 We mentioned the states of well-defined momentum, referred to the playing waves in mynd on the wave number Pe upon momentum upon H bar. We can go back to the problem of too-slit interference and ask ourselves why it is that quantum interference is not observed in Macroscopic objects and assess what experimental set up will be required to see quantum interference with say electrons. I can put some numbers into this experiment. So, as you will recall, we had some gun here that was symmetrically placed with respect to a couple of slits here in an obscuring stream. Screening it fired out particles. Some of the particles got through the holes and they came to here and we said that the amplitude up here would be the sum of the quantum amplitude to arrive at this point would be the sum of the amplitude to go via this root, or to go via this root. These roots, well, the distance from the gun to the two slits by setup is symmetrical, is equal. So any difference in the amplitude that they come there is to do with the change in the amplitude when it goes along this root as against along this root. So let d plus be the distance from the upper slit to this place and d minus the distance to the lower slit, then we can write a formula that d plus by Pythagoras' theorem is going to be l squared plus x minus s squared square root, and correspondingly d minus is going to be the square root of l squared plus x plus s squared, and we make the reasonable conjecture. I mean the right way to think about this is these particles, let's imagine that we've got our gun here has been tuned to emit particles with some well-defined energy. That means that as they go along here, the particles will have some reasonably well-defined momentum because their energy will consist of their kinetic energy. So along here we will have that, that the amplitude is going to be on the order of e to the i p upon h bar times x. So that's a reasonable model for how the amplitude varies with position. That's the new information that we bring to bear on this. So what will be the difference? So the probability to arrive at x, as you recalled, was equal to the amplitude to arrive by the top slot plus the amplitude to arrive at the bottom slot, mod squared, and the argument that this was a plus mod squared plus a minus mod squared plus twice the real part of a plus a minus, and these were about equal, and well, these were the classical probabilities. So these were p plus plus p minus, and then we have this quantum interference term which we want to assess. So what is this quantum interference term? So the interference term is going to be mod a plus mod a minus, neither, not very interesting, and the crucial thing is that this one is going to be e to the i p upon h bar d plus, and the other one's going to be e to the minus i p upon h bar plus, sorry, one of these needs to have a star in it, which means that one of these requires a minus sign. So we're interested, excuse me, in the real part of this, and the real part, so this thing can be written as cos p upon h bar d minus minus d plus, et cetera, et cetera, et cetera. So this is looking like to the probability to go either way through each hole separately times the cosine of p over h bar d plus minus d minus, actually it's easier to do it the other way around, isn't it, minus plus. So what is this difference of distances? This difference of distances from up there, well, let's binomial expand this, and because we can argue that l is going to be big experimentally compared to x, x will be millimeters, l will be a meter or so, so we can binomial expand this and say that this is l, it's about half of l brackets one plus x minus s squared over l squared plus dot dot dot, and the other one is going to be about half l one plus x plus s squared over l squared plus dot dot dot. So when we take the difference of those two, so this is going to be two p plus minus, this probability times the cosine of p upon h bar of the difference is going to be x plus s squared over l minus x minus s squared over two l in fact. So when you take the difference of those two, you will be looking at two x s over l, yes, two x s over l is what that difference will be. So now let's put it, now we need to put in some numbers, right? Suppose we take the energy which is p squared over two m, what do we want to do? We want to, no. So what does this do? This gives us a probability of arrival as a function of x which is going to consist of twice, the sum of these two which are going to be about equal and so about twice this plus twice this thing times this cosine. So it's going to be an oscillating commodity which we'll be doing this and there's some characteristic distance between these minima which, so this we'll call this delta x say, no let's call it big x, I think that's what the, so we call this big x, the distance between the places where it's a minimum because the quantum interference term is cancelling the classical term and this difference is what causes that, the argument of that cosine to become two pi, so we can write that two pi is equal to p over h bar times