 There are the four summarized rules for the quantization, a semi-classful quantization of one dimensional system. The rule itself is very simple, although we went through some pain to get it involving area functions and asymptotic forms, and so on, but the basic rule is really simple, and it has a simple interpretation also. It's said that the area of the ornamented space space, which is otherwise loop-in or PDX, is an integer plus a half times 2 by h bar. 2 by h bar is the same thing as Planck's original constant h. So you can think of the area of one cell, so-called Planck's cells being in the area of h and space space. I'll remind you of the origin of this. Well, let's see. This is sketched in the order of some phase space for a typical oscillator. There'll be things that look like this, and the quantized ornaments are the ones that have an area of 3 plus a half blood cells. So the middle one, the center one, has an area of a half, and then after that, it goes 3 halfs, 5 halfs, and so on. So the area of each little angular region that connects one quantized orbit to the next has an area of one blood cell. Now, I'll remind you that the origin of this formula is that it's a basic idea that as a particle moves around its orbit, it accumulates a phase, which is integral of PDX divided by h bar. This is a very old idea and actually precedes even in a modern quantum. This goes back to the days of the old quantum theory. In any case, you can imagine it's racing out of waves as it goes around its orbit. Now, in addition to that, it loses a phase of minus pi over 2 every time it passes through a turning point. And since we have two turning points in a typical oscillator, that's a total lost phase. And the answer then has to be a integer multiple of 2 pi in order to get phase coherence. I'm repeating some things I said in a previous lecture. So this has to be an equal to 2 in pi. And this condition in this form is the same as the condition of above. It's just rewritten in slightly different language. Here, this has to be a loop in the loop if you're going to go all the way around the orbit. All right. Now, that's the basics of the rule. So I wanted to begin today by making a simple illustration of how we can apply this rule. Let's work with the ordinary harmonic oscillator by zoning this p squared over 2m plus m of omega squared over 2x squared, as I guess you know. The first thing about this classically, we set h equals e. We get an orbit of energy e. And if we plot this in phase space x and p, the equation is that of an ellipse in phase space. So it looks something like this. And if you use, of course, number 12 rule, we're going to need the area of the orbit. Well, the area of an ellipse is pi times the semi-major axis times the semi-minor axis. So let's call this x max here as the maximum position for x. That's the x intercept of the orbit. We get that by setting momentum p equals to 0. So the kinetic energy term goes away and then solving for x. And if you do this, you find that you get twice the energy e divided by m of omega squared. Likewise, let's call this p max here, which is the maximum value of momentum. That's the momentum intercept. We get that by setting x equal to 0 and then solving for p. So that becomes equal to the square root of 2 and e, like this. And thus, the area is, you can calculate the area without doing any integrals. You can get a formula for an ellipse of pi times the semi-major axis, which is the square root of 2 pi, excuse me, 2 and e, times the semi-minor axis, which is the square root of 2 e divided by m of omega squared. Well, maybe I've got the semi-major and minor axis. We've got these reversed, but it won't matter. This is just a product of the two. And we multiply this out. You can see the masses will cancel. And what you get is a 2 pi times the energy e divided by the frequency. Now, like for a summer cell, that has to be equal to n plus a half times a 2 pi h bar on the quantized ordinance. And you can see that we cancel the 2 pi's here and then solve for the energy. And the result is that the energy is equal to n plus a half times h bar omega, which you recognize as the exact answer. So not much trouble for the harmonic oscillation to get the exact answer for summer cell quantization because of the exact initialize. WKB theory does not give the exact wave functions for the harmonic oscillation. It does give the exact energy problem. All right. Now, I'm going to run through a series of examples just because it's so easy to do them and we might as well have some fun just getting some answers. We'll start with some answers so you know when the harmonic oscillator. Let me next do a particle in a box. So here's the axis. Let's say x equals l is the size of the box. The potential energy is that it's zero inside the box but it's infinity outside. So these are hard walls. And the size of the box is gone. Of course, the classical part of the response is back and forth in the box like this. To get the organs in the face space we need to think about face space. So let's plot x and p. Here's the line x equals l again. And the particle is free inside the box. Let's say it has a momentum of p zero just to distinguish the momentum of the particle from the variable momentum p there. So the orbit going across on the right side is a straight line with a constant velocity at momentum p zero. When it gets to the opposite wall it bounces back which just