 In this lecture, we will learn the following things. We will learn about the postulates of quantum mechanics, the inviolable tenets that are the foundations of this branch of physics. We'll also learn about some guidelines that you can employ for wave functions so that you can learn to solve the Schrodinger wave equation. We'll learn about classical analogs of quantum systems that we might want to model, building on what we know already about a classical system, but employing that in the Schrodinger wave equation. And finally, we'll learn about a specific archetypal model, a quantum model of a bound particle known as the particle in a box, or infinite square well model, and we will solve it using the Schrodinger wave equation. Let me remind you first about the one-dimensional Schrodinger wave equation, which I will represent using a shorthand going forward, SWE, much easier to carry you that around than Schrodinger wave equation. The Schrodinger wave equation has a time-dependent statement on the left, and on the right it has a spatially-dependent statement about the wave function, and finally it has a portion here that describes the action of an external force on the particle or system represented by the wave function. The above is the one-dimensional Schrodinger wave equation, and generally speaking it allows for solutions that vary in space and time, and it also allows for forces represented by the underlying potential that gives rise to the force that varies in space and time. This is very complex, so to utilize this equation we will need to do the following. First, we will represent physical situations with a model, and what that usually boils down to, because the timepiece on the left and the spacepiece in the middle are essentially fixed by the form of the equation, is varying the form of the potential V. This describes how the system constrains particles described by the wave function. Now, this effort may involve simplifying assumptions in the aid of creating a simple model of the force or forces that can act on the particle, and these choices, these simplifying assumptions, have consequences that I'll talk about later. We will define the basic rules of quantum mechanics. What are the inviolable tenets of problem solving in quantum mechanics that, if untrue, mean the fundamental dissolution of quantum mechanics? We'll also define some guidelines for how to write down wave functions that will work to solve the Schrodinger wave equation, for instance, in a specific situation. Now, these guidelines may be viable depending on how you approximate physical situations, but don't represent the fundamental failure of quantum mechanics if violated. In other words, poor assumptions on the part of the problem solver, the physicist, are not to be held against the fundamental framework of quantum mechanics. So what are the inviolable tenets of quantum mechanics? Well, these are known as the postulates of quantum mechanics, and I'm going to warn you at the beginning that I am glossing over some of the elegance of these postulates in favor of a bit more wordiness because we don't have the mathematical foundations quite yet in order to take advantage of the more elegant and direct way of stating these postulates. So what are the postulates of quantum mechanics? Well, the first one is that at each specific time, the state of a system, that is, for instance, a particle, or collection of particles, can be entirely represented by a space of functions that are related to the wave function psi. Now, while psi depends on a finite number of things, like spatial position along the horizontal space axis and time, the space of functions that can be related to the wave function and can fully represent the possible state of a system can be infinite in dimension. Now, for our purposes, we will concentrate just on the wave function rather than on this larger notion of a space of functions that can describe a system. A more advanced course will concentrate rather on that space of functions, which has all kinds of properties and rules associated with it. It's called a Hilbert space, and it's named after mathematical physicist David Hilbert. The second postulate is that every observable quantity of a system, for instance, a measurement of momentum or energy, will be represented mathematically by the action of an operator on the state of the system. Now, I'll elaborate more on this a little bit later, but think back to how I waved my hands and derived the Schrodinger wave equation. For example, the total energy is measured in that equation by a time derivative acting on the wave function, and as you'll see, other actions of other derivatives effectively represent operators that measure quantities of the system. These would be the outcomes of doing experiments. And finally, the only possible results of a measurement of an observable are related to characteristic numbers known as eigenvalues of those operators that represent taking measurements. Now, the details behind the word I just used, eigenvalues, demands a much higher level of mathematics than can be assumed for this course at this moment, so I will emphasize this particular postulate much less here in this lecture, and in fact, generally speaking, going forward. Now, a dedicated course in quantum mechanics will develop these ideas using fundamental mathematical frameworks like linear algebra and specifically focused on Hermitian matrices or function spaces with specific emphasis on Hilbert spaces. I'm going to gloss over those aspects and instead try to equip you to solve the Schrodinger wave equation while utilizing the implications of these postulates to get useful information from that equation and from the solutions to that equation. Now, what are some implications of the postulates? Well, from the first postulate of quantum mechanics, here are some of the consequences. For example, there exists a product of states that yields a measure of distance between states. So, for instance, if you want to think about a state of a system represented by a wave function psi, there is some product of that wave function that will give you a real valued measure of the system, effectively the distance between possibilities in the system. And that's in the same way that components of a vector, the x component, the y component, the z component, form a kind of basis over which the full vector itself, represented by putting those components together, is actually composed. So, the distance measure of a vector is given by combining the lengths of the components using the Pythagorean theorem. The Pythagorean theorem in real 3D space provides the framework for defining a distance in that space. Now, in our case, since the wave function represents the probability density of the system, and in one dimension that's the probability per unit length, we need a real valued number to come out of our wave functions. And we've seen this before. It's the product of the wave function and its complex conjugate. And this is interpreted to yield the probability per unit length of outcome. Now, the above interpretation, taking this product at all, between the wave function and its complex conjugate, and the link between that product and the probability density, was actually introduced as an interpretation in the mathematics by physicist Max Born, whose image was featured at the beginning and will be featured again at the end of this lecture. This was the proposal that allows you to take this complex function and get real numbered interpretations out of it so that you can apply it to the real world. Now, to get the probability itself, rather than just the probability density, what we need to do is we need to consider this product in a region of space and multiply by, effectively, the size of that region of space. Now, if we divide the space into an infinitesimally sized bits of very large numbers, then we can sum up the products of this psi-si star quantity with those little bits of distance along the x-axis to get the probability of being within a region of x. And that's simply represented by an integral. If we sum up all of the products of psi-si star and the dx in which we're computing the product psi and psi star, that's just an integral. And that integral yields the probability of outcomes in a region of space. So rather than per unit length, this gives you raw probabilities of something happening in a particular region of space. Of course, you have to set the limits of the integral if you want to define the region of space in which you want answers. Now, another