 It reminds you where we are, we're going to think about it with a bunch of stuff here, so I'll summarize it a little bit more. In the first place, I'm going to use a pointer. We're considering a case in which there is a potential that's rising and you should go to the right so that there's a classically allowed region to the left of the turning point and a classically forbidden region, C, F, R to the right. These are also labeled 1 and 2. X, R here is the turning point itself, and the R just means that this turning point is to the right of the classically allowed region. Down here is the face of this picture. Then, using the WKD solutions that we've worked out so far, in a classically allowed region, which is region number 1, the momentum function is defined like this. It's a real function, real and positive function. The action function, or the function s, which appears in the exponent, is also called an action, is the interface of the integral of pdx. Here's the question of the lower limit, which is, I pointed out, that's kind of just related to the face convention of the wave function. So you can set it for anything for now or about face conventions later. The most obvious point to choose is the turning point itself, and that's what I've done in this integral here. And I've called it s of x, comma, x, r, and indicated that the lower limit integration has to be explicit about that. In any case, using the linear combination of the forward and backward and the right and left point solutions, c to the i s and even right and psi s over h bar, with coefficient c right and c left, this gives us a general linear combination in the classically allowed region. Now in the classically forbidden region, which is also labeled as number 2 here, the momentum is purely imaginary. If I take its absolute value, it just means to drop the i and just deal with this positive square root. And we can find a modified action we call k, which is the integral of this square root dx, the absolute value of p of x dx. And again, the lower limit is based on purely determined normalization in the face of the wave function. And just to make a definite, so we have a definite number, we'll use again the right turning point, x, r here as the lower limit. And then we perform a linear combination of the 2WKB solutions, a growing one and a tapping one with coefficient c growing and c napping. And so we have these two solutions in these two different regions and in each region there's two arbitrary coefficients. Now the problem is to connect together the c right and c left in the classically allowed region with the c growing and c napping in the classically forbidden region. But as I explained last time, you can't do that just by putting the both solutions up to the turning point because there's a certain region around the turning point and both these solutions become invalid. Let me elaborate on that breakdown near the turning point just a little bit. I'll cover this up and bring it back again. The, as you recall, the amplitude is the, well the amplitude squared times the velocity is the same as the, you know, the amplitude, the amplitude squared is the same as the classical probability density which is proportional to one over the velocity v. This gives the amplitude is proportional to one over the square root of the velocity which is proportional to one over the square root of the momentum which is how it's occurring in our formulas. In any case, if we have a situation like I sketched out of a moment ago where we've got a turning point that goes up like this, there's an energy v. Then if we plot the velocity, you see the velocity is positive here but then it goes to zero which becomes the turning point. So one over the velocity or one over the square root of the velocity is going to be a function that's going to do this. It's going to diverge in the turning point. In fact, you can see that the, the, the divergence goes as, well the divergence goes, the amplitude h goes to one over the square root of the velocity and goes to infinity there. Also as I explained last time, this implies that the conditions of the WKB approximation which is that the wavelength should be much less than the scaling breaks down there because near the turning point, lambda actually goes to infinity since the momentum is going to zero. Okay, so no wonder the WKB solutions don't work there. But there's an, you might say, the nominal WKB solution blows up at the turning point. It does also in the class they've forbidden region because there's a whatever square root of the momentum there as well. They both go up at the same place. Now you might ask, what does the real solution do? What is the real quantum mechanics do? Well by the way, the classical probabilities we mean for quantum to be classically and it does diverge there. It diverges as one over the square root of the distance from the turning point which means that the integral of the classical probability is finite. As you expect since it's a, this probability should be for a finite oscillator, a bound state system you can normalize. And that means the amplitude actually goes as one over the distance to the one, one quarter power from the, from the turning point itself. But they diverge. And so there's a question about what does the real way function look like? Does it diverge? No, it doesn't. It's a solution to the Schrodinger equation. I'll show you in a bit what it actually looks like there. But in any case, that's the problem near the turning point. And we fix it up by inventing yet a third solution, again an approximate solution which is valid in the immediate neighborhood of the turning point. We do this by approximating the potential by straight line there, by drawing elements like this. This is going to be valid at least in some region around the turning point. Here's a series of the potential carrying the first order. XR is the turning point. This is just a constant coefficient. Here's the slope of the potential. If you plug it into the Schrodinger equation, it becomes this. It's now Schrodinger equation naturally the linear potential because that's what you replace the real potential by. And as I explained last time, we'll clean this equation up by doing the transformation going from an old variable X