I think it doesn't need to be in the first place, we could call it anything (e.g. \lambda) to start with, but then upon solving the generalized eigenvalue problem, it turns out that \lambda is equal to the expectation value of H with respect to the resulting eigenfunction \tilde{\Phi}. I think the ordering that Trent introduced E first (in magenta, LHS) versus the Lagrange multiplier (in orange, RHS) is what caused confusion for me too initially.
It's really cool how the functional variation in the etymology of the hamiltonian operator is hydrating. At a glance, you would have thought that the wave function would be amphoteric, but it is actually conjugating! The only mistake was that the Scorsese lambdonian integer was mistaken for the atomic number of the hydrating atom. Besides that, good work!
Sure. The coefficient vector, c, is a linear combination of a bunch of basis functions. These basis functions might not be normalized, and they might not be orthogonal. If they're not normalized, then there is "too much" or "too little" electron density. If they're not orthogonal, then basis functions might "overlap" with each other, producing either too much or too little total electron density depending on how they overlap. The overlap matrix S includes of this information, because it contains the overlap integrals of all basis functions. If normalized, the diagonal terms are all 1 (S_ii). If orthogonal, the off-diagonal terms are all 0 (S_ij, i =/= j). So if the basis set is already orthonormal, the S matrix is just an identity matrix, and the answer is the same whether or not it is present. If not, the S matrix corrects for any non-orthonormality of the basis set.
It's analogous to a differential calculus problem where we have a function and we need to find the critical point. The catch is that the function has an extra constraint, requiring us to maintain orthonormal orbitals as we compute the derivative and solve for its zeroes. Besides that, we're just taking a derivative and finding when it is equal to zero.
Why does the Lagrange multiplier have to be the same as Energy (E)? Great videos btw.
I think it doesn't need to be in the first place, we could call it anything (e.g. \lambda) to start with, but then upon solving the generalized eigenvalue problem, it turns out that \lambda is equal to the expectation value of H with respect to the resulting eigenfunction \tilde{\Phi}. I think the ordering that Trent introduced E first (in magenta, LHS) versus the Lagrange multiplier (in orange, RHS) is what caused confusion for me too initially.
It's really cool how the functional variation in the etymology of the hamiltonian operator is hydrating. At a glance, you would have thought that the wave function would be amphoteric, but it is actually conjugating! The only mistake was that the Scorsese lambdonian integer was mistaken for the atomic number of the hydrating atom. Besides that, good work!
What?
Why in this video are we varying the coefficients but in the next video we are varying the spin orbitals?
Sorry, but i did not quite understand what is the overlap matrix. Can you explain me?
Sure. The coefficient vector, c, is a linear combination of a bunch of basis functions. These basis functions might not be normalized, and they might not be orthogonal. If they're not normalized, then there is "too much" or "too little" electron density. If they're not orthogonal, then basis functions might "overlap" with each other, producing either too much or too little total electron density depending on how they overlap. The overlap matrix S includes of this information, because it contains the overlap integrals of all basis functions. If normalized, the diagonal terms are all 1 (S_ii). If orthogonal, the off-diagonal terms are all 0 (S_ij, i =/= j). So if the basis set is already orthonormal, the S matrix is just an identity matrix, and the answer is the same whether or not it is present. If not, the S matrix corrects for any non-orthonormality of the basis set.
Hi, I'm having a hard time linking the concept before 4:10 and after... just before you mention that we need to find E when dE=0
It's analogous to a differential calculus problem where we have a function and we need to find the critical point. The catch is that the function has an extra constraint, requiring us to maintain orthonormal orbitals as we compute the derivative and solve for its zeroes. Besides that, we're just taking a derivative and finding when it is equal to zero.
so fancy :')
First.