In this video, I'd like to explore the validity of the perturbation series expansion that we've been using to correct for the energy and the state of a monol Hamiltonian in the presence of a perturbation. So intuitively by the name, it's clear that perturbation theory should be valid if the energy scales that are associated with our perturbation delta H hat are much smaller than the scales, the energy scales associated with our original monol Hamiltonian. It turns out, however, that this is not a sufficient condition to ensure that our perturbation series expansion that we used in the previous videos is valid. And to see that, we're going to consider a two-state system, a two-state quantum mechanical system with the following Hamiltonian. This has a part that we know the solution to, H0, and a perturbative term. And knowing the solution of the Schrodinger equation for a given Hamiltonian is equivalent to having the Hamiltonian in a diagonal form when it can be represented by a matrix, because this trivially satisfies an eigenvalue equation. And then we have our perturbation. We're going to call the energies at Vc, and they're off the diagonal, meaning that they're coupling the two levels together. And for this to be Hermitian, for the perturbation Hamiltonian to be Hermitian, this term has to be equal to the complex conjugate of this term. This is because delta H must be Hermitian. So we can write this out in a single matrix. And here E1-0 is the energy of the first state, and E2-0 is the energy of the second state. And now the perturbation couples them together in some way. And what we're interested in then is to find the eigenvalues of this new Hamiltonian. That's what it means to solve the Schrodinger equation. Right? So to solve the time independent Schrodinger equation, for example, we want to diagonalize this new Hamiltonian. And to do that, we need to find the eigenvalues and eigenvectors. We're just going to focus on the eigenvalues of this new Hamiltonian. These will be given by following expression. It will be a function of lambda. You'll get two distinct eigenvalues, well denoted by E plus and E minus. And this will be the new energy of the two levels of our system. So this plus or minus differentiates the two energy levels. Okay, so these are the eigenvalues of this new Hamiltonian. Notice that there is no first order term in lambda because there's no diagonal contributions from the perturbation. And for future simplicity, we're going to call this term that occurs over here and over here. We're going to call it delta E naught over two. So this is our exact solution to this new two level system. If we're to think about this in a perturbation series, this will that would be equivalent to being able to expand this radical to all orders and that series converging. So perturbation theory implies in this case a Taylor expansion of these energy eigenvalues converging. So we're going to do that. We're going to see what condition needs to be satisfied for this Taylor series to converge. And for that, we're going to focus on the radical. Which we're going to know by sure, and it's one plus x squared. This will be some function of x. So what we're interested in is the Taylor series expansion of this function squared of one plus x squared. So the Taylor series of that. So there's one plus x squared over two minus x to the four over eight plus x to the six over sixteen minus infinitely many terms. And one of the conditions for this Taylor series to expand to all orders. So for the Taylor series to converge to all orders, we need that x or the absolute value of x has to be smaller than one. I put another way if we substitute back in all of our variables. This won't mean that lambda BC over delta E naught over two has to be smaller than one. So, absolute value over here. Or. And value. Or equivalently that the perturbation has to be smaller than the difference of energy between the two states approximately up to a constant factor. So what this model showed is that the perturbation not only has to be smaller than the energy associated with the original Hamiltonian. And the perturbation also has to be smaller than the difference in energies of the original Hamiltonian.