PRESENTER: Thefollowing content is provided under aCreative Commons license. Your support will help MITOpenCourseWare continue to offer high qualityeducational resources for free. To make a donation orview additional materials from hundreds of MIT courses,visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Today we beginwith the harmonic oscillator. And before we get intothe harmonic oscillator, I want to touch ona few concepts that have been mentioned in classand just elaborate on them. It is the issue of nodes,and how solutions look at, and why solutions havemore and more nodes, why the ground state has no nodes. This kind of stuff. So these are just a collectionof remarks and an argument for you to understand alittle more intuitively why these properties hold. So one first thingI want to mention is, if you have a Schrodingerequation for an energy eigenstate. Schrodinger equation foran energy eigenstate. You have an equation of thefrom minus h squared over 2m. d second dx squared psi ofx plus v of x psi of x equal e times psi of x.Now the issue ofthis equation is that you’re trying to solve fortwo things at the same time. If you’re looking at whatwe call bound states, now what is a bound state? A bound state is something thatis not extended all that much. So a bound state willbe a wave function that goes to 0 as the absolutevalue of x goes to infinity. So it’s a probability functionthat certainly doesn’t extend all the way to infinity. It just collapses. It’s normalizable.So these are boundstates, and you’re looking for boundstates of this equation. And your difficulty isthat you don’t know psi, and you don’t know E either. So you have to solve a problemin which, if you were thinking oh this is just a plaindifferential equation, give me the value of E.We know the potential, just calculate it. That’s not the way itworks in quantum mechanics, because you need to havenormalizable solutions. So at the end of the day,as will be very clear today, this E gets fixed. You cannot getarbitrary values of E’s. So I want to make a couple ofremarks about this equation. Is that there’s thisthing that can’t happen. Certainly, if v of x is a smoothpotential, then if you observer that the wave functionvanishes at some point, and the derivativeof the wave function vanishes at that samepoint, these two things imply that psi ofx is identically 0.And therefore itmeans that you really are not interested in that. That’s not a solution ofthe Schrodinger equation. Psi equals 0 is obviously solvesthis, but it’s not interesting. It doesn’t representthe particle. So what I claim hereis that, if it happens to be that you’resolving the Schrodinger problem with somepotential that is smooth, you can take derivatives of it. And then you encounter thatthe wave function vanishes at some point, and its slopevanishes at that same point. Then the wave functionvanishes completely.So you cannot havea wave function, a psi of x thatdoes the following. Comes down here. It becomes an inflectionpoint and goes down. This is not allowed. If the wave functionvanishes at some point, then the wave functionis going to do this. It’s going to hit at an angle,because you cannot have that the wave function is 0 and needsthe derivative 0 at the same point. And the reason is simple. I’m not going toprove it now here. It is that you have a secondorder differential equation, and a second orderdifferential equation is completely determinedby knowing the function at the point and the thederivative at the point. And if both are 0s, likethe most trivial kind of initial condition, the onlysolution consistent with this is psi equals 0 everywhere. So this can’t happen. And it’s something goodfor you to remember. If you have to do aplot of a wave function, you should never have this. So this is what we call thenode in a wave function. It’s a place where thewave function vanishes, and the derivativeof the wave function better not vanish at that point. So this is one claim thatit’s not hard to prove, but we just don’t try to do it. And there’s another claim thatI want you to be aware of. That for bound statesin one dimension, in the kind of thing that we’redoing now in one dimension, no degeneracy is possible. What do I mean by that? You will never findtwo bound states of a potential thatare different that have the same energy. It’s just absolutely impossible. It’s a very wonderfulresult, but also I’m not going to prove it here. Maybe it will be given as anexercise later in the course. And it’s discussedin 805 as well. But that’s another statementthat is very important. There’s no degeneracy. Now you’ve looked atthis simple potential, the square well infinite one. And how does it look? You have x going from 0 to a. And the potential is 0. From 0 to a isinfinite otherwise. The particle is boundto stay inside the two walls of this potential. So in here we’ve plotted thepotential as a function of x. Here is x. And the wave functions arethings that you know already. Yes? AUDIENCE: Is it true if you havetwo wells next to each other that there’s still no degeneracyif it’s an infinite barrier? PROFESSOR: If there’stwo wells and there’s an infinite barrierbetween them, it’s like having two universes.So it’s not really aone dimensional problem. If you have an infinitebarrier like two worlds that can talk to each other. So yes, then youwould have degeneracy. It’s like saying you can havehere one atom of hydrogen or something in one energylevel, here another one. They don’t talk to each other. They’re degenerate states. But in general, we’re talkingabout normal potentials that are preferably smooth.Most of these things are trueeven if they’re not smooth. But it’s a little more delicate. But certainly twopotentials that are separated byan infinite barrier is not part of what wereally want to