two s over l times x. In other words, we have a formula for the distance between the minima which is two pi, two pi h bar, but h bar is h Planck's constant over two pi, so that's going to be h over p hl, sorry, yes. And we could also write this, I guess, we could also say that, oh no, let's not bother. So let's put some numbers in, let's say that e, the energy is, so in order to get a big value of x we want to take a big value of l, needless to say we want to take a small value of p and of course a small value of s, I've lost a two, yeah you're quite right, so let's take l to be a metre, let's take s to be, I think, yeah what did I do, I think it's, yeah, a micron, because you want to make it as small as you can, but if you make it much smaller than a micron you'll find it difficult to make the whole, using ordinary materials, and in order to get a small value of p we want to take a small value of the energy, but you can't take too small a value of the energy or your particles will be deflected by stray electromagnetic fields and stuff and it'll be difficult to keep any kind of coherence. So let's take 100 ev say as a sort of convenient low speed, if you plug all this stuff into there, so that gives you, what does it give you, a twentieth of a millimetre or something, I think it's, is it 0.06 millimetres, which is obviously perfectly, perfectly observable, yeah. So this such an interference experiment is possible but hard using electrons. If you do the same thing with bullets, when we're not expecting anything to happen, what could we do? We would, we would take that the velocity for a gun is say 300 metres a second, might be a bit faster these days, I'm not sure, but that's a classical, that's, you know, faster than sound, so that's sort of a reasonable ballpark figure. I suppose we took L to be 1 kilometre, 1000 yards per reasonable shooting distance for a rifle, and if we took out the mass to be 10 grams, put it into the same formula, and we discover that X is some ridiculous figure, 10 to the minus 29 metres. So it's obvious that you cannot observe this interference using anything like a bullet, any kind of macroscopic, any, any kind of macroscopic object, because it's going to be vastly bigger itself than the, than the size of the interference pattern. Obviously an absolutely basic requirement for this experiment to work is that the physical size of your particle has to be smaller than the value of X that you derive out of this, so you haven't a hope of measuring this interference. So that's why classically we don't, we are unaware of this interference term, but I would remind you that in the last lecture we recovered classical results, which would explain why cricket balls move as they do, why satellites and so on move as they do by interpreting, we calculated this, we obtained results which recovered classical physics by decomposing the amplitude to arrive into a sum of contributions from states of different well-defined momentum, and these were all interfering with each other, and the classical physics came back as a result of quantum interference, so this quantum interference on the one hand is something which is very hard to observe with classical objects, on the other hand our entire picture of the classical world, a classical world is only recovered through quantum interference. It's not, it's not some esoteric corner of the subject, but it is hard to, it's hard to have it happen in a controlled way. Okay, so we, yeah, we should just, so we've done, we've done the position representation in just one dimension, everything has been with a one-dimensional motion, motion along X, we obviously need to generalize the position representation to three dimensions because we live in a three-dimensional world for whatever reason, and the generalization is, is nice and trivial, we don't need to worry about it, we have, we now have three position operators, X, Y, Z, also known as XI, all right, and we have, of course, three momentum operators, three more operators, PX, PY, and PZ, also known as PI, and we have that every one of these operators commutes with the other one, so we have that XI, XJ is nothing, and every one of the momentum operators commutes, PI, PJ equals nothing, so it is possible to simultaneously know your X coordinate, your Y coordinate, and your Z coordinate, there's a complete set of eigenstates of well-defined states, states where you know all those strings simultaneously, or you can know all three components of momentum, but you can't know there's not a complete set of states for knowing, and so on, and the only other interesting thing we have to have is XI commuted with PJ is IH bar delta IJ, so it is possible to know the X position and the Y momentum, but it's not possible to know the X position and the X momentum, so most of these operators commute with, well, each operator commutes with five of the, sorry, four of the remaining five operators, but it does not commute with its own momentum, that's what each of these position operators, so that's the generalisation there. What else do we have to say? Well, we