 All right, so there is no time to do, we'll probably have two more assignments before our next exam. Alright, so last time we were, we quickly reviewed some elementary solutions to the Schrodinger equation for one-dimensional problems. So we're looking at the energy eigenvalues and eigenstates, that is, the solutions to the independent Schrodinger equation. And so we have motion of a particle along a line, 1D. And there, of course, are two kinds of motion that one has. One has bound motion where the particle is bound between two turning points because at some point it's kinetic energy goes to zero and then it turns around. And then I hit another wall where it's kinetic energy goes to zero and turns around again. Or the particle might be unbound. There's no two turning points that bind bound motion. And in the quantum world there are thus two kinds of energy eigenstates, bound states, which typically we call the energy levels of the system, and then the unbound states, which correspond to situations where the particle scatters off of whatever is inducing the force. And the spectrum thus, the set of energy eigenvalues of this equation come in two flavors, the bounced eigenvalues, which are discrete because of the boundary conditions. They're only allowed, circle-loud solutions. Whereas in the unbound case there is no restriction set by the boundary conditions on the eigenvalue. And thus for the unbound states you have the continuum, we call it the continuum of energy eigenvalues. So when one is solving for something related to the unbound states, when it's not solving for the energy eigenvalues, because all of them are allowed, right? What one is solving for is what is the nature of the eigenfunction, and in particular we want to extract certain aspects about the scattering problem associated that are encoded in the eigenfunction associated with the solution of that energy, okay? And there are many different solutions at that energy. They're degeneracies. So we look at a particular physical context we might be interested in. For example, we might be interested in the problem of I send in waves from the left asymptotically far away as a plane wave, and I want to know what fraction of that say we send it in with a definite energy and far enough away, it's a free particle, so the wave function looks like a plane wave. And then we want to know what fraction or what probability amplitude is there for the wave to be reflected and what is the probability amplitude to be transmitted, right? So how do you solve that? How do you find those positions? So I have a question. That is a question to you. You solve the short equation with a boundary condition. So in this case, you have to find what the wave is inside the valve. You imagine this valve, you imagine that valve. That those boundary conditions involve the continuity of the wave function, the continuity of its derivative, and you solve it and you find those coefficients, okay? And that's a problem that we have in homework, which you've all done because it's due today. All right? We're going to study this kind of problem in more detail in three dimensions next semester where there, again, the problem isn't what is the energy item that is all or allowed. The problem is about something related to the strength of the scattering of a particle off of another particle or off of a scattering center, as we call it. And typically what we're interested in solving for is the scattering system. So we're going to talk about that next semester in greater detail. We might touch on it a little bit later on this semester. For the bound states, what we're interested in is the bound state energies because there are only certain ones are allowed. We're interested in what the energy item values are because not all are allowed. And also typically we're interested in what the eigenfunctions are as well. So we have a few factoids. One is that in 1D there are no, it's impossible to have a degeneracy unless the wave function vanishes in some finite breakage between the two. That is possible. But otherwise the eigenfunctions are not in general. In 1D, that's not true in higher dimensions, higher spatial dimensions. Moreover, the lowest energy level, what we call the ground state, the wave function has no nodes. And each successive bound state has one more node. So those are some little factoids. If the potential is reflection symmetric, then what we can say in that case is that the Hamiltonian is invariant or commutes with the parity operator. And because the Hamiltonian commutes with the parity operator, and because unless under some very bizarre circumstances there are no degeneracies, the eigenfunctions of the Hamiltonian must also be eigenfunctions of parity. So the wave functions corresponding to bound states are parity eigenstates. As I say, they're symmetric or anti-symmetric functions under reflection of the energy. And because of this fact that they go from no nodes to one node to two nodes to three nodes, they alternate. Ground state is an even parity solution, even odd, et cetera, et cetera. Of course, the simplest problem is the particle in a box where we think about hard wall boundary conditions