 Shall we begin? So last term, we discussed the dynamics of a harmonic oscillator, which is a one-dimensional system where the particle is bound by a potential v is a half kx squared. And we found a number of important results. Most obviously, there was quantization of the energy, so discreteness of the energy levels. And very importantly, there was the phenomenon of zero point energy that even when something was in its ground state, it had the uncertainty principle applied to have non-negligible kinetic energy and potential energy indeed. So what we're going to do now is study a number of very simple one-dimensional potentials, motion in a number of very simple one-dimensional potentials, which are themselves very artificial. These are step potentials, so they change discontinuously at some values of x from one value to another value. So they're very artificial, but the merit of them is that we can solve the governing, well, the key equation, the time-independent Schrodinger equation and obtain the states of well-defined energy from which we can, as we saw with the harmonic oscillator, recover the dynamics of the system. We can solve this equation, this important equation, simply for these rather artificial potentials. And we'll discuss at the end which of our results is artificial reflects the artificiality of the potentials and which are generic and are ones that we can believe in. So this is what we're going to start with is the square potential well, which is this structure. This is potential energy being plotted vertically, position being plotted horizontally. Here is the origin. This is going to be a distance a. It's going to be symmetrical. This is going to be a potential level v0. And this is going to be a potential level nothing. So we set the zero as a potential energy to be at the bottom of the well. And then there's when you're more than distance a from the origin, you have some potential energy v0, which is a constant. So this is highly artificial, but let's see what quantum mechanics has to say about motion in here. And in particular, what we're going to look for stationary states. That is to say states of well-defined energy, which we write like that. So these are states of well-defined energy. We're interested in these because they enable us to solve the time-dependent Schrodinger equation trivially. Once we know what all these things are, we can write down. And once we know what all these things are, and we know how to, then we can express any arbitrary initial condition as a linear combination of these things. And we can time evolve it in a simple way as we saw last term. So we're going to find these things. And they're going to have wave functions, which we'll call u of x. So this is x e is the wave function. Now this potential is symmetrical. It's an even function of x. So we have that v of minus x is equal to v of x. So it's an even function. The consequence of that is that we introduced the parity operator last term. We had that p, if you remember, x, p is an operator, the parity operator, which makes out of a state of psi, the state that you would get, which has amplitude to be at minus x, well, what is this? This is the amplitude to be at x when you're in the state that p makes out of psi. And this was defined to be minus x of psi. In other words, it was defined to be the amplitude to be at minus x when you're in the state of psi. So the parity operator makes a state which is the same as the state you first thought of, except if every point is reflected through the origin. We discussed this operator. And because b is an even function of x, we have that pvx pv of psi is going to be equal to minus x v of psi, which because v is a function of position. It's a function of the position operator. This is going to be v of x, x of psi. In other words, it's going to be simply v of x. Sorry, sorry, sorry, sorry. I need a minus x and a minus x. So it's going to be v of minus x times of psi of minus x. Because v is an even function of x, so v of minus x is the same as v of x, this is equal to this is this, and this could be written as p. So what does this mean? This means that because v is an even function of x, it commutes with the operator p. In other words, pv equals 0. Because this is rather a rigmarole, I have to admit, p commutes with v because v is an even function, which is a way of saying that the potential is symmetric about the origin. This is the bottom line. If you have a potential which is symmetric about the origin as an even function of x, then it commutes with the parity operator. The consequence of that is, so the Hamiltonian is, of course, is, as ever, p squared over 2m plus v of x. p squared is an even function. If you write this in the position representation, it's minus h bar squared d2 by dx squared. So this is an even function of x. So this commutes with the parity operator. We've just figured that this commutes with the parity operator. So we have that p comma h. So this implies that p comma h is 0. So the parity operator commutes with a Hamiltonian whose stationary states we would like to find, whose eigenstates, that is to say, the stationary states we would like to find. So what does that imply? When two operators commute, remember the fundamental rigmarole is that this implies there's a complete set of mutual eigenstates of p and h. That is to say, we can, if we wish, look for eigenstates of h, which are states of well-defined parity. And the wave functions of these states of well-defined