PHYS 343 Lecture 20: A Closer Look at the (Quantum) Harmonic Oscillator

nanohub.org. >> So the lecture is on a closer look at the quantum harmonic oscillator. If you have just walked in, I'm guest lecturing for Erica. Um my name is Dr. Ayar Piswas or Shri or Shidya, whatever you want to call me. Just call me uh by whatever whatever um whatever is comfortable for you. Okay. Okay, let's get started. Rec recall the classical harmonic oscillator. Are you able to read fine from the back? So a quintessential example is mass attached to spring. Mass M attached to a spring with spring constant K. As you know potential energy is does this sound right? half kx² okay so potential energy is half kx² Newton's laws d2x dt² is let's start with force is minus kx corresponding to this potential energy so we can write down d2 d2x dt2 is uh minus kx x from here. And the solution x of t is a sin omega t plus b cossine omega t. The familiar simple harmonic oscillator where omega is the classical angular frequency of oscillation which is clearly related to the spring constant through this or omega square is equal to k by m. Everybody more or less with me on this? That's the classical version. Why do we obsess so much over the harmonic oscillator? If you feel like you've seen enough of it in this course, rest assured, if you stick with physics, you'll be seeing a great deal more of the harmonic oscillator. We are just getting warmed up. Why Why do we obsess so much over the harmonic oscillator? I will expect more participation going forward. But since it's early in the lecture, let me give you the answer. Consider a generic potential. So sorry, generic V of X. etc. And let's focus on some local minimum and Consider a parabolic approximation to this generic potential around the local minimum. Why are we considering the parabolic approximation? Well, for those of you who like to think in terms of algebra rather than pictures, let's expand tailor expand V of X around X knot the local minimum. As you know this is constant term plus linear term the linear portion plus half second derivative times quadratic term etc. This is a potential energy. This term is constant as you know. It doesn't do anything interesting because it doesn't change the force. X not is a local minimum. So what's this term? Zero. >> Louder. >> Zero. >> Please clap for this gentleman. [applause] >> And what's your name, sir? >> Pratush. >> Pratush. Thank you. Pratush. Not just for answering correctly for but for answering at all. And what about the second term? It's our first nonzero or non-trivial term and it's quadratic in x - x kn and that is this parabolic approximation. Everybody remember parabola goes as x square parabolic. And that is why we obsess over a simple harmonic oscillator cuz this potential energy form is what you get if you take any generic potential energy and [clears throat] expand around the local minimum. This is the leading order term. Good. So are we a little bit more invested in the simple harmonic oscillator knowing that it's not just some esoteric thing that physicists like to pull out all the time but it actually is practically useful all the time. Okay to summarize any generic v of x for small enough x - x is approximately equal to half second derivative at x not times this quadratic term. So we can identify this second derivative second derivative at x not term with the effective spring constant here. Okay, that's the spirit in which we are approaching the harmonic oscillator. So that was lightning introduction in the classical uh version. But now we go to the quantum version. for quant for the quantum version. As you know, the thing we need from here is the form of E of X [laughter] quantum analog V of X. We had half kx² and we also discussed what omega was the angular frequency in terms of k and m. For the quantum analog it's convenient to write this potential in terms of m and omega. This is how I want it. Yeah. M is mass of course. Omega is now to be understood as the classical angular frequency. It's just notation because you may not always have a spring connected to a mass as the physical instantiation of the potential. Okay, so this is just setting up our language to talk about this harmonic oscillator now within the quantum framework. Shall I proceed? At this point, let me pause and ask you if you know the drill for given a potential, how do you do quantum mechanics with it? I know you do because I've seen the lecture topics that Erica has gone through. But I am going to without writing the prescription down on the board because it won't be directly useful for this lecture just go through the prescription for how you're expected to treat a quantum mechanical system given the potential with you and you can nod with me as we go through each step. Okay, does that sound like a plan? As you know for a generic problem given the functional form of V of X we assume it's a timeindependent potential. In this course you're only going to meet V of X given the