in that case. And there has to be that the wave function has to go to zero at the walls. And then we have simple alternating cosines and sines, right? And the energy eigenvalues are just h squared, h bar squared, two n times the wave vector for the n eigenvalues squared. And I just want to emphasize one point here. Of course, we should be able to use the uncertainty principle just qualitatively to estimate what we expect the ground state energy to be in the problem. We should be able to do that without going through matching boundary conditions just from our intuition about the problem. Just from the uncertainty principle. How would you think I have any suggestions about that? Yes, please. Would the uncertainty and the position be the width of the wall? Right. So the particle is localized. So just qualitatively, qualitatively, for the ground state, we expect the uncertainty and the position is on the order of the width of the box, right? So how would you then estimate what the ground state energy is? Well, the energy is, yeah, go on. Do you know something like using the weight number? Sure, exactly. It's in the weight numbers related to the momentum, right? I mean, just physically, the energy of the system is the kinetic energy and the potential energy, right? Now in here we can say let's call this energy potential zero, okay? So inside this well, the energy is just connected. What would you say then? What can we say with this value of p squared? Well, for the ground state, this is the expected value of p squared for the ground state. Yeah. Go back to the uncertainty principle. Yeah. If we assume that this system is in a minimal uncertainty position, because we're in a box, we already know what the momentum localized in the box must be, right? Uh-huh. It's just H4 to L, right? So we don't have to, we don't know for sure if it's a minimum uncertainty state, but we could say it's on that order and that's a good thing. So we could say just from the uncertainty principle that this is on the order of that. And so what is the energy of the ground state there? Well, this is just the spread in momentum, right? Because the mean momentum is zero. So this is HR squared over 2 mL squared, which is basically what this is. Wait, back to the pie. It's not exactly that because, you know, inside of the uncertainty state and all that, but that is a kind of back of the envelope calculations everyone should be able to do. Okay? Now, what would you say happens, suppose I have a finite square well? And so now this is some finite depth, not infinity. Same thing. What would you say happens to the ground state energy? Does it go up or down relative to the infinite square well with hard walls? Ground states below the V dot where it's cousin on one's graph, they not notice the change in the tetrahedral rule? It will because of tons, right? So what does the ground state look like in this case? Well, it's got a little bit of an eminence in tail, okay? So what happens to the uncertainty in X in this case? Does it go up or down relative to this case? It goes up. It's slightly bigger, right? So the uncertainty in X is now, you know, it's bigger than that. The details are there. So what happens to the uncertainty in P? It goes down. It goes down when it was for the infinite square well. Which means the ground state energy does what? It goes down. So the ground state energy for the finite square well has to be less than the ground state energy for the infinite square well. This brings up an interesting concept that if the wall gets shallower and shallower, the momentum will go to zero. It will get smaller and smaller and smaller. It doesn't go to zero in this case, in fact. There is always one bound state. No matter how shallow the well, there's always one bound state. That's true for this. There's always a solution. No matter how shallow the well is, there's always a solution. It's barely bound, but it's bound, okay? Always one bound state in this case. You can always find a solution to this kind of problem. At least one. In fact, you're looking at the opposite problem in your homework where you have infinitely small width, where you might think the momentum gets infinitely high that it doesn't because of the way the width goes. There's always, you know, this bound state for the delta potential. Okay, now let me ask you one last question about this kind of problem. Suppose I have this kind of well. So this goes up to infinity. It's got a hard well right at the origin, but then a finite well will be not at this distance. Can you tell me what you expect the solutions to that to be like? Well, this is not reflection of symmetric, right? This is not a, you only have sines and cosines inside the well if their eigenstates are parity. This is not a reflection of symmetric potential. But it does have to be, the wave function does have to go to zero right there, I agree. But maybe I could give you a hint. This potential is what I have in the finite well and then I put a wall right in the middle, right? So given the solutions to this, can you tell me what the solutions to this problem are? Does that have to be sined just because we need that node? It's the odd parity