parity will either be even functions of x if the parity is even, or there will be odd functions of x if the parity is odd. So what this means is we can insist or we can look for. So we can look for stationary states with wave functions u of x, meaning, of course, x, e, that are either even functions ux equals u of minus x, or odd functions, i.e. u of x, is minus u of minus x. And this observation, knowing that you're looking for an even function, say, makes it much easier to find that function than if you don't know whether it's even or it's odd. This is a general observation, and we'll just find a concrete example of it in a moment. So here is our potential again, a minus a. What do we have? We have here what is the time-independent Schrodinger equation, that is to say, what is that? That is the equation which says that h of psi is equal to e of psi. In the position representation, what does this important equation look like here? Well, at this point here where the potential is 0, h is p squared over 2m. So at this location here, this becomes p squared over 2m, which is minus h bar squared over 2m d2u by the x squared. So this is, I suppose I should change this, sorry, in this context to u, probably, let's leave it with e. That's what we were calling it, the state e. So this left-hand part now reduces to just this in the position representation, because we only have the kinetic energy at this location. There is no potential energy. And on the right-hand side, of course, we simply have e times u. So we're trying to solve this equation, and we know all about the solutions of this equation. This is just a simple harmonic motion equation of classical physics, essentially. It tells us, so we know that u is cos kx, or u is sine kx, provides solutions of this equation. Cos kx is an even function, so that must be the solution belonging to an even parity state. And sine kx is an odd function of x, so it's an odd parity thing. So this has solutions u of x is equal to either cos kx or sine kx, depending on parity. And we have that k. So when you double differentiate this, you're going to get minus k squared cos kx. So the minus sign deals with that. We're going to have that minus k squared e is equal to this stuff here. In other words, we're going to have that k is equal to, no, this is wrong. Sorry, sorry, I've got this the wrong way up. What am I doing? Sorry, the double derivative is here. So when we double differentiate, we'll get minus h bar squared k squared over 2m. We'll get a minus coming from the differentiation, which will cancel this. And we will have that k squared is equal to 2me over h bar squared square root, well, k is that. So we have determined what the wave functions are of the stationary states in that interval from minus a to a in terms of the energy. What we now need to do is think about the state of affairs here. So this stuff is all true for mod x less than a. What about the case when mod x is greater than a? So another picture. In this zone here, what does the time independent Schrodinger equation look like? It still has kinetic energy minus h bar squared over 2m d2u by the x squared. But now we have potential energy, how much? v0, so v0 times u. So this is the Hamiltonian operator operating on u. And that's equal to e u. And let's now say we're looking for bound states. That's to say states where the energy e is less than the potential energy out here, so that the classically the particle will be confined inside here. So we're going to look for bound states. There are other states too, but let's focus on the bound states. That means that e is less than v0. So the particle classically is not allowed to get out of the well. What happens then? Well then, this equation becomes d2u by the x squared is equal to, if we put this onto this side, since v0 is by hypothesis bigger than u, we have a negative right hand side. And we can cancel the minus signs on the two sides. So if we write it as we have 2m v0 minus the energy over h bar squared times u. So again, we have a double derivative is equal to some constant times the function. But the difference is now that we don't have a minus sign here. So instead of having sinusoidal solutions, we have exponential style solutions. So the solution to this equation is that u is equal to a constant times e to the plus or minus k times x, where big k is the square root of this stuff here. What about this sine ambiguity here? There is a sine ambiguity here because it's the double derivative, which has to be equal to a constant times u. If we go for the minus sign in taking the double derivative, two derivatives, we get down to minus signs. We get a plus obviously. So that's why there's this ambiguity. What do we do about that ambiguity? Well, when x is greater than 0, we want the wave function to decrease as we head off to infinity as x becomes larger and larger. Well, in fact, because we would like to be able to normalize the wave function, we'd like to have the wave function mod squared integrated over all space comes to 1. And that's not going to be possible if we have an exponential divergence. So the consequence of that is that in this zone here, we take that u is proportional to e to the minus kx because x is positive over there, and that means the bigger x gets, the more we move over here, the smaller the wave function becomes. If we're on this left side here where x is negative, then we want to take u goes like either plus or minus e to the plus kx because x is negative in this