means it changes the sign of momentum. So it changes the minus p pinot and goes back this way with a momentum of minus p pinot. Unless the orbit in face space is really kind of a rectangle and it continues to jump from here to there because it's a hard wall. But I'll fill it in like this if it continues to move back up. And you can see that the momentum is a rectangle in the face space. Now here's a tricky part about the particle in the box. And that is that instead of having a phase shift of pi over two at a turning point we have a phase shift of pi. And the reason is that this is a hard wall. Remember where this pi over two came from? It came from taking the potential energy which we look like this and we had the actual energy and then we looked at the wave function in the neighborhood of the turning point. This is what you call a soft wall where the particle its velocity decreases in a continuous manner. And it was some complicated analysis to get this pi over two but once you get it it's a simple rule. In the case of a hard wall you can see that because the boundary conditions are that the wave function has to vanish at the wall and so you minus five means the reflected wave is minus the incident wave. That's needed to make the sum of the two waves be equal to zero at the wall. So here the Borsomical condition is going to be the area instead of being n plus a half it's going to be n plus one times two pi h one. So it's slightly modified in the Borsomical condition. Now as far as the area is concerned however this is quite easy because it's just the area of a rectangle the height is twice p zero and the width is L so this is equal to two p zero times L. However p zero is just the momentum of the particle so that's the square root of two m e using energy and so I put this in and I'll get two times L times the square root of two m e which is n plus one now times two pi h one and so we'll cancel two's here and maybe I'll leave it in the form of p zero from the time being if I do that what we get is that p zero is equal to n plus one times pi h one and then we get the energy e which is p zero square root of two m which is divided by L p zero square root of two m n plus one squared pi squared h bar squared over two m L squared which you recognize as being the correct formula for the energy levels of the particle in the box here again it goes zero one two three so n plus one goes one two three four the ground state is n equals one now not n equals zero alright so that's another simple example now of course I can help this up for example yes I'm sorry I just gave a solution of why we have n plus one half when we calculate the quantization for the area for the square well versus I'm sorry n plus one through the square well versus n plus one half it's because the one half comes from this the n plus one half that applied to the harmonic oscillator came from the fact there's a phase shift of minus pi over two when you go through a turning point and that's why you multiply it times two so it's half of the phase cycle of two pi that's where the one half comes from up here but on a hard wall you're going to phase shift the pi at each wall because the reflective wave and the instant wave have to cancel out oh okay so that's a hard wall gives you a different phase shift and that's why it doubles the amount of phase and it becomes n plus one alright now I'll do another example this will be what is it going to be this will be the rigid rotor so rigid rotor let's do it classically first the rigid rotor has two masses and m1 and m2 they don't have to be the same this is actually a model of a diatonic molecule which we'll come back to in a few weeks I'd say there's the center of mass right there so I can put an origin and a coordinate system here x and y the center of mass and that's introducing as an angle to the center of mass now this is all classical to the time being the moment of inertia is equal to the reduced mass mu times capital R squared where R is the length of the rigid rotor and mu is the reduced mass so it's 1 over mu is equal to 1 over m1 plus 1 over m2 like that now the classical Lagrangian potential energy but for a rigid rotor this is a free rotor there is no potential energy so it's just the kinetic energy and as you know the kinetic energy of a rotor is one half of the moment of inertia times the the derivative of the angle they make frequency squared which all frequency is 5 dot here it's the rate of rotation it's obviously just constant it's just turning around and around let's translate this into Hamiltonian first of all we'll find the momentum which is let's say we'll call it p5 because it's contrary to the angle 5 this is dL d phi dot and that becomes equal to i times phi dot which you easily see by differentiating i times phi dot however is the same thing as the angular momentum of the rotor well now we have a problem because the angular momentum L is not to be confused with Lagrangian L so let me call the angular momentum Lz just to distinguish it but also if we think about the z axis coming out of the board it really is the z component of angular momentum of the thing rotating alright and then the Hamiltonian is equal to so p5 is the same thing the momentum as we say conjugate to phi is the same thing as the z component of angular momentum but the classical Hamiltonian is p5 times phi dot minus Lagrangian and if you work this out you want to express it