consider. OK so these wave functionsstart with m equals 0, 1, 2, 3, and psi ends of x aresquare root of 2 over a sin n plus 1 pi x over a. Things that you’ve seen already. And En, the energiesare ever growing as a function of the integern that characterizes them.2ma squared. And the thing that younotice is that psi 0 has technically no nodes. That is to say thesewave functions have to varnish at the end, becausethe potential becomes infinite, which means a particlereally can go through. The wave functionhas to be continuous. There cannot be any wavefunction to the left. So it has to vanish here. These things don’tcount as nodes. It is like a bound statehas to vanish at infinity.And that’s not whatwe count the node. A node is somewherein the middle of the range of x wherethe wave function vanishes. So this is the ground state. This is psi zero has no nodes. Psi one would be somethinglike that, has one node. And try the next ones. They have more and more nodes. So psi n has n nodes. And the interesting thingis that this result actually is true for extremelygeneral potentials. You don’t have to justdo the square well to see that the groundstate has no nodes. That first excited state was onenode, and so on and so forth. It’s true in general. This is actually avery nice result, but its difficult to prove. In fact, it’s prettyhard to prove. So there is a nice argument. Not 100% rigorous,but thoroughly nice and really physicalthat I’m going to present to you whythis result is true. So let’s try to do that. So here is the general case.So I’m going to takea smooth v of x. That will be of this kind. The potential, here isx, and this potential is going to be like that. Smooth all over. And I will actuallyhave it that it actually goes to infinity asx goes to infinity. Many of these things arereally not necessary, but it simplifies our life. OK, so here is a resultthat it’s known to be true. If this thing grows toinfinity, the potential never stops growing, you getinfinite number of bound states at fixed energies. One energy, twoenergy, three energy, infinite number of bound states. Number of bound states. That’s a fact. I will not try to prove it. We’ll do it for theharmonic oscillator. We’ll see those infinitenumber of states, but here we can’tprove it easily. Nevertheless, whatI want to argue for you is that thesestates, as the first one, will have no nodes.The second state, thefirst excited state, will have one node. Next will have two nodes,three nodes, four nodes. We want to understand whatis this issue of the nodes. OK. You’re not tryingto prove everything. But we’re trying toprove or understand something that isvery important. So how do we prove this? Or how do we understand thatthe nodes– so there will be an infinite numberof bound states, psi 0, psi 1, psi 2, upto psi n, and it goes on. And psi n has n nodes. All right. So what I’m going to do, inorder to understand this, is I’m going to produce what wewill call screened potentials. Screened potentials. I’m going to select thelowest point of the potential here for convenience. And I’m going to call it xequals 0 is the lowest point. And the screen potentialwill have a parameter a. It’s a potentialwhich is equal to v of x if the absolutevalue of x is less than a. And its infinity if the absolutevalue of x is greater than a. So I come here,and I want to see this is thispotential, v of x, what is the screenedpotential for sum a? Well, I wanted coloredchalk, but I don’t have it. I go mark a here minus a. Here are the pointsbetween x and a. Absolute value of x less than a. Throughout this regionthe screened potential is the potential that you have. Nevertheless, for therest, its infinite. So the screenedpotential is this thing. Is infinite there, andit’s here this thing. So it’s just some potential.You take it a screenand you just see one part of the potential,and let it go to infinity. So that’s a screen potential. So now what I’m goingto do is that I’m going to try toargue that you could try to find the bound stateof the screen potential. Unless you remove thescreen, you will find, as you let a go toinfinity, you will find the bound states ofthe original potential. It’s reasonablethat that’s true, because as youremove the screen, you’re letting more ofthe potential be exposed, and more of thepotential be exposed. And the wave functionseventually die, so as the time that you’re very far away,you affect the wave functions less and less. So that’s the argument. We’re going to tryto argue that we’re going to look atthe bound states of the screened potentials andsee what happened, whether they tell us about the boundstates the original potential. So for this, I’m goingto begin with a screen potential in whicha goes to 0 and say that a is equal toepsilon, very small. So what potential so do I have? A very tiny potential herefrom epsilon to minus epsilon. Now I chose the original pointdown here to be the minimum. So actually, the bottom part ofthe potential is really flat. And if you takeepsilon going to 0, well, the potentialmight do this, but really at the bottom forsufficiently small epsilon, this is an infinite squarewell with psis to epsilon. I chose the minimum so that youdon’t get something like this. If it would be apoint with a slope, this would be an ugly thing. So let’s choose the minimum. And we have the screenpotential here, and that’s it.Now look what we do. We say all right, herethere is a ground state. Very tiny. Goes like that. Vanishes here. Vanishes there. And has no nodes. Very tiny. You know the two 0s arevery close to each other. And now I’m goingto try to increase the value of the screen a. So suppose we’veincreased the screen, and now the potential is here. And now we have a finite screen. Here is the potential. And I look at the wave function. How it looks. Here is