zone here, and the negativity of x gives the exponent of the exponential negativity so that the bigger that x becomes in modulus, the more we move over here, the smaller the wave function becomes. And whether we want to take a plus sign, if we're looking for a state of even parity, then we want to take this plus sign. So the wave function over here has the same numerical value as the wave function at the corresponding point over there. If we're looking for a state of odd parity, we take this minus sign so the wave function at negative x becomes minus the wave function at the corresponding point at positive x. So that's where we are so far. What we now need to do, so we've now solved, what have we done? We've solved the time independent Schrodinger equation everywhere except at x equals a and x equals minus a. But we haven't solved it at those points because at those points, if you go a bit to the, for example, at the point x equals a, if you go to the bit to the left of that point, then the wave function is supposed to be cos or sine kx. And if you go a bit to the right, it's going to be this exponential function. But at that point, we must still have the time independent Schrodinger equation satisfied. So what does that require? Well, for the time independent Schrodinger equation to mean anything even, the second derivative of the wave function has to be well-defined at that point because the time independent Schrodinger equation equates the second derivative to some stuff. So what we can say is that at x equals plus or minus a, we need that d2u by dx squared is defined. Or we can't solve the tizy there, satisfy the tizy there. Well, the rate of change, this is the rate of change of the gradient that is certainly not going to be defined if the gradient isn't continuous. So that implies that du by dx is continuous. And that's a non-trivial requirement because at the moment we've got the wave function in pieces and there's no obvious reason, unless we do some engineering on our pieces, why the gradient is defined by one piece on one side of the barrier of the transition should equal the gradient from a completely different function on the other side. So the gradient has to be continuous. The gradient is certainly not going to be even, is not even going to be defined unless the wave function itself is continuous. So we also require same similar reasons that u is continuous at these points. So we have to insist that u of a minus some tiny bit is equal to u of a plus some tiny bit. And we have to insist that du by dx evaluated at a minus a tiny bit is equal to du by dx evaluated at a plus a tiny bit called epsilon. That's what we've got to insist on. What does that amount to? That amounts to here, if we're just to the left of a, we have for the even parity solution, we have that u is equal to cos k x. And x is a minus epsilon, but epsilon's as small as we like. So let's just make it equal to cos k a. That's got to equal the wave function on the right-hand side, which is some constant. We'll call that constant big a times e to the minus big k times x plus a tiny bit. Let's forget about the tiny bit because this is a continuous function. So we require that. That's the continuity of u of x. Similarly, the gradient just to the left of x equals a is given by the derivative of cos k x. So we're looking at minus k sine k a. And that's got to be equal to the gradient of a e to the minus k x just evaluated at x equals a. So that's minus big k a e to the minus k a. And that's the continuity of du by dx. And we also have to satisfy these continuity conditions at x equals minus a. But the nice thing about deciding that you're going to look for a wave function of well-defined parity, either an even function or an odd function, is that it's easy to persuade yourselves. You probably want to sit down quietly and do this afterwards. That if you satisfy these conditions on the right-hand side at x equals plus a of the origin, then you've also satisfied these conditions on the to the left of the origin. These equations suffice to fix up the arrangements on both of the discontinuities. And you don't have to deal with them separately. That's the great advantage of choosing wave functions of well-defined parity. So what do we have here? We have a pair of equations, and we have a number of unknowns. As it stands, we do not know what little k is or big k is. And we do not know what a is, big a is, right? Those are all unknowns. We've two equations, and we need, fundamentally, to determine these unknowns. Most important is to determine little k and big k, because they are related to the energy by formulae, which are there's little k right at the top there is the square root of 2m e over h bar squared. And big k halfway up is 2m v0 minus e. So little k and big k are both related to the energy. And once we found the energy, we'll know what both big k and little k are, and it's the energy we're fundamentally after. So that's what we want to focus on. Big a is of less interest. So let's get rid of big a by dividing this equation through by this equation. Then we will find, so we divide equation 2, basically, by equation 1. And that leads to the conclusion that minus, let's do it down here, minus k tan k a from sine over cosine is equal to minus big k. Nothing, everything else goes because we have an a e to the minus big k a in both equations. And let's ask what this is. Let's try and relate this. So big k and little k are both related to the energy from which it