initial condition capital S of X comm 0. So you're given B of X and initial condition. The problem is to find capital S of X comma T. Capital S is the solution to the time dependent Shreddinger equation. The typical strategy is to solve instead the time independent shreddinger equation which yields a set of solutions small si of x each with associated energy e1 e2 e3 etc. And then our strategy is to write the initial condition capital S of X comm 0 as a sum of terms C1 * small S1 of X plus C2 * small S2 of X and so forth and then to find these coefficients through magic using the FIA trick. And then are you still with me? Should have draw done this drill a few times by now. And then once you have those coefficients, it's trivial. Uh you just tag on the phase factors and get your general solution. Does this sound vaguely familiar? The good news is that we are not going to do this brute force technique for the harmonic oscillator in class today. It can be done. But what I want to walk you through is a diabolically clever alternate way of arriving at the same answer which you will only fully appreciate if you sweat your guts out through the brute force method afterwards. So will at least some of you take that on so that you can appreciate the full beauty of this alternate technique for the harmonic oscillator. Yes. Okay. All right. With that background and having declared that we will not take the brute force route, what we want to do is a clever solution of the time independent shreddinger equation. That's what the rest of the lecture is about. And you've also seen some of this, but I want to walk you through the nuts and bolts in detail. Okay, everybody remember the time independent shreddinger equation? Let me write it down. Let me write it down. So we have the potential energy term time. I'll write it once and then I'll independent shreddinger. It's laborious to write this again and again. So going forward I am just going to write this. Okay. The time independent shinger equation is minus hr² by 2 m d2 small s dx2 plus v of x small s of x equ= e * of x to be determined given here and also to be determined. Everybody with me? This is not a grim topic even though it seems like it right now. it. The material we are going to cover in today's lecture if you were to ask me is could easily go into uh centerpiece in museum of amazing human thought or human achievement in thought. It is an extraordinary set of u ideas that come together to give you the story that I'm telling you today. And even in an hour and change, I am hopeful that you'll feel the wow of human accomplishment here. So stay with me. We'll get there. So here is the time independent shinger equation which I'm going to call this from going forward. The program is to find small s of x given the potential energy term which is quadratic in x for the simple harmonic oscillator. Okay. And corresponding to to s you also need to find e. And typically we expect an infinitum of solutions size correspond with corresponding solutions E. Have you seen this before? For let's say the for what potential have you seen this before? Beside the harmonic oscillate >> well. >> The well. Okay. The infinite well. Yep. So you know the drill. So instead of the brute force technique, now we want to try a clever technique to solve for the infinitum of size and ease allowed for this problem. And to motivate the method of solution, let me first clean up this equation as follows for the simple harmonic oscillator. So to start with we write the time independent shinger equation in a more suggestive form [snorts] 1 by 2 M * P operator squared + M omega X operator the whole square operating on S equals P S where as you know P operator is H bar by I ddx. Is this familiar? And so p² is minus h bar² d2 dx2 and x is just x. We must treat operators with tender loving care gently because operator A time operator B need not be equal to operator B * operator A. Okay. And we'll see this explicitly as we go forward. Is this vaguely familiar to everybody? Okay. So take this hat notation seriously. This is just to remind ourselves that if another operator multiplies it, we must preserve the order of multiplication. So if this is on the left and something else is on the right, we want to keep it that way and not flip the orderly. You know what this operator is? Anybody? Take a wild guess. Yes. >> Indeed. The Hamiltonian operator. So in fact we could have thirstly written the time independent shooting equation as this. The reason this is a suggestive form is because if we were to think about the number analog for what we have now in operator form. Let me tell you what I mean by that. Think about the algebraic identity u 2 + v square. Can anybody factoriize this for me? >> Yes. Go ahead. >> Ei. >> I think you said u plus i plus uus iv. >> That is correct. But I just want to be um slightly different and write it as I u + v * minus i + v which is also equally valid. Right? Everybody with me on factorizing this? In