 In this lecture we will learn the following things. We'll take a look at mechanical and electromagnetic wave equations to inspire our thinking about matter waves. We'll learn how to infer the nature of the wave equation for matter from an exercise involving the conservation of energy. We'll look at the meaning of the waves described by the matter wave equation, the so-called Schrodinger wave equation. And finally we'll look at the limits of absolute knowledge that are imposed by the wave nature of matter. Let's take a peek at waves beginning with classical mechanics. An introductory physics class would have taught you about oscillatory phenomena. And a wave is just another kind of oscillatory phenomenon that can be described by time and space-dependent functions. So in introductory physics we learn that a time varying oscillation along one dimension, for instance a mass on the end of a spring that's bouncing back and forth on a friction the surface or up and down in a gravitational field, can be described as simple harmonic and nature. And this allows us to write a mathematical function involving, for instance, the cosine of frequency and time and an offset from the amplitude being maximal at zero. This is a typical equation you might see in introductory physics to describe an oscillatory phenomenon. Now here omega is a special kind of frequency. It's known as the angular frequency and it's given in terms of the period of oscillation, which is a more familiar concept. The period of oscillation, often denoted by capital T, is simply the time required for one cycle of the phenomenon to conclude. The angular frequency is 2 pi divided by the period. And this essentially means that it's 2 pi times the frequency of oscillation of the phenomenon. Angular frequency is the rate of angular displacement if we were to model the repetitive behavior as going around a circle, completing one cycle of the circle, 2 pi radians, as completing one cycle of the phenomenon. Now for all considerations here, let's set the phase angle, the degree by which we would need to offset the cosine function to get the amplitude to match the initial conditions of our oscillator. Let's set that phase angle to zero. Let's set phi to zero to simplify this equation. If you then extend the phenomenon to two dimensions and imagine a long string, for instance, made from a bunch of tiny little masses, each tiny little mass bound to its neighbor as if by a little spring. And we pluck the string, that is, we displace part of the string vertically, then let it go and it bounces up and down and up and down, the vibration of a string. Now we have a distortion in y that's traveling along x in time. And the solution to that problem looks something like this, that the displacement in y at any position x and at time t is given by some initial y times the cosine of a spatial part, k times x, I'll come back to k in a moment, minus a temporal part, omega t, which we're already familiar with from the equation up here on a simple one-dimensional oscillatory phenomenon. Now what is k? Well, k in this context is known as the wave number. And it's defined by 2 pi, the number of radians in a circle, divided by the wavelength of the phenomenon. So you can think of this as describing the number of cycles per unit distance in the phenomenon, whereas the angular frequency is the number of cycles per unit time. But these functions answer some question. And if they're the answers to a question, what is the question? Well, they are all solutions to a wave equation. That is, an equation that describes how changes in space relate to changes in time. Now the one-dimensional mechanical wave equation, at least the one that tells you about vertical displacements and how they vary as a function of horizontal position in time, is simply given by the second derivative with respect to time of the vertical amplitude y. And that's equal to the constant squared times the second derivative with respect to space of the vertical displacement y. And y, of course, is a function of x and t. So if you try applying this wave equation to the solution on the previous slide, you'll see the following. First of all, the left-hand side is the second derivative with respect to time of the vertical displacement y. Plugging in our function for the vertical displacement, we would get this equation now. The second derivative with respect to time of our description of the vertical displacement versus x and time. Taking one of the derivatives of this function results in us having to do the derivative twice. First of the cosine function and then of the argument of the cosine function. Well, the derivative of cosine is going to be the negative sine function. And the derivative of the argument is going to return a negative omega, a negative angular frequency multiplier. And so we'll be left with this. Omega times the original amplitude y sub zero times a sine function of the original argument. We have to take the time derivative of this one more time. If we do that, we wind up with an additional factor of negative omega out in front of the original cosine function. And so at the end of this, we wind up with an equation that's just negative omega squared times the original function y of x and t. Now let's handle the right-hand side of the wave equation. This is a constant term squared times the second derivative with respect to x position of the displacement y. We plug in our function for y again. Now taking the first of the two spatial derivatives that we have here, we wind up with a function that looks like this. So the first derivative of cosine returns negative sine. And the derivative of the argument multiplied by that gives us a factor of k. And so we wind up with this. And we have to take the spatial derivative of this one more time. And at the