that Lz is not in phi but in terms of phi dot so you get Lz squared right about twice the momentum of inertia so I'll write this again in the box because this is the classical Hamiltonian now what about the quantum mechanics of the rigid rotor here we're going to use a process of quantization this is really ad hoc and kind of hand way it's just following intuition but it works like this the process of quantization is that of going from a classical Hamiltonian to a quantum one the quantization that we've used so far in these recent lectures has been called what is called direct quantization I didn't use that terminology but this is what it amounts to in direct quantization what you do is you take the x's and p's which are the classical variables and you've got a classical Hamiltonian h of x and p and you reintermit x as an operator which is multiplication by x on a wave function side of x and you reintermit p as a minus i h bar d x as an operator acting out of side of x so the wave function lives on the configuration space x and this is transcription of classical variables into quantum operators and then the Hamiltonian comes into the operator and this is called the direct quantization of the direct rules for quantization and what it takes you it doesn't take you to the pet space it takes you to the space of wave functions on the configuration space now without a question whether it's valid or not let's just do direct quantization in these other coordinates now the angular coordinates the angular coordinates phi and the conjugate momentum p phi so let's apply this for the case of the rigid rotor in the case of the rigid rotor this suggests that phi, the classical phi goes over into a multiplication by phi it suggests that the wave function is going to be some wave function defined on the configuration space which is the phi angle space and it suggests that p phi which is the same thing as lz should go over into a minus i h bar d d phi and thus the quantum Hamiltonian is obtained by replacing lz by minus i h bar d d phi and hence the Schrodinger equation should be this minus h bar squared divided by twice the momentum inertia d squared psi d phi squared is equal to e psi phi this is the Schrodinger equation based on direct quantization this is easy to solve and you find that the psi of phi is equal to e to the phi m phi where m is an integer and you can normalize it again and m is equal to the integer positive or negative so it's 0 plus or minus 1 plus or minus 2 and so on like that alright configuration space here is we're going to plot this here for phi here's the wave function side the configuration space only goes out to 2 pi like this and the wave functions are e to the i and phi because they have to be the wave function whatever it is it has to be periodic because 2 pi is the same as s5 physically the rotor returns to its original configuration so wave functions have to be periodic and this is the solution we get by solving Schrodinger equation that's why you have to e to the i to the phi periodic functions that's why you get a given integer because it has to be an integer to make this periodic so this is a different set of boundary conditions than we had at the particle in the box instead of hard walls these are but notice that now there's no turning points the angle just increases and phi keeps on going around increasing and increasing so before a summer cloud condition is going to change again because now we won't have any one half at all we'll just be integer multiple to 2 pi well to go through the details of that let's go back to the classical mechanics which we have to do to apply one summer cloud and let's talk about the phase space for the for the system so the phase space is going to be phi coordinate here and the p phi which is the same thing as an LZ coordinate here and we've got 2 pi the periodic and the angle of the pi like this in period 2 pi and what is the classical motion classical motion is obviously one where just phi dot is constant it just rotates in a constant velocity and it will look like LZ is constant so that means the classical orbit is going to just go straight across like this at some given value of LZ which is the constant LZ for the for the motion unlike the particle of the box it doesn't reflect and come back it just keeps on going this line is the same as this line if you like the phase space it's really similar it just goes around and around and around alright so the classical motion has to be this it has to be the area which is going to be the loop integral of p phi d phi must be equal to an integer which I'll call m instead of n now times 2 pi h bar with no one half let me write this way it's m plus 0 m besides that with no one half because there's no turning points on the other hand this integral is just equal to 2 pi times LZ and so look what you get you get LZ to 2 pi is cancelled it's equal to mh bar which is the correct quantization of the z component of any momentum for the energy you get LZ to use the classical formula LZ squared over 2i which becomes m squared x bar squared divided by twice the moment in inertia and these are the energy levels for the the correct answer the quantized orbits are orbits that go across here for m equals 0 it's right on the the phi line they just call it the one like this and the area between them is a single quant cell right so that's the ridge of river there's an example of the book which I