over to a new variable Z such that Z equals 0 at the turning point. So over here, in this diagram here, this class of the above region is one where Z is less than 0 and the class of the forbidden region is where Z is positive and the turning point itself is Z equals 0. In the constant A of the transformation, we choose to clean up the physical constants in the Schrodinger equation and this turns out to be the choice and if you do like to end up with a differential equation for the wave function in the neighborhood of the turning point which does not involve any constants at all. It's a completely dimensionless form. This is as I explained last time is the standard equation for the varying function which has special functions. These are the two linear independent solutions of this equation and so in the neighborhood of the turning point the wave function is now a linear combination where the coefficient area and coefficient area of these two functions vary and as a result we now have three different regions with three solutions and each solution there's two pairs of coefficients so there has to be a connection and not six coefficients altogether there's got to be connections amongst these coefficients which we need to work out. So how do we do this? Well before I do that let me talk a little bit about the area and vary functions. Of course you'll have to cover this up but if you go to the books on standard functions standard special functions you'll find the chapter on the area and vary functions you'll find these limiting forms for these two functions when z is large and negative or when it's large and positive and this is just what you get out of the books I could derive these formulas for you that hard take a lot of lecture or lecture and a half but for lack of time I won't do that let's just say we read about the books but what you can recognize is that these are asymptotic forms valid in large and small z what you can recognize is they're actually a WKB form you see in the classically allowed region which is z negative you've got cosine and sine terms these are real exponentials of two waves going in the two directions so if that part's easy to see in order to understand these formulas however let's adopt a physical model in which the potential really is a linear function of position an example would be let's say v of z let's say a particle in the gravitational field where the potential is f of g of z then the Schrodinger equation is of course minus h plus 2 and also let's take the energy equal to 0 let's do that so it's minus h plus 2 then you have d squared psi dc squared plus mgz psi equal to 0 you may compare that to the varying equations which is right here and you see except for the constants it's the same equation so apart from scaling the constants the varying variable functions are the exact solution of the Schrodinger equation in the particle in the gravitational field with a 0 energy let's look at that problem from the standpoint of classical mechanics or maybe you might say wkv theory so here's the z axis and so you imagine the particle is climbing in the gravitational field the total energy is 0 so it reaches a turning point at z equals 0 then turns around and comes back so as it climbs in the gravitational field of course its velocity decreases its decelerating that means the momentum is decreasing and that means the wavelength which starts out short gets longer and longer and longer as it comes closer and closer to the turning point also then the velocity is going down it means the amplitude must be going up because as we just have learned the amplitude goes as one of the squared with the velocity so just like in general wkv brands we expect a kind of an impulse like this then there's going to be oscillations in here and as you move further towards the origin the wavelength is longer and finally what it does is it does something like that it doesn't go to infinity but it reaches an axle of here z equals 0 that's the turning point and then there's some penetration where the classical particle wouldn't go at all positive z and you get a decay over here and exponential decay well this is actually a sketch not a bad sketch but this is a way of interpreting what that solution means now as far as the very function is concerned that's the other minimum independent solution and it also has the same envelope like this but the oscillations are 90 degrees out of phase from the very function so I can't really draw them in the board and they get longer wavelength as you get closer to z equals 0 and then what happens is in the classical bin region they diverge exponentially and this is the very function which you would say is the solution that has to be rejected because of boundary conditions for the particle in the gravitational field but nevertheless it gives you again I say by this physical model it gives you an idea of these of these what these asymptomatic forms mean and carrying these carrying these exact solutions out the large negative z or large positive z and the very functions yields these expressions we have here summarized over here for you copy that in the book alright so that's the properties of the very and very functions it's not some bad actually so now if I may reveal my board again here it must be that these WKB solutions side one and side two which you see here will be very the same as we get by taking linear combinations of these asymptotic forms so for example if I look at the classical bin region first which is where z is positive then this exponent which occurs in our WKB solution here which is k of x comma x r over h r must be equal to that exponent of beta of z up there let's see if it is let's show you this so here's what we'll do so the exponent of these asymptotic formula of beta of z which is equal to which you see is two thirds of z to the three halves this must be equal to I'll have a question about that because we want to verify it 1 over h r of k and value of value of value of k to the turning point in the first place they both manage the turning point so that's what we know that this has to be x r on the right hand side now this is to check and see if this is true let's plug in the definition of k this is the integral from x r to x of the square root of twice the mass