The followingcontent is provided under a CreativeCommons license. Your support will help MITOpenCourseWare continue to offer high qualityeducational resources for free. To make a donation or toview additional materials from hundreds of MIT courses,visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: So,finally, before I get started on the newstuff, questions from the previous lectures? No questions? Yeah. AUDIENCE: I have a question. You might have saidthis last time, but when is the first exam? PROFESSOR: Ah, excellent. Those will be posted on theStellar page later today. Yeah. AUDIENCE: OK, so we’reassociating operators with observables, right? PROFESSOR: Yes. AUDIENCE: AndProfessor [? Zugoff ?] mentioned that wheneverwe have done a wave function with anoperator, it collapses. PROFESSOR: OK, so let merephrase the question. This is a very valuablequestion to talk through. So, thanks for asking it. So, we’ve previously observedthat observables are associated with operators–and we’ll review that in more detailin a second– and the statement wasthen made, does that mean that acting on a wavefunction with an operator is like measuringthe observable? And it’s absolutelyessential that you understand that acting on a wavefunction with an operator has nothing whatsoever to dowith measuring that associated observable.Nothing. OK? And we’ll talk aboutthe relationship and what those things mean. But here’s a verytempting thing to think. I have a wave function. I want to know the momentum. I will thus operate withthe momentum operator. Completely wrong. So, before I even tell youwhat the right statement is, let me just get thatout of your head, and then we’ll talk throughthat in much more detail over the next lecture. Yeah. AUDIENCE: Why doesn’t itcollapse by special relativity? PROFESSOR: We’redoing everything nonrelativistically. Quantum Mechanicsfor 804 is going to be a universe in whichthere is no relativity. If you ask me that moreprecisely in my office hours, I will tell you arelativistic story. But it doesn’t violateanything relativistic. At all. We’ll talk about that– justto be a little more detailed– that will be a veryimportant question that we’ll deal with in the last twolectures of the course, when we come back to Bell’sinequality and locality. Other questions? OK. So, let’s get started. So, just to review where we are.In Quantum Mechanicsaccording to 804, our first pass at thedefinition of quantum mechanics is that the configuration ofany system– and in particular, think about a singlepoint particle– the configurationof our particle is specified by givinga wave function, which is a function whichmay depend on time, but a function of position. Observables– and this isa complete specification of the state of the system. If I know the wavefunction, I neither needed nor have access toany further information about the system.All the information specifyingthe configuration system is completely containedin the wave function. Secondly, observablesin quantum mechanics are associated with operators. Something you canbuild an experiment to observe or to measure isassociated with an operator. And by an operator, Imean a rule or a map, something that tells youif you give me a function, I will give you adifferent function back. OK? An operator is just athing which eats a function and spits out another function. Now, operators– which Iwill denote with a hat, as long as I can rememberto do so– operators come– and in particular, thekinds of operators we’re going to care about, linear operators,which you talked about in detail last lecture–linear operators come endowed with a natural setof special functions called Eigenfunctions withthe following property.Your operator, actingon its Eigenfunction, gives you that same functionback times a constant. So, that’s a veryspecial generically. An operator will takea function and give you some other randomfunction that doesn’t look all like theoriginal function. It’s a very special thing togive you the same function back times a constant. So, a useful thingto think about here is just in the caseof vector spaces. So, I’m going to considerthe operation corresponding to rotation around thezaxis by a small angle. OK? So, under rotation aroundthe zaxis by a small angle, I take an arbitrary vectorto some other stupid vector.Which vector is completelydetermined by the rule? I rotate by a smallamount, right? I take this vector andit gives me this one. I take that vector,it gives me this one. Everyone agree with that? What are the Eigenvectorsof the rotation by a small anglearound the zaxis? AUDIENCE: [INAUDIBLE] PROFESSOR: Yeah, it’sgot to be a vector that doesn’t change its direction. It just changes by magnitude. So there’s one, right? I rotate.And what’s its Eigenvalue? AUDIENCE: One. PROFESSOR: One, becausenothing changed, right? Now, let’s consider thefollowing operation. Rotate by small angleand double its length. OK, that’s a different operator. I rotate and Idouble the length. I rotate and Idouble the length. I rotate and Idouble the length. Yeah, so what’s the Eigenvalueunder that operator? AUDIENCE: Two. PROFESSOR: Two. Right, exactly. So these are a veryspecial set of functions. This is the same idea, butinstead of having vectors, we have functions. Questions? I thought I saw a hand pop up. No? OK, cool. Third, superposition. Given any two viablewave functions that could describeour system, that could specify states orconfigurations of our system, an arbitrary superposition ofthem– arbitrary linear sum– could also be a validphysical configuration. There is also astate corresponding to being in an arbitrary sum.For example, if we know thatthe electron could be black and it could bewhite, it could also be in an arbitrary superpositionof being black and white. And that is a statement inwhich the electron is not black. The electron is not white. It is in thesuperposition of the two. It does not havea definite color. And that is exactlythe configuration we found inside our apparatusin the first lecture. Yeah. AUDIENCE: Are those PhiAarbitrary functions, or are they supposedto be Eigenfunctions? PROFESSOR: Excellent. So, in general thesuperposition thank you. It’s an excellent question. The question was are thesePhiAs arbitrary functions, or are they specificEigenfunctions of some operator? So, the superpositionprinciple actually says a very general thing. It says, given any twoviable wave functions, an arbitrary sum, anarbitrary linear combination, is also a viable wave function.But here I want to marksomething slightly different. And this is why I chosethe notation I did. Given an operatorA, it comes endowed with a special set of functions,its Eigenfunctions, right? We saw the last time. And I claimed the following. Beyond just the usualsuperposition principle, the set of Eigenfunctionsof operators corresponding to physicalobservables– so, pick your observable, like momentum. That corresponds to an operator. Consider theEigenfunctions of momentum. Those we know what those are. They’re plane waves withdefinite wavelength, right? u to the ikx. Any function can beexpressed as a superposition of those Eigenfunctions ofyour physical observable. We’ll go over this inmore detail in a minute. But here I want to emphasizethat the Eigenfunctions have a special property that– forobservables, for operators corresponding to observables–the Eigenfunctions form a basis. Any function can be expandedas some linear combination of these basis functions,the classic example being the Fourier expansion. Any function, anyperiodic function, can be expanded as a sumof sines and cosines, and any functionon the real line can be expanded as a sum ofexponentials, e to the ikx.This is the same statement. The Eigenfunctionsof momentum are what? e to the ikx. So, this is the same thatan arbitrary function– when the observableis the momentum, this is the statement thatan arbitrary function can be expanded as a superposition,or a sum of exponentials, and that’s the Fourier theorem. Cool? Was there a question? AUDIENCE: [INAUDIBLE] PROFESSOR: OK, good. Other questions on these points? So, these should not yet betrivial and obvious to you. If they are, then that’sgreat, but if they’re not, we’re going to beworking through examples for the next severallectures and problem sets.The point now is to give youa grounding on which to stand. Fourth postulate. What these expansioncoefficients mean. And this is alsoan interpretation of the meaning ofthe wave function. What these expansioncoefficients mean is that the probability thatI measure the observable to be a particular Eigenvalueis the norm squared of the expansion coefficient. OK? So, I tell you thatany function can be expanded as a superpositionof plane waves– waves with definite momentum–with some coefficients. And those coefficientsdepend on which function I’m talking about. What these coefficientstell me is the probability that I will measure themomentum to be the associated value, the Eigenvalue. OK? Take that coefficient,take its norm squared, that gives me the probability. How do we compute theseexpansion coefficients? I think Barton didn’tintroduce to you this notation, but he certainly told you this. So let me introduce to you thisnotation which I particularly like. We can extract theexpansion coefficient if we know the wave functionby taking this integral, taking the wavefunction, multiplying by the complex conjugate ofthe associated Eigenfunction, doing the integral. And