now that we have our atomic units for molecular Hamiltonian we quickly encounter a problem and that problem is that the Schrodinger equation for an arbitrary molecule can't be solved or rather I should say can't be solved exactly with a finite amount of computing resources in a finite amount of time so for all of our purposes we can't solve the Schrodinger equation exactly for anything other than a hydrogen atom and the reason this is is because we have n charged particles so a typical atom is composed of a bunch of nuclei and our typical molecule or molecular system is composed of a bunch of nuclei and a bunch of electrons so we have a bunch of charged particles there and if we have n total charge particles we'd have n choose 2 or n times n minus 1 over two interacting pairs so there's a quadratic number of those N squared over 2 minus n over 2 and this is what is known in physics as the many-body problem and the many-body problem is typically unsolvable for any number of particles which is greater than 2 you can solve it exactly for two particles which is why we can do it for a hydrogen atom containing one electron and one nucleus but we can't solve it exactly for any molecular system which has more than two charged particles which rules out pretty much every interesting system that we can think of so in in computational terms or in computer science terms this is what you would call an np-complete problem and if you use this asymptotic complexity notation you notice that the difficulty of solving such a problem scales as an exponential in the number of particles so it actually gets exponentially more difficult to solve this problem as the number of particles increases so it's not twice as difficult to solve the problem if you go from 10 to 20 particles it's you know e to the 20 more difficult and then it is with ten all right and we have to solve this so this is where we have to start introducing approximations in order to be able to get a handle on the molecular Hamiltonian and try to get out some approximate solutions which will eventually end up at hartree-fock theory so the first thing we notice is that the mass of the nuclei is much much greater than the mass of an electron so as I noted in the previous video a hydrogen nucleus weighs about 2,000 times more than an electron and most nuclei contain more than just one particle for example a carbon atom would have 12 carbon-12 would be 12 you know whatever other nuclei there are whatever their atomic whatever their mass number is that's going to be you know thousands tens of thousands of times greater than that electron mass so what this is going to mean is that because of that very high mass that our nuclei are going to move very slow relative to our electrons so the set of nuclei is going to move slower relative to the set of all electrons so what we're going to do is approximate that the positions of all the nuclei are fixed relative to the electrons so this means that we're going to treat the kinetic energy of the nuclei as being zero because they can't have kinetic energy if they're not going anywhere if they're fixed they can't move no motion no kinetic energy so this is essentially equivalent to treating the nuclei as classical point particles classical in the sense that they do not have quantum behavior we're going to pretty much concern ourselves with them as just charged particles that exist somewhere in space and then not be concerned much about their wave functions in the vast majority of applications in quantum chemistry alright so there was just some diagrams or I was showing there where our interaction of the charged particles if there is zero you have no pairs of interactions two of them you have one pair three of them you have three pairs four you have six five you have ten as predicted by this in choose to formula alright so how does this approximation called the born-oppenheimer approximation first used by those two scientists what does that do when we apply it to our molecular Hamiltonian so the first thing to note is that our nuclear kinetic energy goes to zero because as I mentioned they're all fixed so the laplacian of them is going to be zero at all points because they're not moving so that goes to zero additionally we have our nuclear-nuclear repulsion term now instead of being spread out over all space where the wave functions of nuclei might exist the nuclei is just a particle which has a specific location in space and so are all the others so that means that Rab is going to be a constant so every as is za and ZB so every term in this pairwise sum between all pairs of nuclei is going to be constant so that means our nuclear repulsion energy is going to be a constant as well and for the rest of the terms we have electron kinetic energy electron nuclear attraction electron-electron repulsion these are still difficult and as of yet we still haven't discussed the way that we can go about solving those so in light of this approximation we can now separate our calculations into what we call an electronic Hamiltonian and obtain an electronic wave function so we're only concerned with the wave functions typically of electrons so our elektra electronic Hamiltonian acting on our electronic wave function gives us our electronic energy times our electronic wave function the eigenfunction there right and then just for good measure indicating what our electronic Hamiltonian is sum over all electrons of their kinetic energy one-half del square and i laplacian operator minus it's attraction so it's a negative sign some overall electron-nuclear pairs charge of the nucleus divided by the distance between the electron and the nucleus and then plus because it's a repulsion term plus sum over all pairs of electrons of 1 over their distance apart the magnitude in which electrons repel one another so that last term there are constant but nonzero nuclear-nuclear repulsion term that's going to end up being a constant so we can add it or not to our electronic energy should we choose I believe in some of the videos in the quantum chemistry playlist I do include this term and the electronic Hamiltonian but since we're going in some more depth here I think I'm typically not including it in the electronic energy in this video or in this chapter so since it's a constant we can do whatever we want with it it is only going to affect our total energy and it is not going to affect whatever our electronic wavefunction is so the nuclear positions determine what that energy is and that's going to be a shift up word or well it's positive so it's always going to be up but it's going to be some shift up in the energy but not changing actually the shape or a function of what those orbitals look like