times the potential energy of d of x minus d of dx now we use the approximation we use to get the area function in replacing the potential energy d of x by d of x at the turning point plus x minus x at the turning point times v prime at the turning point that's just the first two terms of the Taylor series expansion and if you do that then the energy e cancels the v of x r because the energy is equal to the potential at the turning point this is linear in x minus x r times the constant so this turns into when we take on the constants that's 1 over h bar the square root of 2m times v prime of x r and then we've got the integral from the x r to x of x minus x r to the one half power dx we should really be trying this be careful about distinguishing the variable of integration from the upper limit but anyway that's the integral we get in this straight line of approximation and this integral is easy to do it gives you three halves x minus x r excuse me it's two thirds of x minus x r to the three halves power and that's nice because first place you see the two thirds of area you can see where it came from and as far as it's even the three halves which appears in math books x minus x r to the three halves but our linear transformation was that the x minus x r was this quantity a times z so this is a z to the three halves power so that's a to the three halves times z to the three halves and so the z to the three halves came out of the math books in the theory that our integral here there it is and so the only thing that's left over is to show that these remaining constants here which is this thing one well if you go back to the definition of a which is very important here here it is I remember what it is it is h bar squared over 2m v prime x r that whole thing to the one third power that's the a but we need to raise that to the three halves power and you see it gives over all one half of the square root therefore the square root of h bar squared cancels h bar squared 2m v prime cancels that the whole thing works out and this thing is verified this is indeed true these exponents are the same so this exponent of z is the same as the exponent of h bar k for one of our two solutions in the class that we did in the region now what about the amplitude we've got one over z to the one quarter here for the amplitude that came out of the math book how does that compare to the amplitude of our wkb formula we can't expect it to be exactly the same because it's constant it's not supposed to be proportional so our amplitude in the wkb formula is one over the absolute value of p of x square root of that but that's one over the square root of p of x that's what value of p of x is the same as this square root here let me do it this way let me just concentrate on p of x itself absolute value of p of x is equal to this square root but in this linear approximation you see it becomes equal to the square root of 2m times v prime of of xr times x minus xr but x minus xr is the same thing as a times z and so the whole thing becomes proportional to z to the one half power as you can see here just the square root of z but this is not the amplitude what actually appears in the amplitude is the square root of this now we've been denominator so what we get is the square root of the absolute value of p of x that is proportional to z to the one quarter power and that's exactly what comes out of the math books to see it in one quarter so these asymptotic forms that appear in the math books but that's what they are ok so the only thing that's left over to match these solutions together is just the constants you just need to work out what the relation is between the coefficients and multiply these solutions and so that's entirely straight forward to put that together and I'll just summarize what you get when you do that maybe this is a good place to do it because you find that the damping and the growing coefficients c-bamp against c-growing are equal to the square root of h bar over pi times a where a is that funny constant multiplying one half of the c-bary and the top of the c-bary at the bottom this is just a translation between these two types of coefficients this final formula here that I've written down basically connects us from the turning point region to the classic forbidden region and connects this solution number two here with this area and very solution over here finish things we need to go the other way as well we need to connect the turning point region with a classally allowed region this is about in exactly the same manner as I just outlined that's to say we first start by looking at this exponent s and sitting out it relates to our function alpha out there and if you go through the same type of analysis just use a linear approximation for the potential and do the integral you'll find that they're the same except for one thing which is that pi over four right there in fact you can see that has to be because if I evaluate alpha at the turning point it's equal to zero then it's equal to pi over four whereas if I evaluate the function s of xxr at the turning point it's the integral of zero so they're the same except for that added constant now to fix that up I'm going to do something here which is just to amounts to a redefinition of these coefficients cr and cl allow me to add an i pi over four to this form here and over here I'll subtract an i pi over four and all that really does is amount to a absorbing or removing factor even the i pi over four from the cr and cl it just simplifies the algebra so why and if you do that then it becomes easy to find the connection between the right and left coefficients and the area of varying coefficients in range of this cr and cl is equal to one piece of paper here paper this is the one is equal to the one half of the same square root h bar over pi a and then what we get is cr minus ic bar on top of the cr plus ic bar in the bottom we probably aren't so interested in the area in various coefficients we're really interested in the right and left and the drawing of the dam and if we put those together just by eliminating the area in very coefficients in this we can find connections between the right and left and the drawing so if we do this then we need to go on to the next page and this is