that notation is thisround brackets with Phi A and Psi is my notationfor this integral. And again, we’ll still seethis in more detail later on. And finally we havecollapse, the statement that, if we go about measuringsome observable A, then we will always, alwaysobserve precisely one of the Eigenvaluesof that operator. We will never measureanything else. If the Eigenvalues are one,two, three, four, and five, you will never measurehalf, 13 halves. You will alwaysmeasure an Eigenvalue.And upon measuringthat Eigenvalue, you can be confident that that’sthe actual value of the system. I observe that it’sa white electron, then it will remain white if Isubsequently measure its color. What that’s telling you isit’s no longer a superposition of white and black,but it’s wave function is that correspondingto a definite value of the observable. So, somehow the processof measurement– and this is a disturbing statement, towhich we’ll return– somehow the process of measuringthe observable changes the wave function from ourarbitrary superposition to a specific Eigenfunction,one particular Eigenfunction of the operator we’re measuring. And this is called thecollapse of the wave function. It collapses frombeing a superposition over possible states tobeing in a definite state upon measurement. And the definitestate is that state corresponding to the valuewe observed or measured. Yeah. AUDIENCE: So, when thewave function collapses, does it instantaneously notbecome a function of time anymore? Because originallywe had Psi of (x,t).PROFESSOR: Yeah, that’sa really good question. So I wrote this onlyin terms of position, but I should moreprecisely write. So, the question was, does thishappen instantaneously, or more precisely, does it ceaseto be a function of time? Thank you. It’s very good question. So, no, it doesn’t ceaseto be a function of time. It just says thatPsi at x– what you know upon doingthis measurement is that Psi, as a functionof x, at the time which I’ll call T star, at whatyou’ve done the measurement is equal to this wave function. And so that leaves us withthe following question, which is another way of askingthe question you just asked. What happens next? How does the systemevolve subsequently? And at the very endof the last lecture, we answered that–or rather, Barton answered that– by introducingthe Schrodinger equation.And the Schrodinger equation,we don’t derive, we just posit. Much like Newtonposits f equals ma. You can motivate it,but you can’t derive it. It’s just what we mean bythe quantum mechanical model. And Schrodinger’s equationsays, given a wave function, I can determine the timederivative, the time rate of changes ofthat wave function, and determine itstime evolution, and its time derivative,its slope– its velocity, if you will– is one upon I hbar, the energy operator acting on that wave function.So, suppose we measurethat our observable capital A takes the valueof little a, one of the Eigenvalues ofthe associated operators. Suppose we measurethat A equals little a at some particularmoment T start. Then we know thatthe wave function is Psi of x at thatmoment in time. We can then computethe time derivative of the wave functionat that moment in time by acting on this wavefunction with the operator e hat, the energy operator.And we can then integrate thatdifferential equation forward in time and determine howthe wave function evolves. The point of today’slecture is going to be to study how timeevolution works in quantum mechanics, and to lookat some basic examples and basic strategies forsolving the time evolution problem in quantum mechanics. One of the great surprisesin quantum mechanics– hold on just one sec– one ofthe real surprises in quantum mechanics is thattime evolution is in a very specific sensetrivial in quantum mechanics.It’s preposterously simple. In particular, time evolution isgoverned by a linear equation. How many of you have studieda classical mechanical system where the time evolution isgoverned by a linear equation? Right. OK, all of you. The harmonic oscillator. But otherwise, not at all. Otherwise, the equationsin classical mechanics are genericallyhighly nonlinear. The time rate of changeof position of a particle is the gradient of the force,and the force is generally some complicatedfunction of position. You’ve got some capacitorsover here, and maybe some magnetic field.It’s very nonlinear. Evolution in quantummechanics is linear, and this is goingto be surprising. It’s going to lead to somesurprising simplifications. And we’ll turnback to that, but I want to put that yourmind like a little hook, that that’s somethingyou should mark on to as different fromclassical mechanics. And we’ll come back to that. Yeah. AUDIENCE: If a particleis continuously observed as a notevolving particle? PROFESSOR: That’san awesome question. The question is, look,imagine I observe– I’m going to paraphrase–imagine I observe a particle and I observe that it’s here. OK? Subsequently, its wave functionwill evolve in some way– and we’ll actually study thatlater today– its wave function will evolve in someway, and it’ll change.It won’t necessarily bedefinitely here anymore. But if I just keep measuring itover and over and over again, I just keep measureit to be right there. It can’t possibly evolve. And that’s actuallytrue, and it’s called the Quantum Zeno problem. So, it’s the observationthat if you continuously measure a thing,you can’t possibly have its wave functionevolve significantly. And not only is it a cuteidea, but it’s something people do in the laboratory. So, Martin– well, OK. People do it in alaboratory and it’s cool. Come ask me and I’ll tellyou about the experiments.Other questions? There were a bunch. Yeah. AUDIENCE: So after you measure,the Schrodinger equation also gives you theevolution backwards in time? PROFESSOR: Oh, crap! Yes. That’s such a good question. OK. I hate it when peopleask that at this point, because I had tothen say more words. That’s a very good question. So the question goes like this. So this was going to be apunchline later on in the in the lecture but you stolemy thunder, so that’s awesome. So, here’s the deal. We have a rule for timeevolution of a wave function, and it has somelovely properties. In particular– let me talkthrough this– in particular, this equation is linear. So what properties does it have? Let me just– I’mgoing to come back to your questionin just a second, but first I want to set it upso we have a little more meat to answer yourquestion precisely.So we note some propertiesof this equation, this time evolution equation. The first is that it’sa linear equation. The derivative ofa sum of function is a sum of the derivatives. The energy operator’sa linear operator, meaning the energy operatoracting on a sum of functions is a sum of the energy operatoracting on each function. You guys studied linearoperators in your problem set, right? So, these are linear.What that tells youis if Psi 1 of x and t solves the Schrodinger equation,and Psi 2 of x and t– two different functions ofposition in time– both solve the Schrodinger equation,then any combination of them– alpha Psi 1 plus Beta Psi 2–also solves– which I will call Psi, and I’ll make ita capital Psi for fun– solves the Schrodingerequation automatically. Given two solutions ofthe Schrodinger equation, a superposition of them–an arbitrary superposition– also solves theSchrodinger equation. This is linearity. Cool? Next property. It’s unitary. What I mean by unitary is this. It concerns probability. And you’ll give aprecise derivation of what I mean byunitary and you’ll demonstrate that, in fact,Schrodinger evolution is unitary on yournext problem set.It’s not on the current one. But what I mean by unitary isthat conserves probability. Whoops, that’s an o. Conserves probability. IE, if there’s anelectron here, or if we have an object, apiece of chalk– which I’m treating as a quantummechanical point particle– it’s described bythe wave function. The integral, the probabilitydistribution over all the places it could possiblybe had better be one, because it had better besomewhere with probability one. That had betternot change in time. If I solve the Schrodingerequation evolve the system forward for half anhour, it had better not be the case that thetotal probability of finding theparticle is one half. That means thingsdisappear in the universe. And much as my sockswould seem to be a counter example of that,things don’t disappear, right? It just doesn’t happen. So, quantum mechanicsis demonstrably– well, quantum mechanicsis unitary, and this is a demonstrably gooddescription of the real world.It fits all the observationswe’ve ever made. No one’s ever discoveredan experimental violation of unitarity ofquantum mechanics. I will note that there isa theoretical violation of unitarity in quantummechanics, which is dear to my heart. It’s called the Hawking Effect,and it’s an observation that, due quantum mechanics, blackholes in general relativity– places from which lightcannot escape– evaporate. So you throw stuff andyou form a black hole. It’s got a horizon. If you fall throughthat horizon, we never see you again. Surprisingly, a black hole’sa hot object like an iron, and it sends off radiation. As it sends off radiation,it’s losing its energy. It’s shrinking. And eventually it will,like the classical atom, collapse to nothing. There’s a quibblegoing on right now over whether it reallycollapses to nothing, or whether there’sa