psi 0. This ground state psi 0. Well, since this thing in here,the potential becomes infinite, the wave functionstill must vanish here and still must vanish here. Now just for yourimagination, think of this. At this stage, it still moreor less looks like this. Maybe. Now I’m going to ask, as Iincrease, can I produce a node? And look what’s going to happen.So suppose it might happenthat, as you increase, suddenly you produce a node. So here’s what I’m saying here. I’m going to show it here. Suppose up to thispoint, there is no node. But then when I doubleit, when I increase it to twice the size, when I goto screen potential like that, suddenly there is anode in the middle. So if there is anode in the middle, one thing that could havehappened is that you have this. And now look what musthave happened then. As I stretch this, this slopemust have been going down, and down, and down, untilit flips to the other side to produce a node here. It could havehappened on this side, but it’s the same, sothe argument is just done with this side. To produce a node youcould have done somehow the slope here musthave changed sine.But for that to happencontinuously, at some point the this slope must have been 0. But you cannot havea 0 and 0 slope. So this thing can’tflip, can’t do this. Another thing thatcould have happened is that when weare here already, maybe the wave functionlooks like that. It doesn’t flip at theedges, but produces something like that. But the only way thiscan happen continuously, and this potential ischanging continuously, is for this thing atsome intermediate stage, as you keep stretching thescreen, this sort of starts to produce a depression here. And at some point, to gethere it has to do this.But it can’t do this either. It cannot vanish and havederivative like that. So actually, as youstretch the screen, there’s no way toproduce a node. That property forbids it. So by the time you go andtake the screen to infinity, this wave function has no nodes. So that proves it that theground state has no nodes. You could call thisa physicist proof, which means– not inthe pejorative way. It means that it’sreasonable, it’s intuitive, and a mathematician workinghard could make it rigorous. A bad physicist proof isone that is a little sloppy and no mathematician couldfix it and make it work. So I think this is a goodphysics proof in that sense. Probably you canconstruct a real proof, or based on this, avery precise proof. Now look at excited states. Suppose you take now herethis screen very little, and now consider the thirdexcited state, psi three. I’m sorry, we’ll call this psi2 because it has two nodes. Well, maybe I should do psi 1. Psi 1. One node. Same thing. As you increaseit, there’s no way to create anothernode continuously. Because again, you haveto flip at the edges, or you have todepress in the middle. So this one will evolveto a wave function that will have one node inthe whole big potential. Now stayed does thatstate have more energy than the ground state? Well, it certainly begins witha small screen with more energy, because in the square wellpsi 1 has more energy. And that energy shouldbe clear that it’s not going to go below theenergy of the ground state. Why? Because if it went below theenergy of the ground state slowly, at some point forsome value of the screen, it would have the sameenergy as the ground state. But no degeneracy is possiblein one dimensional problems. So that can’t happen. Cannot have that. So it will alwaysstay a little higher.And therefore with one node youwill be a little higher energy. With two nodes willbe higher and higher. And that’s it. That’s the argument. Now, we’ve argued by thiscontinuous deformation process that this potential notonly has these bound states, but this is n nodes and En isgreater than En prime for n greater than n prime. So the more nodes,the more energy. Pretty nice result,and that’s really all I wanted to sayabout this problem. Are there any questions? Any? OK.So what we do now isthe harmonic oscillator. That’s going to keep us busyfor the rest of today’s lecture. It’s a very interesting problem. And it’s a most famous quantummechanics problem in a sense, because it happens to be usefulin many, many applications. If you have anypotential– so what is the characteristic ofthe harmonic oscillator? Harmonic oscillator. Oscillator. Well, the energy operator is psquared over 2m plus, we write, one half m omegasquared x squared where omega is thisomega that you always think of angular velocity,or angular frequency. It’s more likeangular frequency. Omega has units of 1 over time. It’s actually put 2pi overthe period of an oscillation. And this you know fromclassical mechanics. If you have a harmonicoscillator of this form, yeah, it actually oscillateswith this frequency.And E is the energyoperator, and this is the energy of the oscillator. So what defines an oscillator? It’s something in which thepotential energy, this term is v of x. v of x is quadratic in x. That is a harmonic oscillator. Then you arrange theconstants to make sense. This has units ofenergy, because this has units of length squared. 