end of this, we wind up with an equation that's negative the constant squared times the wave number squared times the original function y of x and t. Now setting these two things equal to each other, as would be required by the wave equation, we find out that the function y of x and t drops out of both sides of the equation, leaving us with this simple relationship between the angular frequency squared, the constant squared, and the wave number squared. And if we take the square root of all of this, we see that we have c equals omega over k. And this is one of the velocities that's present in mechanical waves. The speed of the mechanical wave is given by the ratio of the angular frequency and the wave number. Now this is a very quick tour of a solution to the wave equation and how you can see that it does solve the wave equation and how when you plug it in, it returns a relationship between frequency squared, speed squared, and wave number, which is related to wavelength squared. A dedicated waves course would spend a lot more time on this, motivating the derivation of the wave equation itself from a simple model of a vibrating string or something like that, motivating how one sets up and solves that equation, and then showing you what relationships emerge from solutions under different conditions. Here, I am merely trying to motivate some thought process about wave equations and the resulting relationships that can be derived from the application of those wave equations to their solutions. So sticking with mechanical waves for a moment, let's think a little bit about the energy that's contained in that wave. So again, our model here is a mechanical distortion of a physical medium. And that medium that I have in mind here might be a string made from many little bits of mass all hooked together as if by little springs, each with a spring constant and so forth. So if we model a string that way, we can think about the string as having a total mass, capital M and a total length, capital L. And the little bits of mass it's made from are all equal in size and uniformly distributed along the length L. And so this string has a uniform linear mass density given by the Greek letter mu, which is mass divided by length big M divided by big L. No matter what chunk of the string we look at, every chunk will have the same mu because it's a uniform distribution of mass. And so we can always relate mu to the mass in that chunk and the length of that chunk. If we then vibrate the string such that a given part of it at some time t in location x will have a small mass M and that mass will have a vertical velocity Vy, that velocity will oscillate transverse to its length just as the displacement oscillates transverse to its length. That tiny little chunk of the string will have a length dx, a differential of x, and a little mass M that can be related to the length dx by the linear mass density. So the little m divided by the little dx would be mu because it's a uniform distribution of mass. So that means that m is equal to mu dx. And every place we see m we can replace it with this product and vice versa. The kinetic energy of that little chunk in a moment of its motion as the string vibrates will be defined by its mass and its velocity at a given moment in time t. So taking the classical definition of kinetic energy, we're thinking about a mechanical wave here, so let's think classically for a second. We have the little bit of kinetic energy possessed of by that little bit of mass is going to be one-half times its mass, which is mu dx, times its transverse velocity squared, Vy squared. Well, Vy is just the derivative of the displacement in the y direction with respect to time, dy dt, and we're going to square that. So we wind up, if you plug in that derivative and do that as we did on the previous page, we wind up with this equation for the little bit of kinetic energy possessed of by that little bit of mass that makes up the string. So this is the term for the little bit of kinetic energy possessed of by that little chunk of mass. Now, because it's hooked to its neighbors by springy things, each mass is linked to the next by, you could imagine, a little spring with a spring constant. The potential energy stored in that same chunk of mass will depend on the elasticity of the string. That is the stiffness of the little springs that you could imagine hold one chunk of mass to the next. So thinking of the string this way is concocted of a whole bunch of little masses M connected to their neighbors by little springs with spring constants kappa. Then as an introductory mechanics for oscillatory phenomena, masses on a spring, you can relate the angular frequency squared to the ratio of the spring constant and the mass. That is to say, the spring constant is related to the mass times the angular frequency squared for an oscillating mass on the end of a spring with spring constant kappa. So the little bit of potential energy that's stored at that location at x in that little mass M is just one-half times the spring constant times the displacement from equilibrium squared. Well, that's just going to be a little chunk of potential energy held by that little bit of mass. One-half times kappa times the displacement squared. We go ahead and substitute for kappa with M omega squared and we can substitute for M with mu dx. And then finally we can put in our equation for the displacement and that now involves the cosine and that whole thing is squared. So we have kinetic energy. We have potential energy. Let's look at the total energy possessed of by this little bit of mass M. So that little bit of mass M will have total energy de composed of kinetic and potential added together at any moment in time. We have the expressions for those two things, dk and du. And you'll notice that if you