follows that I could express big k as a function of little k. So let's do that. This is minus the square root of 2m v0 minus e over h bar squared, which is equal to minus square root. Now, 2m e over h bar squared is actually from, we maybe just lost it, unfortunately. Let's bring it back into focus right at the top there. It's 2m e over h bar squared is in fact k squared. So 2m e over h bar squared is k squared. So what we want to do is write this as 2m v0 over h bar squared minus k squared. So here we have an equation now that this equals this, which has only one unknown, namely k. So the left side is a function of k. The right side is a function of k. Any values of k for which these two sides are equal are provide solutions to our time-independent Schrodinger equation and provide wave functions for stationary states, for states of well-defined energy. To solve this equation, the way to go is to divide through by this, well, obviously cancel the minus signs, divide through by k, both sides. And this then becomes tan kA is equal to the square root of 2m v0 over h bar squared k squared minus 1. And it's good now to multiply the top and the bottom of this by a squared and write this as the square root of w over kA squared minus 1, where w is 2m v0 a squared over h bar squared. Why do I want to do that? kA is obviously dimensionless because this is a wave number. The wave function was sine kx. The argument of sine must always be dimensionless. So this is dimensionless. That's obviously dimensionless. Therefore, this must be dimensionless. You can explicitly check that this is dimensionless. So the reason I've defined w is that this thing is dimensionless. It's a dimensionless measure of the depth and width of the potential well. It depends on the depth of the potential well. It depends on the width of the potential well. And it's the dimensionless, the mathematics is telling us that this is how you quantify, how you know what kind of potential well you've got. You've got a very deep one or a very shallow one. So how are we going to solve this equation here? Well, the way to go is to plot both sides of the equation graphically. So the tangent is to plot both sides of the equation graphically and see what you get, see at what points they meet. So if this is a plot, this is kA being plotted this way, then if I plot tan kA, it starts off at 0 and rises like this. And when kA becomes equal to pi over 2, it zooms off to infinity. That's what tangents do. And down here it goes symmetrically. What does this thing do? When kA is nothing, that thing is obviously infinity because it becomes w over nothing. Square-rooted. So this is the left-hand side. I better write that down. So this here is the left-hand side. It's tan kA. The right-hand side is coming down from infinity. And it's going to go to 0 when kA is equal to, or k squared is equal to w. So this right-hand side is cruising down from infinity. And it's going to go to 0 when kA is equal to w. 0 when kA is root w. And from this we see, it's obvious now, that no matter what the value of w is, so as you change the width and depth of the potential, you move this point to the right or the left. But no matter where you put it from nowhere to infinity, it will cross. There will be this intersection here. These two curves always cross. And where they cross gives you a value of kA, and therefore a value of k, which is a solution to the time-independent Schrodinger equation. So what we've just learned is that this well always has a bound state. It doesn't matter how shallow the well is or how narrow the well is, it always has a bound state. Because these two curves always cross. This tangent on the left has other branches. There's also a branch. So tangent comes along here, goes off to plus infinity. And then somewhere down here, when kA is equal to pi upon 2 plus a bit, it becomes minus infinity, or it comes in from minus infinity and repeats itself. So this is the LHS second branch. And this second branch may or may not cut this curve of the right-hand side again. So if we made kA smaller than here, this is where kA, this is the place pi, if we would make this less than that, we wouldn't get a second solution. But if we have kA bigger than this, we do get a second solution. So it may have other even parity stationary states. So depending on the depth of the potential, we have 1, 2, 3, 4, et cetera, stationary states of even parity. We've only dealt with even parity case. If we want to look at the odd parity states, let's just begin to do this. But this is basically an exercise for the problems. Then our conditions are a wave function for the odd parity state. We've lost it somewhere, but it's a wave function for the odd parity states is the sine kx right up there. So in the middle, it's sine kx. At the edge, it's still a e to the minus kx. So our continuity conditions become that sine kA is equal to a e to the minus kA. And that's the continuity of the wave function itself at x equals a. The derivative of this is going to give me k cos kA is equal to minus big kA e to the minus kA. If we divide this equation by this equation analogously to what we did before, we are going to get k cotangent of kA is equal to minus big k. So now we have only one minus sign. Previously, we had a pair of minus signs, which cancelled. Now we have only one minus sign because we differentiated a sine. We didn't differentiate a cosine. And correspondingly, we have a cotangent instead of a tangent. So this equation can be