that spirit we will attempt to factoriize the Hamiltonian. Okay. But can we do that given that P and X are operators and may not elegantly allow us to write it in this kind of neat factorized form? That's the question. So take a deep breath because that's what we are going to do for the next several minutes. Attempt to factoriize this Hamiltonian in the spirit of that algebraic identity. So that is the So we are now in the hard slog part of the hike. Scenic views will come but you have to stay with me through this. Is the motivation clear? We are going to attempt to factoriize this Hamiltonian which has these two quadratic terms which have operators in them in the spirit of that algebraic identity. Okay, let's get started. To sort of clean up notation, I want to introduce. Let's examine operator E plus defined as stay with me and I'll motivate why we are doing this some factors in the front 2 h bar m omega under square root times this is the important part minus this I P operator plus N omega X operator and it's analog A minus operator which is same prefactor in front which at the moment can [snorts] just take down as something that we have put in the front to make a future result look nice. Okay, but that's not the important stuff. times the analogous plus IP plus M omega X in the spirit of this relation. Okay. And so you won't be surprised that the next thing we want to examine is the product A minus operator time A+ operator which at least in spirit looks like this. Are you with me? Okay, hard coming up. Take a deep breath. Here's the definition of a plus comes with a negative sign here. Here's the definition of a minus comes with a positive sign here. We want to multiply in the order. This times that. Okay, a minus * a plus. So, let's do that. The preactors under square root multiply with each other. Getting rid of the square root 1 by 2 h m omega time let's write it out ip + m omega x * - i + m omega x. So straightforward to do the multiplication provided you remember these are operators and need to be handled with care and so what stays on the left stays on the left and what stays on the right stays on the right. Okay. So this becomes p² + m omega square + m² omega 2 x² those are the easy terms and now come the terms we have to handle with care i - i m omega - i m omega * x * p minus p * x Is my algebra sound? You're getting the same thing. This thing, as you may have seen before, is written in shorthand form as this. where this notation stands for something that's referred to as the commutator. Have you heard this before? And so to proceed further, we need to find the commutator of X with P. which even though I expect you know from before or you've seen before, I'm going to quickly evaluate for you just you just to get you warmed up with operators. Okay, do I belabor the point? Handle operators with care. uh to find out what this is. If you were to calculate this, I would highly recommend sticking in a test function here to evaluate what it's equal to. And in the end, the test function will cancel off. So let's do that. This is x * h bar / i dx of f minus h bar / i x * f. So that's h bar / i * x df dx minus from here you're going to get two terms. So that's x df dx canceling off with this minus f. Quick reminder i² is -1. So minus I is the same as 1 by I. We have a 1 by I here and a minus here. So we are left with I H bar * F compare. So we end up with commutator of x with p this i h bar the famous canonical commutation relation with me so far. This should be old stuff. And the reason that we got into this is because we are trying to evaluate this product because we are trying to write the Hamiltonian in a factorized form in the spirit of this relation. And we were stuck at a point where we needed the commutator of X and P. The hard slog part of the hike is almost done. Scenic views are coming up. So please stay with me. So now we take this and stick this in here and see what happens to this product A minus A+. Continuing a minus A+ is therefore equal to what we had before 1 by 2 H bar M omega times this stuff just what we had 4 minus i by 2 h bar. Just opening up this bracket, you have a 1x 2 h bar m omega here and a i m omega here. M omega's cancel off minus i and 2 h bar minus i 2 h bar times the commutator of x p. This will be a lot more fun if you actually work it out with me. If I'm talking at you I don't know maybe me don't feel pressure but scribbling along with me you may have a lot more fun. Okay. So we are going to substitute for this. We just derived what this is. This is IH bar I H bar. Kind reminder I square is -1. I I - sign. So we can clean this up. What do we get? >> 12. through a derivation like this. It is a relief to know that somebody's paying attention. So, please follow with me, follow along with me and yell out answers as I ask for them. We are almost at the end of this one. Okay, of this one. So, we conclude A minus A+ which is what we started out with is P² + M omega X² / 2 M + half. What's the