Finite square well energy eigenstates
PROFESSOR: Find our final solution, we just have to match the equations. So Psi continues at x equals a. And what do we have? Well from two, you have cosine of ka. And from four you would have a equals e to the minus ka. This is the value of this-- so the interior solution at x equals a must match the value of the exterior solution of k equals a. Psi prime must be continuous at x equals a as well. Well what is the derivative of this function? It's minus the sine of this. So it's minus k sine of kx, that becomes ka, is equal to the derivative of that one, which is minus Kappa A e to that minus Kappa little a. Two equations, and how many unknowns? Well there's A and some information about Kappa and k. And the easiest way to eliminate that is to divide them. So you divide the bottom equation by this equation. So what do we get? Divide the bottom by the top. Minus k and and the minuses cancel, we can cancel those minus signs and you get k tan ka is equal to Kappa....