what we get this we get c right c left it's equal to this matrix which is minus i over two i over two here and then a one and a one and a multiplying c drawing and c band and for reference I'll write down the inverse of this c drawing and c band and take the inverse of that two by two matrix and this becomes i minus i one half one half multiplying on to c right and c left and these final matrices are the main result of this immediate stage of calculation these are what are called connection rules because they allow you to connect together the wk solutions in the two regions classically allowed and classically forbidden having gone through this intermediate and very invariant which not drops out because we have eliminated those problems alright so that's the main result of this now having gotten those connection rules let me now apply them to a practical problem let's see the diagram I want to show you is right here this is still the same picture we used here let's consider a case where we have a potential that rises like this and say it never comes back down again and you're doing a scattering, one dimensional scattering problem we send in a wave like this, it goes in as an incident wave it hits this potential barrier and reflects it comes back out it's just a reflective wave going like this now a large negative x here, this potential goes to 0 so it's really three particles this is typical in scattering situations because you want your particle in a region that's far away from where the potential is active this is only 1d but anyway the general solution is going to be obviously a linear combination of the incoming or the incoming wave and the reflective wave and that's exactly these two ways right and left going away so let's analyze this problem and see what the solution of the Schrodinger equation is well the way to do it is to use the connection rules I'll erase this stuff here and I'll go back to the right board here in the first place in the classically forbidden region which is over here the wave function has to die out it has to be dampened out we can't have it blowing up because of the verge of infinity and that doesn't satisfy the common conditions besides this in the classical forbidden region it's got to be dampened and as a result of these two coefficients going and damping the growing coefficient must be 0 and as for the damping coefficient which is the only one that's left also, alright, so then I can erase this here and as far as the damping coefficient which is the only one that's left you can see it's equivalent to a normalization in phase of psi so let's just take the damping coefficient of 1 we can re-normalize it later on if you want to just one will work for now so on the basis of boundary conditions what we do is we say that C growing is 0 and C damping is equal to 1 now then by this matrix right here this connection rule we just do the multiplication so that C right is equal to C left is equal to 1 and the damping is 1 now this is 0 1 and the result is that we find that psi in region 1 of x is 1 over the square root of e of x and then what we have is e to the i s of x how the x are over h bar plus i pi over 4 plus e to the really plus the complex conjugate because the second term is the complex conjugate at first and we also get the solution in region 2 it's 1 over the square root of the absolute value of p of x which is real in that region times either the lightest k of x coming x of r over h bar we're probably mainly interested in the solution in the classical amount of region because that's where the particle is that's where the particle is going to come back so let's look at this solution here in the first place you can see that the solution is real because it's the sum of the terms of the complex conjugate in fact it can be written as 2 over p of x times the cosine of s of x of r over h bar plus pi over 4 there's a couple of lessons in this it is a simple problem there's a couple of lessons in it in the first place there's no quantization of energy there's nothing in the solution that requires the energy to take on a particular value that's because we're in the unbound positive energy it's in the unbound spectrum it's unbound states and so it's a continuous spectrum and the energy can range from zero to infinity it's a problem like this the second thing to notice is that the solution there's only one solution we attain the solution in other words is non-degenerate I argued or showed you that in one-dimensional problems if the weight function is forced to go to zero at either one or the other of the infinities or even at a hard wall it doesn't have to be infinity then the solutions are non-degenerate and that's just what we see here another point that I made was that in one dimension the non-degenerate energy eigenfunctions can be chosen to be real that is to say if they come out complex it's only because they're multiplied by some overall complex number and immediately see this as real yes would you like that, Ryan? would you replace the exponential with the cosine? is that p dx? excuse me, yes it's a square root of p of x, thank you is that what you're asking? yeah, square root of p of x so anyway, all these features of these one-dimensional problems that I mentioned are revealing themselves here while we're writing this abstract as many lessons as I can on the scattering problem there's several interesting things allow me to write this solution in a slightly different way by factoring out the heat of the i pi over four this is just a a criminal change the heat of the i pi over four is going to all face back eventually right about the square root of p of x and then what's left over is e to the i over h bar s of x comma x r this is the right travel of wave let me write the left traveling wave as a coefficient of r times e to the minus i s of x comma x r over h bar and r here is by factoring out e to the i pi over four that was e to the minus i pi over four of here but that goes into minus pi over two in fact with this factor out and so r is equal to e to the minus i pi over two it's the same thing as minus i it's a pretty base factor now the idea here is that this is the incident wave that we send in and here's a reflected wave coming back and the reflected wave is multiplied by