first term? together. >> It's got this. It's got P square. Sorry, my first term. I should be clear. What's this term? >> All right. >> Louder. >> Hamiltonian. >> The Hamiltonian. The P square by 2M part was the kinetic energy part. And this is the potential energy part. So together the Hamiltonian. So we have rewritten the Hamiltonian in terms of the product a minus a plus as a factor. >> Yeah, I'm searching for it actually. Oh, from here. [snorts] So we've rewritten the Hamiltonian in terms of the product a minus a plus as the Hamiltonian is h bar omega time a + a minus minus we have a minus sign problem. Okay, this is what happens when we try to do too many steps in one shot. Give me a moment to remind you what we had. We had - i by 2 h bar here. - i by 2 h bar. And then we multiplied that by i h bar. And so we got rid of this and we got a plus sign. Ah my bad. We are doing fine. Okay. We are doing fine. A minus a + min - half. Okay. [snorts] So let's summarize what we have found because we'll use this again and again. The Hamiltonian is a minus a + -/* h bar omega. Analogously instead of starting with the product a minus a + we could have started with a + a minus following through the derivation in exactly the same way. you would have actually gotten this commutator with a different sign so and so forth. So you will find you could also write this as h by omega * a + a minus plus half. So there's a minus here and you would get a plus here if you did the complimentary derivation with a + a minus. Still with me? Okay. What's the big deal? Well, before we start unpacking all the wisdom that's contained here, and again, I'm hoping to convince you um I'm hoping to convince you in the next 35 minutes that this is an extraordinary accomplishment in the Museum of Human Thought. So, stay with me. This is there's a lot to unpack here. But the next thing I want to calculate before we actually unpack that result is the commutator of a minus and a plus. We are almost there because the commutator as you know by now is just this difference. Are you with me? And from here you can do the math in your head. What is it? Just one. It's to get this nice result of one rather than some yucky factors on the right hand side that we chose these three factors before. So this result that the commutator of a minus and a plus is one will turn out to be useful. So, let's put that up here. And I just remembered what I wanted to point out to you earlier. What units does H bar omega have? Energy. Indeed. Hamiltonian units of energy. H bar omega units of energy. What is the unit of the product A minus A plus? Stare at the formula. energy on the left hand side for Hamiltonian energy for H bar omega what's left over must be dimensionless so the product a minus a plus is dimensionless which also squares with commutator is one because one is dimensionless so dimension dimensionally all of this makes sense we haven't dropped any factors by accident All right. Now that we have these, we are ready to start constructing pictures. So this is just cosmetic but stylishly you can combine these two relations into one as Hamiltonian equals h omega time a + a minus + half. This is one of the relations. And then you can write the second relation right here. And if you see a formula written like that, that means either you consistently read the top line or you consistently read the bottom line. Yes. So sometimes you'll see those two formula written like this. And that's what that means. So everybody with me still on the science. Okay. So for the rest of what comes we will have to make clever choices of which of these two equivalent formula for the Hamiltonian we want to use for the time independent shreddinger equation. [snorts] Okay so both are equally valid but one route is the scenic route and the other one is the slog route. So I'll motivate you through the use of this formula again and again over what comes next to appreciate better what the a minus a plus operators do for us. Here is the crucial argument in this lecture claim. If satisfies the time independent shingle equation with energy value E. It's a lot of words that basically says if hi equals e si that's what we have so far then here's the claim. So does a + s but with energy value [music] E plus H bar omega that is if this if S satisfies the time independent shinger equation with energy E. Then A+ type satisfies the time independent shingle equation with energy E plus H bar omega. So if this then this. As you can intuit this is one of a pair pair of analogous results. We could write the analogous result for a minus. And let me also write that statement. And since that proof will be very similar to this one, I'm going to do one of the proofs for you and leave the other proof for you to try at home. The treat this treatment is taken out of the marvelous textbook by David Griffiths. Are any of you familiar with this book? And I use the second edition. I believe the edition that's out there right now is the third. I