little granule nugget of quantum goodness. [LAUGHTER] We argue about this. We get paid to argue about this. [LAUGHTER] So, but here’s the funny thing. If you threw in a dictionary andthen the black hole evaporates, where did the informationabout what made the black hole go if it’s just thermalradiation coming out? So, this is aclassic calculation, which to a theorist says, ah ha! Maybe unitarity isn’t conserved.But, look. Black holes, theorists. There’s noexperimental violation of unitarity anywhere. And if anyone ever didfind such a violation, it would shatter the basictenets of quantum mechanics, in particular theSchrodinger equation. So that’s something we wouldlove to see but never have. It depends on yourpoint of view. You might hate to see it. And the third– andthis is, I think, the most important– is thatthe Schrodinger evolution, this is a time derivative. It’s a differential equation. If you know theinitial condition, and you know the derivative,you can integrate it forward in time. And they’re existence anduniqueness theorems for this. The system is deterministic. What that meansis that if I have complete knowledge of thesystem at some moment in time, if I know the wave functionat some moment in time, I can determineunambiguously the wave function in all subsequentmoments of time.Unambiguously. There’s no probability, there’sno likelihood, it’s determined. Completely determined. Given full knowledge now, Iwill have full knowledge later. Does everyone agreethat this equation is a deterministicequation in that sense? Question. AUDIENCE: It’s also local? PROFESSOR: It’s all– well, OK. This one happensto be– you need to give me a betterdefinition of local. So give me a definitionof local that you want. AUDIENCE: The time evolutionof the wave function happens only at apoint that depends only on the value of the derivativesof the wave function and its potentialenergy at that point.PROFESSOR: No. Unfortunately,that’s not the case. We’ll see counterexamples of that. The wave function–the energy operator. So let’s think aboutwhat this equation says. What this says is thetime rate of change of the value of the wavefunction at some position and some moment in time is theenergy operator acting on Psi at x of t. But I didn’t tell you whatthe energy operator is. The energy operatorjust has to be linear. But it doesn’t have to be–it could know about the wave function everywhere. The energy operator’s a mapthat takes the wave function and tells you whatit should be later. And so, at this level there’snothing about locality built in to the energyoperator, and we’ll see just how bad that can be. So, this is relatedto your question about special relativity, andso those are deeply intertwined. We don’t have thatproperty here yet. But keep that in yourmind, and ask questions when it seems to come up. Because it’s a very,very, very important question when we talkabout relativity.Yeah. AUDIENCE: Are postulatessix and three redundant if the Schrodinger equationhas superposition in it? PROFESSOR: No. Excellent question. That’s a very good question. The question is, look, there’spostulate three, which says, given any two wave functionsthat are viable wave functions of thesystem, then there’s another state whichis a viable wave function at some moment in time,which is also a viable wave function. But number six, the Schrodingerequation– or sorry, really the linearityproperty of the Schrodinger equation– so it needs to bethe case for the Schrodinger question, but it sayssomething slightly different. It doesn’t just say that anyany plausible or viable wave function and anothercan be superposed. It says that, specifically,any solution of the Schrodinger equation plus any other solutionof the Schrodinger equation is again theSchrodinger operation.So, it’s a slightlymore specific thing than postulate three. However, your question isexcellent because could it have been that theSchrodinger evolution didn’t respect superposition? Well, you could imaginesomething, sure. We could’ve done adiffer equation, right? It might not have been linear. We could have had thatSchrodinger equation was equal to dt Psi. So imagine this equation. How do we have blown linearitywhile preserving determinism? So we could have added plus, Idon’t know, PSI squared of x. So that would now bea nonlinear equation. It’s actually refer to as thenonlinear Schrodinger equation. Well, people meanmany different things by the nonlinearSchrodinger equation, but that’s a nonlinearSchrodinger equation. So you could certainlywrite this down. It’s not linear.Does it violatethe statement three that any two statesof the system could be superposed togive another viable state at a moment in time? No, right? It doesn’t directly violate. It violates the spirit of it. And as we’ll seelater, it actually would cause dramatic problems. It’s something we don’tusually emphasize– something I don’t usually emphasizein lectures of 804, but I will makea specific effort to mark the places wherethis would cause disasters. But, so this is actuallya logically independent, although morally–and in some sense is a technically related pointto the superposition principle number three. Yeah. AUDIENCE: For postulate three,can that sum be infinite sum? PROFESSOR: Absolutely. AUDIENCE: Can youdo bad things, then, like creating discontinuouswave functions? PROFESSOR: Oh yes.Oh, yes you can. So here’s the thing. Look, if you have two functionsand you add them together– like two smoothcontinuous functions, you add them together–what do you get? You get another smoothcontinuous function, right? Take seven. You get another. But if you take aninfinite number– look, mathematicians are sneaky. There’s a reason we keepthem down that hall, far away from us. [LAUGHTER] They’re very sneaky. And if you give them an infinitenumber of continuous functions, they’ll build for you adiscontinuous function, right? Sneaky. Does that seemterribly physical? No. It’s what happens when yougive a mathematician too much paper and time, right? So, I mean this lessflippantly than I’m saying it, but it’s worth being alittle flippant here. In a physicalsetting, we will often find that there areeffectively an infinite number of possible thingsthat could happen. So, for example in this room,where is this piece of chalk? It’s described by acontinuous variable.That’s an uncountableinfinite number of positions. Now, in practice,you can’t really build an experimentthat does that, but it is in principlean uncountable infinity of possible positions, right? You will never get adiscontinuous wave function for this guy, becauseit would correspond to divergentamounts of momentum, as you showed on theprevious problem set. So, in general, we will oftenbe in a situation as physicists where there’s thepossibility of using the machinery– themathematical machinery– to create pathological examples. And yes, that is a risk. But physically it never happens. Physically it’sextraordinarily rare that such infinitedivergences could matter. Now, I’m not sayingthat they never do.But we’re going to be verycarefree and casual in 804 and just assume that whenproblems can arise from, say, superposing an infinitenumber of smooth functions, leading potentiallyto discontinuities or singularities, that they willeither not happen for us– not be relevant– or they willhappen because they’re forced too, so forphysical reasons we’ll be able to identify. So, this is a veryimportant point. We’re not provingmathematical theorems. We’re not trying to be rigorous. To prove a mathematicaltheorem you have to look at allthe exceptional cases and say, thoseexceptional cases, we can deal withthem mathematically. To a physicist, exceptionalcases are exceptional. They’re irrelevant. They don’t happen. It doesn’t matter. OK? And it doesn’tmean that we don’t care about the mathematicalprecision, right? I mean, I publishpapers in math journals, so I have a deep lovefor these questions.But they’re not salient formost of the physical questions we care about. So, do your best to try notto let those special cases get in the way of your understandingof the general case. I don’t want you tonot think about them, I just want you notlet them stop you, OK? Yeah. AUDIENCE: So, inpostulate five, you mentioned that[? functions ?] in effect was a experiment thatmore or less proves this collapse [INAUDIBLE] But, so I read that itis not [? complicit. ?] PROFESSOR: Yeah, so as with manythings in quantum mechanics– that’s a fair question. So, let me make a slightlymore general statement than answering thatquestion directly. Many things will– how tosay– so, we will not prove– and experimentally you almostnever prove a positive thing. You can show that a predictionis violated by experiment.So there’s always goingto be some uncertainty in your measurements,there’s always going to be some uncertaintyin your arguments. However, in the absenceof a compelling alternate theoreticaldescription, you cling on to what you’ve got it aslong as it fits your data, and this fits thedata like a champ. Right? So, does it prove? No. It fits pretty well,and nothing else comes even within the ballpark. And there’s noexplicit violation that’s better than ourexperimental uncertainties. So, I don’t know ifI’d say, well, we could prove such athing, but it fits. And I’m a physicist. I’m looking for things that fit. I’m not a metaphysicist. I’m not trying to give yousome ontological commitment about what things are trueand exist in the