1 over time squared. Length over timeis velocity squared times mass is kinetic energy. So this term hasthe units of energy.And you good with that. And why is this useful? Because actually in any sortof arbitrary potential, rather general potentialat least, whenever you have a minimum wherethe derivative vanishes, then the second derivativeneed not vanish. Then it’s a goodapproximation to think of the potential at the minimumas a quadratic potential. It fits the potentialnicely over a good region. And therefore when you havetwo molecules with a bound or something oscillating,there is a potential. It has a minimum at theequilibrium position. And the oscillationsare governed by some harmonic oscillator. When you have photonsin space time traveling, there is a set ofharmonic oscillators that correspond to photons. Many, many applications. Endless amount of applicationsfor the harmonic oscillator. So we really want tounderstand this system quantum mechanically. And what does that mean? Is that we really wantto calculate and solve the Schrodinger equation.This is our first step inunderstanding the system. There’s going tobe a lot of work to be done even once we havethe solutions of the Schrodinger equation. But the first thingis to figure out what are the energyeigenstates or the solutions of the Schrodingerequation for this problem. So notice that herein this problem there’s an energy quantity. Remember, when you havea harmontonian like that, and people say so what isthe ground state energy? Well, have to find theground state wave function. Have to do things. Give me an hour, I’ll find it. And all that. But if you want anapproximate value, dimensional analysis will do it,roughly what is it going to be. Well, with this constant howdo you produce an energy? Well, you rememberwhat Einstein did, and you know that h baromega has units of energy. So that’s an energy associatedwith Lagrangian energy like quantity. And we expect thatthat energy is going to be the relevant energy.And in fact, we’ll findthat the ground state energy is just one half of that. There’s another quantitythat may be interesting. How about the length? How do you construct alength from these quantities? Well, you can startdoing m omega h bar and put powers and struggle. I hate doing that. I always try to find someway of doing it and avoiding that thing. So I know thatenergies go like h over h squared overm length squared. So I’m going to callthe length a quantity a. So ma squared. That has units of energy. And you should rememberthat because energy is b squared over 2m,and b by De Broglie is h bar over sub lamda. So h bar squared, lambdasquared, and m here, that’s units of energy.So that’s a length. On the other hand,we have another way to construct an energyis with this thing, m omega squared length squared. So that’s also m omegasquared a squared. That’s another energy. So from this equation Ifind that a to the fourth is h squared over msquared omega squared. And it’s a littlecomplicated, so a squared is h bar over m omega. So that’s a length. Length squared.I don’t want to takethe square root. We can leave itfor a moment there. But that’s important becauseof that’s a length scale. And if somebody would askyou in the ground state, how far is thisparticle oscillating, you would say probablyabout a square root of this. Would be a natural answerand probably about right. So OK, energy andunits is very important to begin your analysis. So what is theSchrodinger equation? The Schrodingerequation for this thing is going to be minus hsquared over 2m, d second psi, dx squared plus the potential,one half m omega squared x squared psi is equal E psi.And the big problem is I don’tknow psi and I don’t know E. Now there’s so many elegantways of solving the harmonic oscillator. You will see those next lecture. Allan Adams will be back here. But we all have to gothrough once in your life through the direct, uninspiredmethod of solving it. Because most of the timeswhen you have a new problem, you will not come up witha beautiful, elegant method to avoid solving thedifferential equation. You will have to struggle withthe differential equation. So today we struggle withthe differential equation. We’re going to just do it. And I’m going to do it slowenough and in detail enough that I hope youfollow everything. I’ll just keep acouple of things, but it will be one linecomputations that I will skip. So this equation is some sortof fairly difficult thing. And it’s complicated andmade fairly unpleasant by the presence ofall these constants. What kind of equation is thatwith all these constants? They shouldn’t be there,all this constants, in fact.So this is the first step,cleaning up the equation. We have to clean it up. Why? Because the nicefunctions in life like y double prime is equal tominus y have no units. The derivatives create no units. y has the same units of that,and the solution is sine of x, where x must have no units,because you cannot find the sine of one centimeter. So this thing, we shouldhave the same thing here. No units anywhere. So how can we do that? This is an absolutelynecessary first step. If you’re going to becarrying all these constants, you’ll get nowhere. So we have to clean it up. So what I’m goingto try to see is that look, here ispsi, psi, and psi. So suppose I do thefollowing thing, that I will cleanup the right hand side by dividing by somethingwith units of energy. So I’m going to dothe following way. I’m going to divide allby 1 over h bar omega. And this 2 I’m goingto multiply by 2. So multiply by 2over h bar omega. So what do I achievewith that first step? I achieve thatthese 2s disappear.Well, that’s not too bad. Not that great either, I think. But in the right hand side,this has units of energy. And the right hand side willnot have units of energy. So what do we get here? So we get minus. The h becomes an h alone over–the m disappears– so m omega. The second psi the x squared. The 1/2 disappeared,so m omega over h bar x squared psi equals2 E over h bar omega psi. It looks actuallyquite better already. Should agree with that. It looks a lot nicer NowI can use a name for this. I want to call thisthe dimensionless value of the energy. So a calligraphic e. It has no units. It’s telling me if I find someenergy, that that energy really is this number, this purenumber is how many times bigger is e with respectto h omega over 2. So I’ll write this now as e psi. And look what I have. I have no units here. And I have a psi.And I have a psi. But things haveworked out already. Look, the same factorhere, h over m omega is upside down here. And this factor hasunits of length squared. Length squared times d dlength squared has no units. And here’s 1 overlength squared. 