pull out all the multiplicative factors, you'll be left with the same coefficients multiplying a sine term squared and the same coefficients multiplying a cosine term squared. So you wind up being able to pull all those multiplicative factors one-half times and why not? Omega squared and mu and dx out in front of a sum of a sine squared and a cosine squared. And there's a trigonometric identity that comes into play. Sine squared plus cosine squared is one. And so the sine and cosine functions vanish from the total energy of this little chunk and all that defines its little bit of energy that it possesses at any moment in time is that the total energy of that little chunk is constant. It may be divided differently between kinetic and potential but the total energy of that chunk of mass that makes up that string that's vibrating is constant and it's given by this number here. And again it depends on angular frequency, linear mass density, the length of the element, the initial displacement of any element of the string and so forth. This is just the energy stored in this little piece of mass M at a location x in space and t in time. Now note that the total energy depends on the square of the angular frequency. The presence of the omega squared multiplier tells us something about the number of time derivatives or the product of the number of time derivatives that had been present in the original equation for total energy. Remember we had to square the time derivative of the wave function that solves this mechanical wave equation and that yielded an omega squared term in all of this. So you can see that there are shades of the number of derivatives left over as sort of vestigial elements of the energy equation for this little bit of mass M. So here are the key takeaways from this look at mechanical waves. The mechanical wave equation relates the second derivative with respect to space and the second derivative with respect to time. You would derive its form in a dedicated class on waves but we don't have time for that here. Nonetheless I want you to take away the big lessons from this. Now the solutions to the wave equation when acted upon by the derivatives in the wave equation yield squares of the angular frequency omega and the wave number k. Recall we had an equation relating omega squared k squared and the speed of the wave squared. The energy equation for the wave or a part of the wave is sensitive to the number of time or space derivatives in the underlying equation and these manifest as multipliers like omega squared. And one might think about the presence of the squares of these quantities like omega or k as indicative of the underlying wave equation that you needed to have solved in order to get these solutions in the first place. Now let's take a quick look at waves and electromagnetism. This is the next classical wave equation that was discovered in the history of physics and it's derived from Maxwell's equations for electric and magnetic fields. So the wave equation that results describes the propagation of oscillating electric and magnetic fields in empty space for instance although it's not limited to only empty space and that wave equation can be written as follows. That the speed of light in empty space squared times a spatial derivative squared minus a time derivative squared all of this acting on an electric field vector is equal to zero. Well again notice that like the mechanical wave equation we've got second derivatives in space and second derivatives in time all acting on a solution E vector whose form we don't necessarily know beforehand. But if we solve the equation we find out that the solutions to the oscillating electric field components look very similar to the mechanical waves in that they have a vector amplitude instead of a scalar amplitude a cosine function a spatial piece and a temporal piece of the argument of the cosine function. Now I should note that yes there is an identical equation for magnetic fields that can be derived from Maxwell's equations but you can think of it as a bit redundant it describes the action of the oscillating magnetic field but can be related through the mathematics of the solution to the electric field and so if you can remember the electric field wave equation which I'm not asking you to do but will come in handy in a dedicated course on electromagnetism later you can very quickly work out what the form of the magnetic field wave equation is and relate the electric field to the magnetic field although they are in independent directions of one another the field strengths are not independent of each other. This is an interesting problem in that electromagnetic waves are two component waves they have an oscillating electric component and an oscillating magnetic component they have two kinds of information that are stored in the wave and this is a theme that we'll return to later when we look at matter waves. Now applying the wave equation written here to the solution written here similarly yields quadratic multipliers of k squared and omega squared so for instance we find that the speed of light squared will be equal to the angular frequency squared divided by the wave number squared and this latter relationship turns out to be a direct consequence of the massless nature of light that we learn from Planck's relationship for energy and momentum and wavelength and so forth and the special relativistic relationship between energy and momentum for massless particles. This allows its speed to be related directly to its frequency and wavelength with no other multiplicative or additive factors involved. So let's revisit that the relationship between frequency and wavelength through a wave can be directly related to the energy present in the radiation quantum the photon. The photon is the particle-like