world, right? That’s not my job. OK. So much for our review. But let me finallycome back to– now that we’ve observedthat it’s determinist, let me come back tothe question that was asked a few minutesago, which is, look, suppose we takeour superposition.We evolve it forward for sometime using the Schrodinger evolution. Notice that it’s time reversal. If we know it’stime reverted, we could run itbackwards just as well as we could run itforwards, right? We could integratethat in time back, or we could integratethat in time forward. So, if we know the wavefunction at some moment in time, we can integrate it forward,and we can integrate it back in time. But, If at somepoint we measure, then the wavefunction collapses. And subsequently, thesystem evolves according to the Schrodingerequation, but with this new initial condition. So now we seem tohave a problem.We seem to have–and I believe this was the question that was asked. I don’t remember who asked it. Who asked it? So someone asked it. It was a good question. We have this problemthat there seem to be two definitions of timeevolution in quantum mechanics. One is the Schrodingerequation, which says that thingsdeterministically evolve forward in time. And the second is collapse,that if you do a measurement, things nondeterministicallyby probabilities collapse to some possible state. Yeah? And the probabilityis determined by which wave function you have. How can thesethings both be true? How can you have two differentdefinitions of time evolution? So, this sort offrustration lies at the heart of muchof the sort of spiel about the interpretationof quantum mechanics.On the one hand, wewant to say, well, the world is inescapablyprobabilistic. Measurement comes withprobabilistic outcomes and leads to collapseof the wave function. On the other hand, whenyou’re not looking, the system evolvesdeterministically. And this sounds horrible. It sounds horrible toa classical physicist. It sounds horrible to me. It just sounds awful. It sounds arbitrary. Meanwhile, it makes itsound like the world cares. It evolves differently dependingon whether you’re looking or not.And that– come on. I mean, I think we can allagree that that’s just crazy. So what’s going on? So for a long time, physicistsin practice– and still in practice– for along time physicists almost exclusivelylooked at this problem and said, look,don’t worry about. It fits the data. It makes good predictions. Work with me here. Right? And it’s really hard toargue against that attitude. You have a set of rules. It allows you to compute things. You compute them. They fit the data. Done. That is triumph. But it’s deeply disconcerting. So, over the last, Idon’t know, in the second or the last quarter,roughly, the last third of the 20th century,various people started getting moreupset about this.So, this notion of justshut up and calculate, which has been enshrinedin the physics literature, goes under the name of theCopenhagen interpretation, which roughly says,look, just do this. Don’t ask. Compute the numbers,and get what you will. And people have questioned thesanity or wisdom of doing that. And in particular,there’s an idea– so I refer to the Copenhageninterpretation with my students as the cop out,because it’s basically disavowal of responsibility. Look, it doesn’t makesense, but I’m not responsible for making sense. I’m just responsiblefor making predictions.Come on. So, more recently has comethe theory of decoherence. And we’re not going totalk about it in any detail until the last couplelectures of 804. Decoherence. I can’t spell to save my life. So, the theory of decoherence. And here’s roughlywhat the theory says. The theory says,look, the reason you have this problem betweenon the one hand, Schrodinger evolution of a quantumsystem, and on the other hand, measurement leadingto collapse, is that in the case of measurementmeaning to collapse, you’re not really studying theevolution of a quantum system. You’re studying the evolutionof a quantum system– ie a little thing that you’remeasuring– interacting with your experimentalapparatus, which is made up of 10 to the 27th particles,and you made up of 10 to the 28 particles.Whatever. It’s a large number. OK, a lot more than that. You, a macroscopic object,where classical dynamics are a good description. In particular,what that means is that the quantum effectsare being washed out. You’re washing outthe interference of fringes, which is whyI can catch this thing and not have it split intomany different possible wave functions and where it went. So, dealing with that is hard,because now if you really want to treat the systemwith Schrodinger evolution, you have to study thetrajectory and the motion, the dynamics, of every particlein the system, every degree of freedom in the system. So here’s the question thatdecoherence is trying to ask. If you take a system whereyou have one little quantum subsystem that you’retrying to measure, and then again a gagillionother degrees of freedom, some of which you careabout– they’re made of you– some of which you don’t,like the particles of gas in the room,the environment.If you take that wholesystem, does Schrodinger evolution in the endboil down to collapse for that singlequantum microsystem? And the answer is yes. Showing that takesome work, and we’ll touch on it at the end of 804. But I want to markright here that this is one of the most deeplyunsatisfying points in the basic storyof quantum mechanics, and that it’s deeplyunsatisfying because of the way that we’re presenting it.And there’s a muchmore satisfying– although still younever escape the fact that quantum mechanicsviolates your intuition. That’s inescapable. But at least it’s not illogical. it doesn’t directlycontradict itself. So that story is thestory of decoherence. And if we’re verylucky, I think we’ll try to get one of my friendswho’s a quantum computing guy to talk about it. Yeah. AUDIENCE: [INAUDIBLE]Is it possible that we get twodifferent results? PROFESSOR: No. No. No. There’s never any ambiguityabout what result you got. You never end up in a stateof– and this is also something that decoherence issupposed to explain. You never end up in a situationwhere you go like, wait, wait. I don’t know. Maybe it was here,maybe it was there. I’m really confused. I mean, you can getup in that situation because you did abad job, but you don’t end up in thatsituation because you’re in a superposition state. You always end up when you’rea classical beast doing a classicalmeasurement, you always end up in some definite state.Now, what wavefunction describes you doesn’tnecessarily correspond to you being in a simple state. You might be in a superpositionof thinking this and thinking that. But, when you think this,that’s in fact what happened. And when you think that,that’s in fact what happened. OK. So I’m going to leavethis alone for the moment, but I just wanted to markthat as an important part of the quantum mechanical story. OK. So let’s go on to solvingthe Schrodinger equation. So what I want to dofor the rest of today is talk about solvingthe Schrodinger equation. So when we set about solvingthe Schrodinger equation, the first thingwe should realize is that at the end of theday, the Schrodinger equation is just somedifferential equation. And in fact, it’s a particularlyeasy differential equation. It’s a first order lineardifferential equation. Right? We know how to solve those.But, while it’sfirst order in time, we have to think about whatthis energy operator is. So, just like the Newtonequation f equals ma, we have to specifythe energy operative before we can actually solvethe dynamics of the system. In f equals ma, wehave to tell you what the force is beforewe can solve for p, from p is equal to f. So, for example. So one strategy to solvethe Schrodinger equation is to say, look, it’s justa differential equation, and I’ll solve it usingdifferential equation techniques.So let me specify, forexample, the energy operator. What’s an easy energy operator? Well, imagine you had a harmonicoscillator, which, you know, physicists, that’s your goto. So, harmonicoscillator has energy p squared over 2m plus M Omegasquared upon 2x squared. But we’re goingquantum mechanics, so we replace theseguys by operators. So that’s an energy operator. It’s a perfectlyviable operator. And what is the differentialequation that this leads to? What’s the Schrodingerequation leads to? Well, I’m going to put theih bar on the other side. ih bar derivative with respectto time of Psi of x and t is equal to p squared. Well, we remember thatp is equal to h bar upon i, derivativewith respect to x. So p squared is minus h barsquared derivative with respect to x squared upon 2m, orminus h bar squared upon 2m. Psi prime prime. Let me write this as dx squared. Two spatial derivatives actingon Psi of x and t plus m omega squared upon 2xsquared Psi of x and t. So here’s adifferential equation. And if we want to know howdoes a system evolve in time, ie given some initialwave function, how does it evolve in time, we just takethis differential equation and we solve it. And there are many tools tosolve this partial differential equation. For example, you couldput it on Mathematica and just use NDSolve, right? This wasn’tavailable, of course, to the physicists atthe turn of the century, but they were less timidabout differential equations than we are, because theydidn’t have Mathematica. So, this is a verystraightforward differential equation