1 over length squaredtimes length squared. So things have worked out. And we can nowsimply say x is going to be equal to au,a new variable.This is going to be your newvariable for your differential equation in which is this thing. And then thisdifferential equation really has cleanedup perfectly well. So how does it look now? Well, it’s all goneactually, because if you have x equals au, d dx bychain rule is 1 over a d du. And to derivativesthis with respect to x it’s 1 over a squaredtimes the d second du squared. And this thing is a squared. So actually youcancel this factor. And when I write xequals to au, you get an a squared times this. And a squared times this is 1.So your differentialequations has become minus thesecond psi du squared, where u is a dimensionlessquantity, because this has units of length, thishas units of length. No units here. You have no units. So minus d seconddu squared plus u squared psi is equal to e psi. Much nicer. This is an equationwe can think about without being distractedby this endless amount of little trivialities. But still we haven’tsolved it, and how are we going to solve this equation? So let’s again thinkof what should happen. Somehow it should happenthat these e’s get fixed. And there is some solutionjust for some values of e’s. It’s not obvious at this stagehow that is going to happen. Yes? AUDIENCE: [INAUDIBLE]. PROFESSOR: Here for example,let me do this term. h bar over m omega is minus,from that equation, a squared. But dx squared is 1 overa squared d du squared.So a squared cancels. And here the x is equala squared times u, so again cancels. OK so what is the problem here? The problem is that most likelywhat is going to go wrong is that this solution forarbitrary values of e’s is going to diverge atinfinity, and you’re never going to be ableto normalize it. So let’s try to understandhow the solution looks as we go to infinity.So this is the firstthing you should do with an equation like that. How does this solutionlook as u goes to infinity? Now we may not be able to solveit exactly in that case either, but we’re going to gaininsight into what’s happening. So here it is. When u goes to infinity,this term, whatever psi is, this term is muchbigger than that, because we’re presumablyworking with some fixed energy that we still don’tknow what it is, but it’s a fixed number and,for you, sufficiently large. This is going to dominate. So the equationthat we’re trying to solve as u goes to infinity,the equation sort of becomes psi double prime– primeis for two derivatives– is equal to u squared psi. OK, so how do we getan idea what solves this is not all that obvious. It’s certainly not a power of u,because when you differentiate the power of u, youlower the degree rather than increase the degree. So what function increasesdegree as you differentiate? It’s not the trivial function. Cannot be a polynomial. If it could beeven a polynomial, if you take two derivatives,it kind cannot be equal to x squared times a polynomial. It’s sort of upside down. So if you think aboutit for a little while, you don’t have anexact solution, but you would imaginethat something like this would do it, an eto the u squared. Because an e tothe u squared, when you differentiate with respectto us, you produce a u down. When you one derivative. When you takeanother derivative, well, it’s more complicated,but one term you will produce another u down.So that probably is quite good. So let’s try that. Let’s try to see if wehave something like that. So I will try something. I’ll try psi equals 2. I’m going to trythe following thing. e to the alpha u squared over2 where alpha is a number. I don’t know how much it is. Alpha is some number. Now could try thisalone, but I actually want to emphasizeto you that if this is the behaviornear infinity, it won’t make any differenceif you put here, for example, somethinglike u to the power k.It will also beroughly a solution. So let’s see that. So for that I haveto differentiate. And let’s see what we get. So we’re trying to see how thefunction behaves far, far away. You might say welllook, probably that alpha should be negative. But let’s see whatthe equation tells us before we put anything in there. So if I do psi prime,you would get what? You would get oneterm that would be alpha u times this u tothe k into the alpha u squared over 2. I differentiatedthe exponential. I differentiatedthe exponential. And then you wouldget a term where you differentiate the power. So you get ku to the kminus 1 into the alpha u squared over 2. If you take a secondderivative, well, I can differentiate theexponential again, so I will get alphau now squared, because each derivativeof this exponent produces a factor of alpha u.U to the k into thealpha u squared over 2. And a couple more terms thatthey all have less powers of u, because this termhas u to the k plus– already has u to the k plus 1. And this has u to the k minus 1. They differ by two powers of u. So for illustration,please, if you want, do it. Three lines, you should skipthree lines in your notebook if you’re taking notesand get the following. No point in me doingthis algebra here. Alpha u squared over 2. Because actually it’snot all that important. Over alpha 1 over u squaredplus k minus 1 over alpha squared 1 over u to the fourth. That’s all you get. Look, this is alphasquared u squared times psi times these things. 