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BARTON ZWIEBACH: --that has served, also, our first example of solving the Schrodinger equation. Last time, I showed you a particle in a circle. And we wrote the wave function. And we said, OK, let's see what is the momentum of it. But now, let's solve, completely, this problem. So we have the particle in the circle. Which means particle moving here. And this is the coordinate x. And x goes from 0 to L. And we think of this point and that point, identify. We actually write this as, x is the same as x plus L. This is a strange way of saying things, but it's actually very practical. Here is 2L, 3L. We say that any point is the same as the point at which you add L. So the circle is the whole, infinite line with this identification, because every point here, for example, is the same as this point. And this point is the same as that point. So at the end of everything, it's equivalent to this piece, where L is equivalent to 0. It's almost like if I was walki...

BARTON ZWIEBACH: Hydrogen atom is the beginning of our analysis. It still won't solve differential equations, but we will now two particles, a proton, whose coordinates are going to be coordinate of the proton, subbing Xp for the proton, and momentum of the proton. And there's an electron. And there's the coordinates, the three coordinates of the electron, and the three components of the momenta of the electron. And these are your canonical variables. This means that the components of this object satisfy the standard commutation relations. That is-- I have to write it like the following. They considered the coordinates of the proton, the i-th component. And the momentum of the proton, the j-th component, that's equal to ih bar delta ij. You see, we used to code for its x, y, and z, and momenta Px, Py, Pz. You could've called it X1, X2, X3, momenta P1, P2, P3. And in that way, you can use a Kronecker delta over here. So, these are the commutation relati...

BARTON ZWIEBACH: Do normal wave analysis to demonstrate that indeed these things should not quite happen. So for that, so ordinary waves and Galilean transformations. So when you have a wave, as you've probably have seen many times before, the key object in the wave is something called the phaze of the wave. Phaze, the phaze. And it's controlled by this quantity kx minus omega t. k being the wave number, omega being the angular frequency and we spoke about. And the wave may be sine of that phaze or cosine of that phaze or a linear combination of sines and cosines, or E to this wave, any of those things could be your wave. And whenever you have such a wave, what we say is that the phaze of this wave is a Galilean invariant. Invariant. What it means is that two people looking at this wave, and they look at the point on this wave, both people will agree on the value of the phaze, because basically, the reality of the wave is based on the phaze, and if you have, for e...

BARTON ZWIEBACH: Today's subject is momentum space. We're going to kind of discover the relevance of momentum space. We've been working with wave functions that tell you the probabilities for finding a particle in a given position and that's sometimes called coordinate space or position space representations of quantum mechanics, and we just talked about wave functions that tell you about probabilities to find a particle in a given position. But as we've been seeing with momentum, there's a very intimate relation between momentum and position, and today we're going to develop the ideas that lead you to think about momentum space in a way that is quite complimentary to coordinate space. Then we will be able to talk about expectation values of operators and we're going to be moved some steps into what is called interpretation of quantum mechanics. So operators have expectations values in quantum mechanism-- they are defined in a particular wa...

BARTON ZWIEBACH: Finite square well. So this brings us also to a little common aside. So far, we could find every solution. Now we're going to write the equations for the finite square well, and we're not going to be able to find the solution. But we're going to understand the solution. So you're going to enjoy a little-- mathematicians usually say it's the most important thing, understanding the solution. Finding it, it's no big deal. But we're physicists as well. So we sometimes have to find the solutions. Even if we don't understand them very well, it's nice to find them. And then you're going to use numerical methods, and this is the part of the course where you're going to be using numerical methods a lot. Here is the finite square well, and now we draw it symmetrically. Here it is. Here is x. We're drawing the potential V of x. It extends from a to minus a. It's 0. This is the 0 of the potential. It's here. And...

BARTON ZWIEBACH: Let's talk now about entanglement. So we talk about entanglement when we have two non-interacting particles. You don't need a strong interaction between particles to produce entanglement, the particles can be totally non-interacting. Suppose particle 1 can be in any of these states-- u 1, u 2. Let's assume just u 1 and u 2. And particle 2 can be in states v 1 and v 2. And you have these two particles flying around, these are possible states of particle 1 and possible states of particle 2. Now you want to describe the full system, the quantum state of the two particles. States of the two particles. Two particles. Well, it seems reasonable that to describe the state of the two particles that are not interacting, I should tell you what particle 1 is doing and what particle 2 is doing. OK, so particle 1 could be doing this. Could be u 1. And particle 2 could be doing v 1. And in a sense, by telling you that, we've said what everything is doing...

BARTON ZWIEBACH: De Broglie, as we discussed last time, we spoke about waves. Matter waves. Because people thought, anyway light is waves so the surprising thing that would be that matters are waves. So a free particle with momentum p can be associated to a wave-- to a plane wave, in fact-- plane wave-- with wavelength lambda equals Planck's constant over p. So this wave is what eventually becomes a famous wave function. So de Broglie was writing the example or trying to write the example of what eventually would become wave functions, and the equations for this wave would become the Schrodinger equation. So really, this is a pillar of quantum mechanics. You're getting there when you talk about this wave. So Schrodinger's equation is a wave equation for these matter waves and this plane wave eventually will become the wave function and there is a Schrodinger equation for it. So it's a wave of what? he was asking-- de Broglie had little idea what that wave ...

BARTON ZWIEBACH: We want to understand now our observables. So we said these are observables, so can we observe them? Can we have a state in which we say, what is the value of Lx, the value of Ly, and the value of Lz. Well, a little caution is necessary because we have states and we have position and momentum operator and they didn't commute and we ended up that we could not tell simultaneously the position and the momentum of a state. So for this angular momentum operators, they don't commute, so a similar situation may be happening. So I want to explain, for example, or ask, can we have simultaneous eigenstates of Lx, Ly, and Lz? And the answer is no. And let's see why that happens. So let's assume we can have simultaneous eigenstates and let's assume, for example, that Lx on that eigenstate phi nought is some number lambda x phi nought, and Ly and phi nought is equal to lambda y phi nought. Well, the difficulty with this is essentially-- well, we co...