implication of the postulates, and this one comes from the second postulate of quantum mechanics, come in two forms here. So, first of all, this postulate was about measurements and how they're represented mathematically using operators. So measuring the energy or momentum of a state will be represented by the application of a mathematical operator on the state of a system. An operator is a mathematical object that takes in a function and spits out information. So a derivative is a good example of that. A derivative acting on a function tells you information about the slope of the function at a particular point, for instance, at any point along the function, for example. In our case from the Schrodinger wave equation, we might realize already that measuring the total energy E of a system is accomplished by applying an operator. We can call that E with a little hat over it to denote that it's a special thing, it's a special mathematical thing. Applying this operator E hat to the wave function psi is supposed to yield the energy of the system, for example. So let's go ahead and write that down. The operator E hat acting on psi, well, that's supposed to return the energy. And in the Schrodinger wave equation, the thing that returns the energy is actually this thing, i times h bar times the partial derivative of the wave function with respect to time. And this, by definition, because it's an operator acting on a wave function, will return a measured value from the wave function. In this case, it's supposed to be the total energy. Now, I haven't proved anything here. Let me show you that if you apply this operator to the free particle wave function, you get sensible answers out of it. So let's go ahead and try the simple case of a free particle psi. And I've written down here the wave function that we figured out for this in a previous lecture. So we go ahead and write the operator i h bar times the partial derivative with respect to time acting on psi. I'm going to skip some steps here, but if you act with this derivative on the wave function, you should get back something that looks like this, i h bar times the quantity a times negative omega, the angular frequency, times this quantity negative sine plus i cosine. Now, I've left out the arguments here. They're implied. This is a compact notation. This is sine of kx minus omega t and cosine of kx minus omega t. I've just compactified the notation for convenience. Now, if you multiply the minus signs together so that they cancel each other out and you pull the omega out in front, you'll notice that if you then distribute the i into the sine and cosine piece, you wind up just getting psi back again. So the action of this operator on psi returns the function psi at the end times a number. Well, what is that number? The number turns out to be h bar omega. So we acted with this mathematical operator on psi and we got back a number, h bar omega, both of which are real valued, times the wave function. We have effectively returned a number from the function. Well, what number have we returned? Well, from the de Broglie postulates, that's just the energy of a matter wave, h bar omega in angular notation, or hf in our old Planck's constant and frequency notation. Now, I will let you figure this out for yourself, but momentum is similarly measured using a different operator. We can denote that as p hat, the operator that returns momentum from a wave function and it's given by negative i, h bar, and the partial derivative of the wave function with respect to space. In this case, in one dimension, it's just dx. So go ahead, pause the video if you like, apply this operator to our wave function for the free particle, and see that it returns the momentum h bar k from the wave function, otherwise leaving the wave function intact. This is what operators do. They act on functions, they return numbers, and the original function intact. This is how you represent measurement in a physical system mathematically in quantum mechanics. Now, I mentioned that there are postulates, which are inviolable tenets of quantum mechanics. If any one of them is ever proved to be wrong, then quantum mechanics is no longer on firm footing and has to be, its assumptions basically have to be revisited. Now, I want to talk about guidelines on writing down good solutions, wave functions. Now, these are guidelines for writing down well behaved as physical as possible solutions to the Schrodinger wave equation. Now, they can be violated. Why? Because not all ways in which you would implement a physical system in the Schrodinger wave equation using a wave function to solve the problem will exactly and accurately reproduce every tiny detail of a physical situation. I'll comment more on this in a moment. Now, the violation of any of these guidelines, if you find out that you have to abandon one of them to solve a problem, it's okay. Don't panic. It really says more about the mathematical model you have chosen to implement in quantum mechanics rather than about what's going on with quantum mechanics itself. The Schrodinger wave equation and the way that you get information from the wave function, whatever it is, is not to be blamed for your choices in approximating the forces on a problem. And that's really what this boils down to, is that when you approximate systems using certain design choices, and I'll comment on this in a bit, there come risks with that that may violate some of these guidelines. And I'll give you examples of these below. So, one of the guidelines for generating a good wave function solution, and it supplies a nice constraint that lets you then try to solve for unknowns in a wave function, is to assume that the probability of finding the particle or particles represented by the wave function psi anywhere at all in space is 100%. That is, if you imagine a system with a particle in it, being acted upon by forces, the probability of finding the particle somewhere in all of space is 100%. It's a good guideline. It has its flaws, and I'll comment on those in a moment, but it's not a bad starting assumption for solving a problem. Now, we can represent this mathematically by something known as a normalization condition, norm being in reference to the value of 1 or 100% that is imposed as a constraint on the probability of finding the particle anywhere at all in space. If we take that integral of the measure of the wave function psi psi star over all space, in one dimension this would be from negative infinity to positive infinity, the integral of this measure psi psi star over all space must be equal to 1. That's a constraint that we can impose. Now, when might it be violated? Well, we've studied unstable particles in this course. The muon is a good example. Radioactive nuclei are another good example. Some particles, even when trapped, don't stick around forever. So, if a particle is unstable and can decay into other things, then the probability of finding that original particle, say an unstable radioactive nucleus, is not 100% at some later time. It may have completely vanished from the universe and having been replaced by some other energy and matter. Now, it is possible to accommodate particle decay in the wave function, but that's to be left for a more advanced class. Another guideline is that the wave function must be continuous, and it must be continuous both in the function itself and in the first derivative of the function. In other words, the wave function must be smooth. Even if it transitions between one region governed by one force and another region governed by another force at the boundary of the transition, the function value should be the same on either side of the boundary, and the derivative should be continuous across the boundary, no kinks, no jump discontinuities. Those things in reality correspond to places of infinite momentum, for instance, which are supposed to be unphysical. So, that's what we're trying to avoid here. We're trying to avoid solutions with seriously bad behavior in them that at some point will cause us trouble because it's fundamentally unphysical. Now, when is this guideline violated? Well, let's say we can't be bothered to represent a force that has a very short spatial action. It's a very narrow region of space where the force acts, and so we choose to represent it by imagining just a potential that's an infinitely deep spike at that point in space. So, we just make it an infinitesimally thin, infinitely deep spike that represents the force that's acting on the particle. That's not a very physical model