aspect of light's behavior and the electromagnetic wave is the full wave description of light's behavior. So for example from above and from our previous look at electromagnetic waves we know that the speed of light can be related to the wavelength and the frequency of light as follows wavelength lambda times frequency f. Now to get angular quantities shoehorned into this equation like omega and k what we can do is we can multiply lambda f by a clever number one so I'm gonna multiply by 2 pi divided by 2 pi. If I group the 2 pi in the numerator with f I get omega the angular frequency 2 pi f. If I group the 2 pi in the denominator with lambda I get k. k is just 2 pi over lambda. So I wind up with the equation that the speed of light is omega divided by k the angular frequency divided by the wave number. Now recalling that Planck's relationship for the energy and frequency of light related by Planck's constant is e equals hf. We can play that same game and shoehorned angular quantities in here by multiplying hf by a clever number one 2 pi divided by 2 pi. Grouping the 2 pi in the numerator with f the frequency in the numerator and taking h and dividing it by the remaining 2 pi in the denominator we wind up with this compact equation for the energy as related to the angular frequency. h bar is to denote h over 2 pi and it is known as the reduced Planck's constant. Remember that Planck's constant has a value of 6.626 times 10 to the minus 34. So all you do is take that divide by 2 pi. 2 pi is approximately 6 and so you wind up with a number that's about 1 times 10 to the minus 34. It's very convenient for all of these angular wave concepts to carry h bar around rather than h and 2 pi. So it's very nice to define this reduced Planck's constant notationally as an h with a line through the vertical part of the h. Now the momentum of the quantum is given by p equals h over lambda and again if you insert a clever one in that you'll find that this is equal to h bar times k the wave number. What's nice about this is it kind of puts e and p for an electromagnetic wave on equal footing. e is h bar omega, p is h bar k. These are a little easier to remember than the h over lambda and h times f thing. At least I find them more convenient once you feel more comfortable with angular frequency and wave number as angular concepts of oscillatory motion. Recalling that the speed of light for our electromagnetic waves is given by omega over k that is the angular frequency divided by the wave number if we substitute in for those two quantities with their energy and momentum expressions we recover the special relativistic relationship for light as a massless phenomenon. e equals p times the speed of light. So we've exactly recovered the Einstein energy relation for light a massless phenomenon. So let's take an overview of the wave equations and classical mechanics and electromagnetism. They involve second derivatives of both space and time. So you see both d squared dx squared and d squared dt squared in these wave equations. Now we could have inferred that from the results of the wave equations as their application results in squares of time and spatial frequencies. So for instance omega squared and k squared appear in equations that result from the application of the wave equations. The energy equations tell us the proportionality of frequency to wavelength as well as other useful information like that. And all of this kind of leaves you wondering if we have a mechanical wave equation and an electromagnetic wave equation where's the matter wave equation? Where's the evidence for that from the history of science up to the early 1900s? That's the problem. Since its presence could not be inferred directly from previous measurements in the same way that Maxwell's equations were inferred from Coulomb's law and Gauss's law and other things like that, Ampere's law, and ultimately when composed together in the form of Maxwell's equations led to the wave equation for light. And in the same sense that considering a string as a bunch of masses bound by springs that are you know tugged up and down and then caused to vibrate by being stretched will lead you to the wave equation where where's the exercise that leads to the wave equation for matter? And that didn't really exist up through the 1920s, the early 1920s. So the question I would put to you is is it possible given other equations that we could infer from what we know its form? So if we know things about particles and waves like wavelength and frequency and energy and momentum and we know the relationships between those things, can we figure out the wave equation using all the information we have from atomic spectra, the black body radiation spectrum, the photoelectric effect, Compton scattering, and all that other stuff? Can we figure that out? So one can glimpse the hints of the underlying but unseen matter wave equation. Sort of like seeing a shadow cast on a wall by a complex object that's out of your line of sight, but whose shadow is projected onto a wall giving you hints about the real shape of the thing you can't see. And we can get that glimpse of the shadow of the matter wave equation by considering the conservation of energy for a particle that's acted upon by an external force. Now such a particle in classical physics would have of course a kinetic and a potential energy with specific forms depending on the force involved in the problem. Now conservation of energy would require no matter what that the total energy of that particle is going to be the sum of its kinetic and potential pieces. Now sticking to classical physics for a moment because employing special relativity to derive the rules of matter waves involves a whole skill set of mathematics that really can't be expected of you at this stage of a university