cannot imagine that something profound has changed for what I've told you. Okay. So the analogous result which I'll or the analogous claim which I'll also write down is well hold on let's do this proof first and then I'll write down the analogous claim. So let's start with h * a + which is what we have here. We want to show that it's equal to this given that s and e given that 8 s equals e si. So starting with this we can unpack using an appropriate choice of formula for Hamiltonian from one of these two and I am going to choose this one and I'll tell you why in just a moment. So I chose the top line. So this is what I have substituted for the Hamiltonian. And this is as before. Remember we have operators. So we treat them gently and keep the order of multiplication the same. This is on the right hand side and we want to bring it in. So let's do that. H bar omega time a + a minus. And now we are going to stick in the a+ on the right hand side plus a + 2 from the second term. With me so far? And now we pull out the A+es to the left. That's a legit operation. Are you with me? I'm not changing the order of multiplication. So this is h bar omega * now we pulled out the a+ to the left times what's left over here is this a minus a + once again half. Now I want to flip the order of this but I have to pay a cost and that cost is given to me by the commutator because the commutator. So let's unpack this relation again this relation is telling us a minus a + minus a + a minus is 1. I want to trade a minus a plus for a plus a minus. So I have to rewrite a minus a plus as 1 + a + a minus. Are you with me? Almost there. This is h bar omega time a + time flipping the order a + a minus + + 1 and then the half from before and we got this from the commutator which we unpacked as this Almost there. H bar omega is just a number regular number not an operator. So we can freely pass it through operators. Are you with me? So we can bring this in here and then take it inside. So what we in fact have is Hamiltonian. Sorry, we have to keep track of this a+. So, a + time the h bar omega passes through and this term combined with this term is once again just the Hamiltonian. read the top line and then what's left over is h bar omega multiplying the one I'll write it down and then you can take a deep breath and stare at it and then what we had from before is the s almost there do you agree with my algebra any questions is h bar omega is just a number. So we passed it through and multiplied and identified two of those terms with the Hamiltonian identifying with the top line of that equation again. What was left over was 1 * h bar omega which gives me that second factor of plus h bar omega inside the bracket. And now we do the clever thing of multiplying this with this again to get number time I rather than operator time s. So but this is a number so we can bring this here. So this is just e + h bar omega time a + s proved. All of a sudden we have arrived at physics from this extremely dry algebra. And the physics is this. If satisfies the time independent shing equation with energy value e, so does a plus s but with energy value e plus h bar omega. And for this reason, the a+ operator is referred to as a raising operator and is one of a pair of ladder operators. Have you heard these terms before? And the other analogous result is the proof. Um this is one of a pair of analogous results. And the other result is as you can possibly guess. If Satisfies the time independent stringer equation with energy value E then so does A minus S but now with energy value E minus H bar omega And for this reason, the proof is very similar to what we just did. You would just use the other formula for the Hamiltonian to do it conveniently. Are you with me? And I strongly recommend going home and trying this. What's the worst thing that can happen? You'll probably not watch a video you would have watched otherwise. and lose 20 minutes doing this instead. Since we are doing ancient math uh for fun today, you might as well close this out and try these things out at home at least once. We are almost at something very beautiful. We are ready to construct a picture. What these two results are telling us is that somehow through magic if you happen to have one state sigh and the corresponding energy E for this problem. So you just happen to have this then you can solve the full problem because you can then construct the next state with the next allowed energy value by just operating the raising operator once and the next state by operating the raising operator another time with corresponding energy value E + 2 H bar omega and so forth. So on this side you can climb the ladder through a plus and on this side you can climb down the ladder by operating the lowering operator and correspondingly lowering the energies as follows. So you could climb down the ladder with a lowering operator. This is fantastic but problematic. Why is this problematic? Are you saying that the problem is that