for a force. It may be convenient if the size of the potential is very small with respect to other features of the system, but it has consequences, and one of those consequences is that the first derivative need not be continuous in that particular case. Now, the question you always have to ask yourself is, when you accept approximations like this, you are trading convenience for physicality. You're getting something that's easier to solve, but at the risk of not being very physical. You always have to come back and ask yourself, can a physical force really be infinite in its strength? Can it really have an infinitely deep potential? Well, probably not, but maybe for a situation you're trying to solve, it's a perfectly acceptable approximation. We make approximations in life all the time when trying to solve hard problems, and there's no shame in that. What I'm trying to get you to separate in your mind is that there are approximations that you state outright, the normalization condition, the continuousness, the smoothness of the wave function. Those are assumptions that we impose. When they fail, it's maybe because we made some approximations to the system that actually aren't very physical. We can't hold that against quantum mechanics. That's us as model builders trying to represent the universe, being a little short-sighted in our choices, even if they're convenient. Now, let's think about physical situations, and let's start by a purely classical situation that's interesting, and we'll come back to a little bit later in the course. Imagine a particle that is free to move on a track. The track has been bent into a parabola, or a u-shape, or something like that, and the particle can roll along the track without friction. There's no dissipative forces in the situation whatsoever. We put the particle up on the track, and then we let it go, and we imagine gravity pulling it down, so it accelerates, and when it gets to the bottom of the track, it has the highest kinetic energy it's going to have in zero gravitational potential energy, and then it rolls at the other side of the track, and eventually it slows to a stop, and there it has no kinetic energy and lots of potential energy, and then it rolls down again, and the whole thing kind of repeats itself. This is an excellent classical model for a something that's trapped in a system. The particle's free to slide up and down the sides of the u-shape until it trades all of its kinetic energy for potential energy, but its total energy, the sum of its kinetic and its potential energies, never changes. There's no dissipative forces, and so if it doesn't start out off the track, then it will never escape the track. It's trapped there. It's bound. So this is what is known as a bound state. This is a trapped particle that in a classical situation could never escape its constraints. It couldn't ever get out of the situation unless acted upon by an external force that removes it from the trap. Now there's a quantum analog to this, and a quantum analog of this is the trapping of an atom in a system from which it ostensibly is supposed to be unable to escape. You know, for instance confining electrons using the Coulomb force around the nucleus of an atom. That's a confinement situation, and we already know that at the scale of the sizes of atoms, that can't be solved without invoking quantum mechanics at some level, putting in the matter wave hypothesis or something like that. Now there are other ways to confine an electron, and I'll talk a little bit about that in a moment. Many of these confinement methodologies are used to design all kinds of interesting custom systems, including custom electronics. The confinement of an electron in extremely small enclosure leads to strong emergence of its wave behavior and all kinds of interesting phenomena result from that, including the possibility of the matter wave escaping its constraints, because the wave is not 100% and fully confined in this system. Some of it leaks out, and that means there's a non-zero chance that the particle could escape from the system. So in modern electronics, this is taken advantage of via the manufacturing of things, for instance, like quantum dots. Quantum dots in three dimensions, for example, are enclosures that hold maybe hundreds or thousands of electrons in a small volume, and in doing so confines them and creates an extremely highly tunable situation where the exact energies of the electrons can be tightly controlled. And this is of great advantage in making light emitting systems where, for instance, you need different chemicals or different engineered structures, each of which, if acted upon by light, like ultraviolet light, would then re-radiate that light at specific target wavelengths. And you can already see this has immense applications for television sets and video screens and all kinds of precision equipment for imaging or for representation or visualizing of information. By tuning the size of the confinement cell itself, you can change the properties of the quantum dot. So for instance, just by tuning the cell size, you can completely alter what light the electrons would emit if acted upon by external photons. And this allows you to create custom atoms. And that is a remarkable feature of modern technology made possible by confining electrons, in this case, to small volumes. Now, these above classical analogies and actual quantum applications lend themselves to being studied by starting from an overly simplistic model. And this is known as the particle in a box or infinite square well model. I'll show you a picture of this in a moment and expand on this a little bit more. But we're going to model the trapping of a particle like an electron in a very small system, as if it's in a deep well, infinitely deep, that it can, in principle, never escape from. Now, of course, that's not very physically plausible. You can't really make an infinitely deep well. There are ways you can approximate that, but it's never perfect. The idea here is to make the barrier for getting out of this as insurmountable as possible under normal conditions. So what we're going to do is we're going to write this model down mathematically. We'll be able to write potentials at different points in space that represent the barriers, the place where the particles free to move, and so forth. And we're going to then solve this using the Schrodinger wave equation. And of course, the classical analog of this is not the U shaped track thing I led with at the beginning, but rather a cart moving on a nearly frictionless air track that's horizontal, bouncing back and forth and back and forth between the ends of the track where there are walls. Now, the cart, once you start it moving, will just bounce back and forth and back and forth, almost free from dissipated forces. It can never get outside the walls of the track, but while it's moving between the walls, it acts as if it's a completely free particle in one dimension. It's only when it gets to the walls that it has to turn around and go back. So that's an excellent classical analogy to the infinite square well or particle in a box problem that we're going to set up here using the Schrodinger wave equation. Now, here is a mathematical picture of an infinite square well. Of course, I've compressed infinity onto one page, so let me walk you through this a little bit. The left wall of the box is located at x equals zero. The right wall of the box is located at x equals capital L, the length of the box. Now, the box volume, the space in between the walls, is where the particle is free to travel absent any external forces. But the particle upon reaching L or zero encounters an infinitely high barrier represented by an infinitely tall potential, v equals infinity. It cannot pass through the barrier, and it cannot in principle be found anywhere outside of zero and L. This potential is infinite at all places, beyond L and below zero. So this is the picture that represents the idealized infinite square well or particle in a box problem. We're going to take advantage of the fact that there are very clear regions here, regions where the energy is infinite for the potential, here on the left, here on the right, and regions where the potential is zero, representing no external force acting on the particle between zero and L on the x-axis. So let's begin to apply the postulates and the guidelines and the Schrodinger wave equation to try to figure out, first of all, what do the wave functions of a particle trapped in this box look like, and