to solve,and we’re going to solve it in acouple of lectures. We’re going to study theharmonic oscillator in detail.What I want to emphasize foryou is that any system has have some specified energy operator,just like any classical system, has some definiteforce function. And given that energyoperator, that’s going to lead to adifferential equation. So one way to solve thedifferential equation is just to go ahead andbrute force solve it. But, at the end ofthe day, solving the Schrodinger equationis always, always going to boil downto some version morally of solve thisdifferential equation.Questions about that? OK. But when we actually lookat a differential equation like this– so, say we havethis differential equation. It’s got a derivativewith respect to time, so we have to specifysome initial condition. There are many ways to solve it. So given E, givensome specific E, given some specificenergy operator, there are many ways to solve. The resultingdifferential equation. And I’m just going to markthat, in general, it’s a PDE, because it’s gotderivatives with respect to time and derivativeswith respect to space. And roughly speaking,all these techniques fall into three camps. The first is just brute force. That means some analog ofthrow it on Mathematica, go to the closet and pullout your mathematician and tie them to thechalkboard until they’re done, and then put them back. But some version of abrute force, which is just use, by hook or bycrook, some technique that allows you to solvethe differential equation. OK. The second isextreme cleverness. And you’d be amazed howoften this comes in handy. So, extremecleverness– which we’ll see both of thesetechniques used for the harmonic oscillator. That’s what we’ll do next week. First, the bruteforce, and secondly, the clever way of solvingthe harmonic oscillator. When I say extremecleverness, what I really mean is a more elegant useof your mathematician. You know somethingabout the structure, the mathematical structure ofyour differential equation. And you’re going touse that structure to figure out a good way toorganize the differential equation, the good wayto organize the problem. And that will teach you physics. And the reason I distinguishbrute force from cleverness in this sense isthat brute force, you just get a list of numbers.Cleverness, you learnsomething about the way the physics of thesystem operates. We’ll see this at workin the next two lectures. And see, I really shouldseparate this out numerically. And here I don’t just meansticking it into MATLAB. Numerically, it canbe enormously valuable for a bunch of reasons. First off, thereare often situations where no classic techniquein differential equations or no simple mathematicalstructure that would just leap to the imaginationcomes to use. And you have some horribledifferential you just have to solve, and youcan solve it numerically. Very useful lesson, anda reason to not even– how many of y’all are thinkingabout being theorists of some stripe or other? OK. And how many of y’allare thinking about being experimentalists ofsome stripe or another? OK, cool. So, look, there’s thisdeep, deep prejudice in theory against numericalsolutions of problems. It’s myopia. It’s a terrible attitude,and here’s the reason. Computers are stupid. Computers arebreathtakingly dumb. They will do whateveryou tell them to do, but they will not tell youthat was a dumb thing to do.They have no idea. So, in order to solve aninteresting physical problem, you have to firstextract all the physics and organize theproblem in such a way that a stupid computercan do the solution. As a consequence, you learnthe physics about the problem. It’s extremely valuable tolearn how to solve problems numerically, and we’re goingto have problem sets later in the course inwhich you’re going to be required tonumerically solve some of thesedifferential equations. But it’s useful becauseyou get numbers, and you can checkagainst data, but also it lets you in the processof understanding how to solve the problem. You learn thingsabout the problem. So I want to mark that as aseparate logical way to do it. So today, I want tostart our analysis by looking at acouple of examples of solving theSchrodinger equation. And I want to start by lookingat energy Eigenfunctions. And then once we understand howa single energy Eigenfunction evolves in time, once weunderstand that solution to the Schrodingerequation, we’re going to use the linearityof the Schrodinger equation to write down a general solutionof the Schrodinger equation.OK. So, first. What happens if we have asingle energy Eigenfunction? So, suppose our wave functionas a function of x at time t equals zero is in a knownconfiguration, which is an energy EigenfunctionPhi sub E of x. What I mean by Phi sub E of x isif I take the energy operator, and I act on Phi sub Eof x, this gives me back the number E Phi sub E of x. OK? So it’s an Eigenfunction of theenergy operator, the Eigenvalue E. So, suppose ourinitial condition is that our systembegan life at time t equals zero in this state withdefinite energy E. Everyone cool with that? First off, question. Suppose I immediatelyat time zero measure the energyof this system. What will I get? AUDIENCE: E. PROFESSOR: Withwhat probability? AUDIENCE: 100% PROFESSOR: 100%, because thisis, in fact, of this form, it’s a superpositionof energy Eigenstates, except there’s only one term. And the coefficientof that one term is one, and the probabilitythat I measure the energy to be equal to that value isthe coefficient norm squared, and that’s one norm squared.Everyone cool with that? Consider on the other hand, ifI had taken this wave function and I had multiplied itby phase E to the i Alpha. What now is the probabilitywhere alpha is just a number? What now is the probabilitythat I measured the state to have energy E? AUDIENCE: One. PROFESSOR: It’s still one,because the norm squared of a phase is one. Right? OK. The overall phasedoes not matter. So, suppose I have thisas my initial condition. Let’s take awaythe overall phase because my life will be easier. So here’s the wave function. What is theSchrodinger equation? Well, the Schrodingerequation says that ih bar timederivative of Psi is equal to the energyoperator acting on Psi. And I should be specific. This is Psi at x at time t,Eigenvalued at this time zero is equal to the energy operatoracting on this wave function.But what’s the energy operatoracting on this wave function? AUDIENCE: E. PROFESSOR: E. E onPsi is equal to E on Phi sub E, whichis just E the number. This is the number E,the Eigenvalue E times Psi at x zero. And now, instead of havingan operator on the right hand side, we just have a number. So, I’m going to write thisdifferential equation slightly differently, ie timederivative of Psi is equal to E upon ih bar,or minus ie over h bar Psi. Yeah? Everyone cool with that? This is the easiest differentialequation in the world to solve. So, the time derivativeis a constant. Well, times itself. That means thattherefore Psi at x and t is equal to E to the minus iET over h bar Psi at x zero. Where I’ve imposed the initialcondition that at time t equals zero, thewave function is just equal to Psi of x at zero. And in particular, I knowwhat Psi of x and zero is. It’s Phi E of x. So I can simply writethis as Phi E of x. Are we cool with that? So, what this tells me isthat under time evolution, a state which is initiallyin an energy Eigenstate remains in an energyEigenstate with the same energy Eigenvalue. The only thing that changesabout the wave function is that its phasechanges, and its phase changes by rotating witha constant velocity. E to the minus i, theenergy Eigenvalue, times time upon h bar. Now, first off, before wedo anything else as usual, we should first check thedimensions of our result to make sure wedidn’t make a goof. So, does this makesense dimensionally? Let’s quickly check. Yeah, it does. Let’s just quickly check. So we have that the exponentthere is Et over h bar. OK? And this should have dimensionsof what in order to make sense? AUDIENCE: Nothing.PROFESSOR: Nothing, exactly. It should be dimensionless. So what are thedimensions of h bar? AUDIENCE: [INAUDIBLE] PROFESSOR: Oh, no, thedimensions, guys, not the units. What are the dimensions? AUDIENCE: [INAUDIBLE] PROFESSOR: It’s an action,which is energy of time. So the units ofthe dimensions of h are an energy timesa time, also known as a momentum times a position. OK? So, this has dimensions ofaction or energy times time, and then upstairswe have dimensions of energy times time.So that’s consistent. So this in fact is dimensionallysensible, which is good. Now, this tells you avery important thing. In fact, we justanswered this equation. At time t equalszero, what will we get if we measure the energy? E. At time t prime– somesubsequent time– what energy will we measure? AUDIENCE: E. PROFESSOR: Yeah. Does the energychange over time? No. When I say that, what Imean is, does the energy that you expect tomeasure change over time? No. Does the probability thatyou measure energy E change? No, because it’s just a phase,and the norm squared of a phase is one.Yeah? Everyone cool with that? Questions at this point. This is very simpleexample, but it’s going to have a lot of power. Oh, yeah, question. Thank you. AUDIENCE: Are we going to dealwith energy operators that change over time? PROFESSOR: Excellent question. We will later, but not in 804. In 805, you’ll discussit in