lect17.ppt_notesit is for sure man man man
- 1. finite potential well “I don't like it, and I'm sorry I ever had anything to do with it.”—Erwin Schrödinger, on quantum mechanics.
- 2. 5.9 Finite Potential Well Our new toy worked very well for the particle-in-a-box. Where do you find infinite square wells in real life? Answer—nowhere that I know of.* Our next test of Schrödinger’s equation ought to be a step up in realism. In real life, particles do exist inside of potentials, only the walls are not infinitely high. So our next problem will be a finite potential well. *A black hole might come close, except its potential is not square.
- 3. E=U x E x=0 x=L L Our finite potential well has length L and walls of height U. There are three regions: I, II, and III. I II III In regions I and III, Schrödinger’s equation is 2 2 2 2m + (E - U) = 0 , x which we can re-write as 2 2 ψ - ψ 2 a = 0 , x 2m (U - E) α = . where - + On the next slide we will assume E<U so that a will be real.
- 4. E=U x E x=0 x=L L I II III - + Consider a particle with E < U. Solutions in region I and III are ax -ax ax -ax I III = Ce + De and =Fe + Ge . Why call the coefficients C, D, F, G? You’ll find out later. 2 2 ψ - ψ 2 a = 0 , x 2m (U - E) α = .
- 5. E=U x E x=0 x=L L I II III ψ ax -ax I = Ce + De ψ ax -ax III = Fe + Ge These solutions are real, and not complex exponentials, so they are not “wiggly” waves. C, D, F, and G are coefficients to be determined using the boundary conditions of the problem. Regions I and III extend to x = - and x = +. Because the wavefunction must be finite everywhere, the coefficients D and F must be 0, so that ψ ψ ax -ax I III = Ce and = Ge . Both solutions decrease exponentially as we move away from the barrier walls. - +
- 6. ψII 2mE 2mE = A sin x + B cos x , E=U x E x=0 x=L L I II III In region II, Schrödinger’s equation is 2 2 2 2m + (E - U) = 0 , x This equation has the same kind of solutions as we had for the particle in the (infinite) box: except that now B 0 because has an amplitude at each barrier. Now you see why we called the region I and III coefficients C, D, F, G. A and B were reserved for region II, and E is reserved for energy. - +
- 7. ax -ax I II III 2mE 2mE = Ce = A sin x + B cos x = Ge There are 5 “pieces” of information we want: the coefficients A, B, C, and G, and the energy E. We have 5 conditions: continuous at x = 0 and x = L (2 boundary conditions), continuous at x = 0 and x = L (2 more boundary conditions), and normalization of . These 5 conditions give 5 equations. Five equations, five unknowns, the rest is “just” mathematics. Here is a Mathcad document that explores the finite square well.
- 8. It’s a bit difficult to see the extent of the well in the Mathcad plots, so I have reproduced some of them here. (The well goes from -1 x 1.) for n=1 for n=2 for n=3 Notice how the wave function tails extend inside the barrier- -there is a finite probability of finding the particle there. Know this for the test! Longer wave function tails mean longer wavelengths and therefore lower momenta and energies. A particle inside a finite box can have a lower energy than a particle in an infinite box.
- 9. Here’s a comparison of wave functions for infinite and finite square wells, n=1, 2, and 3. Be able to tell me which wave function goes with which well. How can you tell?
- 10. Here’s a comparison of the probability density functions for infinite and finite square wells, n=1, 2, and 3. (X-axis scale is changed from previous slide.) Which probability goes with which well. Why? Which plot corresponds to n=2? How can you tell? What is the meaning of the red shaded areas?
- 11. Are you trying to tell me there is a probability of finding this brick stuck halfway through an impenetrable wall? Well…not exactly… …but I am telling you there is a probability of finding the brick somewhere inside the impenetrable wall! (You’re supposed to imagine the brick inside the wall, please.)