what are the allowed energies of the particle in this configuration? Does it have any energy that can be conceived of in nature, or is it constrained by the system to some set of quantized energy states, and it can only exist in those states? Let's explore the problem and see what happens. So we're going to begin by setting up the Schrodinger wave equation and preparing to solve the problem. And this is a good point to pause and reflect on a basic property of unknown functions and how to solve for them. And this is known as separation of variables. So first of all, let me begin by rewriting the Schrodinger wave equation in a slightly more convenient form for what's going to happen next. This is actually more conventional to the way the equation is typically written in books and so forth. Now the minus sign that I've been carrying around in the Schrodinger wave equation has been affiliated with the left side of the equation, the one with the time derivative in it. What I'm going to do is I'm just going to multiply the whole equation by a minus sign. So that effectively moves the minus sign to the right hand side. But in order to keep the potential from having a minus sign in front of it, which may confuse things later about, well, which way does the potential point? Is it infinite down or infinite up? I don't like to have a stray minus sign floating around in the equation. I'm just going to redefine negative v of x and t to just v of x and t. We're just going to absorb that minus sign into the definition of the potential. I know you're probably unhappy with this, but get used to this because this is a fairly standard mathematical trick for moving signs around in an equation. As long as you're clear about what you did, that you absorbed a minus sign into the definition of this thing, v of x and t, no harm, no foul. But basically what this means is we have a positive time term on the left, we have a negative space term here in the middle, and we have our positive potential term here on the right. Now let's proceed with this problem by making a few assumptions. And one of them is a pretty sensible one. First of all, let's assume that the force is time independent. That is that we don't have v of x and t, it's just v of x. And it stays whatever it is at any time t. So we don't have to worry about a time variation in the potential. This basically means we don't have a time varying force. And that makes this problem a lot easier to solve. Now this is applicable to the problem I'm trying to solve right now because I didn't say anything about the walls of the potential moving back and forth. That's something you could introduce into the problem if you wanted to make your life difficult. Real situations may have walls that move back and forth. But we're not going to worry about that here. I could imagine having a potential that's infinite at some times, but not at others. That's also a time dependence. I don't want to have that here. I want my infinite walls to be infinite at all times and fixed in space so they're not moving. So great. I have a time independent potential. That's certainly true for our case that we're studying here. Now the other thing I want to assume, and this may or may not be true, but in this case it won't do any harm to assume this, is that the wave function can be separated into the product of two independent pieces. A time dependent piece and a time independent piece. Now this trick is classic in mathematics. When you have a function of several variables, it is very common if there's nothing that rules this out at the beginning to assume that the variables are separable in the function and that you could rewrite the function as a product of individual functions, one depending on space, one depending on time. And that's what I'm going to do here. This is again known as separation of variables. It may not work in all problems, but let's see the implications if it turns out to be true in a specific problem. So this gives me the freedom to write psi of x and t as two pieces multiplied times each other, little psi of x and phi of t. So little psi of x is the space dependent part of the wave function and phi of t is the time dependent part of the wave function. How is the Schrodinger wave equation transformed by assuming that the wave function has this separability feature? Well, let's work it out. So the implication of this then is that we would plug this little psi of x and phi of t into our Schrodinger wave equation and if you do this, you'll get something that looks like this. On the left hand side, the time derivative only cares about the time dependent part of the function. So we can move the little space dependent part out with i h bar, no harm, no foul. It's not dependent on time. The time derivative would treat it as a constant. Now on the right hand side, similarly, the time dependent part of the wave function is unaffected by the space derivatives. So only little psi of x remains to the right hand side of the second derivative with respect to x, because it's the thing whose form we don't know and we know that, you know, d dx is going to do something to it, but we don't know what it's going to do yet. And then finally we have the potential part over here where we just written out the product of little psi of x and phi of t. Now if I divide through by the wave function, psi of x times phi of t, I wind up moving all the time dependent stuff to the left hand side. The little psi of x cancels out on the left hand side and I'm just left with 1 over phi of t times the time derivative of phi of t and i h bar. And on the right hand side, I wind up cancelling out the phi of t that was in the space dependent part and I only have little psi of x now. I only have the space dependent stuff here, oh and also here, because when I divide the whole equation by the wave function psi of x and phi of t disappear from the potential term. This is great. I've got all the time dependent stuff on one side and I've got all the space dependent stuff on the other. I've isolated the time stuff from the space stuff and in principle I've gone a long way toward making this a much easier to solve equation. Now let's write that one more time just to help keep us in focus here. So we wind up with the time dependent piece on the left, the space dependent piece over here on the right including the term with the wave function and the term with the potential. Now I want you to think about the left hand side. Think about this thing over here. Ignore the rest of it for a second. Just focus on this stuff. Imagine I have a function phi of t and I'm going to do all this stuff to it. Well let's imagine evaluating the function at one specific spacetime coordinate. Let's call it x1, t1. Okay so in that case the left hand side is going to take on some value at t1. Now imagine again evaluating the left hand side of this equation at a different spacetime coordinate, x2, t1. So different location in space, same location in time. What happens to the left hand side of this equation? Has its value changed? Well it's being evaluated at t1 again. I'd better get the same value back. So that means that at these two different spacetime coordinates, x1, t1, and x2, t1, the left hand side remains constant. This is an equation. If the left hand side remains constant under that situation, the right hand side also must remain constant under that situation. And if that feels a little mind bending, pause and reflect on that for a second. If we consider the left hand side at two different spacetime coordinates where only the spatial component changes, we know that the left hand side doesn't change in value. But because this is an equation neither can the right, even though we're evaluating at different space coordinates. So this implies that the left hand side and the right hand side are constant at all spacetime coordinates. So that must mean the following statement is true. That the time dependent part is equal to the space dependent part is equal to a universal constant of some kind for the specific problem we're considering. And this thing, capital C, is known as the separation constant. It's effectively a consequence of having separation of variables be a feature of your solution in the Schrodinger wave equation. Not only does it make it easier to try to figure out what the wave function is, but it also makes the Schrodinger wave equation effectively equal to a constant value. So let's see what we can learn about the separation constant from the time dependent part of the wave function, phi of t. Now