1 plus 2 k plus 1 overalpha 1 over u squared. So when u goes toinfinity, your solution works, because thesething’s are negligible. So you get a numbertimes u squared.That is the equation you aretrying to solve up there. And therefore, you getthat the equation if alpha squared is equal to 1. And that means and really thatalpha can be plus minus 1. And roughly thissolution near infinity, probably there’s two solutions. This is a second orderdifferential equation, so even near infinity thereshould be two solutions. So we expect as u goesto infinity psi of u will be some constant Atimes u to the k times e to the minus u squared over 2. That’s where alphaequal minus 1. Plus Bu to the k into theplus u squared over 2. And what is k? Well, we don’t know what is k.It seems to work for all k. That may seem a littleconfusing now, but don’t worry. We’ll see other thingshappening here very soon. So look at what has happened. We’ve identified that mostlikely your wave function is going to look likethis at infinity. So we’re going to want tothis part not to be present. So presumably we’re goingto want a solution that just has this, becausethis is normalizable. The integral of any power timesa Gaussian is convergence.So this can be normalized. The Gaussian falls sofast that any power can be integratedagainst a Gaussian. Any power however bigdoesn’t grow big enough to compensate a Gaussian. It’s impossible tocompensate a Gaussian. So we hope for this. But we want totranslate what we’ve learned into sometechnical advantage in solving thedifferential equation, because, after all, we wantedbe insight how it looks far way, but we wanted to solvethe differential equation. So how can we usethis insight we now have to simplify the solutionof the differential equation? The idea is to changevariables a little bit.So write psi of u to beequal to h of u times e to the minus u squared over 2. Now you’re going to saywait, what are you doing? Are you making anapproximation now that this is what isgoing to look far away? Or what are you putting there? I’m not makingany approximation. I’m just saying whateverpis is, it can always be written in this way. Why? Because if you havea psi of u, you can write it as psi of uover e to the minus u squared over 2 times e minusu squared over 2.Very trivially thiscan always be done. As long as we saythat h is arbitrary, there’s nothing,no constraint here. I have not assumeanything, nothing. I’m just hoping that I have adifferential equation for psi. That because this is a veryclever factor, the differential equation for h will be simpler. Because part of the dependencehas been taken over. So maybe h, for example, couldbe now a polynomial solution, because this producthas been taken care.So the hope is that bywriting this equation it will become anequation for h of u, and that equationwill be simpler. So will it be simpler? Well, here again thisis not difficult. You’re supposed to pluginto equation one– this is the equationone– plug into one. I won’t do it. It’s three lines ofalgebra to plug into one and calculate theequation for h of u. You should do it. It’s the kind of thing thatone should do at least once. So please do it. It’s three, four lines. It’s not long. But I’ll just write the answer. So by the time yousubstitute, of course, the e to the minusu squared over 2 is going to cancelfrom everywhere. It’s very here. You just need totake two derivatives, so it becomes a secondorder differential equation. And indeed, it becomesa tractable differential equation. The second h, du squared minus2u dh du plus e minus 1 h equals 0. OK, that is our equation now. So now we face the problemfinally solving this equation. So before we start,maybe there’s some questions of whatwe’ve done so far. Let’s see. Any questions? Yes? AUDIENCE: Do you haveright there in the middle would be– this equationis linear, so can we just [INAUDIBLE] minus u squared over2 and you stuck it to that u to the k. PROFESSOR: It’s here? This thing? AUDIENCE: Yeah. Could you then just power serieswhat’s going on at 0 with those u to the k terms [INAUDIBLE]? PROFESSOR: No. This is the behavioras u goes to infinity.So I actually don’t knowthat the function near 0 is going to behavelike u to the k. We really don’t know. It suggest to you thatmaybe the solution is going to be near 0 u tothe k times some polynomial or something like that. But it’s not that, becausethis analysis was just done at infinity. So we really have no informationstill what’s going on near 0. Other questions? Yes? AUDIENCE: So is k some arbitrarynumber or is it an integer? PROFESSOR: At this moment,actually, it doesn’t matter. Is that right? Doesn’t matter. The analysis that we did heresuggests it could be anything. That’s why I just didn’tput it into h or u. I didn’t put it becausewould be strange to put here a u to the k.I wouldn’t knowwhat to make of it. So at this moment,the best thing to say is we don’t know what it is,and maybe we’ll understand it. And we will. In a few seconds, we’llsort of see what’s going on. OK, so how does onesolve this equation? Well, it’s not atrivial equation, again. But it can be solvedby polynomials, and we’ll see that. But the way wesolve this equation is by a power series expansion. Now you could doit by hand first, and I did it whenI was preparing the lecture yesterday. I said I’m going tojust write h of u equal a constant a0 plusa1u plus a2u squared plus a3u cubed. And I plugged it in here. And I just did thefirst few terms and start to see what happened. And I found aftera little thinking that a2 is determinedby a0, and a3 is determined by a1once you substitute.It’s not the obvious when youlook at this, but that happens. So when you face aproblem like that, don’t go high powerto begin with. Just try a simple seriesand see what happens. And you see a little pattern. And then you can do a moresophisticated analysis. So what would be a moresophisticated analysis? To write h of uequal the sum from j equals 0 to infinityaju to the