pieces. So there is a solution that looks like this, which would have been like that. So the odd parity solution of the finite well has exactly the right boundary conditions to put a hard wall right in the middle. So this problem has all the odd parity solutions, all the ones that were reflection symmetric guys. So I'd rather just do it a little bit higher. Up here there's another guy, right here. Yeah. I don't think you could start the well where you only have the even parity solution then. That's not symmetric. Impossible. Okay, so if you have even symmetric, but this of course is odd symmetric. I'd be a little bit careful. V solutions are not odd parity with respect to the center of this well. Right? They're not, they're V solutions. This well does not have reflection symmetric, but they are the odd parity parts of this on that half line. There are no ones on any part of the line which are just the even solutions that doesn't exist. I don't know if it's an artifact in the picture, but was there like an adult potential in the middle of that one that you drew or no? It's just like, I mean you were just drawing the other half of the symmetric well. It wasn't like a delta function with potential infinity right at zero. No, a delta function potential is not an infinite part of the wall. Okay, so it's just... This thing is a wall. There's a potential here, and then it goes up to infinity. And this goes on forever. Right, okay. The delta function is just a spike. The delta function would have been this goes back down, and then it's a spike. I think I was confused by the axis. Right, and then it could turn around. That's that. Okay, last question about this do hit. I said in this problem, no matter how shallow the potential is, there is always one bound state. You could always find a solution that would satisfy the boundary conditions. What about this guy, just going up to infinity? The ground state always even though? No, the ground state always has no nodes. It's only even if the potential itself is reflection symmetric. This is not a reflection symmetric potential. These eigenfunctions are not eigenfunctions of parity. Yeah, so there the ground state would be the lowest the u2. Yeah, and as I lower the depth of this potential, is it always possible to have a bound state? No, no, why not? Because this guy disappears. And this guy is not this guy. So because of that problem, as I try to squeeze this guy in, it just can't make the boundary conditions. There's just no way to do it. So it's not the case that every finite square well always has a bound state. This guy there is a minimum depth in order to minimum depth, such that this solution, once I lower it too much, boop, pops out of the well. There's just no bound states at all. This case with the hard wall of the origin is important because it relates to the motion in 3D where this axis is now the radius from the origin. If I think about the problem in 3D where I separate coordinates, it says variable coordinates, an R only exists on the half line from zero to infinity. And so when we talk about, as we will, about bound states in 3D and scattering in 3D, there's finite numbers of bound states, has important implications for, say, scattering in 3D. All right. Very good. So that was our lightning review of piecewise onset potentials in life. And what we're going to, now I'll turn our attention to for about a week or so, is one of the most important problems of a well in 1D, the simple harmonic constant. I didn't care so much about the simple harmonic constant. Well, there are a number of reasons why we care about the simple harmonic constant. One is that this is, you know, one of the most important problems. Well, firstly, if I had any binding potential, or I would say most any binding potential, not all, but most, has finite curvature. So I might have, you know, a binding potential, say, for a molecule that looks like this as a function of position. But near the minimum of the potential, this thing is approximately quadratic. So as long as the particle is not moving too far up the potential, near the equilibrium point, it undergoes simple harmonic. So when we talk about, for example, vibrational spectra of molecules, to good approximation, often we can approximate that by harmonically bound atoms bound together in the intercomic potential that binds them. So that's one reason. Another important reason is, of course, so in field theory, we talk about quantum fields where fields, the degrees of freedom are in the normal mode. So if I have waves on a strain, waves moving on, surface of water, I can always decompose them into normal modes. And those normal, the dynamics of those normal modes are the dynamics are those of a simple harmonic constant, which I will always call the SHL. So when we talk about field theory, which we maybe will do at the very end of the academic year, we talk about the modes of the field and they are described by simple harmonic constant. So that's a whole lot of important reasons. There's one other important reason that we'll see is that it's a natural forum for formulating quantum mechanics in the phase space, as I say, in the coordinates of x and p