used to have a wave function of psi being a function of scalar X, we now, it's trivial, the argument of the wave, we can now label a complete set of states by X, Y and Z, so we can write that there's a, we have states of well-defined position which are labelled by a vector now, vector position X, because there are three, this is an eigenstate of the X operator, it's an eigenstate of the Y operator and an eigenstate of the Z operator, so we need three eigenvalues written inside here to describe what this is, it is, that's at a mathematical level, at a physical level, this is the state of being at the location position vector X, correspondingly our wave functions become functions of X, Y and Z because they become these complex numbers, right, that's still a complex number, this complex number, but it's a function of X and Y and Z, the locations of the particles, similarly we have states of well-defined momentum, UP, we have states, yeah, yeah, UP of X, which is X, P, so now we have, here we have PX, PY and PZ because we have a state of well-defined momentum which is labelled with all three components of momentum, so we have this function of a complex of three, sorry, this complex function of three variables X, Y and Z labelled by the momentum, this is just an identical notation, whereas in single, when we were doing this in one dimension we found that this was E to the I, P over H bar times X not vectors, now that's a vector, that's a vector, and whereas on the bottom we used to have H bar to the one-half, now we have H bar to the three-halves reflecting the fact that there's an X component to this, a Y component to this, and a Z component to this, so this wave function of a state of well-defined momentum has now become a plain wave whose wave surface is a normal to the vector P, and that's what it is, it's easy to check that that stuff works, it's a very straightforward generalisation of what we did before, and I think that's all we have to say, oh you know, not quite, we also want to say what the momentum operator P looks like, so previously we had an X, P, X, P, sorry, it's not what I'm talking about, yeah, X, P, of Psi was minus IH bar was introduced by this formula here, D by the X of X of Psi, all right, that was what we did in one dimension, that generalises in three dimensions very straightforwardly to X, P, Psi, so that's become a vector, that's become a vector because we have to write down what it is for PX, PY and PZ, this is really going to be a shorthand for three formulae and it's going to be minus IH bar gradient operator on the function of three variables, function on space, this one here, all right, the wave function, so this is a vector reflecting the fact that that's a vector, this is just a label which appears on both sides of the equation, that's what this formula generalises to that formula, I don't think we need to be detained about that any longer, before we leave the position representation it's good to do a useful result which falls into our laps now because of what we've already done called the virial theorem, which is a theorem, it's a result in classical physics which you may not have met, I don't know but you in a way should have met, did you cover the virial theorem in classical mechanics anywhere, no anyway, so it's nothing quantum mechanical about the virial theorem, it has a classical counterpart but it's going to fall into our laps because it's got this powerful machinery, so do you recall if we are in a stationary state that is to say a state in which the result of measuring energy is certain then all expectation values for such a state are constants, that's why we call it a stationary state, it's going nowhere, so every expectation value for a stationary state, for a state of well-defined energy is independent of time, so we want to exploit that result, so for a stationary state, this is just recalling what we already had, it was a consequence of Ehrenfest theorem, for a stationary state we have that dbd time of EqE equals 0 for all operators Q, it doesn't matter what observable you stuff in there, as long as the observable is defined in a way that is independent of time, so it's something like position momentum, angular momentum, whatever, it has a vanishing rate of change or with respect to time, it's a constant, so we now apply this result to Q is equal to x dot p, so then we have that 0 is equal to dbdT, sudden moment of doubt, so I want to apply this to x dot p E, and let's divide that by Ehrenfest theorem is x dot p comma h, so Ehrenfest theorem tells us that this rate of change which vanishes is equal to this here, and now let's take, suppose we're dealing with a particle which has kinetic energy and potential energy, so we'll take the Hamiltonian to be of that form, which is a pretty useful form, and stuff it in there, and we're going to have that 0 is equal to E dot p comma p squared over 2m plus v, closed square brackets, so now we need to work out what this commutator is, and this is where a little bit of, so this is where we get a bit of practice in using the three-dimensional