j. Then if you take aderivative, because we’re going to need thederivative, dh du would be the sum fromj equals 0 to infinity. j times aju to the j minus 1. You would say thatdoesn’t look very good because for j equals0 you have 1 over u. That’s crazy. But indeed for j equals 0,the j here multiplies it and makes it 0. So this is OK. Now the term that we actuallyneed is minus 2u dh du. So here minus 2u dh du wouldbe equal to the sum from j equals 0 to infinity, and Iwill have minus 2jaju to the j.The u makes this j minus 1 j,and the constant went there. So here is so far h. Here is this otherterm that we’re going to need for thedifferential equation. And then there’s thelast term that we’re going to need for thedifferential equation, so I’m going to go here. So what do we getfor this last term. We’ll have to takea second derivative. So we’ll take– hprime was there, so d second h du squaredwill be the sum from j equals 0 of j times j minus1 aju to the j minus 2. Now you have to rewrite thisin order to make it tractable. You want everythingto have u to the j’s. You don’t want actually tohave u to the j minus 2. So the first thingthat you notice is that this sum actually beginswith 2, because for 0 and 1 it vanishes. So I can write j times jminus 1 aj u to j minus 2. Like that. And then I can say let jbe equal to j prime plus 2. Look, j beginswith 2 in this sum.So if j is j prime plus 2,j prime will begin with 0. So we’ve shifted the sum so it’sj prime equals 0 to infinity. And whenever I have a j Imust put j prime plus 2. So j prime plus 2. j prime plus 1 aj primeplus 2 u to the j prime. Wherever I had j, Iput j prime plus 2. And finally you say j orj prime is the same name, so let’s call it j. j equals 0. j plus 2. j plus 1. aj plus 2 uj. So we got the seriesexpansion of everything, so we just plug into thedifferential equation. So where is thedifferential equation? It’s here. So I’ll plug it in. Let’s see what we get. We’ll get some from jequals 0 to infinity. Let’s see the secondderivative is here. j plus 2 times j plus 1 ajplus 2 uj, so I’ll put it here. So that’s this secondderivative term. Now this one. It’s easy. Minus 2j aj and the uj is there. So minus 2jaj. Last term is just eminus 1, because it’s the function thistimes aj as well. That’s h. And look, this wholething must be 0. So what you learn is thatthis coefficient must be 0 for every value of j. Now it’s possible to–here is aj and aj, so it’s actuallyone single thing. Let me write it here. j plus 2 times j plus1 aj plus 2 minus 2j plus 1 minus e aj uj. I think I got it right. Yes. And this is the same sum. And now, OK, it’s a lot ofwork, but we’re getting there. This must be 0. So actually that solves foraj plus 2 in terms of aj. What I had told you that youcan notice in two minutes if you try it a little. That a2 seems to bedetermined by a0. And a3 seems to bedetermined by a2.So this is sayingthat aj plus 2 is given by 2j plus 1 minus eover j plus 2 j plus 1 aj. A very nice recursive relation. So indeed, if youput the value of a0, it will determine for you a2,a4, a6, a8, all the even ones. If you put the value of a1, itwill determine for you a3, a5. So a solution is determined byyou telling me how much is a0, and telling me how much is a1. Two constants, two numbers. That’s what you expect froma second order differential equation. The value of thefunction at the point, the derivative at a point. In fact, you arelooking at a0 and a1 as the two constants thatwill determine a solution.And this is the value of h at 0. This is thederivative of h at 0. So we can now writethe following facts about the solutionthat we have found. So what do we know? That solutions fixedby giving a0 and a1. That correspond to thevalue of the function at 0 and the derivativeof the function at 0. And this gives one solution. Once you fix a0, you get a2, a4. And this is an evensolution, because it has only even powers. And then from a1, you fixeda3, a5, all the other ones with an odd solution. OK. Well, we solve thedifferential equation, which is really,in a sense, bad, because we were expectingthat we can only solve it for some values of the energy. Moreover, you have a0,you get a2, a4, a6, a8. This will go on foreverand not terminate. And then it will be aninfinite polynomial that grows up and doesn’tever decline, which is sort ofcontradictory with the idea that we had beforethat near infinity the function was going to besome power, some fixed power, times this exponential.So this is what we’re lookingfor, this h function now. It doesn’t looklike a fixed power. It looks like it goes forever. So let’s see whathappens eventually when the coefficient, thevalue of the j index is large. For large j. aj plus 2 is roughly equalto, for large a, whatever the energy is, sufficientlylarge, the most important here is the 2j here, the j and the j. So you get 2 over j aj. So roughly for large j,it behaves like that. And now you have to askyourself the question, if you have a power seriesexpansion whose coefficients behave like that, howbadly is it at infinity? How about is it? You know it’s thepower series expansion because your h was allthese coefficients.And suppose theybehave like that. They grow in that wayor decay in this way, because they’re decaying. Is this a solutionthat’s going to blow up? Or is it not going to blow up? And here comes animportant thing. This is pretty badbehavior, actually. It’s pretty awful behavior. So let’s see that. That’s pretty bad. How do we see that? Well you could do itin different ways, depending on whetheryou want to derive that this is a badbehavior or guess it. I’m going to guess something. I’m going to look at howdoes e to the u squared behave as a power series. Well, you know as apower series exponential is 1 over n u squared to the n. Here’s n factorial. n