together. We already saw a taste of this, which I intended we would do with homework in the bigger function, but we'll get a little deeper into that discussion as we go along. All right. So let's do some more monotonic. We all know it, of course. So the Hamiltonian, let's talk about this problem first classically. The Hamiltonian, kinetic energy and potential energy. The potential energy of the system, well, for a simple harmonic oscillator, the force is, you know, the spring constant and x. That's a simple harmonic oscillator and it is a natural resonance frequency that's better with the capital over n and the potential energy is one half capital, right? And with that, we are always almost always by the return of the oscillation frequency, right? So we have a harmonic potential and we typically take the potential energy to be zero at the origin. The equations of motion, Hamilton's equations of motion, regions of motion. x dot is p over n and p dot is a minus the gradient of force this equation, okay? Right, okay, fine. Now, if we're going to get somewhere in this problem ultimately, and this is something generic that I encourage you to think about always when you're getting first presented with a new problem versus a new one, that might be new in some ways. The first thing you should do is to find out what are the characteristic characteristics scales of the problem? That's the first thing you should do in analyzing any physics problem. Any physics problem has associated some characteristic units, some characteristic scales, and one should always express all physical quantities in terms of those characteristic scales. So now I'm going to go on my diet or I'm against SI units, which has to happen once a year. If you want to have a standardized set of units, it's fine if you're an engineer and you want to make sure all the screws fit into the space level the right way. But this doesn't mean you should measure everything you're doing in SI units all the time, because there are other scales. There are natural units. We're talking about high-energy field theory at the Grand Unified Scale. We're talking about the plan scale. We don't talk about things in terms of kilometers. If we're talking about atomic physics, there's the more rakingness. There's characteristic scales in the problem, which is why Fahrenheit is so much better than Celsius. Why in the world do I care where water boils when I'm talking about the weather? Hot is 100, not whenever. And I should measure my size in feet because a foot is how big I am. What's a centimeter? What does a centimeter have to do with my height? So that's why it makes much more sense in my view. So, come on, Cameron. I've said it a few times. Metric is stupid for measuring. All right, so... That's all his decision. All right. So, what are these characteristic scales? So, the way you go about... I mean, one way to do it is sort of systematically is to say the following. Let's say there's a characteristic... So there's a... Let's call it Xc, the characteristic length, Pc, the characteristic momentum, and dc, the characteristic energy. There's some characteristic scales of the problem. So I'm going to define dimensionless. The position I'll call capital X is the position in units of the characteristic. And the momentum coordinate in phase space is dimensionless. And I'll call the dimensionless Hamiltonian. So let's then plug that in over here. And what it says is dimensionless Hamiltonian is equal to the characteristic momentum squared over the characteristic energy and the mass times 1 half e squared plus the mass omega squared, the characteristic length squared over the characteristic energy. Just factoring out all the units. Now, where you put that factor of 2 is the mean of my existence. I have many things, but that's one of them. Because it just drives you crazy when you see there's a square root of 2 to float around. But so I'm going to not put the factors of 2 in here. So one way to define the characteristic units is to say, well, this is of order 1. That just defines the relationship between what I'm calling the characteristic energy and the characteristic momentum and the characteristic energy. So I'm going to define the characteristic units, define the scales. The characteristic energy, whatever it is, is whatever the characteristic momentum over which it's got to be the same. So this is the relationship between these this sets the scale. It doesn't define them yet because I have two equations with three unknowns. I need one more thing. We'll have that soon. So with that set, we can say the following. Let's look at what the dimensionless position in units of the characteristic scale is what is its time derivative as a function of time. So that's equal to the characteristic momentum over mass over there divided by the characteristic scale times, according to that, that's the scheme of putting all that together. You get this. Similarly, p dot is equal to minus m omega squared the characteristic this over the characteristic momentum x, which is minus 4. So in dimensionless units, we have the dimensionless Hamiltonian is 1 half x squared plus p squared and x dot in a dot plus omega p and p