generalisation, we obviously have two things to work out, we've got a commutator of x comma p with p squared, so let's work that out, x comma p with p squared, now we write that in components, we write x dot p, sorry x dot p, did I say comma x dot p comma p squared, this x dot can be written as a sum over j, j equals 1 to 3 of x i p i, sorry x j p j, so that's just a way of writing that, and now I have a sum pk, well p squared k, so now I'm summing over k as well, right, p squared is px squared plus py squared plus pz squared, now we can work out this using our rules for a commutator, we had that rule that a comma b, sorry ab comma c was equal to a comma c b plus ab comma c, p, we know that pk commutes with pj that's been written down up there, so that commutator vanishes, that's this one here in some sense, sorry there's this one here in some sense, and so what we're left with is, so we have this double sum, we're going to have xj pk pj, sorry let's squared, squared comma, no no, no comma, so that's what we get, so this has to be commuted with that, that's what I've written down I hope, and then there should in principle be another term this commuting with this but that vanishes because pj commutes with pk for all for all j and k, so we have to work out what this one is now, and we can use the same rule if we're being pedantic, we would say this is x on pk pk, so we would say that this is xj pk pk plus pk the commutator of x and pk, I'm using the same rule and that all has to be multiplied by pj, the same because I'm now writing p squared as pk pk, but this is ih bar, this is ih bar, so these two terms actually contribute the same thing, this becomes two ih bar times delta jk times pk times pj, and I'm sorry I've lost track of the sum sign, here we have a sum sign, we're summing over j and we're summing over k, sum over j and you get nothing because of this delta jk unless j is equal to k, so this becomes pk pk summed over k, but pk pk summed over k is the same thing as p squared, so this is two ih bar p squared, that's what the commutator is of x dot p with p squared, now let's write, now let's do the x dot p commutator with v which is itself a function of x of course, these things all ought to have hats really but one gets difficult to write down enough things, well what we want to do is write this thoroughly in the position representation, in the position representation x dot p is minus ih bar x dot gradient, right that's what this becomes in the position representation on v which becomes a function of x, just so this is in the position representation, so what does that mean, that means ih bar minus ih bar brackets x dot gradient working on v minus v x dot gradient, and this is an operator statement so it's waiting for you to put in the function of your choice of psi on the right, right, there's a virtual function there for it to work on that's what that's the meaning of this v x dot gradient, and this x dot gradient v doesn't mean extra gradient only v, it means of everything that is to the right of it including your psi, so when you use this x dot v on v times alone you'll get a term and then you will still have to use the x dot v on the psi, but the result of using the x dot v on the psi will be killed by here an x dot v on psi, so what this is equal to is minus ih bar x dot gradient of v, that's all that survives this is the action of nabla the gradient operator on the potential itself, the operation of the action of the gradient operator on the wave function that's virtually sitting here is cancelled by this contribution here, so we now have we can put these results back into what we had up there so what we had was naught is equal to yeah is equal to the sum of these of these commutators is equal to e x dot p comma p squared um e plus e x dot p v, that's just summarizing where we stand, this we've discovered to be this is 2 ih bar, this commutator turned out to be p squared so it becomes the expectation oops there should have been over 2m on this shouldn't there because it was the Hamiltonian p squared over 2m uh yeah this p came from the Hamiltonian where it was p squared over 2m this v came from the Hamiltonian where it was just v so we have over 2m uh no let's leave that alone of p squared over 2m e plus um we figured out that this one was uh minus ih bar um so we want to cancel uh what we this is the expectation value of the kinetic energy clearly right p squared over 2m is the kinetic energy this is the expectation value of it um so canceling the ih bar we can say that 2 times the expectation value of the kinetic energy is equal to this stuff that's as far as we can go in general um but consider now very important cases have that v of x is proportional to mod x to the alpha so for example for a simple harmonic oscillator we're about to discuss the potential energy goes like x squared alpha is 2 if we were doing with the coo dealing with the Coulomb interaction the potential energy potential energy goes like 1 over radius so it would be v of r is proportional to 1 over r um well this mod x is r so alpha would be minus 1 so we