think I'm going to skip but I'll let you read it in the book this is the bouncing ball that Zachariah talks about and so this is the idea you drop the ball and it just bounces up and down so it's a particle in the gravitational field except it looks like this I plot the potential energy let's say the potential energy V of x is equal to the energy axis for a gravitational field so this is the potential energy and so the current potential energy is a straight line like this except if it's a balancing ball and we put a hard wall at x equals 0 and then for some energy you like the particle bounces back and forth in this turning point in that one here you've got one hard turning point for the phase shift from the pi and a soft one in the phase shift from the pi over 2 so now you've got an n plus 3 quarters actually anyway I'll let you think about this one you can read about it in the book yes the thing that makes the horizontal equation and all of these examples actually the right answer for the potential energy yeah the reason these answers are all coming out the exact so far is because the potential is the most quadratic polynomial actually in this case it doesn't come out exactly but approximations to the roots of the area functions you can see in this table before leaving this I want to mention something about multi-dimensional cases the Planck's cell in one dimension as I mentioned is 2 pi h bar that's the area of a single cell if you're at multiple dimensions it has to be a power of this it's a level of 3D it can take a 3-dimensional problem but this is the volume of a Planck's cell in phase space Planck's cells are in phase space so this is actually the 6-dimensional phase space that's positioned in a particle in 3 dimensions the basic rule is that each quantum state occupies a single Planck's cell and I'll show you something let's call the region R let's say that it's volume is capital B let's say that inside this region what we've got is a free particle but now in 3 dimensions so it's Hamiltonian is p squared or 2M but I want to emphasize that the momentum is now a 3 vector like this so this is a particle in a box except we let the box have an arbitrary shape and it's in 3D alright then now let's let N of E be the number of the eigenstates let's name it E naught N of E naught is the number of energy eigenstates with energy which is less than or equal to E naught we can call this the number function the number of eigenstates with an energy less than or equal to an energy I'm going to apply the Planck's cell rule to this and what we'll obtain is an estimate of N of E naught as a semi classical estimate so according to this idea that a single Planck's cell corresponds to a single Planck's state what we need to do is to integrate the phase space volume which is now 6 dimensions d cubed x d cubed p over all 3 positions 3 momentum variables of the region where the Hamiltonian is less than or equal to E naught the value of E naught over here and then we need to divide this by the Planck's cell in 3 dimensions which is 2 pi h bar cubed this is a 6 dimensional integral now it looks like it's hard to do but actually it's not difficult at all because as far as the x in this breaks up into an x integral which is just taken over the region r because that's the only place and then it's times the momentum integral 3 dimensional momentum integral which is taken over the region where let's say the magnitude of p is less than or equal to p naught where p naught is the magnitude of the momentum corresponding to E naught so the E naught is equal to p naught square root of 2 m or in other words p naught is equal to the square root of 2 m E naught so this is easy because the first integral just gives you the volume of the region and the second integral you just use the formula for the volume of the sphere it's 4 thirds pi times p naught cubed and if we put this together we get the the estimate for the number of energy eigenstates and it's equal to the volume to be divided by 2 pi h bar cubed times the 4 thirds pi and then for p naught p naught cubed I'll write this as 2 m E naught to the 3 halves of power and this gets you the answer I can clean this up by collecting powers of pi and stuff like that but this is the basic answer I'll show you the structure of the formula I'm sure you've seen this formula before you use this in statistical mechanics when you want to count states usually they make it a box and they count the number of plane wave solutions of a given energy less than a given energy and you end up with this formula all I'm doing here really is to show you that underwise those counting states is the basic fact that in the base space every state occupies a cell of 2 pi h bar 3 particles although 3 particles makes it easy to do the integral you can just do an applied or any system it could be a good estimate for this number function if you differentiate this to get dN dE this is what you call the density of states the differentiation is easy this is the number of states per unit energy interval which appears in solid state physics in all of the places alright so those are all examples of aspects of the force of a cell okay so this is the all I want to say about WKB theory and semi-classical methods for an answer are there any questions about this before we go on? how do train points work and multiple dimensions? well that's more complicated and when I write a formula down like