dot. So what's the solution to this equation? Well, if I take the second derivative of this p double dot, so what we get is that x double dot plus omega squared x minus x is 0 and p double dot plus omega squared p. Of course, that is the differential equation of the simple harmonic oscillator as we know it. And the solution is there is two initial conditions. As usual, x dot is omega p dot, so this is p dot is 0 times omega, so we have the following solutions. x of t is x of 0 plus omega t plus p of 0 sine omega t and p of t is equal to p of 0 plus omega t and p dot is minus omega x so this is minus x dot. So the trajectory in phase space written in terms of these dimensionless canonical coordinates is what? It's a circle. Indeed, it's a circle and the circle has an amplitude here which is the radius of the circle which is what? Well, it's equal to the square root of x 0 squared and p 0 squared. Is this circle going clockwise or counterclockwise? So what's happening to p as a function of time? p is decreasing so it's going in this direction, right? It's going clockwise. So this phase this is phase space but I can also think about this as a complex space. This is a phasor and its phase is changing as e to the minus i. That's why in physics we always choose e to the minus i. Because that's the direction it's going in phase space. The phase is decreasing. Well, this gives us a hint that we can define this in terms of complex amplitudes. Whenever you have signs and code signs you can always define that in terms of the real or imaginary parts of some complex amplitude. And that complex amplitude so this is the trajectory in phase space it's a circle. The energy of the system by the way is what? 1.5 is constant. This is the energy at all times. So this says that we should define a complex amplitude and this is where that student factor 2 becomes confusing. What is in here? Let me define alpha. It's a complex number it's a function of time whose real part is related to x and the imaginary part is related to p. If it were just that it would just be that vector but we're going to put in the square root of 2 inside there so that x of t is the real part of alpha kind of term. And p of t is the imaginary part of alpha. So this is where p of t is square root of 2 there. I'm just saying that that's p. You're right. If you're actually right it does that must be the case because that's what's constant. It is over root 2 because that's where the energy is constant. Right? Because that is what you get right? Right. And so that's exactly why we're putting it this way. So the energy of the system is in fact equal to alpha star alpha squared A squared. That's equal to 0. So of course it's true at any time that this case alpha is constant. What is alpha of t? Well let's plug that in. Let's plug in these guys. We have x of t and p of t. When you plug that in you get the following assignment for you to try that. This is what we call so this is A into the i phi where A is the square root of the magnitude of alpha that's the square root of and phi is the argument. It's the art tangent of the imaginary part over the real part. So what we have is the solution to the motion of the particle in phase space alpha of t is equal to alpha of 0 e to the minus i omega t where we have alpha of 0 is x plus i t at time equal 0 over 2 and the position of the particle at time t is the real part of alpha and the momentum is the imaginary part. We always think about harmonic motion as the motion of a phaser in phase space that just rotates at the angular velocity omega crop wise. Projected on the x axis we have harmonic motion back and forth at frequency omega projected on the momentum axis we have 90 degrees out of phase harmonic motion. So if we have two harmonic motions that are 90 degrees out of phase with one another that's a circle. Of course we know that the momentum is 90 degrees out of phase with momentum because when the kinetic energy goes to 0 potential energy is maximum when potential energy goes to 0 the energy is maximum. It's called classical physics. However now we want to quantize the phi right there is that like the initial displacement? That's exactly what it is. So it's just the initial if we go back to this picture let's say at time equal 0 we have some initial we have our spring it has some initial displacement and then we have a ball p hammer and we give it a kick. So we have some initial momentum and some initial position. This is phi 0 and then the guy rotates. So this phi is the initial phase in the space plane and then at a time oscillates the question is excellent. Alright quantize what it means is that now X and P are not no longer commuting numbers they're non-commuting operators. So our dimension full X and P satisfy the canonical commutation relation. So we have a new constant in the column before we only had two constants and that completely defined the problem. Now we have a new constant H bar. So we can define all of our characteristic scales now. What is the natural union? What we should have is that the characteristic of the problem should be H bar. Put that together with what we have over there. So we had EC is equal to H bar over M