a coefficient complex number and this complex number is called the reflection amplitude it's the definition of it and so as I say it's just the molecular factor giving in a reflected wave there's something that comes out of the unit called capital r which is the square of the reflection amplitude and this is called the reflection probability which in this case since the reflection amplitude is a base factor it's just equal to one so the fact that the reflection probability is equal to one means physically that since this barrier goes up to infinity that any particle that hits the barrier must bounce back so therefore the probability of getting a particle back where you came from all the particles have to come back there's no possibility of penetrating the barrier coming out to another side somewhere there's no tunnel in there this goes on up forever alright we can understand these reflection amplitudes and the probability is a little better if you stand for the complexes so let me remind you of some things about complexes which I spoke of in the last lecture and I'm clearly out of shock except for 20 to 30 pieces here the the first place the probability equation in one dimension says that v rho dt plus dj dx is equal to 0 but this is a time-independent problem so everything the density in particular is independent of time so is j so this term it just turns into dj dx equal to 0 or to say that j as a function of x is actually a consequence independent of s however j's definition is that it's psi star times minus ih bar with the mass m dx applied to psi and that's not quite that j is a real part of this this is the probability of flux in one dimensional problem now we have two waves here an incident wave and a reflected wave and we can get the flux of those two things separately if we compute this for the incident wave what do you get? well I'm pressing around that what do you get? well the edx brings down a derivative of the phase s which is s prime but the psi star psi is going to involve the square of the amplitude which is one over the square of p of x and so that turns into a p and in combination those is equal to one because p is equal to s prime in fact what you find is that it's only this mass term is the only thing that survives you find the incident flux of one over n and sure enough it's independent of x as it has to be anyway this is what it is now what about the reflected flux and this problem is I'll just show the reflection amplitude as a phase factor so it's square is one let's consider maybe we have a more general problem where r is any complex number if you go through the same calculation you'll find that this is equal to the answer of the value of r squared divided by n which is the capital r times the incident flux so this is why capital r is called the reflection probability it's the any fact it's the probability that a particle which is sent in will be reflected in this problem this square of this r is equal to one in this very simple case alright so that's the lesson about reflection of amplitudes and probabilities there's yet another lesson that I'd like to make regarding this very simple solution and it goes back to this phase picture of the motion particle maybe I have enough room to say what I need to say here the action function s it's called an action it's divided by h parts the phase of the wave function is basically the integral of p dx that's the definition of it now a useful way of thinking about this action function is to imagine as a classical particle is going around the orbit as a function of time it carries a little integrator with it it knows what its momentum and position are and it just integrates this as a function of time we create a function let's call it s of t as a function of time along the orbit and that's equal to the integral of the final time t of the momentum as a function of time times dx dt which of course is the velocity as a function of time dt it's the same integral but just expressed of time as an integrator so let's say our particle carries an integrator like this for this kind of problem let's choose t equals 0 to be the turning point itself and let's take the lower limit of this integral to be 0 so just to get a definite lower limit now when you're on the other half of this orbit here the momentum is positive and the dx is positive too because you're moving forward so that means that s is increasing in this part of the orbit when you get to be t equals 0 and you go down this way it's negative but the dx is negative also and so s is still increasing if you actually plot s of t as a function of time it looks like this it's actually negative when negative time kind of goes like that the point is it's increasing everywhere on the other hand the s of t is obviously closely related to the s of x comma xr which is appearing in the x corner here and the question is what is the relation between them and the answer is this is that s of t is equal to s of x comma xr when t is negative and it's equal to minus s of x comma xr when t is positive and the result is is that it's this main part of the phase which you see here and here and the only point of this is is that it's merely a conceptual one of these phases and both of these terms is being due to the same particle doing the same integration and just continuing on around from the upper branch down to the lower branch in some sense these are the same functions if they're thought of as being a function along the orbit instead of as being a function of x they're right however there is this phase shift this pi over 4 here plus and minus pi over 4 or if I write it in this reflection form pi over 2 so in other words when the particle bits around to the lower side the overall phase is really not s of t but it's s of t minus pi over 2 compared to what it was in the upper branch what's happening here is is that the WKB solution breaks down in the neighborhood of the training points so these formulas and these nice s's are even valid in this region so this integration of this particle doing the carry along the phase and kind of tracing the wave along as it goes when it gets into this region it doesn't even make sense to talk about the WKB wave anymore the whole concept breaks down but nevertheless the particle continues