career. We're going to stick with purely classical low velocity matter. Even small matter like electrons we're going to have to consider moving at relatively low velocities not very close to the speed of light. Now obviously that doesn't cover the full domain of phenomena of small particles like that but it'll get us going and it will allow us to solve a great number of problems that are actually within our grasp once we figure out the matter wave equation. So using classical physics we can write the kinetic energy as one-half mv squared and we're going to leave the potential energy unspecified I'm not going to worry about what the force is that's acting on this let's just say it has a potential energy u for now and leave it at that. Now we have relationships for matter waves between total energy and frequency and total momentum and wavelength but we don't have momentum in this equation so let's get momentum into this equation and the way we do that is we multiply yet again by a clever number one so if we insert a number one in the kinetic energy equation that is just m divided by m then we get an m squared v squared in the numerator and mv is just momentum so we wind up with momentum classical momentum squared in the numerator divided by twice the mass of the particle plus its potential energy u. So we have our kinetic term now expressed in terms of momentum and we still have our potential energy term and they're summed together to get the total energy e. So let us now inject de Broglie's postulates into this equation that is e equals hf which is equal to h bar omega and p equals h over lambda which is just equal to h bar k again employing all these nice angular quantities and if we do this we now obtain the shadow cast on the wall by the matter wave equation and that is h bar omega equals h bar squared k squared over 2m plus u. You see it? The single power of omega on the left side indicates to us that a shadow is cast here by a single time derivative that's acted on some solution to the underlying wave equation. We don't see the solution and we don't see the wave equation but we see the result of applying those two things and that is a single power of omega on the left side. The k squared on the right hand side implies that there's a second derivative with respect to space in the wave equation acting on the solutions to that equation whatever they may be. So we have a single time derivative and a second space derivative that result in k squared and omega. So let's go ahead and take that equation with our hypotheses our hunches about what the underlying wave equations form might look like and let's try inserting those hunches into this equation above assuming that an appropriate derivative has acted on an unknown solution to yield a single omega or a k squared. So if we do that if we take our hypothesis about the number of derivatives acting on some unknown function that solves the wave equation yielding this relationship we wind up with the following equation. On the left h bar times the first time derivative of an unknown solution which I'm denoting with the Greek letter psi and it's a function of space and time we're only considering motion in one dimension right now. On the right hand side we have h bar squared over 2m times the second derivative with respect to space of that same function psi of x and t and of course we have the potential energy of the the matter wave still tacked on to the right hand side over here and I'll return to that a little bit more later. So this looks promising it has all the hallmarks of a wave equation but it's different from Maxwell's equations or a mechanical wave equation in one key way. It has a second derivative in space but only a single or first derivative in time. This will have implications for the kinds of functions that can solve such equations and the solutions to this as I've said are denoted by the capital Greek letter psi as a function of x and t in one dimension. So let's explore solutions to this equation and as we do this we'll find that we are missing at least one key piece of the underlying equation. We've guessed at the form of the object casting the shadow on the wall and we may have guessed incorrectly. So let's begin by guessing the form of the solutions to our equation and then plug them in and see if we recover our energy conservation statement. So to simplify matters let's consider for now free particles that is particles free from external forces and that is most simply expressed by setting u to zero. The particle has no potential energy associated with it we're only considering motion at a constant velocity so that's a fixed kinetic energy which then relates to the total energy of the particle. Okay well to solve our wave equation we need a kind of function that when acted on by a derivative transmutes into another version of itself. So for instance in the old wave equations the mechanical wave equation the electromagnetic wave equation we had second derivatives acting on the solutions. Sines and cosines were great for that because after two derivatives they returned to their original selves. So that's what we did for traditional waves we used sines and cosines. So let's take a guess let's guess that psi of x and t is just one of our mechanical wave solutions a cosine of kx minus omega t. Though we're not doing anything original here we're just taking mechanical waves getting inspired by them and blindly applying that idea here. So let's write down our wave equation that we've guessed. H bar times the first derivative of our solution with respect to time. H bar squared over 2m times the second space derivative of our solution. Let's go ahead and plug our guess at the solution in. All right so we plugged our function in on both sides now and then go ahead and work out the derivatives and