this we're kind of blowing off the one half we're just thinking that N is large enough so one half won't matter I think that's the simplest answer the multi-dimensional problems are more complicated the classical mechanics are more complicated some of them are inimitable and some of them are not they're chaotic and that affects the whole picture in the days of the old quantum theory people didn't understand that distinction although they knew about integral systems because most of the ones that are soluble it turns out that integral systems have multiple periods one dimensional or somerville rule that's written down already gets applied to each of the periodic motions in the integral system and so in the period from about 1911 when Moore worked out his results for the hydrogen atom Moore really did a one-dimensional problem even though hydrogen atom is 3D somerville generalized this to integral systems in more than one dimension which is why his name gets attached to it as well and so then you get these quantization conditions for each of the degrees of freedom that's the quantization rule in more detail where you take into account turning points and stuff this is a similar approach in which I'm actually ignoring k-shifts but it's an interesting result because it shows you that the number of states and the density of states is proportional to the volume independently of the shape of the container which is something that's going to be quite hard to get by learning the other way because you know the wave functions for a box but if you make it a lobby-shaped thing nobody knows what the wave functions are so this is still correct it's only an asymptotic result but it's still correct okay in the days of the old funk period they didn't understand these spacious returning points because they didn't know about wave functions so they got some answers wrong by this one half the fact that a harmonic oscillator is n plus a half instead of just n was something that took a while but anyway the answers they got were in many cases very good so if everybody knew they were on the right track but they didn't really understand what this quantization mission meant alright so I'd like to turn now to a new topic which is the the elementary harmonic oscillator which we all have seen before so I expect a good deal of this will be reviewed and there's no eraser in there could I cover one up there's no eraser no I can't take it wrong basically what an elementary harmonic oscillator is I say I'm sure I know you to you but I'm going to go through the Rax algebraic derivation of the eigenstates and eigenfunctions of the eigenvalues of the harmonic oscillator which I'm sure you've seen before also but it's an interesting argument and I would encourage you to pay attention to the details of the CFL logic flows Dirac's algebraic treatment of the harmonic oscillator is the simplest and the most famous example of the solution of a quantum mechanical problem by algebraic means what that means is we work with the properties of the operators and their commutation relations but without attempting to solve the Schrodinger equation and we don't do it by solving differential equations and worrying about wave functions based on the harmonic oscillator and also an eigenveminim theory this is really by far the most effective and efficient way of dealing with problems involving harmonic oscillators which you know the one dimensional harmonic oscillator is a Hamiltonian which is p squared over 2M plus 0 to the power of omega squared over 2x squared this is expressing the potential energy potential energy is a this is expressing the potential energy in terms of frequency instead of the spring constant as I expect you to know this is frequently just a small amplitude approximation to a true potential real potentials are never really x squared because they don't go to infinity they have to flatten out somewhere so typically the potential energy is only this quadratic form of the potential energy is only the approximation to a real potential near the bottom of the well another dimension is if you have a multidimensional problem for example if you have two dimensions x and y and the potential energy as a function of x and y has a minimum somewhere like this imagine there's a bowl sitting down like that you can expand the true potential without the minimum and carry out the second order you're going to get a quadratic form this is a symmetric quadratic form and it can be diagonalized by a lithogonal transformation and the result is that the Hamiltonian of a multidimensional Hamiltonian will break up into a sum of one dimensional harmonic oscillators generally with different frequencies the frequencies of the so-called normal loads are usually not the same they may be the same but they're not always but the point is is that the quadratic approximations of the potential in multidimensions can be reduced to the problem of a one dimensional harmonic oscillator looking like this so in solving the one big problem we're actually solving a multidimensional harmonic oscillator as well I only drew this with two coordinates x and y but in some cases the number of coordinates is much larger than that the famous example is the lattice vibrations in a solid in which case the number of coordinates is equal to the number of atoms times 3 and so it could be an enormously large number in that case the diagonalization of the quadratic