we are only allowed to climb rungs >> um which are at which are h bar omega or multiples of h bar omega different from each other. That's not the problem. That is in fact the quantum in quantum mechanics which is that the energy is quantized and so there are these discrete discrete allowed levels which for the quantum harmonic oscillator it so happens are equaced. So that's not the problem. The problem is that yes >> we we don't have a ground state like >> why do we need one? So like maybe you'll have end up with like a negative energy and that's not like >> you're close but what's wrong with that? >> Almost there. And there's a very cute argument which I want to give you in 2 minutes which is quite useful to understand why what you're saying is true. Okay. So theorem E must exceed the minimum value of V of X for any quantum mechanical problem. E must exceed the minimum value of V of X for normalizable wave functions. Can you prove this to me in a couple of sentences? In the interest of time, I'll give you the proof. This is a deep statement. We are making a general statement about any potential and then we'll also apply it to this problem which will get us to what this gentleman told us. Pratush rewrite the time independent shreddinger equation through trivial rearrangement as d2 dx2 is 2 m by hr² * vx minus e * s. Okay, that is s prime is positive number time this thing time. Let's interpret this equation in a figure. Since we have second derivative is positive number times this time the function if e less than v mining if this term in the brackets is positive then sp prime and s must have the same sign. Are you with me? I'm just reinterpreting the time independent shinger equation. Since this is a positive number, if e is less than v min, that would make this positive. This and this will have the same sign. What does that mean in figures? In figures, that means that s must be like this, always bending away from the x-axis. Do you see why? Consider let's say this portion. If s is increasing with x and its slope is increasing with x then you get a function of that form. So s and s doublep prime both have the same sign here and the rest of the shapes follow from symmetry arguments. So let's step back. If e is less than v min s has to always curve away from the x-axis. But what's the problem with such a sigh? If it never curves back down as x goes to infinity, you cannot normalize this. Right? S has to go to zero as x tends to infinity for it to be normalizable. And so we have proved our theorem. E must exceed the minimum value of vx for normalizable s. Are you wowed? You barely needed high school math to make this argument, but it's deep physics. And so we conclude that we have a problem with this ladder because our potential energy was half m omega² x². What's the minimum value for this? 0. The minimum value of our potential for the harmonic oscillator is zero. So surely I cannot indefinitely keep climbing down this ladder because it has to be kept at the bottom. There has to be a lowest energy state with energy greater than zero for this theorem to hold. What gives? Its energy must be greater than zero. must exceed. >> Yeah. So that's why >> Oh, sorry. You were saying say that again. >> No. How do we fix the ladder is the question. >> The ladder allows you to climb down indefinitely. That seems legit. We just derived this. But theorem says you can't have uh well- behaved wave functions if E is less than V min. How do we reconcile the two? Our ladder picture was constructed from theorems which did not guarantee that the size we would construct by raising and lowering the original well- behaved normalizable size would themselves be normalizable. [snorts] Let me say that again. You can start with well- behaved normalizable S and as you apply the raising operator. The theorem doesn't guarantee that the new wave function you have create or new solution to the time independent shooting equation you've created is itself normalizable. It only says that it solves the time independent shooting equation. So normalization is not guaranteed. Are you with me? By the same token, as you climb down the ladder, you have to check if the new solutions you're creating are normalizable. And what this theorem is saying is that at some point you will hit a non-normalizable solution which will correspond to the which will give you not correspond because it's which will give you the lowest rung of the ladder. Let's see how that works. the lowest rung of the ladder in will in fact okay maybe there's one more point that's worth unpacking before we do this how do you what does it mean for s to be non-normalizable what are the options how can you end up with non-normalizable size if size identically zero okay because then you can't make the area under the curve for mod square equal to one which is what we need for decently behaved probability densities. So one option is that size is identically zero that's non-normalizable. Another option is that mod s² is