributions). The Hartree-Fock METHOD finds the optimum wave function for that operator, and it then evaluates the expectation value of the CORRECT Hamiltonian FOR that (approximate) wave function. The CI method involves building an IMPROVED (but still approximate, unless the CI considers all possible excitations (full CI) and uses an infinite basis set) wave function constructed from the Hartree-Fock reference, and evaluating THAT with the correct Hamiltonian. Hope that helps...
@@ChemProfCramerThank you for the detailed explanation but I don't quite understand the term "correct Hamiltonian." are you using perturbation theory to get expectation values?
@@Zephyrus006 For electronic structure theory the correct Hamiltonian is nuclear repulsion, electron-nuclear attraction, electron-electron repulsion, and electron kinetic energy. It's trivial to write it down, it's just impossible to solve the many-body problem that would deliver the correct wave function analytically. In HF theory, you modify the Hamiltonian operator to replace the electron-electron repulsion term with an electron-mean-field repulsion term. THAT can be solved for a wave function. The expectation value (of the correct Hamiltonian) FOR that wave function, which is NOT an eigenfunction of the correct Hamiltonian, will be an upper bound on the correct energy.
CI will lower energy of bonding MOs (and will increase anti-bonding), therefore making them more "stable", so as Chemist and going with chemical intuition :) I can guess that more stable bond is shorter bond, shorter bonds are stronger and needs more energy for dissociation, example is H2, very short and strong bond. Also, you said pick a molecule, so the answer is suppose to be same for every molecule, so I went with very easy H2 approach for my answer :D.
Note that CI doesn't change the one-electron orbitals -- rather, it changes the MANY-electron wave function by reducing the weight of the HF Slater determinant and adding in some weight for determinants defined by excitations from HF occupied orbitals to HF virtual orbitals. And, since the HF occupied orbitals are usually BONDING (lower in energy) and the virtual orbitals are usually ANTIbonding (by needing to be orthogonal to the occupied orbitals), adding weight for determinants with less bonding and more antibonding actually LENGTHENS bonds in the structure that is optimized at the CI level. This generally DOES improve agreement with experiment, i.e., HF overbinds.
@@ChemProfCramer Yes, I understand know. I was looking only from HF determinant perspective, but HF determinant is HF determinant and was mistake by me thinking that CI will alter it, also not considering that in CI there is weight on (doubly) excited determinant in case of H2 (antibonding) and weight less than 1 for HF determinant is another mistake. I have clearer picture now. Thank you for clarifications. Usually I use only DFT in my work and I don't like when I need to check which functional is good for that particular problem, but what I like in WFT methods that there is well defined path on how to improve the results (WF).
Can u tell us when to use RASSCF and CASSCF: like what type of molecules use RASSCF and CASSCF?
CAS implies a full CI (all possible excitations) within a given active space, so the number of configuration state functions (CSFs) gets enormous very quickly. RAS "restricts" how many electrons can leave the occupied window and how many can enter the virtual window. This permits many more orbitals to be included but at the cost of possibly missing some CSFs that need more excitations to describe. Something with a large, conjugated pi system, for example, or multiple metals with active d orbitals, would present too many active space orbitals for a practical CAS but might be accessibly with a well-designed RAS.
Thank you professor that was very helpful