form with the potential energy amounts to obtaining the normal loads in the vibration of the solid actually this could be considered a classical problem finding those normal loads but once you find them then each one of them becomes a single one dimensional harmonic oscillator which is applied by the methods I would have said in the moment probably the it's actually applicable to multidimensional problems also by the way this is a mechanical oscillator but the harmonic oscillator Hamiltonian occurs in the number of contexts where it's not a potential energy problem for example the particle moving in a uniform magnetic field is a Hamiltonian which can be transformed into the form of a harmonic oscillator another very important example is examples is what happens in field theory in which one field theory which the most of the free field whether they be electromagnetic field or a field describing massive particles can be decomposed into normal loads and each one of those becomes a harmonic oscillator in cases like that the excitations of the given motor are identified with particles and we speak for example with photons the quantization of the electromagnetic field which will very similar to the quantization of sound waves, the phonons and the solid which otherwise is just the normal most of the vibrations in the solid in any case this is just a way of saying a harmonic oscillator has many many applications in physics alright so let's go back to the one dimensional harmonic oscillator and do this there are three constants that appear here a mass and a frequency of omega and it's convenient to avoid colliders and constants to choose units such that all three of these constants are equally one the equivalent way of speaking to this is to say that if you take a hidden value of mh bar and omega you can construct a fundamental unit of length of this equal to the square root of h bar divided by m omega and similarly the fundamental unit of momentum which is the square root of mh bar times omega this fundamental unit of energy which is h bar omega time which is one over omega and so on etc and by choosing units like this it's equivalent to make choosing units of each of these this length momentum energy time that's what are all equal to one of the units that we choose this is going to make the harmonic oscillator now become simply one half mass x square it's a conventional form now to convert back to real units you'd need to multiply or divide by these scale factors to get the actual units to one by the way I can only do this for one dimensional oscillator if I had two harmonic oscillators these frequencies were different I can't set both frequencies equal to one then I have to I can still set in an h particle to one but not to two anyway for one oscillator we can do this this is a quantum oscillator let's say at this point so the commutation relations of the position momentum are that their commutator is i h bar except for setting h bar to the one so it's just i in Dirac's method he begins by defining a set of new operators one is a which is defined as x plus i p divided by the square root of two and at submission contrary at a dagger which is x minus i divided by the square root of two these are the first one is called the annihilation operator or it's also called the lowering operator and the second one is called a creation operator or a raising operator the terminology annihilation and creation was typically applied to photons the quantization of electromagnetic field but the terminology is growing now these two operators are not permissioned so they remember that so they don't diagonalize them on the other hand they have some nice properties in particular let's define a new operator which is a product of a dagger times a well just from the line above this is one half of x minus i p times x plus i p if those were c numbers instead of operators you just both line together and you get one half of x squared plus p squared which is obviously the annihilation but because these are operators there's also a commutator that occurs you see x times i p minus p times i x divided by two so you get i over two times the commutator of x with p so this first term is the annihilation and the second term is i over two times i so it's one half so we can summarize this about only one half over the other side by saying that h is equal to n plus a half the other operator by the way is called the number operator here in the name and the result of this is that the Hamiltonian is the number operator plus a half the number operator has some properties here is that first of all the number operator which is a permission which is obvious if I take the submission conjugate it just goes into itself and it's also the non-negative definite so yeah it is one revision two non-negative definite the fact that it's non-negative definite it means that the eigenvalues of the number operator are non-negative they can be positive or zero it's easy to show that it's non-negative definite because if I take an arbitrary state and I damage it around here recall the definition of the non-negative definite operator is this expectation value has to be non-negative for all choices of phi let's see what this is this is phi a dagger a phi like this this however is the square of the state a phi and the square of the state is equal to zero because that's one of the possible of the of the definition of the definition of the