equals 0 to infinity. Now these two n’s, u to the2n, these are all even powers. So I’m going tochange letters here, and I’m going to work with jfrom 0, 2, 4, over the evens. So I will write u to the j here. And that this correct, becauseyou produce u to the 0, u to the 2, u to thefourth, these things.And j is really 2n,so here you will have one over jover 2 factorial. Now you might say, j over2, isn’t that a fraction? No, it’s not a fraction,because j is even. So this is a nice factorial. Now this is thecoefficient, cj u to the j. And let’s see how thiscoefficients vary. So this cj is 1 overj over 2 factorial. What is cj plus 2 over cj? Which is the analogueof this thing. Well, this would be 1 overj plus 2 over 2 factorial. And here is up there,so j over 2 factorial. Well, this has one morefactor in the denominator than the numerator. So this is roughlyone over j over 2 plus 1, the last value of this. This integer is justone bigger than that. Now if j is large,this is roughly 1 over j over 2,which is 2 over j. Oh, exactly that stuff. So it’s pretty bad. If this seriesgoes on forever, it will diverge likee to the u squared.And your h will be like e to theu squared with e to the minus u squared over 2 is goingto be like e to the plus. u squared over 2 is goingto go and behave this one. So it’s going to doexactly the wrong thing. If this seriesdoesn’t terminate, we have not succeeded. But happily, theseries may terminate, because the j’s are integers. So maybe for some energiesthat are integers, it terminates, andthat’s a solution.The only way to get a solutionis if the series terminates. The only way it canterminate is that the e is some odd number over here. And that will solve the thing. So we actually need to do this. This shows the energy. You found why it’s quantized. So let’s do it then. We’re really donewith this in a sense. This is the most importantpoint of the lecture, is that the seriesmust terminate, otherwise it willblow up horrendously. If it terminatesas a polynomial, then everything is good. So to terminate you can choose2j plus 1 minus e to be 0. This will make ajplus 2 equal to 0. And your solution, yourh of u, will begin. aj will be the lastone that is nonzero, so it will be ajtimes u to the j, and it will go down like ajminus 2 u to the j minus 2. It will go down in steps of2, because this recursion is always by steps of two. So that’s it. That’s going to be the solutionwhere these coefficients are going to be fixed bythe recursive relation, and we have this.Now most peoplehere call j equal n. So let’s call it n. And then we have 2nplus 1 minus e equals 0. And h of u would be an u tothe n plus all these things. That’s the h. The full solution is htimes e to the minus u squared over 2 as we will see. But recall what e was. e here is 2n plus 1. But he was the true energydivided by h omega over two.That was long ago. It’s gone. Long gone. So what have youfound therefore? That the energy,that’ we’ll call en, the energy of thenth solution is going to be h omegaover 2 2n plus 1. So it’s actually h omega, andpeople write it n plus 1/2. Very famous result. The nth level of the harmonicoscillator has this energy. And moreover, theseobjects, people choose these– yousee the constants are related by steps of two. So just like you could startwith a0, or a1 and go up, you can go down.People call these functionsHermite functions. And they fix the notation sothat this an is 2 to the n. They like it. It’s a nice normalization. So actually h ofn is what we call the Hermite function of u sub n. And it goes like 2 to then u to the n plus order u to the n minus 2 plusn minus 4, and it goes on and on like that. OK, a couple thingsand we’re done. Just for reference,the Hermite polynomial, if you’re interestedin it, is the one that solves this equation.And the Hermitesub n corresponds to e sub n, which is 2n plus 1. So the Hermite solutionfrom that the equation is that the Hermite polynomialsatisfies this minus 2u d Hn du plus 2n. Because en is 2n plus 1. So it’s 2n Hn equals 0. That’s the equation forthe Hermite polynomial, and interesting thing to know. Actually, if you want togenerate the efficiently the Hermite polynomials, there’ssomething called the generating function. e to the minus zsquared plus 2zu. If you expand it ina power series of z, it actually gives youn equals 0 to infinity. If it’s a power series of z,it will be some z to the n’s. You can put a factor heren, and here is Hn of u. So you can use yourmathematic program and expand this in powers of z. Collect the various powers ofu that appear with z to the n, and that’s Hn It’s the mostefficient way of generating Hn And moreover, if you wantto play in mathematics, you can show thatsuch definition of Hn satisfies this equation.So it produces the solution. So what have we found? Our end result is the following. Let me finish with that here. We had this potential, and thefirst energy level is called E0 and has energy h omega over 2. The next energy is E1. It has 3/2 h omega. Next one is E2 5/2 h omega. This polynomial isnth degree polynomial. So it has n zeros,therefore n nodes. So these wave functions willhave the right number of nodes. E0, the psi 0,will have no nodes. When you have psi 0, theHn becomes a number for n equals zero. And the whole solutionis the exponential of u squared over 2. The whole solution, infact, is, as we wrote, psi n Hn of u e to theminus u squared over 2. In plain English,if you use an x, it will be Hn u with x overthat constant a we had. And you have minus xsquared over 2a squared. Those are your eigenfunctions. These are the solutions. Discrete spectrum, evenlyspaced, the nicest spectrum possible. All the nodes are there. You will solve this in amore clever way next time.[APPLAUSE] .
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