remember we don't know what phi of t is yet, we just know that this left hand side of the Schrodinger wave equation is constant. And so whatever this stuff is on the left, it's equal to a number that doesn't change in time or space. So here's what we've learned. If we can separate our wave function into space and time pieces that are just multiplied by each other, the following statement must be true. So we can try to guess now at the functional form of the solution to this time dependent portion of the Schrodinger wave equation. And notice now I've switched to using not partial derivatives but full derivatives because there is no space dependence on the left hand side of this equation. The partial derivative doesn't care anymore about any of the other parts of the wave function because there's only time in this wave function. So we can switch to using full derivatives for this. Now here is a tactic that you would learn in a differential equations course, and it may feel very hand-wavy, but it's based on intuition about mathematical functions. And once you build intuition about mathematical functions, you become very good at seeing how to solve new equations. And this is a differential equation. It involves the function and a derivative of the function. This is by definition a differential equation. Its solution is a function. And you can use knowledge of mathematical functions to try to infer what the solution form might look like. We're going to do that right now. So this is something that you'll begin to build up as a skill in this class. So whatever the functional form of phi of t is, it must be a function that, first of all, when acted upon by a first derivative returns a copy of itself. How do I know that? Well, I've got one over the function on the left hand side of this equation. The derivative of the function with respect to time multiplied by one over the function has to cancel out. I mean, after all, there is no phi of t on the right hand side. There's no time dependence on the right hand side. Phi of t must vanish in this transaction. So whatever phi of t is, when you act on it with a first derivative, it has to return a copy of itself. Now it may be modified by constants, but nonetheless the function itself remains intact and be cancelled out. Now, second, the argument of the function had better contain time, but also the constants i, h bar, and c. How do I know that? Well, only c appears on the right hand side. So this derivative, then multiplied by one over phi, then multiplied by i times h bar, must eliminate i times h bar. So whatever is in the argument of the function of phi of t, it's got to pull some constants out when you take the derivative. And i and h bar had better come out of that, or at least one over i and one over h bar had better come out of it. And oh, by the way, c had better come out of it too, because there is no c on the left hand side of this equation. So whatever the derivative does to phi, it also better yield the product involving c as well. And finally, after all that's been said and done, when the results of the derivative are then divided by phi of t and the function is completely gone, and i and h bar are completely gone, only c can be left behind. Then we have the equation c equals c. That's how we'll know we've solved this correctly. I gotta say, I played around with exponential functions a great deal of my life, even in college when I was learning calculus for the first time and not being very good at it. I really embraced the exponential function. Why? Because it's got a really easy to remember derivative. The first derivative of the exponential function with respect to something that it depends on, returns the exponential function times a bunch of constants, whatever happened to be in its argument. That sounds exactly like the function we're looking for. First derivative returns a copy of it, you get some constants liberated in the process. This sounds like the one we're looking for. So let's guess at the form of the solution. I'm going to go ahead and guess that phi of t equals e to the negative i c over h bar times time. You'll see the value of the negative sign in a moment. Go ahead and plug that into the Schrodinger wave equation. We have i h bar times 1 over e to the minus i c over h bar t times the first derivative with respect to time of the function e to the negative i c over h bar t. So why did I guess e to the negative something? Well, when I take the derivative I'm going to get the exponent down less the time. So I'm going to wind up with negative i c over h bar multiplying i h bar. i times i is negative 1. So I'm going to have a lingering minus sign in this if I don't put a minus sign in somewhere. So you have to put negative i in here to cancel out the minus sign that you get from squaring i ultimately. Try this out yourself. Go ahead and verify that this works. Play around with it. Alternatively, I could have guessed that this was e to the c over i h bar times t. But 1 over i is equivalent to negative i. So it's the same solution. And if you're not convinced of that, check that for yourself. We had it as an in-class exercise earlier in the course. So I promise you that if you work through all of this, you'll find that indeed in the end this thing winds up just being equal to c, which is what it was supposed to be equal to in the first place, QED. We've proved that this is a viable solution to this particular part of the problem, just the time-dependent part of the Schrodinger wave equation. Well, what is this constant? We haven't really answered that question yet, but we can figure it out now. The argument of the exponential function must be dimensionless. In other words, negative i c over h bar times t can't have units. Now as a result of this, the only things that carry units are t, time, seconds, and h bar, which has units of joules times seconds. These are the units of angular momentum. So t over h bar has units of time divided by angular momentum, seconds over joules seconds, which just leaves 1 over joules. Therefore, c has got to have units of energy, joules, so that we get joules divided by joules in the exponential's argument. So the constant turns out to be the total energy of the system, h bar omega. That's why it doesn't matter where in time or where in space you observe the wave function. You always get the same number back from this equation, and that's because this part of the Schrodinger wave equation returns the total energy of the system, and that energy is constant because there are no forces that can dissipate energy in these problems. So we just wind up having an identity here that the constant thing, c, is energy, and it's equal to h bar omega, the total energy of the matter wave in this case. Now what about the space dependent part of the solution? Well this is a bit trickier. Let's see what we can learn with what we've done so far. So from the exercise with the time dependent solution, we've learned the following that this thing on the left hand side of the Schrodinger wave equation is equal to a constant, which is the total energy of the system, and that also must be equal to the right hand side of the Schrodinger wave equation, which only contains the space dependent parts of the solution. So all we have to do is we already know what the solution is going to be for phi of t. We've written it down. It's that exponential function. Now we just have to figure out what psi of x is, and that is going to be much more bound up in what's going on with the potential, because the potential affects what happens in space. Now let's also consider what will be the probability density of these separable solutions. To remind you, probability density is the wave function times its complex conjugate, a la Max Born's hypothesis or conjecture about how to get real probability values out of the wave function, which is complex. So if we plug in our wave functions, what we find is that we have the space dependent part times the time dependent function. We have the complex conjugate of the space dependent part times the complex conjugate of the time dependent function. e to the negative something times e to the positive something is just e to the zero. e to the zero is one. So all we're left with here is a very interesting identity. The probability density of the wave function over both space and time is equal to just the probability density of the space part. The time part drops out of this calculation and doesn't play a role in probability density. So the probability density is independent of time