can say that alpha equals 2 is simple harmonic motion alpha equals minus 1 is Coulomb there are you know you can think of other power laws uh which are relevant in this case so then if we ask yourself what is x dot gradient of um of v well that's going to be so so we'll say that this is equal to some constant a times x to the alpha what is this going to be it's going to be um alpha mod x to the alpha minus 1 times x dot the gradient of mod x and the gradient of mod x is uh uh the gradient of mod x is x the unit vector x so it's at the vector x over mod x so this is equal to sorry this is a we have a x to the alpha minus 1 times x dot x over mod x so here this mod x is going to make this an alpha to the minus 2 but from this x dot x we're going to get a mod x squared so this is going to be and i've lost sorry this was an alpha there was also an a unfortunately yes sorry we need an a and an alpha um this is going to be alpha times a x to the alpha which is alpha times v so if v has a power law dependence on distance from the origin then x dot grad v is simply alpha times v so when we put this result back into that formula that back into this statement here we have that twice the ke expectation value is equal to alpha times the expectation value of the potential energy so that's our kepler formula in the case of simple harmonic motion alpha is 2 and kinetic energy is equal to potential energy in the case of Coulomb interaction where alpha is minus 1 you have that the potential energy is minus twice the kinetic energy which is to say that the particle has lost two units of energy uh radiate in falling in from infinity into a bound orbit it's lost two units of energy uh one unit's been sent off to infinity in radiation or something and one unit is used as kinetic energy uh of its orbit so that's the this is a virial theorem so now we opened a new chapter as it were by talking about harmonic motion the harmonic oscillator is the single most important dynamical system in physics most of field theory most of condensed of quantum field theory most of condensed matter physics is fiddling with uh more or less with harmonic oscillators which uh which are which are decorated in some way so the basic physics is that of the harmonic oscillator and it's it's worth just taking a moment to understand why harmonic oscillators are all over the place the universe physicists a fundamental position of physicists well physicists like to represent the universe is a collection of harmonic oscillators and this is partly because physicists uh may be brighter than some other people but they're still pretty stupid we have a quite a small bag of tricks and a harmonic oscillator is is is a trick that we we have and it's an incredibly useful trick for this reason that if you plot force in some direction versus displacement from a point of equilibrium you will get a curve which does something like this um the force vanishes uh at an equilibrium at a point of equilibrium of a system the force on it obviously vanishes so if you do a plot of force versus distance you'll get a curve something like this passing through zero at the point of equilibrium which i happen to put at the origin of x but you know that's by construction clearly and the general idea is that most of the time you can rep sorry that's meant to go through the through the origin most of the time you can represent this uh to some to a good approximation you can say that f of x uh is about equal to kx plus order of x squared or whatever so to lowest order approximation because f has to vanish at a point of equilibrium in the neighbourhood of the point of equilibrium f is going to be proportional to x and if f is and if we neglect this if this is small then we have harmonic motion for displacements these small displacements around here so this is why harmonic oscillators are ubiquitous a very incredibly an incredibly valuable model we can apply we can use to understand many many systems because many systems for small displacements uh almost all systems for small displacements look like a harmonic oscillator okay so let's agree what the Hamiltonian of this thing should be the Hamiltonian of our harmonic oscillator should be p squared over 2m plus a half kx squared right because if the force is going like that you integrate it up this becomes the potential energy and this is of course the kinetic energy we're familiar with that already it's better though to write this in in a different way to anticipate results that are to come and to write this as p squared plus m omega x squared over 2m so uh and of course omega squared is k over m so it's easy defining omega squared to be k over m it's easy to write this formula like that and that's how I want to write it if you want to reproduce this formula just think about dimensional analysis we want to have p squared over 2m because it's the kinetic energy we're always saying that and here I want something that's proportional to x squared and has the dimensions of momentum and obviously omega x has dimensions of speed so m omega x has dimensions of momentum so that's