more detail. Nothing dramatichappens, but you just have to add more symbols. There’s nothing deep about it. It’s a very good question. The question was,are we going to deal with energy operatorsthat change in time? My answer was no, not in804, but you will in 805. And what you’ll find isthat it’s not a big deal. Nothing particularlydramatic happens. We will deal with systems wherethe energy operator changes instantaneously.So not a continuousfunction, but we’re at some of them you turnon the electric field, or something like that. So we’ll deal withthat later on. But we won’t develop atheory of energy operators that depend on time. But you could do it,and you will do in 805. There’s nothingmysterious about it. Other questions? OK. So, these states– astate Psi of x and t, which is of the form eto the minus i Omega t, where Omega is equalto E over h bar. This should look familiar. It’s the de Broglie relation,[INAUDIBLE] relation, whatever. Times some Phi Eof x, where this is an energy Eigenfunction. These states are calledstationary states. And what’s the reason for that? Why are they calledstationary states? I’m going to erase this. Well, suppose this is my wavefunction as a function of time. What is the probabilitythat at time t I will measure the particleto be at position x, or the probability density? Well, the probability densitywe know from our postulates, it’s just the norm squaredof the wave function. This is Psi at x t norm squared. But this is equal tothe norm squared of e to the minus Psi Omega t PhiE by the Schrodinger equation.But when we takethe norm squared, this phase cancelsout, as we already saw. So this is just equal toPhi E of x norm squared, the energy Eigenfunction normsquared independent of time. So, if we happen to know thatour state is in an energy Eigenfunction, thenthe probability density for finding the particleat any given position does not change in time. It remains invariant. The wave function rotatesby an overall phase, but the probability densityis the norm squared. It’s insensitive tothat overall phase, and so the probabilitydensity just remains constant inwhatever shape it is. Hence it’s calleda stationary state. Notice its consequence.What can you say aboutthe expectation value of the position asa function of time? Well, this is equalto the integral dx in the state Psi of x and t. And I’ll call this Psisub E just to emphasize. It’s the integral of the x,integral over all possible positions of theprobability distribution, probability of xat time t times x. But this is equalto the integral dx of Phi E of x squared x. But that’s equal to expectationvalue of x at any time, or time zero. t equals zero. And maybe the best way to writethis is as a function of time. So, the expectation valueof x doesn’t change.In a stationary state,expected positions, energy– thesethings don’t change. Everyone cool with that? And it’s becauseof this basic fact that the wavefunction only rotates by a phase undertime evolution when the system is anenergy Eigenstate. Questions? OK. So, here’s a couple ofquestions for you guys. Are all systems alwaysin energy Eigenstates? Am I in an energy Eigenstate? AUDIENCE: No. PROFESSOR: No, right? OK, expected position of myhand is changing in time. I am not in– so obviously,things change in time. Energies change in time. Positions– expected typicalpositions– change in time. We are not inenergy Eigenstates. That’s a highlynongeneric state. So here’s another question. Are any states ever trulyin energy Eigenstates? Can you imagine anobject in the world that is trulydescribed precisely by an energy Eigenstatein the real world? AUDIENCE: No. PROFESSOR: Ok, therehave been a few nos. Why? Why not? Does anything reallyremain invariant in time? No, right? Everything is gettingbuffeted around by the rest of the universe. So, not only are thesenot typical states, not only are stationarystates not typical, but they actually neverexist in the real world. So why am I talkingabout them at all? So here’s why. And actually I’mgoing to do this here. So here’s why. The reason is this guy, thesuperposition principle, which tells me that ifI have possible states, I can buildsuperpositions of them. And this statement–and in particular, linearity– which saysthat given any two solutions of theSchrodinger equation, I can take asuperposition and build a new solution of theSchrodinger equation.So, let me build it. So, in particular,I want to exploit the linearity of the Schrodingerequation to do the following. Suppose Psi. And I’m going tolabel these by n. Psi n of x and t is equalto e to the minus i Omega nt Phi sub En of x, where Enis equal to h bar Omega n. n labels the various differentenergy Eigenfunctions. So, consider all the energyEigenfunctions Phi sub En. n is a number which labels them. And this is the solution tothe Schrodinger equation, which at time zero is justequal to the energy Eigenfunction of interest. Cool? So, consider these guys. So, suppose we havethese guys such that they solve the Schrodinger equation. Solve the Schrodinger equation. Suppose these guys solvethe Schrodinger equation. Then, by linearity, wecan take Psi of x and t to be an arbitrary superpositionsum over n, c sub n, Psi sub n of x and t.And this will automaticallysolve the Schrodinger equation by linearity of theSchrodinger equation. Yeah. AUDIENCE: Butcan’t we just get n as the sum of theenergy Eigenstate by just applying that andby just measuring that? PROFESSOR: Excellent. So, here’s the question. The question is,look, a minute ago you said no system is truly inan energy Eigenstate, right? But can’t we put a systemin an energy Eigenstate by just measuring the energy? Right? Isn’t that exactly what thecollapse postulate says? So here’s my question.How confident areyou that you actually measure the energy precisely? With what accuracy canwe measure the energy? So here’s the unfortunatetruth, the unfortunate practical truth. And I’m not talking aboutin principle things. I’m talking about it in practicethings in the real universe. When you measure the energy ofsomething, you’ve got a box, and the box has a dial,and the dial has a needle, it has a finite width,and your current meter has a finite sensitivityto the current. So you never truly measurethe energy exactly. You measure it towithin some tolerance. And In fact, there’s afundamental bound– there’s a fundamental bound onthe accuracy with which you can make a measurement, whichis just the following. And this is the analog ofthe uncertainty equation.We’ll talk aboutthis more later, but let me just jumpahead a little bit. Suppose I want tomeasure frequency. So I have somesignal, and I look at that signal for 10 minutes. OK? Can I be absolutely confidentthat this signal is in fact a plane wave with the givenfrequency that I just did? No, because it couldchange outside that. But more to thepoint, there might have been smallvariations inside. There could’ve beena wavelength that could change on a timescale longer than the time that I measured. So, to know that thesystem doesn’t change on a arbitrary– thatit’s strictly fixed Omega, I have to wait a very long time. And in particular, how confidentyou can be of the frequency is bounded by the time overwhich– so your confidence, your uncertaintyin the frequency, is bounded in thefollowing fashion. Delta Omega, Delta t is alwaysgreater than or equal to one, approximately.What this says isthat if you want to be absolute confidentof the frequency, you have to wait anarbitrarily long time. Now if I multiply thiswhole thing by h bar, I get the following. Delta E– so this isa classic equation that signals analysis–Delta E, Delta t is greater than orapproximately equal to h bar. This is a hallowed timeenergy uncertainty relation, which we haven’t talked about. So, in fact, it ispossible to make an arbitrarily precisemeasurement of the energy. What do I have to do? I have to wait forever. How patient are you, right? So, that’s the issue. In the real world, we can’t makearbitrarily long measurements, and we can’t isolate systemsfor an arbitrarily long amount of time. So, we can’t put things ina definite energy Eigenstate by measurement. That answer your question? AUDIENCE: Yes. PROFESSOR: Great. How many peoplehave seen signals in this expression, thebound on the frequency? Oh, good. So we’ll talk about thatlater in the course.OK, so coming back to this. So, we have our solutionsof the Schrodinger equation that are initiallyenergy Eigenstates. I claim I can take an arbitrarysuperposition of them, and by linearity derivethat this is also a solution to theSchrodinger equation. And in particular, whatthat tells me is– well, another way to saythis is that if I know that Psi of x times zerois equal to sum over n– so if sum Psi of x–if the wave function at some particularmoment in time can be expanded as sum overn Cn Phi E of x, if this is my initial condition,my initial wave function is some superposition, then Iknow what the wave function is at subsequent times.The wave function bysuperposition Psi of x and t is equal to sum overn Cn e to the minus i Omega nt Phi n– sorry, thisshould’ve been Phi sub n– Phi n of x. And I know this has tobe true because this is a solution to the Schrodingerequation by construction, and at time t equalszero, it reduces to this. So, this is a solution tothe Schrodinger equation, satisfying this condition atthe initial time t equals zero. Don’t even have todo a calculation.So, having solved theSchrodinger