graphically solved. That's the exercise. And we find that we may get 0, 1, 2 solutions for k. And for every solution, we have an odd parity. We have an odd parity stationary state. Let's have a look at the uncertainty principle in this example. Just to remind you, last term, quite early on, we showed we considered the case. So this is last, this is a summary, last term. We considered the case where psi of x is proportional to e to the minus x squared over 4 sigma squared. So we considered a wave function whose spatial form was a Gaussian such that when you mod squared this, in order to get the probability distribution, you found that it was a Gaussian with dispersion sigma. So this thing here is the expectation of x squared. That's what sigma squared is from that formula. And what did we find? We found, fundamentally by doing a Fourier transform, that if that's what the wave function looks like, then the probability, well, the amplitude to find, so this is x of psi, the wave function looks like that, then the amplitude to get a certain momentum, if you would make a momentum measurement, was looking like e to the minus p squared over 4 sigma p squared, where this thing becomes the expectation value of p squared, so the variance. The expectation value of the momentum in this state is 0. The expectation of the square of the momentum is going to be this sigma p squared. And what we found was that sigma times sigma p is equal to h bar on 2. This was a concrete example of the uncertainty principle, where the smaller the uncertainty in x is, the bigger the uncertainty in the momentum. And correspondingly, the smaller the momentum in p is, the bigger the uncertainty in x has to be because the product is always the same in this particular Gaussian example. So we have some idea that the uncertainty in x times the uncertainty in p, we say to ourselves, is inherently never smaller than this number h bar upon 2. It may be bigger than h bar upon 2 easily. It often is bigger than h bar. It usually is bigger than h bar on 2. But this is kind of as small as it gets. So that was what we did last term. So let's have a look at it in this particular case. So let's look at the ground state. Remember, oh yeah, another point to remind you is that the ground state wave function of the harmonic oscillator. So the ground state of a harmonic oscillator actually has a wave function, which is this Gaussian. At the time that we did this calculation, this Gaussian was just picked out of the air. But for the ground state of the harmonic oscillator, actually does have this thing here. This fits this example. So let's have a look at the wave function of the ground state of this that we have here. What you might argue to yourself, you might say, OK, so what we can say is that p is less than or on the order of, well, p squared over 2m, the energy, is less than v0. Because it's a bound. So it's less than v0. So what we can say is that p squared is less than 2m v0. Yeah, sorry, that's meant to be a less than. Seems reasonable, doesn't it? The particle can't have any more kinetic energy than the energy it requires to escape, because we know it's bound. The next thing you might say is, well, so what's x squared? What's the uncertainty in x? Well, you might say to yourself, well, look, this particle is trapped in this potential well that goes from plus a to minus a. So what's the uncertainty in x squared? Well, this must be on the order of a squared. Seems reasonable, doesn't it? Less than on the order of. I'll put in a factor 2 or 4, or just to be sure that the less than is holding. Because the thing is trapped by a potential well that extends only from plus a to minus a. So it has a range of 2a that it can run in. So we say x squared is definitely less than 2a squared. It's almost certainly, it's clearly less than that. Significantly less than that. So what does that give me? Oh, and this square of the momentum is, so we have that x squared, p squared, I mean, I can put expectation values around that too, I think, is less than on the order of, whatever it is, 8m v0a squared. But that is w, isn't it? Somewhere, yeah, look, w, this is basically w because that was 2m v0. So this v0a squared is something like 4m v0a squared is h bar squared. If I've done this right, because I'm getting an answer, which I'm times w, yes, exactly. That's right, times the dimensionless number w, right. The point is that w could be made as small as you like. We've agreed, and you still have a bound state. So this naive argument suggests that we're going to violate the uncertainty principle and we can make this as small as we like. So what's the problem? This argument's bogus. What's bogus about it? It's not all in the well. As you make w smaller and smaller, let's draw what the wave function looks like. So if we have, let's have a nice big value of w. A big value of w means a nice spacey well with right high walls. And then we will have a ground state wave function, which is a cosine and some little bits like this. And indeed, the probability to find the particle outside the well outside plus or minus a will be not very much. But as you make this smaller and smaller and smaller, you bring this in, making a smaller, and or you lower these, making this curvature smaller. So that curvature of that wave function is a reflection of the kinetic energy, which is a reflection of phi naught. We come to that. We have a nice, we have a narrow, a titchy-witchy thing like this. Then you have barely a, you