why you know that enables you to recover that quickly from this and that's the way to go for practical purposes so there's our Hamiltonian and we're trying of course to solve the stationary states are the key to understanding dynamics because they have this trivial time evolution and by by decomposing any initial condition into a sum of of stationary states into a linear superposition of stationary states then evolving the stationary states we find out how any arbitrary initial condition evolves in time so that's why we want these stationary states I've said that before and I'll say that again so we want to find states of well-defined energy this is the problem we want to solve and this is a completely generic situation in physics first of all you think about your physical system on the grounds of physics you write down the Hamiltonian then the next thing you do is you find the damn stationary states because that once you've got those you can do anything you want pretty much so that's what we're trying to solve the the the way to do this is the the proper way to to find these states uh so we need to find the energies that that uh are possible and we need to find the corresponding states and the way to do this is to introduce some new operators let's introduce a which is m omega x plus ip over the square root of two m omega or h bar omega why do I write that down well basically because I know where I'm going but just to give you some sense of direction the general idea here is that we want to factorize that that's the general idea we want to factorize the Hamiltonian into you know it's a quadratic expression it seems kind of reasonable to factorize it uh if these were if these weren't operators because these are operators sorry and in future I'm not going to even attempt to put hats on operators right these are operators despite the absence of hats it's just too difficult to remember to put the hats on and takes too much time and cronups never do but these are operators now if if they but if they weren't operators this and its complex conjugate would factorize that so that's the drift okay let's write down its complex this of course is an operator and it's not an observable it's not a Hermitian operator it's what is its dagger a dagger is Hermitian adjoint is this thing dagger which is itself because x is a Hermitian operator on its own dagger so it's m omega x plus this thing dagger p is its own dagger but i has the the dagger of i the Hermitian adjoint of i is minus i so this is minus ip over of course this on the bottom is a real number so it's its own it's its own complex conjugate so here we have two operators and the general idea is they're going to factorize h or they almost do whatever that's the plan um and this is called annihilation operator and this is a creation operator and we'll we'll the reason they have these names will emerge but it is that if you use this on a state this operator increases the excitation of our harmonic oscillator and this oscillator reduces the excitation of our harmonic oscillator and since in quantum field theory particles are excitations of the vacuum this thing creates a particle because it creates an excitation which is a particle and this thing destroys a particle because it destroys an excitation so what we next do is work out what a a dagger a is because the idea was it that this product would be more or less the Hamiltonian so what exactly is it let's get this right m omega x minus ip m omega x plus ip over two m h bar omega now when we write this out we have the obvious terms we have p squared and we have m omega x squared so let's write those down that's p squared plus m omega x squared all over two m h bar h bar omega two m omega h bar whatever um and um and then we have some additional terms which would class it would cancel in classical algebra but don't now because we have an x we have an m omega x times ip and here we have an m omega x on the sorry have an i minus ip times an m omega x so the additional term is an m omega x m omega i x comma p and it's again over two m h bar omega right so this is the Hamiltonian over h bar omega and this is an ih bar which and the i's make a minus one with this so this is going to be minus a half and everything else will cancel right because we'll we've got an m omega here and we're going to get an h bar from there so the rest cancels so I should have I should have explained sorry I wanted to factorize this and this on the bottom this normalizing factor on the bottom is put in it's not really essential but it's very convenient and it's put in in order to make this dimensionless so just to check that that's true um h bar has dimensions of position times momentum right so it has the dimensions of position times momentum so what we have here is m x sorry m omega x which we've agreed has dimensions of momentum times p which has dimensions of momentum then we take the square root so this on the bottom has dimensions of the whole square root of momentum and therefore cancel the dimensions of what's on the top so it's dimensionless that's the purpose of