equation once for energy,Eigenstates allows me to solve it forgeneral superposition. However, what I justsaid isn’t quite enough. I need one more argument. And that one more argument isreally the stronger version of three that we talkedabout before, which is that, given anenergy operator E, we find the set ofwave functions Phi sub E, the Eigenfunctionsof the energy operator, with Eigenvalue E. So, given the energy operator,we find its Eigenfunctions. Any wave function Psi atx– we’ll say at time zero– any function of x canbe expanded as a sum. Specific superposition sumover n Cn Phi E sub n of x. And if any functioncan be expanded as a superposition ofenergy Eigenfunctions, and we know how totake a superposition, an arbitrarysuperposition of energy Eigenfunctions, and findthe corresponding solution to the Schrodinger equation. What this means is, we can takean arbitrary initial condition and compute the full solutionof the Schrodinger equation.All we have to do is figure outwhat these coefficients Cn are. Everyone cool with that? So, we have thus, usingsuperposition and energy Eigenvalues, totally solvedthe Schrodinger equation, and reduced it to the problemof finding these expansion coefficients. Meanwhile, these expansioncoefficients have a meaning. They correspondto the probability that we measure theenergy to be equal to the correspondingenergy E sub n. And it’s just the normsquared of that coefficient. So those coefficientsmean something. And they allow us tosolve the problem. Cool? So this is fairly abstract. So let’s make it concreteby looking at some examples. So, just as a quick aside. This should sound an awfullot like the Fourier theorem. And let me comment on that. This statement originally wasabout a general observable and general operator. Here I’m talkingabout the energy. But let’s think about aslightly more special example, or more familiar example. Let’s consider the momentum.Given the momentum, we canfind a set of Eigenstates. What are the set ofgood, properly normalized Eigenfunctions of momentum? What are the Eigenfunctionsof the momentum operator? AUDIENCE: E to the ikx. PROFESSOR: E to the ikx. Exactly. In particular, oneover 2 pi e to the ikx. So I claim that, for everydifferent value of k, I get a different value of p,and the Eigenvalue associated to this guy is p isequal to h bar k. That’s the Eigenvalue. And we get that by actingwith the momentum, which is h bar upon i, h bar timesderivative with respect to x. Derivative withrespect to x pulls down an ik times the same thing.H bar multiplies thek over i, kills the i, and leaves us with an overallcoefficient of h bar k. This is an Eigenfunctionof the momentum operator withEigenvalue h bar k. And that statementthree is the statement that an arbitraryfunction f of x can be expandedas a superposition of all possibleenergy Eigenvalues. But k is continuouslyvalued and the momentum, so that’s an integraldk one over 2 pi, e to the ikx timessome coefficients.And those coefficientsare labeled by k, but since k is continuous, I’mgoing to call it a function. And just to give it a name,instead of calling C sub k, I’ll call it f tilde of k. This is of exactlythe same form. Here is the expansion–there’s the Eigenfunction, here is the Eigenfunction, hereis the expansion coefficient, here is expansion coefficient. And this has a familiar name.It’s the Fourier theorem. So, we see thatthe Fourier theorem is this statement, statementthree, the superposition principal, for themomentum operator. We also see that it’s truefor the energy operator. And what we’re claiminghere is that it’s true for any observable. Given any observable, youcan find its Eigenfunctions, and they form a basis on thespace of all good functions, and an arbitrary function canbe expanded in that basis. So, as a last example,consider the following. We’ve done energy. We’ve done momentum. What’s anotheroperator we care about? What about position? What are theEigenfunctions of position? Well, x hat onDelta of x minus y is equal to y Delta x minus y. So, these are the states withdefinite value of position x is equal to y.And the reason this is trueis that when x is equal to y, x is the operator thatmultiplies by the variable x. But it’s zero, exceptat x is equal to y, so we might as wellreplace x by y. So, there are theEigenfunctions. And this statementis a statement that we can representan arbitrary function f of x in a superposition ofthese states of definite x. f of x is equal to the integralover all possible expansion coefficients dy delta xminus y times some expansion coefficient. And what’s theexpansion coefficient? It’s got to be a function of y. And what functionof y must it be? Just f of y. Because this integralagainst this delta function had better give me f of x. And that will only betrue if this is f of x. So here we see, in somesense, the definition of the delta function. But really, this is a statementof the superposition principle, the statement that anyfunction can be expanded as a superposition ofEigenfunctions of the position operator.Any function can beexpanded as a superposition of Eigenfunctions of momentum. Any function can beexpanded as a superposition of Eigenfunctions of energy. Any function can beexpanded as a superposition of Eigenfunctions of anyoperator of your choice. OK? The special cases– the Fouriertheorem, the general cases, the superposition postulate. Cool? Powerful tool. And we’ve usedthis powerful tool to write down a generalexpression for a solution to the Schrodinger equation.That’s good. That’s progress. So let’s look at someexamples of this. I can leave this up. So, our first example is goingto be for the free particle. So, a particle whoseenergy operator has no potential whatsoever. So the energy operatoris going to be just equal to p squared upon 2m. Kinetic energy. Yeah. AUDIENCE: When you sayany wave function can be expanded in terms of– PROFESSOR: EnergyEigenfunctions, position Eigenfunctions,momentum Eigenfunctions– AUDIENCE: Eigenbasis,does the Eigenbasis have to come from an operatorcorresponding to an observable? PROFESSOR: Yes. Absolutely. I’m starting withthat assumption. AUDIENCE: OK. PROFESSOR: So, again,this is a first pass of the axioms ofquantum mechanics. We’ll make this more precise,and we’ll make it more general, later on in the course, as wego through a second iteration of this.And there we’ll talk aboutexactly what we need, and what operators areappropriate operators. But for the moment, thesufficient and physically correct answer is,operators correspond to each observable values. Yeah. AUDIENCE: So are the set ofall reasonable wave functions in the vector spacethat is the same as the one withthe Eigenfunctions? PROFESSOR: That’s anexcellent question. In general, no. So here’s the question. The question is,look, if this is true, shouldn’t it be that theEigenfunctions, since they’re our basis for thegood functions, are inside the space ofreasonable functions, they should also bereasonable functions, right? Because if you’re goingto expand– for example, consider two dimensionalvector space.And you want tosay any vector can be expanded in a basis of pairsof vectors in two dimensions, like x and y. You really want to makesure that those vectors are inside your vector space. But if you say thisvector in this space can be expanded in termsof two vectors, this vector and that vector, you’rein trouble, right? That’s not goingto work so well.So you want to make surethat your vectors, your basis vectors, are in the space. For position, the basisvector’s a delta function. Is that a smooth, continuousnormalizable function? No. For momentum, thebasis functions are plane waves thatextend off to infinity and have support everywhere. Is that a normalizablereasonable function? No. So, both of thesesets are really bad. So, at that point you might say,look, this is clearly nonsense. But here’s an important thing. So this is a totallymathematical aside, and for those of you whodon’t care about the math, don’t worry about it. Well, these guysdon’t technically live in the space ofnonstupid functions– reasonable, smooth,normalizable functions. What you can show is thatthey exist in the closure, in the completion of that space.OK? So, you can find asequence of wave functions that are good wave functions,an infinite sequence, that eventually thatinfinite sequence converges to these guys, eventhough these are silly. So, for example, for theposition Eigenstates, the delta function is not acontinuous smooth function. It’s not even a function. Really, it’s some god awfulthing called a distribution. It’s some horrible thing. It’s the thing that tellsyou, give it an integral, it’ll give you a number.Or a function. But how do we buildthis as a limit of totally reasonable functions? We’ve already done that. Take this function witharea one, and if you want, you can round this out bymaking it hyperbolic tangents. OK? We did it on one ofthe problem sets. And then just make itmore narrow and more tall. And keep making it more narrowand more tall, and more narrow and more tall, keepingits area to be one. And I claim that eventuallythat series, that sequence of functions, convergesto the delta function. So, while this function isnot technically in our space, it’s in the completionof our space, in the sense that we take aseries and they converge to it.And that’s what you need forthis theorem to work out. That’s what you needfor the Fourier theorem. And in some sense,that observation was really thegenius of Fourier, understanding thatthat could be done. That was totallymathematical aside. But that answer your question? AUDIENCE: Yes. PROFESSOR: OK. Every once in awhile I can’t resist talking about thesesort of details, because I really like them. But it’s good to know thatstupid things like this can’t matter forus, and they don’t. But it’s a very good question. If you’re confused aboutsome mathematical detail, no matter how elementary, ask. If you’re confused, someoneelse the room is also confused. So please don’t hesitate.OK, so our first example’sgoing to be the free particle. And this operator can bewritten in a nice way. We can write it as minus–so p is h bar upon iddx, so this line isminus h bar squared upon 2m to the derivativewith respect to x. There’s the energy operator. So, we want to solvefor the wave functions. So let’s solve itusing an expansion in terms of energyEigenfunctions. So what are theenergy Eigenfunctions? We want to find thefunctions E on Phi sub E such that this isequal to– whoops. That’s not a vector. That’s a hat– such as this isequal to a number E Phi sub E. But given thisenergy operator, this says that minus h bar squaredover 2m– whoops, that’s a 2. 