two, only having dependence on what happens spatially to the wave. So for situations where the space and time dependent portions of a wave function can be separated from one another, we need only worry about solving one equation and that is the time independent Schrodinger wave equation. It's the space derivatives of the wave function plus the action of the potential on the wave function. That whole thing will be equal to the total energy of the system times the wave function. And all I did to get this was I multiplied the right hand side of this equation up here by psi of x. If you multiply through by psi of x, you get exactly this equation that I've written down here. Now let's return to our model of the infinite square well, the particle in a box problem. We're considering a particle of mass m trapped inside the infinite square well. So as I said before, when the particle is located between x equals zero and x equals l, there is no potential. The force vanishes. The particle is free in that region. So v of x is equal to zero. Its energy in that region, e, will be entirely in the form of kinetic energy, p squared over 2m. Now because the walls of the potential are represented by x equals zero and x equals l, infinite potential energy, the wave function has got to vanish at x equals zero and x equals l. Why? Well if the wave function has any non-zero value at x equals zero and x equals l, then it will also have infinite energy. Because if it has any energy at that point, because v is infinity at that point, e is infinity. And that just isn't physical. So in order to avoid having infinite total energy for this particle, potential plus kinetic, it is forbidden to find the particle in any other region of the problem than between zero and l. Its wave function must vanish at zero and vanish again at l, never to be seen above l and never to be seen below zero. Otherwise the particle will have infinite total energy and that is a completely unphysical consequence of a model. We have to at least impose physicality on this thing at some point, and this is where I choose to do it. So let's articulate a strategy, a general strategy for solving problems at this point using the Schrodinger wave equation. If the wave function can be separated into a time-dependent and a space-dependent part, we already know the function that describes the time-dependent part. We've got it. The space-dependent part however depends on the spatial distribution of the potential. So begin by writing down the Schrodinger wave equation that includes the potential used to model the situations. That may mean breaking the problem into different regions. So go ahead and separate the problem into any regions that can be treated distinctly from other regions. For example, there's the region below zero or at zero where there's an infinite wall, there's the region between zero and l where there's no force whatsoever, and then there's the region at l and above of infinite potential again, a solid barrier that the particle cannot pass. Those are three regions in the problem and at the boundaries of those regions we have to match up the wave functions from region to region to region. Identify trial functions that might solve the Schrodinger wave equation in each of the regions and determine the values of unknown parameters in the functions that you guessed using constraints. So for instance, employ the guidelines of good wave function behavior, conservation of probability, smoothness, the functions have to match up at boundaries, the first derivative should be continuous, those are good guidelines and try to employ them to solve for unknowns in your trial solutions. These are all constraints and you can see if these constraints will yield values for the things that you inserted as unknown parameters in your trial solution. And I'll show you an example of this in a moment. Okay, so let's go ahead and get going on the particle in a box test solution. We only need to worry about the space dependent portion of the Schrodinger wave equation, since if the wave function can be separated into space and time dependent portions, we already know the time dependent part. I've said this before, I'm repeating it again. The time dependent part of the Schrodinger wave equation is equal to a constant, the total energy of the system. We use that to then solve for the form of phi of t, it's e to the negative iet over h bar. So that's it. We have solved for the time dependent part of the wave function. We don't know what the allowed energies e of the system are yet, but we'll get there, but once we know that we know everything about this function. So we only need to concern ourselves with solving the space dependent part, which I've rewritten here. It's got the second derivative with respect to space of psi, it's got the potential multiplying psi, and it's got e times psi of x on the right hand side. So staring at this equation for a moment, we can already see that solving this equation is going to demand a function that after two space derivatives returns a version of itself. Why? We ultimately need psi to drop out of this equation completely so that we can solve for the energies of the system. In order to satisfy this equation then we need a function that when acted upon by a second derivative with respect to space returns a version of itself. Well that sounds like a job for a sine or a cosine, and really any sine or cosine function would do. So I'm going to begin by guessing that a sine function will do the trick. Why? Well remember that whole business about the fact that the wave function in our problem needs to vanish at x equals zero? The sine function already does that for you. At an argument value of zero, sine returns zero. So I'm guessing here a little bit intelligently guided by past experience and the fact that I know that one of my goals is to make the wave function vanish at x equals zero in our problem. So I'm going to begin by guessing that psi of x is in the region between zero and l, inclusive of those boundaries, a sine function. It's an unknown constant a, I don't know what a is, times a sine of an unknown constant kappa, I don't know what kappa is, times x. And everywhere else in the problem, because otherwise the matter wave would have infinite total energy, it's got to vanish. It's got to be zero everywhere else. It's got to be zero above l, it's got to be zero below zero in x. So we don't know kappa, it's an unknown coefficient multiplying x, but out of good form you should assume that there's an unknown coefficient there because something's got to cancel out the value of x and the sine function. The argument of sine must be dimensionless. And a is an overall and potentially complex, meaning a complex number, multiplicative constant whose form is also unknown. But I promise you by employing the guidelines of solving these problems, using the constraints in the system, boundaries, the good behavior of the functions, conservation of total probability, we can figure these things out. So let's begin by attacking the question, what is kappa? Let's plug our test function into the time independent Schrodinger wave equation, and let's do so inside the infinite square well between the walls where v of x is zero everywhere, so that greatly simplifies the Schrodinger wave equation. And let's see what happens. So here is the simplified form of the Schrodinger wave equation inside the box. v has vanished. All we have to do is evaluate the second derivative with respect to x of psi of x, the space-dependent part only, which we've guessed at the form of. So let's go ahead and plug that guess into the equation. Now what I'm going to do is I'm going to move the negative h-bar squared over 2m to the right-hand side of this equation, so I wind up with negative 2m e over h-bar squared. Now I'm going to act with my derivatives. I do the first derivative with respect to x, and then the derivative again of that with respect to x, and I get a times negative kappa squared times sine of kappa x. So I've got a sine of kappa x back again, but now I have this constant negative kappa squared that's left over from doing the two derivatives. No big deal. I cancel out a sine of kappa x from both sides. It's gone. There's no more space dependence. We have some unknown constant equal to some other constants. Whatever e is, we'll figure that out. So all I have to do is cancel the minus signs and take the square root of both sides of the equations, and I've solved for kappa. Kappa is the square root of 2m e divided by h-bar. Great, but