on going and after a while it comes back out again and when it does it's now you can continue the integration if you get a wave that's going along here except there's been a phase shift when it was in this region where the overall s function didn't make sense as a phase minus pi over 2 as we say it suffered a phase shift with minus pi over 2 in the crossing through the training points so this is a basic rule as the WKB theory is that the wave function suffers a phase shift with minus pi over 2 in the crossing and turning point okay now and that minus pi over 2 comes out of this elaborate analysis of carry and carry functions okay now that's all I want to say about this scattering problem let me now turn to a more complicated problem which is an oscillator let's suppose we have a potential energy as opposed to the x which has some kind of a well and try to draw it so it's steep on the one side and shallow on the other because I don't want to make it a harmonic oscillator it doesn't have to be a harmonic oscillator but let's say it's an oscillator like this now there's two turning points there's a left one I'll call xl and a right one I'll call xr and if we plot the order of the phase space x and v we have the same xl and xr coming down and then we order it with something like this it kind of does could be kind of an edge shape object like this in an oval shape I don't want to draw this on the list because the general is not on the list it's called logically a circle however and it part of it goes around like this now let's take an initial position I want to first analyze this problem just from an intuitive standpoint let's take an initial position on this orbit and we have a classical particle and it's going to go around and around and as it does let's imagine carrying out this integral of pdx to accumulate a phase of a wave which is going along like this it passes through a turning point something's a phase shift of minus pi over 2 and gives a wave that's going back it passes through a secondary point minus pi over 2 and then continues on back and eventually comes back to where it started from now the wave that it's carrying with it better be in phase with the wave that it started with or else you get destructive interference and there's no wave function at all so what is the total phase the total phase of going around the orbit actually the phase is integral of pdx over h bar so the total phase of going around the orbit the total phase it's really the total delta phase change in the phase relative to the initial position it's equal to you might be better in the corner room here it's equal to 1 over h bar integral now around the entire loop of pdx that's the accumulated action function s but then we have to subtract twice by ever 2 because of the two turning points now this total phase is going to be an integer multiple of 2 pi because as I say otherwise you will have destructive interference and so if I bring this this is equal to pi of course and I bring it over to the other side multiply by h bar what we get is this is the integral of pdx is equal to m plus 1 half times 2 pi h bar and the integral of pdx around the loop of the area of the orbit in the phase space and this is a quantization condition it says that only when the area takes on this special value of a half integer then plus a half times 2 pi h bar then we have a phase coherence this is the semi-class conversion of the quantization condition and one can see that it's really the area of the orbit it's quantized it's quantized in half integer multiples of 2 pi h bar it's quantized in original constant h sometimes called a plot cell it's the fundamental area of the phase space and in a certain sense one can say that that every quantum state occupies this fundamental area of h of phase space if I plot the quantized orbits there's a special class of orbits with one that has an area of a half that's the beginning of n equals 0 and in successive orbits have the property that the area of the angular strength separating one orbit from the next is equal to a single plot cell so in an oscillator like this you can imagine the quantized orbits moving out like this notice that the minimum orbit does not have an area of 0 it has an area of 1 half this is related to the uncertainty principle the fact that you couldn't have a part of this stationary in the bottom of the well because then the adult would be infected this takes care of this that takes care of the uncertainty principle the ground state alright so this is an intuitive understanding of what's called the Borsamert-Hill quantization condition for one dimensional oscillators there's an according to the in classical mechanics it's called the action of the orbit actually the word action gets used in several sentences in classical mechanics as they all have dimensions of action there are different things this is called I and it's just a definition it's 1 over 2 pi times the area of the orbit this is how it's inclined to the one dimensional problem so if I apply both sides of this equation by 2 pi we can say that for a quantized orbit the action of I in the sense is equal to n plus 1 half times h bar in fact let me put an n on it and this is the exact the quantized value of the action of the orbit the action is just proportional to the area so if the action increases the orbit is getting bigger and bigger moving onwards the way I drew the potential that means the energy is increasing also so the energy of the orbit is a function of the action and classically you can write a function of this energy as a function of the action like this the class of the the action as it takes on all calories is quantization of Borsamert-Hill there's a result in classical mechanics all of the classical mechanics is worth mentioning which is the frequency of the motion is actually equal to the derivative of the energy with respect to the action I don't approve this but that's an important result in any case once the actions are quantized you can use the classical relation connecting energy and action to find the quantized values of energy that's to say that e sub n is equal to the classical energy action relationship evaluated by n the orbit of this area is equal to n plus a half times h these then are the sort of classical