2m– Phi prime prime ofx is equal to E Phi of x.Or equivalently, Phi primeprime of x plus 2me over h bar squared Phi of xis equal to zero. So I’m just going tocall 2me– because it’s annoying to write itover and over again– over h bar squared. Well, first off,what are its units? What are the unitsof this coefficient? Well, you could do it two ways. You could either do dimensionalanalysis of each thing here, or you could just know that westarted with a dimensionally sensible equation,and this has units of this divided by length twice. So this must have towhatever length squared. So I’m going tocall this something like k squared,something that has units of one overlength squared. And the generalsolution of this is that phi E E of x– well, thisis a second order differential equation that will have twosolutions with two expansion coefficients– A e to the ikxplus B e to the minus iks. A state with definite momentumand definite negative momentum where such that E is equal to hbar squared k squared upon 2m. And we get that just from this. So, this is the solution of theenergy Eigenfunction equation.Just a note of terminology. People sometimescall the equation determining an energyEigenfunction– the energy Eigenfunctionequation– sometimes that’s refer to as theSchrodinger equation. That’s a sort of cruelthing to do to language, because the Schrodinger’sequation is about time evolution, and this equationis about energy Eigenfunctions. Now, it’s true that energyEigenfunctions evolve in a particularly simpleway under time evolution, but it’s a different equation.This is telling you aboutenergy Eigenstates, OK? And then morediscussion of this is done in the notes, which I willleave aside for the moment. But I want to do one moreexample before we take off. Wow. We got through a lotless than expected today. And the one last exampleis the following. It’s a particlein a box And this is going to be importantfor your problem sets, so I’m going to go ahead andget this one out of the way as quickly as possible. So, example two. Particle in a box. So, what I mean byparticle in a box. I’m going to take a systemthat has a deep well. So what I’m drawing hereis the potential energy U of x, where this issome energy E naught, and this is theenergy zero, and this is the position xequals zero, and this is position x equals l. And I’m going to idealizethis by saying look, I’m going to be interestedin low energy physics, so I’m going to just treatthis as infinitely deep.And meanwhile, mylife is easier if I don’t think about curvybottoms but I just think about thingsas being constant. So, my idealizationis going to be that the well isinfinitely high and square. So out here thepotential is infinite, and in here thepotential is zero. U equals inside, betweenzero and l for x. So that’s my systemparticle in a box. So, let’s find theenergy Eigenfunctions. And again, it’s the samedifferential equations as before. So, first off, before weeven solve anything, what’s the probability that Ifind x less than zero, or find the particleat x greater than l? AUDIENCE: Zero. PROFESSOR: Right,because the potential is infinitely large out there. It’s just not going to happen. If you found it there,that would correspond to a particle ofinfinite energy, and that’s not going to happen.So, our this tellsus effectively the boundarycondition Psi of x is equal to zero outside the box. So all we have todo is figure out what the wave function is insidethe box between zero and l. And meanwhile, whatmust be true of the wave function at zero and at l? It’s got to actuallyvanish at the boundaries. So this gives us boundaryconditions outside the box and at the boundaries xequals zero, x equals l. But, what’s our differentialequation inside the box? Inside the box, well,the potential is zero. So the equation is thesame as the equation for a free particle. It’s just this guy. And we know whatthe solutions are. So the solutions can bewritten in the following form.Therefore inside thewave function– whoops. Let me write this asPhi sub E– Phi sub E is a superposition of two. And instead of writingit as exponentials, I’m going to write itas sines and cosines, because you can expressthem in terms of each other. Alpha cosine of kxplus Beta sine of kx, where again Alpha and Betaare general complex numbers. But, we must satisfy theboundary conditions imposed by our potential at xequals zero and x equals l.So from x equalszero, we find that Phi must vanish when x equals zero. When x equals zero, thisis automatically zero. Sine of zero is zero. Cosine of zero is one. So that tells us thatAlpha is equal to zero. Meanwhile, the condition that atx equals l– the wave function must also vanish– tells usthat– so this term is gone, set off with zero– thisterm, when x is equal to l, had better also be zero. We can solve that bysetting Beta equal to zero, but then our wavefunction is just zero. And that’s a reallystupid wave function. So we don’t want to do that. We don’t want toset Beta to zero.Instead, what must we do? Well, we’ve got a sine,and depending on what k is, it starts at zero andit ends somewhere else. But we need it hit zero. So only for a veryspecial value of k will it actually hit zeroat the end of l. We need kl equals zero. Or really, kl isa multiple of pi. Kl is equal– and wewant it to not be zero, so I’ll call it n plus1, an integer, times pi. Or equivalently, k is equalto sub n is equal to n plus 1, where n goes from zero to anylarge positive integer, pi over l.So the energyEigenfunction’s here. The energy Eigenfunctionis some normalization– whoops– a sub nsine of k and x. And where kn is equal tothis– and as a consequence, E is equal to h bar squaredkn can squared E sub n is h bar squared kn squaredover 2m, which is equal to h bar squared– just pluggingin– pi squared n plus 1 squared over 2ml squared. And what we found issomething really interesting. What we found is, firstoff, that the wave functions look like– well,the ground state, the lowest possible energythere is n equals zero. For n equals zero, this isjust a single half a sine wave. It does this. This is the n equals zero state. And it has some energy,which is E zero. And in particular, E zerois not equal to zero. E zero is equal to h bar squaredpi squared over 2ml squared. It is impossible fora particle in a box to have an energy lowerthan some minimal value E naught, which is not zero.You cannot have lessenergy than this. Everyone agree with that? There is no such Eigenstatewith energy less than this. Meanwhile, it’s worse. The next energy iswhen n is equal to 1, because if we decreasethe wavelength or increase k a little bit, we getsomething that looks like this, and that doesn’t satisfyour boundary condition. In order to satisfyour boundary condition, we’re going to have toeventually have it cross over and get to zero again. And if I could only draw–I’ll draw it up here– it looks like this. And this has anenergy E one, which you can get byplugging one in here. And that differs by one,two, four, a factor of four from this guy. E one is four E zero.And so on and so forth. The energies are gaped. They’re spread awayfrom each other. The energies are discrete. And they get further andfurther away from each other as we go to higherand higher energies. So this is alreadya peculiar fact, and we’ll explore some ofits consequences later on. But here’s that I wantto emphasize for you. Already in the firstmost trivial example of solving aSchrodinger equation, or actually even beforethat, just finding the energy Eigenvalues and the energy ofEigenfunctions of the simplest system you possibly could,either a free particular, or a particle ina box, a particle trapped inside a potentialwell, what we discovered is that the energyEigenvalues, the allowed values of the energy, are discrete, andthat they’re greater than zero.You can never have zero energy. And if that doesn’tsound familiar, let me remind you of something. The spectrum of light comingoff of a gas of hot hydrogen is discrete. And no one’s everfound a zero energy beam of light coming out of it. And we’re going to make contactwith that experimental data. That’s going to be partof the job for the rest.See you next time. [APPLAUSE] .
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