what is e? We still haven't answered that question. What are the allowed energies of this system? So let's see if we can tackle that question next, and then we'll finally know what e is. So to answer that question, what are the allowed energies e of the wave function of the particle in the box, let's consider the guidelines on good wave function solutions. They should be smooth, either by being continuous in the function itself, but also possibly in the first derivative. So in our case, for the infinite square well or particle in a box problem, this means that at the left boundary, x equals 0, we need the following to be true. That if we plug in x equals 0, the wave function is 0. And the good news is, as I said before, this one is automatically satisfied by our test solution, thus why I picked a sine function and not a cosine function. Now, here's where things get a little trickier, but this turns out to be a benefit. It must also be true at the right boundary, x equals capital L, that the function vanishes. That is, a times the sine of kappa L must be equal to 0. That's what it means for that function to return to 0 and b0 forever after that, after the right-hand boundary of the infinite square well. How do we satisfy this? Well, think about the properties of the sine function. The sine function is 0 at what places in a circle? At 0 degrees, at 180 degrees, at 360 degrees, etc. So where is it 0? It's 0 at 0. It's 0 at pi. It's 0 at 2 pi, and so forth. Well, we've already employed 0 as a solution to our problem. Let's focus on when it's multiples of pi that are non-zero integer multiples of pi. So this equation will be satisfied so long as the product of kappa and L is always equal to pi times an integer n, where n is 1, 2, 3, etc., all positive integers. So this turns out to be our quantization condition. This is the thing that's going to ride in and give us only allowed certain energies of the system. So let's go ahead and plug that in. If we plug that in, we find out that the square root of 2m e over h bar times L, that's kappa, so kappa times L is pi times n. Solve for e. What are the allowed energies of the system, e? Well, they will be given by the following equation. The nth allowed energy of the system, so e subscript n, is pi squared, n squared, h bar squared, over 2m L squared. Rather elegant. All of these things are just numbers. The energies of the system are fixed by two parameters, the mass of the particle, and the width L of the square well. Otherwise, it's just integer multiples of pi squared h bar squared over 2m L squared. Of course, it's n squared, not n, but you get the idea. For n equals 1, n equals 2, n equals 3, you can calculate all of the allowed energy levels just knowing pi squared h bar squared over 2m L squared. This is remarkable. You can see why it is that a quantum dot, which is a very good approximation to an infinite square well system, is effectively a custom atom. We can tune the allowed energies of our custom atom by putting the massive particles in there that we have available. They better be stable and last a long time. Electrons are a good candidate for that, and then tuning the size of the well, L, so that we get the energies, e, that we want out of the system. That's it. We can build a custom atom by doing this, using a larger system than just an atom. This is amazing. Okay. Well, we have, we have another question left here, and that is, what is a? We have the solution a sine of kappa x. We figured out kappa, we figured out the allowed energies, but what is this constant that multiplies the sine function? Well, to answer this question, let's take advantage of the guideline that a good physical solution satisfies conservation of probability. In other words, the integral of psi, psi star over all space must be equal to one. This is just the statement that the sum of all probability densities must yield a 100% chance of finding the particle somewhere in space. So, if we go ahead and plug our solution in and noting that psi of x equals zero, if x is less than zero or x is greater than L, and actually psi of x is also equal to zero at x equals zero and x equals L, which we're going to use to our advantage in a moment, we can proceed to try to figure out what this integral gives us. So, we plug in our trial solutions. We've got the space-dependent piece, psi of x, and we've got the time-dependent piece, phi of t. Notice that we wind up with a sine of kappa x, e to the negative iet over h bar, and then the complex conjugate of that here, a star, a could be complex, we don't know, so we're just going to denote its complex conjugate with a star, sine of kappa x, well we know everything inside the sine function is real, so that function remains untouched, and then e to the positive iet over h bar. Remember, complex conjugation is just the act of taking i and replacing it with negative i, wherever you find it. Well, again, we can move the exponential functions around so that they multiply each other. There's nothing in the way to prevent us from putting them together and multiplying them, and then we just wind up with e to the zero, which is one. So, the time dependence completely drops out of this, and this reaffirms the fact that the probability density could have just been computed from the space-only part of the wave function, little psi of x, little psi of x star. So, we're left with this equation, the integral from zero to l is just a a star sine squared kappa x dx, and that integral has got to be equal to one to satisfy conservation of probability. Now, note that the limits of the integral are zero to l, not negative infinity to positive infinity. Well, that's because I took advantage of the fact that this wave function vanishes everywhere except between zero and l. So, it doesn't matter if we do the integral from zero to l or negative infinity to infinity, the integral has to be one in either case because the the integral is outside the region zero to l or zero by definition. So, we can use a standard technique to solve the integral, we can look it up in a reference, whatever you like. So, this would be the result of doing that activity. The integral that I desire to figure out the form of is just going to be equal to a times a star times this quantity involving x over two minus one over four kappa times the sine of two kappa x, and now this needs to be evaluated at the end points, zero and l, and then we can go ahead and and get a relationship between a, a star, and a number. So, rewriting that here, let's go ahead and plug in the end points. Well, zero is going to result in this whole thing being zero. So, the only end point that matters is l. So, we wind up with l over two minus one over four kappa sine of two kappa l. Well, this looks a little scary at first until you remember that at x equals l or any integer multiple l of the function by construction this thing has to go to zero. So, remember this is just an integer multiple of kappa l, and any integer multiple of kappa l yields zero for the sine function by construction because that's another behavior we've imposed on this wave function. So, in that case we can just replace this less pleasant looking equation with a times a star times l over two equals one, and then we can solve for a times a star. It's just two over l. Well, that's not so bad. In other words, the product of the normalization constant with its complex conjugate is a real valued number, two over l, the width of the square well. So, for sure we know that the measure of a squared, the magnitude of a squared, is two over l. From this we know that the absolute value of a itself is the square root of two over l. Now, this leaves us with two choices for what a is. A can either be only real valued, it could just be root two over l, or it could be entirely imaginary, it could be i root two over l. Go ahead and try multiplying a times a star using this choice, or a times a star using this choice. You'll see you get the same answer. There's an ambiguity here. Now, remember the wave function itself has no physical meaning. So, there's no physical consequence to picking either a purely real or a purely imaginary coefficient, a. So, it's typically convention in this problem to simply select the real valued one. A equals the square root of two over l. i times root two over l is just as valid, but there's no physical consequence to picking one or the other. So, maybe it's a little easier to pick the real valued one. That's certainly what I'm going to do here, and that's the convention. So, here's the full solution