or some of the energy eigenvalues for the problem they're not exactly general they're only in approximation but often times it's a very good approximation now so that is the the oscillation treated from an intuitive standpoint by just counting phases let me now do it from a more sophisticated standpoint in order to do that I'm going to need I'm going to need a new set of connection rules because we need to worry about not this board takes care of turning points looks like this we need to take care of the other turning point so let me go to this board I already got the connection I won't do this anymore let's take this situation now where we have the potential energy is decreasing like better the potential energy is decreasing like this let's say there's a total energy view like this and now there's a turning point here that I'll call xl meaning that it's to the left let's call this region 3 here in this space in region 4 in this space then what we have is that psi in the class of region 3 of x let's write this as the square root of s of the volume of momentum times coefficient c growing k of x, xl now I'll refer this to the left turning point now about the right one over h bar plus c down to either the minus k of x, xl over h bar and I should say that growing in down to your refer to the behavior of the wave function has to move to the right so a growing wave function is going up like this growing wave function is the one that's allowed by boundary conditions if you want to go to the zero the damp ones when it's damping coming in like this means it's going up if you go backwards that's when it's not allowed but anyway the general solution is that the wave function looks like this and then in side 4 which is a class but allowed region that's why this is one of the square root of p of x let's make it a c right e to the i s of x, xl over h bar plus a c left e to the minus i s of x, xl over h bar like that those are the left one locations now the procedure for finding the connections is exactly the same as before the algebra is just the same as before but there's some minus signs of stuff in there so there's no point in going through it again instead I'll just summarize what the connection rules are but I should say here these are the connection rules in which the let me just draw a picture here you might be best to make a picture so here's the x r here and this set of connection rules the picture looks like this this is a potential region here p of x but another picture here of p of x and now a potential region is going down like this and here's a turning point p of p which is xl and so I'm going to just need to site for you the matrices which I'm going to dig them up here in my notes when I will do that these notes here yes, here's what they are so what we get is cg, cd is this matrix one half, one half minus i I don't need to copy this down cr, cl cr, cl the inverse matrix is one i over two one minus i over two cg, cd these are the connection rules and then the other direction so both of these are connection rules that they work at turning points that are either on the right or on the left so let's say let's now let's now use this to solve this oscillator problem on this board here let's solve it in a somewhat rigorous manner instead of achieving this intuition and not have a particle of keyless space as it goes around the orbit so let's do this let's draw our potential energy like this these, so here's the left turning point that must have gone somewhere else because I remember doing it here's the here's the turning points down base to base picture x and p and you get an orbit that goes around like that let's call these regions one two and three not the same number I used before here the one and three are classed and the two is classed in and out and let's write side region one let's write it this way one of the square root of the absolute value of p of x times a growing coefficient of column A g I'll call it coefficients of these regions A, B and C A in this region, B in this region C in this region so let's say A growing e to the k of x, comma, x, l over h bar let's say and h bar side region two is going to be one of the square root of p of x and now I'll use the version of the wave function in which the integral is the action integral that's referred to the left turning point this is, oh, I forgot something one of the guys explaining that you get different connection rules of the turning points on the left what I forgot to mention is just like we did in the case of the turning point of the right it was convenient to introduce space shifts of pi over 4 into these exponents with the s's in order to make a degree of the phases that are under the standard forms of varying functions here it turns out I want to put a minus i pi over 4 on this side plus i pi over 4 on this side and that's going to be necessary to make these formulas correct if I don't do that then there's extra factors of e to the i pi over 4 and that's the most matrices of the block that's here now so here's the side 2 in this region this is let's call it a d coefficient to the right e to the i over h bar s of x coming s l minus i pi over 4 plus d right e to the minus i pi over h bar s of x coming s l plus i pi over 4 however there's another expression for side 2 that we're referring to the action to the right turning point there's actually two expressions here because we had two turning points so both cases get a formula of the way to the left so let's call it a gr prime this is another way of moving to the right but the action is referred to to the right turning point now and the sine of the pi over 4 is the opposite let's call it a b r prime e to the minus i the right turning point minus i pi over 4 and then finally in region 3 let's write this as one of which is a classic for good in region let's write it as one of our absolute value of p and let's call this c growing e to the k of x comma x r over h bar plus c down to e to the minus k of x comma x r over h bar like that and the algebra of doing all this is really a glorified version of what you've done in undergraduate courses where you have square mode potentials this is slightly more complicated but once you've got the matrices it's just the algebra working through the matrices so here