definition alright so it's non-negative definite so the Hamiltonian is equal to this number operator plus a half it's just a trivial change of constant shift the result is that the eigen states of the Hamiltonian states of the number operator are identical and only the eigenvalues are different the eigenvalues of h are just the eigenvalues of n plus a half so we're going to diagonalize n and find the spectrum of the eigen eigen states of n first and then h follows through that curriculum okay so we'll do this in the following way the first thing I want to notice in regard to finding the eigen states and eigenvalues of the n of the number operator n the first thing to mention is that in a one-dimensional problem potential goes to infinity like this on both sides this means that the eigenfunctions have to die off exponentially at both positive and negative acts and as a result it means that the eigenfunctions are non-degenerate they're also discrete they're known as a discrete spectrum of a non-degenerate as we showed earlier in that lecture I gave on the some properties of one-dimensional systems so let's denote an eigenstate of the number operator with a simple nu where nu is the eigenvalue so that n acts on nu and brings out eigenvalue nu now I'm using the simple nu here because at this point for all we know the eigenvalue is the n number positive negative fraction integer or anything else at this point however it's easy to see and the nu has to be actually cannot be negative that's because n is a non-negative definite operator you see from just the line of part so whatever the eigenvalues are they cannot be negative numbers now next let's consider the state nu and I know as soon as this is normalized 2 we'll assume these eigenstakes are normalized now let's consider the state we get when we apply the operator a the annihilation of lowering operator to the eigenstate of the number operator let's take that and let's let the number operator act on it this is the same thing as a dagger a times a times nu now allow me to move the a dagger past the a for which I need commutation relations between a and a dagger which I meant to mention before it doesn't go over here and do it from the commutation relations between x and p it's straight forward without those between a and a dagger so the population and you find the commutator a with a dagger is equal to 1 and so this a dagger a this becomes the same thing as a a dagger times a acting on nu with a minus sign of just a acting on nu however the a dagger a is in the n acts on nu of nu itself so the right hand side is nu minus 1 on point a acting on on state nu like this so we can summarize this we have number operator n acting on a nu is equal to nu minus 1 acting on a nu thus we can say nu here is a an eigenstate of n the number operator with eigenvalue and so this statement says that if we apply the number operator to a nu we get a factor of n minus 1 therefore a nu is an eigenvector of n with eigenvalue nu minus 1 except for one slight wrinkle we have to worry about the possibility that a nu could be equal to 0 because if it is equal to nu nu is normalized but a nu could be 0 and if a nu is 0 then this equation just says 0 equal to 0 and it doesn't give us anything new so to make a precise statement we can say that if a nu does not vanish then it is an eigenstate of the number operator with an eigenvalue nu minus 1 now this raises a question which is does a nu equal to 0 when does this happen when does this happen when does this happen the answer to this will be the basic factor of the space is that a vector is vanish if it only if it is square vanishes so I am going to remove the question mark and say that this condition is true if it only if I take the square of it I get nu a dagger a nu and this is equal to 0 this is how it was the same thing as nu sandwiched from the number operator which is the same thing the same thing as nu times the norm of the eigenstate assuming it is normal I say it is just nu itself so what we see is the answer to win is that a-actuant nu is equal to 0 if it only if nu is equal to 0 and we put a little box around that that is the main conclusion of this so to go back to the formula at the top of the board we will say it again and the light of the new information we will say that if nu is not equal to 0 then a nu is an eigenstate of n with eigenvalue mu minus 1 nu is equal to 0 this equation just says 0 is equal to 0 it doesn't say anything now if we get to something similar with a racing operator let's consider n acting on a dagger at nu this is the same thing as a dagger a with this n acting on nu and now allow me to move this a to the right of that a dagger so here is the commutator of a dagger which is up there and so this gives us a dagger times a dagger times a n acting on nu plus a dagger acting on nu again this is these factors are n n acting on nu brings out nu and the result is that it is nu from this terminal point to this terminal nu plus 1 times a dagger acting on nu so now we have the statement that nu is an eigenstate and not directly with eigenvalue nu so we apply a dagger to it we get a nu state which of it does not vanish is an eigenstate of n with eigenvalue nu plus 1 let's go through the same thing does this state vanish when is the possibility of a dagger nu there actually a dagger nu equals 0 if it only happens square as it equals 0 nu times a a dagger nu however by the commutation