Quantum Chemistry 8.3 - Linear Variational Method

[@chloanthite]

THANK YOU these videos are a blessing and a public service

[@ppppppporing]

Incredible! Very clear explanation!

[@TMPChem]

Thanks, Eric.

[@andyjiang5096]

Very well explained!

[@gabe7478]

So on the right side of your page: I going from the purple-writing-line to the green-writing-line, what happens to the 1/()? I see that you substituted in your values for the 1st derivatives (with respect to c1), but then it looks like you left out the 1/() But otherwise, these are absolutely incredible videos. They are vastly superior to how my P-chem professor is teaching. I've never been so at a loss for what is going on in a class, and your videos really help me out.

[@TMPChem]

Hi Matthew. Notice how that light purple line is equal to zero. Notice also that the factor of 1/ is factored out from the rest of the right side of the equation. Whenever k * x = 0, it is also true that x = 0, by dividing both sides by k. Here k = 1/, and x is the rest of the line. I then substitute in the values for each derivative from the previous yellow and light blue lines, yielding the following green line as a result.

[@lordofutub]

So (just curious), What exactly is the use of these approximations of molecular wavefunctions? Since they're just approximations, won't they be far off due to the complexities of chemical mixtures? My apologies if this is a stupid question.

[@TMPChem]

As with any approximation, the key is understanding the strengths and limitations of the method. For some applications, the approximation is great and we can use it without reservation. For others, the answer is of insufficient accuracy and other methods must be used or appended on top as a correction. For molecular wavefunctions, the typical base method is a "self-consistent field" (either Hartree-Fock [HF] or density functional theory [DFT]), which very much looks like the linear variational method. For some things, like orbital shapes and electron density maps, HF and/or DFT is usually good enough. For other cases where higher accuracy is necessary (non-covalent interactions, electron affinities, open-shell systems, etc.) and we apply successive corrections on top of HF (like MP2, CI, or CC [see video 9.11]). It's not so much the "complexity" of the chemical system that matters, but how well the method at hand approximates the physics of electron-electron interactions, and what types of interactions are present in the system at hand.

[@solsticetwo3476]

lordofutub - I agree with you. Why to think that such a complex function (e.j. H atom for n=4) is going to be fixed by a finite number of basis function? Image for a multi electron atom or even a molecule. It seems that after such a complex and elegant formulation of the Shrodinger equation the problem ended up being a trivial minimization problem.

[@anshulkapila4962]

Thanx dear for these videos

[@joannel1330]

for the derivation of on the left hand side at the top in blue should it be 2c1s11 +2c2s22?

[@TMPChem]

Hi Joanne. Those lines look correct to me. Remember that we must expand the product of two binomials into the four terms listed below it.

[@therealrictuar]

i dont think that you need the reciprocal factor on the top right hand side in the purple color. the derivative with respect to energy is 0, which the reciprocal was multiplied by, so you are scaling a value that is equal to 0. obviously, the math still works the way that you did it.

[@TMPChem]

Correct, which is why that term disappears in the next line where the terms inside the bracket are now equal to zero without the reciprocal. That line just serves to show that we have fully isolated dE/dc1 as intended in that section.

[@panyakorntaweechat8586]

Are basis functions orthonormal? Sij = 1 when i =j and Sij = 0 when i not equal j ?

[@TMPChem]

In general, we cannot assume that the basis functions are orthogonal. The elements of the S matrix may be any finite real number value (or complex, as long as the S matrix is still Hermitian). The only restriction is that their absolute value must be less than infinity and the diagonal elements must be non-zero. The formulas presented here will work for any S matrix which follows those criteria. The next video on the secular determinant shows how this method gets a lot easier to use and interpret when we do have an orthonormal basis set, especially for the case of 2 basis functions.

[@panyakorntaweechat8586]

Thank you.