to the infinite square well problem. We have constructed a model of a system where there's an infinitely insurmountable barrier at zero and an infinitely insurmountable barrier at l. A particle is free in between these two things, but confined at the boundaries. The wave function must vanish below zero and at zero and at l and above l, because otherwise the particle would have infinite energy owing to the fact that it's in a region of infinite potential. This then constrains our solutions to the following form. We guessed at functions for the space dependent part that would solve the space dependent Schrodinger wave equation, and we found out that this would just be equal to the square root of two over l times the sine function of pi n x over l, and that's going to be for zero less than or equal to x less than or equal to l. And you'll see this function vanishes very nicely at the boundaries zero and l. And it has to be zero absolutely everywhere l. So, outside this region between the walls, the function is just defined to be zero everywhere, otherwise we would wind up with infinite energies. So, the only energies we're concerned with are the ones that have to do with the particle that's confined between the walls of the barrier, essentially. And those energies are given by this elegant equation, n squared, where n is integer labeling the lowest one to this next highest two and so forth up energy state, times pi squared times h bar squared over two m l squared. So, once you know the mass of the particle, you're confining. And once you know the width of the region in which it's confined in one dimension, you know it's energies, you know all of its allowed energies at that point. And that gives you an immense amount of control over a system, whether that's incredibly obvious at the beginning or not. So, as I said before, this effectively allows you to create a customized atom. If you can create a situation that looks like the mathematical picture on the left-hand side by creating a semi-impenetrable barrier in which electrons are trapped, you can create a custom atom in three dimensions with tunable atomic energy levels, simply tuned by changing either the mass of the particle that's trapped or the size of the container in which it's trapped. So, these can be represented on the right-hand side as follows. So, let's focus on the blue for the time being. The blue lines represent the different allowed energies, the lowest energy being n equals one, the second lowest being n equals two, etc., n equals three, n equals four are all shown here. Now, the vertical axis is the energy of a given state of the system, and each energy corresponds to a different shape of the real and imaginary parts of the wave function. So, what we're seeing here is the real part of the wave function in space. Now, of course, the exact shape here is not measurable. Only probabilities can be determined from this, and we'll do that as an exercise outside of a lecture video. We'll determine probabilities of finding particles at different places in the well. It's not uniform everywhere. There's definitely places in space where you're more and less likely to find the particle. We'll get to that as an exercise in class. But what I want to point out here is that you can actually visualize the real part of psi of x for each of these allowed energy states. Here you have a single node standing wave. Here you have a two-node standing wave. Here you have a three-node standing wave. These look a lot like, for instance, the standing waves that might develop on a guitar string where it's confined at both ends, and the string is not free to move past its ends. In fact, that's exactly what this looks like. It looks like the fundamental and the harmonics of a plucked guitar string. And, well, it's not really an accident. There is a classical analog to this system and exactly the model I just mentioned. The energies are tunable by the mass of the particle and the size of the system. And again, that actually sounds a lot like tuning a musical instrument, changing the properties of the string, for instance, the thickness or density of the string, or changing the length of the string, as you would do in a piano, to get a different frequency to result from striking it with a hammer. So this actually has a very nice analogy to musical instruments, but here is realized in a quantum mechanical system where you trap a particle in an insurmountable barricade. So to review, we used a strategy for solving this problem and it's a strategy that will work essentially for solving every problem going forward with the Schrodinger wave equation for this course. Write down the Schrodinger wave equation that includes the potential used to model the situations. Try to separate the problem into any regions that can be treated distinctly from other regions, regions where the potential changes value, for instance, is a good rule of thumb. Try to identify trial functions that might solve the Schrodinger wave equation in each region. If you can separate variables in the solution, then you have a huge advantage because the time solutions are already known, you just have to solve for the spatial part of the solutions. Determine the values of unknown parameters in the functions using the constraints in the system. Employ the guidelines of good wave function behavior, conservation of probability, smoothness of the wave function, and match the wave functions at different boundaries in your problem between the regions. Make sure they're smooth across the boundaries both in the function and in its first derivative, again as a guideline, and see if that helps you to constrain unknowns in the problem. For instance, we used conservation of probability to figure out what the multiplying coefficient, a, was. And we used the solution itself plugged into the Schrodinger wave equation to get at another unknown kappa, which turned out to be related to the allowed energies of the system. We figured out the allowed energies of the system by using the boundary conditions of the function at the edges of the potential, at the walls of the potential. All of these basic skills executed in some order over and over and over again, in general, will help you to solve any quantum mechanics problem using the Schrodinger wave equation. Some problems are so difficult you really have to not use paper and pen. You need to use a computer, but the problems that we'll do in this class will be all solvable on pen and paper, pencil and paper, so that you can learn to develop the underlying skills, which could then be applied in more complex situations later that might need more advanced tools to solve them. So to review, in this lecture we've learned the following things. We started with the postulates of quantum mechanics just to lay the groundwork for the inviolable tenets of quantum mechanics that if ever violated mean you need to start over again and figure out what's more fundamental than quantum mechanics. We then learned about some guidelines for wave functions that you can use to practically speaking solve the Schrodinger wave equation for a specific situation. Those are guidelines, they may not always be obeyed, we looked at cases where they might not be, but in general if you start from those and try them and see what happens you will make a lot of progress in solving problems. We then introduced some classical analogs of quantum systems like trapping particles in a classical sense versus trapping them in an atomic or quantum mechanical sense, and then we thought about how to build up a representative mathematical model of a trapped particle situation and that led us to the infinite square well or particle in a box model which we then exercised the postulates and guidelines on in order to try to come up with test solutions to the Schrodinger wave equation. And from that we learned basically that if you can approximate a particle in a box system in the real world you can create a custom one or two or three-dimensional atom for instance a quantum dot being a one-dimensional analog of a particle in a box but in the real world that have highly tunable properties and you can see how elegant and simple those properties actually are controlled only by the mass of the particle and the size of the bounding box. This is really the beginning of an elegant journey through a fantastic world revealed by quantum mechanics and we're going to build on these tools going forward to continue to try to solve the Schrodinger wave equation in different physical models of the real world.