the that's the purpose of the horrible square root so we find that this product which is dimensionless is equal to the Hamiltonian divided by h bar omega which has the dimensions of energy because h bar also has the dimensions of energy times time omega of course has dimensions of one over time it's a free is the frequency of the oscillator so this is has dimensions of energy minus a half which is obviously dimensionless so we have indeed almost factorized we have the statement now that h can be written as h bar omega which carries the dimensions times a dagger a plus a half we've almost factorized h just there's that there the next thing we want to do is calculate the commutator a dagger comma a yes we've just got time to do this a dagger a of these two operators so of course we will have a one over two m h bar omega as a factor on the bottom because each of these a's brings in its own square root and then we will have the commutator of m omega x minus ip on m omega x plus ip now we we have this breaks down into four commutators in principle there's the commutator of this with this and the commutator of this with this the commutator of this with this obviously vanishes because x commutes with itself and the commutator of this with this is so we're going to have an m omega i times x comma p that's the commutator of this with this and now we have to deal with these with these terms this produces a non-negotiable commutator with that we're going to have minus m omega i times p comma x and then we'll have the commutator of p with self which will vanish if i swap those two over then clearly i changed the sign in front and then this becomes a plus x comma p it becomes this thing all over so that cancels this and this whole caboodle is going to equal i x comma p over h bar because we're going to get a two these two terms are going to add together to make us a two which cancels with this and the m omega is clearly go x comma p is itself equal to i h bar so the i's make a minus one the h bars cancel and this is equal to minus one so these two operators have non vanishing commutator actually equal to minus one yeah well we seem to still have time to to nail this this problem i think so let us suppose we have got a state of a station restate let us now apply the operator a dagger to both sides of this equation right then this is just that i can value it's only a number so i can then write e a dagger e is equal to a dagger h e that's obvious i would like to swap these over so i jolly will do i say this is equal to h a dagger plus a dagger commutator h so this commutator puts in what i'm supposed to have and takes away what i'm not supposed to have but have previously written down but we know what h is in terms that we have that h is equal to there it is h bar omega a dagger a so let's use that so this is h a dagger plus commutator of a dagger and h turns out to be a dagger a plus a half closed brackets h bar omega to carry the dimensions close that close that and stick in our e that we first thought of so all i have done is replace h by an expression we already derived um yep now i have to take the commutator of a dagger with this and with this the commutator of a dagger with the half clearly vanishes because a half is just a number not an operator um the commutator of a dagger with itself van vanishes so so when we do the commutator with this product there should in principle be two terms but only one of them survives and that term is that term is uh this sticks uh this stands idly by while the a dagger works on that and then i have an h bar omega h bar omega close brackets close brackets e but we just worked this thing out and found that it's minus one right a dagger a turned out to be minus one so this is equal to um h a dagger e minus h bar omega a dagger e just to remind you of it mind us what we had on the left what we had on the left was e a dagger e this is just a restatement of what's been at the top so we take this a dagger e and we obviously join it onto that a dagger e and we discover that h on a dagger e is equal to h working on this the cap that you get by using a dagger on e is equal to e plus h bar omega of a dagger working on e what does this tell us it tells us that we have out of a state which had energy e we have constructed a state by multiplying by a dagger which has energy e plus h bar omega so this means that a dagger e is equal to a constant normalising constant not discussed times e plus h bar omega a new stationary state this is an incredibly powerful result because it immediately follows that we have states if we can find a state e we can immediately generate e plus h bar omega by using this a dagger beast and also if we use a dagger on this it follows we're going to get e plus two h bar omega and we're going to get another if we use a dagger on this we're going to get e plus three h bar omega so we're going to get a whole infinite series of states of ever-increasing energy simply by applying a dagger again and again and again so what remains is to find what e what number e is and that we will do first thing tomorrow