you should find the following conclusion that we wind up with a positive sine function on the left now multiplied by omega. We get a minus sign from the derivative of the cosine but we also get a minus sign from the derivative of its argument with respect to time. So this winds up being net positive on the left side but over here the two derivatives of the cosine yield an overall minus sign. So we have a positive sine function and a negative cosine function. We can't cancel out the sines and cosines on either side. It doesn't recover our original energy conservation expression. It doesn't work to solve the equation. The left side and the right sides don't give us what we would have expected based on where we had derived this from which was the conservation of energy. And if you try just a sine function it will similarly fail. So what if instead we combine sine and cosine functions? What if we add together sines and cosines? Because when you take the first derivative of something that's a sine plus a cosine you'll wind up with a sine and a cosine in the result and similarly with the second derivative. Maybe a superposition, an addition of sine and cosine will do the trick. All right let's go ahead and try that. Now I'm trying the barest simplest superposition. I'm assuming that they have the same multiplicative coefficient out in front a whatever that is and otherwise it's a cosine of the same argument and a sine of the same argument all added together. When you're playing around with solving equations whose solutions are not known to you a priori that is with prior knowledge beforehand guesses like this will get you through the process and you should always try to start with the simplest guess and work your way up in complexity. So for instance it may be these coefficients aren't supposed to be the same but don't start by assuming that. Try assuming them and then work your way up to a more general set of solutions as you get more comfortable with solving the problem. So we write down our guess at the wave equation again. We plug in our new choice of the solution. We work through the derivatives and we'll get the following equations. Now what I want to do is I want to reshuffle the term order on the left side. I want to get the cosine first and then the sine second so I'm just going to move these terms around without changing anything about the equation and this is the final form of the equation I get. I get a negative cosine and a positive sine and I get a negative cosine and a negative sine over here. I can't cancel these functions out. I can't recover the energy conservation expression we started from. It's a lot closer than we were with just cosine but it's still no good. We've got problems with the plus and minus signs and all of this. It's a mess. What's the real problem we keep running into here? Well we keep generating stray minus signs from the single derivative of only the cosine on the left side. The derivative of the sine gives us something positive but the derivative of the cosine gives us negative sine function and that's what's really hurting us here. Our goal at this point is if we're going to figure this out we've got to find a way to get rid of this minus sign we keep picking up and at this point it helps to remember that there are other kinds of numbers than real numbers in the world. So everything I have done up till now is predicated on the assumption that these solutions and perhaps even the wave equation itself can only be composed from real valued numbers you know like 1.1 or 2.3 or pi or negative 75.6. Those are all numbers that can manifest in the real world. If somebody says look I'm you know I'm going to give you negative 76 dollars it means that they're going to take 76 dollars away from you, right? That has real consequences. Negative numbers are real things in the world around us but there are other kinds of numbers that don't have physical meaning in the world around us and it's important to remember that and they fall under at least one class of these numbers is a category known as imaginary numbers and in particular it's helpful at this point to remember some of the behavior set of the archetypal imaginary number i which effectively serves the role of being the number one in the imaginary number set. So let's pause for a moment and revisit imaginary numbers which presumably you have seen in some context prior to this course. Let's recall that the imaginary number i is a special kind of number one but with no physical interpretation. So i is defined by the question what squared equals negative one and the answer to that is i and i's value would be the square root of negative one which is nonsensical. If somebody told you you know give me i dollars you would not know what to do with that because you don't know how to calculate the square root of a negative number and then turn that into a real dollar value that you then hand to that person. Now this number can exist in a mathematical universe where it has plenty of self-consistent rules that don't violate any of the axioms of mathematics that you're that you're playing with. And in fact i doesn't violate any axioms of mathematics at all so it's perfectly mathematically tenable even if it's not physically realizable around us in the world. Its existence mathematically has consequences though. So for example you know going back to the question that leads to i you can take i and multiply it by itself that's just i squared. If we plug in for i squared we have the square root of negative one times the square root of negative one and by definition that has to yield negative one. And so i squared equals negative one i is the answer to that question that you know the square of what number gives me negative one. But you'll notice that i squared has the ability to add a stray minus