is M omega squared times the characteristic scale of length. You put all that together and you get the following. The characteristic energy scale is exactly how you would expect it. H bar omega. The characteristic momentum scale is whatever the mass is and the characteristic energy scale is squared that's the square root of H bar and omega. And the characteristic length scale is the momentum scale over H bar. So just from this kind of dimensionless analysis analysis of scales we have a kind of sense of what we expected. From the same picture of a particle in a box and just thinking about the characteristic scales from the outside. Oh, there's a length so there's a momentum on certainty. We kind of can guess what order, what's the ground state energy of the harmonic oscillator approximately. And what is the width in position momentum and here they are. And we'll see if, of course, you know that's starting close. We'll get to those. So that's a kind of analysis of every good physicist. So, our Hamiltonian again, we're going to define a dimensionless position in momentum. It's equal to whatever the position operator in units of this characteristic length scale. And a dimensionless momentum operator give them my scaling. The dimension full momentum operator by that scale. And the commutator of dimensionless x with the dimensionless p relative to these scales is i of h bar divided by the product of those scales which is i. Hamiltonian operator kinetic operator and momentum operator plus sorry, kinetic operator and potential energy. In scaling out the units is h bar of omega times x squared plus p squared over. Notice the thing about the harmonic oscillator and this goes to the point I made about phase space plus the potential energy is quadratic and the kinetic energy is quadratic. It's symmetric in x and p which what makes it kind of a natural form for talking about phase space and quantum mechanics. Alright. Now, what about this, we have a non-permission operator an important role here and that is the quantized version of the complex amplitude. The complex amplitude is a complex number classically quantumly it's a non-permission operator. So I'm going to define the operator a as 1 over root 2 x plus i p. It's just the quantized version of the complex amplitude. I could do this in the Heisenberg picture or the Schrodinger picture I don't care. Okay, that operator remember that a does not equal a dagger. A dagger is x minus i p. A is not a normal operator meaning it does not commute with its adjoint. Let us calculate the commutator of a with a dagger. Plugging it in. So I get backering out the root 2s. So let's do it. x can be x but it doesn't commute with p. Right? So this is equal to minus i over 2 the commutator of x with p and then plus i over 2 the commutator of p with x. And the commutator of x with p in dimensionless units is i. Thank you very much. So i times minus i is plus 1 and this is negative i and negative i times i is so what we get is the canonical commutation relation relative to the quantized amplitudes alpha and alpha star no longer commute. Or a and a dagger no longer commute and the commutator is a famous one. If we look at our Hamiltonian and we plug in a and a dagger over here we get the following. This is equal to a 4 omega a dagger a plus a a dagger over 2. Classically we said somewhere here oh thank you that the Hamiltonian which is alpha star alpha quantumly because these guys don't commute you have to symmetrize it. So you have the equivalent of alpha star alpha plus alpha alpha star over 2 I mean I don't do that by fiat that's how it comes out when you plug in a and a you plug in x and p in terms of a and a dagger remember x is the real part which is alpha plus alpha or the Hermitian part I should say times root 2 and p is the anti-Hermitian part which is a minus a dagger over 2i times root 2 so you plug these guys into this guy and you get this which if I use the commutator a with a dagger put it the other way I get a dagger a plus a 1 so I get this hr omega a dagger a things you have all seen before but in a different light alright so as you know the equivalent of the complex amplitude squared which is the magnitude of alpha squared quantizes a dagger a this operator we call n so n is the quantum version of the magnitude of the amplitude squared related to energy of the oscillator so n is defined to be a dagger a this is Hermitian we take the adjoint so this is a Hermitian operator and thus it has eigenvalues and eigenvectors Hermitian we can always diagonalize so there exists a set of eigenvectors with eigenvalues I'll call lambda to start I guess Hermitian implies that ok moreover we know some important commutation properties of n n commuted with a well that's a dagger a commuted with a how do you do that yeah you pull out the one that the first guy with that commutation that's ok you put that guy on that side so that's a dagger a a and this is a negative one and the commutator between n and a dagger do the same trick or rune pull out the a dagger you're left with the commutator of a with a dagger that one's a plus one so with that said we can solve for the eigen values of n and thus the eigenvalues