have an almost a straight line across here. I can't draw it well. Almost a straight line across here. And then enormous, long, exponential decays. And the particle is almost certain not to be in the potential well. That's a very remarkable conclusion to come to, right? That you can trap a particle with a potential well, which it's almost, it has very little probability of actually being in. But that's what the theory says. Something else we can show is we can say, we can argue as follows. Supposing we've given some other potential well, right? Here is some other potential well. Sorry, it's meant to be an even function of x, right? So it's meant to be the same on both sides, symmetrical. And somebody asks you, so does this potential well have a bound state? Then your reason as follows. Let us inscribe a nice square potential well. There are many square potential wells you can inscribe into it, right? But there's one. Then you can argue that this potential well is narrower and shallower than the one you were given. So less likely to trap a particle. But this potential well has a bound state. So this well is shallower, narrower, but it has a bound state. At least one bound state, because it's a square well and we understand about square wells. So your reason that this wider and deeper well also has a bound state. So we're making an inference from this very special, rather artificial, square well potential about all one-dimensional square wells. What happens now, let's conclude with a special case, an important special case, which is the infinitely deep well. Whoops. So now we let v0 become arbitrarily large and we have a well that looks like this. It just goes up and up and up and up and up and up and up and up. What happens then? Well, let's go to our graphical solution. What have we done? We've left a constant. I mean, we've left a at some values. And we've allowed v0 to become arbitrarily large, which means we've made w arbitrarily large. What's the implication of that for our solution of this equation? So we're solving this equation for w arbitrarily large. So we're saying that the roots are going to become where tan kA, the even parity roots, are going to be where tan kA equals infinity. And we know what that means. We know that kA must be pi upon 2, 3 pi upon 2, et cetera. So we've made w goes to infinity, which implies that the governing equation, the equation that determines k, becomes tan kA equals infinity, which implies that kA is equal to either pi upon 2 or 3 pi upon 2, et cetera, et cetera, et cetera. And all of these values are going to be OK. Graphically, what's happening is that this curve, right? This was the curve of the right-hand side for finite w, is going to just become a line along here at infinity. And it's going to cross these various branches of the tangent at infinity, which means that these pi upon 2, 3 pi upon 2 points. And it's going to cross every single one of them all the way out. So we're going to have an infinite number of even parity stationary states. And they're going to have those values of kA. And when you study the corresponding problem for the odd parity states, which we just vaguely discussed here, it's going to be the same deal. We're going to be solving an equation which says that cot kA is going to be infinite. And that means that kA is going to be pi, 2 pi, 3 pi, so on. So these are going to be the even parity states. And then we will find that kA is equal to pi, 2 pi, 3 pi, et cetera, for the odd parity states. So the infinitely deep well will have an infinite number of solutions. And what do the wave functions look like? Well, the wave functions are going to be cos kx, where kA is equal to some number of odd pi by 2s. So the wave function of the even parity states is going to vanish. This is going to be cos kx. And the wave function is going to vanish here and here by virtue of these conditions there. And the same thing will happen for the odd parity states. For example, the first odd parity state is going to have a wave function which looks like this. It's going to be sine kx, need a better piece of chalk. It's going to be sine kx, where kA is equal to a number of pi's, and therefore the sine vanishes. So it's going to vanish. So this is a concrete example, but it leads to the general, it inspires the general principle that a wave function vanishes adjacent to a region of infinite potential. Physically, what's happened is that this big k has grown bigger and bigger and bigger and bigger and bigger. And therefore this exponential has grown deeper. The curvature of it has grown larger and larger and larger and larger. So as we raise the walls, the transition at the edge of the well goes from being sort of nice and easy with a small value of k through a bigger value of k to ultimately an infinite value of k, which allows it to discontinuously go from a finite slope right around to no slope. And that's why it's a general property of these solutions of the time-independent Schrodinger equation that as you approach the edge of an infinite, a region of infinite potential, the wave function vanishes in anticipation of that infinity. So another strange aspect, another physically strange thing that the theory is telling us is that the wave function, the particle has negligible probability of being found in the neighborhood of this region where it's strictly forbidden. It anticipates the fact that it's going to be forbidden. You won't find it even near this dangerous place. OK, it's time to stop.