sign where none would have been present before. And i is the number one in the imaginary number world. It is the unit on which you can build all other numbers, integers for instance, in imaginary space. So the presence of extraneous minus signs when trying to solve equations using functions as we've been doing with our guests at the matter wave equation could actually be an indicator of something that we are trying to use real valued numbers and solutions only. But maybe the problem we're trying to solve is too complicated to only admit real numbers that it requires the ability to store additional information that real numbers alone cannot accommodate. Those can be accommodated by complex numbers. These are numbers that contain both real and imaginary components. So for instance the complex number z is made from two real numbers x and y, but y is multiplied by the number i. And so i, y is imaginary, x is real. This is a combination of a real and an imaginary part. This is what is known as a complex number. And it stores twice as much information as a single real number because it's got this extra component over here. And if this is reminding you a lot of vectors like a vector z being equal to a component along the x axis x and the y component y, that's good because a lot of the basic ideas of vectors translate into complex numbers and give us some confidence about how we can get useful information out of complex numbers. So let's explore complex solutions to this equation. So let's start by trying a guess at a complex solution. It's got a real valued part, a cosine, and it's got an imaginary part, a i sine. So all I've done is I've added the number i to the sine part of my solution. So again here's my guess at the wave equation. I plug in my solution, I take my derivatives, and I get the following results. Now again I've got sine and cosine out of order on the left side. So if I shuffle them around to get cosine first and sine second and try to map that onto the right, I see that I still have a problem. I've got negative i cosine and negative cosine here. I've got a sine and I've got negative a i sine over here. I can't just naively cancel these things out. That doesn't really work. So I have a problem. I still can't get this to work out. If I were to try to move an i or a negative i from the left side to the right side, I'd still wind up with a stray i where I don't want one. And that isn't going to work for this problem. So it's ridiculously close to working out. We're so temptingly close to solving this problem right now, but something is still missing. Some salt is missing from our soup here that we used to try to mimic the recipe for the matter wave equation. So let's see if we can get one more opportunity to think about our assumptions by moving some minus signs around to see if we can find a clue that will resolve this little puzzle. So I'm starting with the last equation from the previous slide here. I haven't done anything to it yet, but what I'm going to do is I'm going to pull the minus signs on the right hand side out as an overall multiplier and then multiply both sides by negative one. So I'm basically moving the minus sign here to the left side of the equation. So this is the net effect of doing it. Notice all the terms over here on the right are now positive. The minus sign that would have been on the right is now moved to the left and overall multiplies these terms. And if I go ahead and do that, if I multiply it through, I see that I now have i cosine here and negative sign here, and I have cosine here and i sine here. And that has a kind of weird rhythm to it. And that's the clue. The difference between the function on the left hand side and the function on the right hand side boils down simply to a missing factor of negative i on the left. If we had originally guessed that the wave equation was negative i h-bar times a time derivative of the wave function, all of this would work out. We would actually get the left hand side being equal to the right hand side. So if we had just done this at the beginning, if we traded our original guess at the left hand side of the wave equation which was all predicated on our bias for real numbers, h-bar times the first time derivative, if we traded that for negative i h-bar times the first time derivative, then we'd solve our problem. And let me show you that. So to wit, let's revisit our guess at the form of the wave equation. I'm leaving the solution completely intact. That's still the same as what I had a moment ago. And what I've done is I've changed the left hand side of our guess at the wave equation to be negative i h-bar times the first time derivative of psi. I've done nothing to the right hand side. Plug in the solutions. Do the derivatives. Play the game again where we shuffle the terms on the left hand side to get cosine first and sine second. Move the minus sign on the right over to the left and distribute that into these terms. Now multiply the negative i into the function. The negative i multiplied by this i gives me negative times negative one, which is one, positive cosine. The negative i multiplied by the sine function cancels out the minus sign here and leaves me with positive i times sine. Cosine, i sine. Cosine, i sine. The left and the right sides level out and the function is now the same on both sides and it can be cancelled out mutually on both sides of the equation. So, Bazinga as a famous tv character might say at a moment of revelation. We've done it. We've cracked the underlying form of the matter wave equation. Let's prove that by seeing if it returns to us the energy conservation relation that we started from. We've only shown here that we have a wave equation that when acting on our guess at the solution to a free particle wave returns the solution with a bunch of