of h by the way go back to this picture of the harmonic gospel here what can you tell me about the spectrum the set of eigenvalues of the harmonic gospel here a continuum this potential as a harmonic potential goes off to infinity means of all the states all the in a harmonic potential everything is a bound state there are no unbound states it's true but they're not it's not a continuum it's disappeared so all that you have only bound states there are no unbound states but a harmonic gospel so the entire spectrum of the harmonic gospel of the potential is a square well but not like the finite square well the finite square well has both discrete and continuous eigen all right back to where we were all right so lemma lemma if I apply a to this eigenvector of n it is also an eigenvector of n with eigenvalue lambda n minus one similarly if I apply this this guy is an eigenvector of n with eigenvalue how do we do that well it's easy to check that's what we're deriving that's what we're just deriving indeed that's what it means so acting on this is equal to the commutator of n plus how do you use the commutator you write it up in the other order and you add the commutator and we just said that this was equal to minus a so this then says that this is even since this is an eigenvector with eigenvalue lambda n this is lambda n minus one by the same argument using the other commutator this is lambda n plus one so it increases the eigenvalue by one or lowers it now what are the eigenvalues well to show that that fact they are the integers non-negative integers use the following fact note and the positive operator what happened to the a there you're ending on a dot a you're going to get a lambda minus one times a u where the a comes from sorry there shouldn't be any dagger for me so I activated n on this I thought that you made the a by accident but you see the a is there for the following reason when I do the commutator nx on that it gives me lambda so then I get lambda minus one alright so what I'm saying here is that n is a positive operator how do we know that well to be a positive operator it means the expectation value of this is greater than or equal to zero for all how do we see that well if we plug in for this equals that but this is equal to the norm squared of that and that's a positive number or a non-negative number I should say it can be zero well what it says is that if this is zero it's annihilated it's the only possible way that's why a is known as the annihilation operator what state does it annihilate well how can we see that what we said is that if we apply a to one of the eigenstates of the number operator we get another eigenstate with eigenvalue decreased by one if I do it twice the same thing I get an eigenvector with eigenvalue that's decreased by two units if I do it three times three units if I do it n times it's decreased n units however however because n is a positive operator it has positive eigenvalues which means that at some point this screws up because this will start being negative which means there's got to be a minimum n below which I can't lower anymore there must exist a minimum value of lambda it can't go to minus infinity because this is a positive operator what this implies is that there exists a minimum case such that a does not lower it to any lower eigenvalues the only possible way to be consistent so if that's true then what is the eigenvalue of the number operator on this state I'm calling 0 well that's a dagger a acting like u0 but u0 is annihilated by a so this is 0 so that implies that lambda0 is 0 lambda1 is one more and lambdan is n QED thus n are the natural numbers and thus we call n the number a and a dagger are played a role here of annihilation and creation or the ladder operator that is to say so by the way for shorthand from now on we'll just label this head by it's label n so this has got to be proportional to n-1 and a dagger gives us something proportional it raises and lowers and a on load n equals 0 finally what is the proportionality constant so let's say you want to know what that proportionality constant is what we can do is multiply it by its adjuvant this is a normalized vector so this is 1 this is the number operator so what is this expectation value it's just n right so that helps me that this proportionality constant is some phase times the square root of n and by convention we choose that to be that by convention there's nothing that fixes that phase we just choose it by convention so what we have is that this is equal to square root of n n-1 and if you go through the same algebra this is the square root of n-1 finally if you do this over and over again what you could show is that n is equal to will you raise this n times on the ground state and then you know the most energy state of n is the ground state of our monocosal n equals 0 the energy eigenvalues so h acting on n is e to the n n according to that is h or omega n plus a half n equals 0 1, 2, dot dot dot ground state of course you know what that says is that the energy eigenvalues equally space in the energy separated by h or omega and the zero point energy the energy of the ground state is h or omega over