for example we'll start from the left and move to the right the boundary conditions in the left are that the coefficient of the damping term of wave function is going down like this has to be 0 because that's the one that's blowing up this x goes to minus infinity so we have to have a d as equal to 0 and I'll use my eraser to get rid of it and as far as a g is concerned the coefficient of the growing term let's set it equal to 1 which is just an arbitrary normalization now once we've got that we can find the coefficient of v r let's see this is v r and v l this is v l I'm right and left we can get v r and v l by using the upper version of the connection rules the symbol C's get replaced by A's and B's but this is going to be v r and v l is going to be that matrix 1 y over 2 1 minus y over 2 multiplying on 2 I'll remind you that A down to is equal to 0 and k growing is equal to 1 so we're going to have 1 0 and the result of this is v r equals v l equals 1 so erase them because they're equal to 1 and this is another formula the first version of the site now these two expressions for the wave function in the class of the allowed region 2 have to be equal to each other there's only one solution there so moreover and the two left side of the wave has to be equal to so this entire term must be termed below and that allows us to solve for v r prime let me show you how this is done this gives us e to the i over h bar s of x comma x l plus i pi 4 that's this upper term is equal to v r prime times e to the i over h bar s of x comma x r referred to the right turning point which should have been a minus prime before but plus i prime before now s of x comma x l its definition is it's the integral from x l to x of p d x that can be written as the integral from x l to x r of p d x plus the integral from x r to x of p d x of which the last term here is s of x comma x r these two actions that went well to the two different turning points are related to one of them just like a constant a changeable or limited just added constant and what is that constant? it's this integral here if you look at the p d x p is a function of x integrated from the left turning point area so that integral here is actually equal to one half of the area of the orbit but the area of the orbit is equal to 2 pi times the action so this is equal to pi times the action i and so the result is that if I make this a product of two things like this v minus prime pi over 4 this first phase here is equal to e to the i over h bar and s of x comma x r which is the right turning point times e to the i over h bar times pi pi now you see that this phase here cancels a bad phase that's the x dependence which had to go away because the v r prime is a constant and the rest of it is just solving for v r prime so what we find is v r prime is equal to just bring this e to the minus pi over 4 on the other side e to the i pi over h bar times the classical action i minus i pi over 2 in that manner we determine the physical condition of v r prime now as far as v l prime is concerned it will be the complex conjugate of v r prime because the line above is real the line below has to be real also so if you have v l prime it's equal to e to the minus i pi of capital i over h bar plus i prime which is really boxed out because that gives us two more conditions now finally the only thing left to do is to use the next connection rules for a right turning point to get the convergence cg and cd so gosh I'm afraid I'll have to erase this just to make room so what we can have then is that the analogy we can use this connection rules for a turning point in the right so cg is going to be equal to i times v r prime I'm using this matrix but I'm reinterpreting c r and c l is the prime v r of v l minus i v l prime if you look at that i times v r prime v i is going to cancel i times v r prime so what we get is that this is a cosine of pi i and as far as c gamut is concerned it's one half and now it's going to be this matrix one half and some of the two so it becomes one half turns it into a cosine it becomes a cosine this was actually twice the cosine turns into a cosine over h bar pi i over h bar minus pi i over 2 however the coefficient of the growing solution in this class which has been the region of the far right must be zero boundary conditions so this we set equal to zero and so that means that pi i over h bar which is the angle inside the cosine has to be equal to pi over 2 or 3 pi over 2 or 5 pi over 2 those are the places where the cosine is zero which is the same thing as n plus a half times pi and if we then multiply this out what we get is i into the n plus a half times h bar and that is the optimization condition of the action of the orbit essentially the area of the orbit same one that we got earlier using the institutive rule of the particle this is a retardation of course some of the conditions using the action of this we get something else too because if I plug in i equals n plus a half h bar into here this thing becomes the cosine of n pi which is equal to minus 1 to n so we get an explicit solution to see growing coefficient of zero so I'll take it out and then cv is minus 1 to n when we do the quantization so in addition to getting the quantization conditions this is more detailed analysis because this is the benefit of getting the wave functions as well the wave function in the class of the allowed region might be the most interesting one it's really the second line here that I can write it this way is twice the length of the square root of p of x times the cosine of s of x, comma, x, l which is grown to the left turning point over h bar minus pi of the 4 and this is actually one of the useful results of the ukv theory is that it is when we get the approximation to the wave function of the oscillator in the class of the allowed region in the class of the presentation 1 whatever the square root of epsilon of the value of p of x e to the minus pi of x, comma, x, l then in side 3 of x is equal to 1 over the square root of epsilon of p of x times minus 1 to the n e to the minus k of h bar of x, comma, s, r of h bar so there's a complete set of solutions to problem with this oscillator oh I'm sorry, I didn't watch it I didn't read it out ok, anyway, please just do the study