multiplying coefficients and so we can cancel a function out of both sides. But we don't know that we've recovered the conservation of energy equation we started from. So, let's see if this all works out. Let's review what we did. Starting from the total energy of a free particle, which is h bar omega equals h bar squared k squared over 2m, we constructed a wave equation that had the right time and space derivatives in it to return these factors of omega and k squared. We played with oscillatory solutions and we learned that only complex functions will satisfy an equation like this. And from this we inferred a missing imaginary number from our original guess at the equation. We should have had a negative i lurking on the guess of the left side of our equation the whole time in order for this thing to have viable solutions for the free particle that work out. The solution guesses that we made for free particles are of this form. They're complex. They have a real part and an imaginary part. And plugging them into the wave equation yields this relationship. H bar omega equals h bar squared k squared over 2m, which was precisely the energy conservation equation for a free particle that we began with. So, it's entirely a self-consistent exercise. And that should be deeply mathematically satisfying, even if you're not completely comfortable with the process by which we arrived at that. But I promise you that this is not the first time, assuming this is the first time you've ever seen this kind of strategy done for solving an unknown equation with unknown solutions. This is not the first time you'll bust out this trick in your life if you ever have to solve hard problems. So, this trick is actually useful even if it feels a little clunky at first. And I hope it conveys to you the sort of incredible exercise that must have been required originally to derive this wave equation in the 1920s. We're doing this with the benefit of a century of hindsight, but our predecessors did not have the benefit of this much hindsight. So, while I am able to look at resources and come up with ways of explaining the wave equations form to you at this level of a physics course, the people that were involved in trying to write down the matter wave equation in the 1920s did not have the benefit of this much hindsight, and so they were struggling with immense difficulty in a different mathematical landscape than we are in now. So, putting back the potential energy piece of this, this is the full one dimensional what is known as Schrodinger wave equation. We have another function that we can tack on to the right-hand side here, v of x and t, which is known as the potential term. That function acting on the wave function psi would return the potential energy of the matter wave, in this case u. And this is one of the most important equations in history where I've added back the potential energy term to complete the equation for a particle in one dimension. The one-dimensional Schrodinger wave equation is one of the most revolutionary insights into the universe in the history of our species, named after Ervin Schrodinger, the first person in 1926 to fully determine the form of the matter wave equation. Now, he was doing this using a whole different set of mathematical algebraic and calculus constraints. That is, this is not how he derived this equation, but this is sufficient at this level to motivate where an equation like this might come from. He was using mathematical guidelines to infer the nature of the equation that we simply don't have the mathematical foundation for at this stage of a physics course to do. Now, you might look at this equation and think, well, this is horrible, I hate time derivatives, I hate space derivatives, and there's two space derivatives and one time derivatives, and these functions are awful, and there's imaginary numbers, and that. Okay, that might seem daunting to you, and perhaps it is. But there's actually a much more difficult part to this equation, and that is that the hard part of the Schrodinger wave equation is the solving of the equation to specific situations. The real challenge of this equation is not this equation itself, although it doesn't look very pleasant, I know. But rather in the finding of solutions, the size of x and t to this equation given different potentials, v, and so forth. Now, we've effectively solved the free particle case, and we'll explore the solutions to that for the rest of this lecture. But if a potential energy term is present, so in other words, if there are forces in a problem, and you can't ignore them, and you have to include them, and those result in potential energy or changes in potential energy for a particle in a problem, you have to completely rework this equation and find the correct solutions that satisfy the equation with the correct potential term added on. And that is much more difficult, and that is effectively what we'll spend the rest of this course learning how to do for different situations that map on to the physical world. Now, I should say that solving this equation for different situations is what has allowed us to understand semiconductors. It is part of what allowed the revolution in microelectronics to happen in the first place. Solving this equation for atomic system leads to an understanding of where chemistry comes from, and specifically doing this on a grand scale is the heart of physical chemistry. Solving this equation for information systems is what results in quantum computing and quantum information, which is a hot subject these days and is one of the many technological frontiers of our species. I cannot understate how important the Schrodinger wave equation is for all the foundations of the world we live in today, but also all the potential for the great discoveries of tomorrow.