relation a a dagger nu is equal to a dagger a plus 1 which is equal to n plus 1 so we can say that a dagger it vanishes if and only if nu plus 1 equals 0 but remember nu is greater than or equal to 0 so nu plus 1 is always greater than 0 and the result is is that a dagger nu never vanishes so we're going to write this a dagger nu a dagger nu is not equal to 0 always so we unbox this part of this well I know so what we've got is that so let's do it this way let's write it this way and say that in happening on a dagger nu it's equal to nu plus 1 a dagger nu and to put it in words a dagger nu is always if nu is an eigenstate of n with an eigenvalue nu then a dagger nu is an eigenstate but eigenvalue nu plus 1 always there are no restrictions alright now from this it follows that the quantity nu the eigenvalue has to be an integer because it's supposed to be a nu not an integer then by applying an a an a squared and a cubed to the nu we get eigenvalue which are nu minus 1 nu minus 2 and we can bring this down up to the point by applying enough a operators we're going to get up to the point where nu minus k is less than 0 well the eigenstates of n don't have any negative eigenvalues so this seems to be a contradiction and the only fact the only way to escape the contradiction is an integer because what happens when you apply these a operators is you're eventually going to get down to the state 0 when you apply n of them and then when you apply the next one a to it instead of getting some state minus 1 you get just a 0 all together that's in this condition here a x and a 0 state here is a 0 so therefore the only consistent solution these sets of relations is that nu is actually an integer which is replaced by n which is an integer and in fact it takes on the value of 0, 1, 2 and so on it has to take on all these values all these non-negative integers because if any one of them occurs you can get all the others by applying raising and lowering operators so if there's any eigenstate at all which there has to be then this is the easily allowed values of integers and so thus we have the spectrum that is the integer 0, 1, 2, 3 and so on and from that it follows meaning that the energy eigenvalues are equal to n plus a half because the Hamiltonian is equal to the number of operators now now I'm going to worry about so now the states so now we'll call the eigenstates n and x on n is equal to n this is the basic relation we're working with here we also have a acting on n is proportional to n minus 1 and we have a dagger acting on n is proportional to n plus 1 summarizes some of the things we learned so far let me call in the first equation we call it a let me call the proportionality constant cn let's say a n actuated it's not proportional to the constant cn times n minus 1 cn is a general complex number to find it we can square both sides of the square the square of the left hand side we get n sandwiched around the a dagger a the square of the right hand side we can guess the value of c n squared times the norm of n minus 1 but we're assuming these states are normalized so that norm is equal to 1 on the left hand side we have a number operator in the middle here and the result is that cn squared is equal to is equal to n just to call the number n therefore cn is equal to a phase factor e to the i let's call it alpha n times the square root n like this so we can now write out some of these relations these annihilation relations here if we apply a to the state 1 we get e to the i alpha 1 times the square root of 1 times 0 if we apply a to the state 2 the annihilation operator we get e to the i alpha 2 times the square root of 2 acting on 0 a acting on 3 we get e to the i alpha 3 I'm just writing down some of these so you can see the pattern this is actually not the case 1 squared and so on this is the information we have so far from these creation annihilation relations now at this point let's take these states 1, 2, 3 and so on and redefine them so as to absorb these phases alpha 1 alpha 2 and alpha 3 and if we do that then we've established phase dimensions for the states 1, 2, 3 and so on for these phases e to the i alpha go away in the process of doing that phase convention phase from state 0 what we are in fact doing is linking the phase conventions for all the excited states to the phase conventions for the ground state and so for that convention I can erase these phase factors and now there's only one phase convention for the eigenstates and that's the basic convention for the ground state 0 and if we do this then I'll draw the result from the little box here it is the result that I think you know is that a acts on n because this is square root of 20 times n minus 1 by a completely similar argument working with the raising operators we find that a dagger acting on n is the square root of n plus 1 times n plus 1 these are relations I'm sure you've memorized for an undergraduate exam at the time so there they are as it turns out it's not necessary to introduce any additional phase conventions for the imaginary equations the phase changes that I made over here are all you need to do and then this imaginary equation automatically follows the effect of these phase conventions is to make the matrix elements of the raising of lowering operators in the energy eigenbasis to make those matrix elements real and that's the most convenient choice to be doing another one and that's all