Square potentials (finite or infinte, well or barrier) are used in intro Quantum Mechanics courses. My question is: What does the word 'square' signify?
My other question: In the infinite square potential well model, the wavefunction outside the well is zero. Two reasons are usually put forward: 1. V is infinite. So Vu is infinite unless u = 0. (I don't understand what's still wrong with Vu being infinite.) 2. The particle can't be in a region of infinite potential if it lacks infinite energy. (Why not? Why should classical principles dictate quantum rules? Moreover, in the finite square well model, there is a finite probability of the particle being within the potential barrier? Isn't that a violation of the classical principle we are using in the above argument?) Therefore, u = 0. Thanks for any help.
Square refers to the appearance of the plot of V versus x. Energy is conserved in both classical mechanics and quantum mechanics. Assuming that the particle inside the well has finite energy, and the potential outside the well is infinite, then when the particle is outside the well, it will have infinite energy, which will violate the conservation of energy. Vu represents the potential energy of a particle outside the well. The other term in the Schroedinger equation represents the kinetic energy. Assuming that the total energy is finite and positive, if Vu were infinite, the kinetic energy would have to be negative infinity (steady state), which is impossible.
There is nothing of the plot of V versus x that reminds me of a square, actually! But isn't that exactly what happens in quantum tunneling? How does the particle get the extra energy to jump into a potential barrier. Isn't that in violation of the principle of conservation of energy? Howcome Vu is potential energy? I thought V was? How can the other term represent the kinetic energy? How can positive and negative infinity add to give a finite positive number? What does 'steady state mean'? Thanks for all the help!
You're not using any classical principles. Since you've apparently solved the finite square well, why don't you just let [tex]V \rightarrow \infty[/tex] and see what happens? You're thinking about it classically. How do you know its potential energy (i.e. the location) and the kinetic energy (i.e. the momentum) at the same time?
Square refers to the fact that the potential jumps discontinuously from one value to another, so that the graph of V(x) has a "square" corner. Compare to e.g. a harmonic potential well, whose graph is a parabola.
When the potential is finite, the "well" forms three corners of a rectangle. I wouldn't worry too much about the fact that it doesn't form a square. Its just a name. I may be wrong here, but as I remember, the time that a particle spends outside the well (delta t) is such that the Heisenberg uncertainty principle is obeyed. In other words, the energy deviation times the time interval is less than Planck's constant/2. A better way to say it is that they are the potential energy and kinetic energy terms in the Schroedinger equation. Its a limiting process. For example, as n grows larger, (n+1)^2- (n^2+2n) equals 1. Each term on the left grows larger without bound. It means no change in time. Thanks for all the help![/QUOTE]