infinity. It turns out in this case we get the lowest rung of the ladder by looking for non-normalizable size which are identically zero. So in fact that gives us the following relation. If you try to climb down below the ground state, you hit a non-normalizable solution. Is the argument clear? So, see, we have not only fixed the ladder, but in fact completely solve the problem. You don't see it yet, but you will in two minutes. To construct this ladder, I needed to magically have E and S given to me and then I could construct the rest of the ladder. But now I have a relation that is going to give me the ground state. So I will have when I'm done solving this the lowest rung of the ladder and the corresponding energy and then through this process I can construct everything else because we already know how to do that. So no magic needed to get this pair. We have this. Okay. So just a little bit algebra left to unpack this and then we are going to finish constructing the ladder and solve the problem and go home. Here is a minus. [snorts] We need to operate it on s not to figure out what s not is conveniently the right hand side is zero. So we can forget about this prefactor. So we just have i * p operator which as you know is h bar / i * ddx operating on s not plus m omega x operator is just x is zero. This is a trivial differential equation that all of us can solve without any help from computers, right? You'll see in a moment. This is just d sin over sin equals - m omega / h bar * x * dx. I've just rearranged. This gives us sin not of x goes as you're going to get x² over two here and then we have to exponentiate because you'll have a logarithm here and so s goes as e ^ of - x² with the preactor m omega by 2 h bar which is the gausian. So you can look up integral tables to figure out the prefactor or just do the normalization yourself to figure out the prefactor. Doesn't matter what it is. Some prefactor which is given by normalization. We have s not and the ladder stops at s not. So we've cleaned up this picture. [snorts] This is now s and we know how to construct everything else provided we have e not. How do we find e not? Almost there. We go back to our dear friend the shreddinger equation and make a wise choice for the Hamiltonian again from the relations I've erased. We in fact choose h bar omega * a + a minus plus half operating on s equals e sin not you can check this in your notes that I got this right and the reason I chose this is that conveniently a minus operating on the ground state gives us zero that's how we identified the ground state correct Right? So this term drops out and you're left with h bar omega * half * sin not equals e sin. Taa we are done. E K is half H bar omega. So we have solved the full problem. For completion, let me write down what happens when you take the pair s not and e not e not and successively construct the first excited state and the next excited state and the next excited state and so on using the raising operator. You end up with as you can directly read off from our ladder picture, en is n + half h bar omega corresponding to n applications of the raising operator to saut. for which we have an explicit formula. That's your s n with the caveat that you must remember to normalize at each rung of the ladder. And that's just the normalization constant solved. What we have d Go ahead please loudly. [applause] Once you go back home and process this and internalize this, I want you to appreciate at least some of the many ways in which we have derived some profoundly disturbing results about the harmonic oscillator and see how you would close the gap between here and what we did in our lightning introduction with what the classical harmonic oscillator What's oscillating? It was very clear in the classical case what was oscillating. Where are the oscillations? Energy is quantized. Well, that's why it's quantum mechanics. What does that mean? What does it mean that the ground state has this nonzero energy half h bar omega? There are several physics courses over which these profoundly disturbing facets of this quantum result can be unpacked. But there are a few that you can try at home already. Are you familiar with the earnfest theorem? Is that how I say it? Let me write it down in case you pronounce it differently. I am sure you have seen this in a problem even if you haven't learn to identify it by that name. It is a result that tells you how you recover the classical picture from the quantum mechanical version. I want those of you who feel motivated to take what you've learned today and extract the maximum physics for all the sweaty algebra we've done to look up Earnfest theorem and see where the oscillations can be recovered from. How do you recover the classical oscillator from those? One more thing to try at home and then I'll stop. You can now construct explicit expressions for well let me just stop with Earnfest theorem. You can clap again and we'll call it job well done. [applause]