Optical potential so far, we assumed that the potential v(x) is real. since the schrödinger equation is already complex, we can ask what happens if we add an imaginary term to it-this is called an 'optical potential'. in class, you derived that the total probability always satisfies p(t)0. evaluate this derivative again, but without assuming that v(x) is real. you will see that it doesn't vanish anymore. now assume that the imaginary part of v(x) is a constant -vo. find a differential equation for p(t). assuming that p(o) 1, solve it to find p(t) in terms of vo and note: we will not use this in the rest of this class, but you will see it again in nuclear physics, where it's used to model particles that can be absorbed or decay. can you see why?