 We're looking at one national WKB theory, so we want to solve the short-term equation in one dimension, pretty simple equation, with some energy keys. We're looking at energy eigenstates. If you imagine a potential energy scaling that's called capital L, and I've kind of sketched here what that means, we're interested in a situation in which the buoyant wavelength of the wave culture, which we call lambda, is much less than the scale. So we're interested in the condition of the wavelength of WKB theory as a lambda that's much less than L. The lambda, of course, is relatively tuned in momentum by the de Broglie relation that's 2 pi h bar divided by t. By use of de Broglie relation, one can say that the lambda is small if h bar is small. And frequently, the WKB approximation is described by saying that the valence of h bar is small. It isn't really dimensionally correct to say that. If you want to put it in dimensionally correct language, you can say that the wavelength is much less than the scale of a potential. Nevertheless, the WKB expansion isn't, in a sense, the formal power series expansion of powers of h bar. Now, by the way, the momentum that comes from the Broglie relation is now a function of position because the wavelength of this wave actually changes as you move around. Now, the need to interpret this will have a function. Yes, that's something you should figure out. Pay attention to. In any case, this leads by looking at a picture like this. This leads to the WKB-onzox of the wave function, which is that it's a slowly varying amplitude, and it's an rapidly varying phase. Slowly, there may be a very little scaling of the potential that you see by this amplitude I've drawn here. And a rapidly varying phase means that in a short change of x, the phase undergoes some sort of thermal oscillations. In fact, in a single label, which is regarded as a short ring here, the phase changes by, of course, by 2 pi. If you think of h bar as being small, the rapidly varying phase is achieved by the fact that since you've got this h bar in the denominator here, the small change in s produces a large change in the phase. In any case, by refining the change in s equal to 2 pi, delta x equals lambda at the end of the hour last time we derived this relation that said that s prime of x, the derivative of the function that appears in the phase, is equal to this momentum function p of x that comes from the de Broglie relation. Now, pay attention to this function p of x because it's being defined here in terms of the local de Broglie relation of the wave length in a common way. But it also has a classical interpretation which is going to come out in a minute. So these two things will be connected together. It's also, as I said, a derivative of s. Alright, so I think that takes us up to where we were at the end of the hour last time. Now, this wkb-onsatz, so the name of the next step is to use this wkb-onsatz in effect, plug it in the Schrodinger equation and give equations to the A and S and then we've got the solution in this approximation. The wkb-onsatz by itself is not a systematic expansion in powers of h-bar. It's rather just something we look down by integration by looking at a picture like this. However, we can put it into the form of a systematic expansion which I think clarifies the structure of the theory in the following way. First of all, let me just rewrite the derivative of k-b-onsatz by bringing the amplitude A up into the x-colon. So I'll write it as e to the i under h-bar times k-b-onsatz plus the logarithm of A. That's no change, of course. But when we do this, you can notice that when Mel appears in the x-colon, he's starting to look like a power series in expansions in powers of h-bar. So the first term goes as h-bar to the minus one and the second term goes as h-bar to the zero. And so what this suggests is if we replace this by a systematic expansion that looks like this, e to the i over h-bar times the series, which I'll write up this way, let's say w-zero at most order plus h-bar times w of x plus h-bar squared times w of x plus dot, dot, dot like this. So if we replace the wave function psi by this expansion in x-colon, you can see the idea is that we're making a power series expansion of the logarithm of the wave function in powers of h-bar, where we're thinking of h-bar as being small. Now, why don't we expand the wave function itself if you're going to study this in the classical limit? You'd think h-bar is small, you'd want to expand psi in powers of h-bar. The answer is that when h-bar goes to zero, if your whole classical formula is fixed, psi does not approach a limit at all. Instead, it just starts oscillating more and more rapidly because of the void wavelength getting shorter and shorter. So it's not suitable to expand psi in powers of h-bar. But instead, we're going to expand the logarithm of psi factoring out the 1 over h-bar, which is what takes care of the rapid oscillations. So in any case, this is what we get and it's a systematic expansion. Let me just call this an entire x-colon and let me just call it w of x like this in effect relating psi to its log w in this manner. So I'll copy it again here to say psi of x is equal to e to the power of h-bar times w of x, this becomes a definition of w now. So the next step what I want to do is to take this version of psi and put it into the Schrodinger equation and get an equation for w. Oh, by the way, maybe before I do that, let's notice that comparing this series here to w of zero is the same as s. So let's write that down. w of zero of x is the same as s of x. It's just another notation for it. And w of one is equal to the logarithm of a except we have to multiply by i's. So w of one of x is equal to as you'll see minus i times the volume of a. And then there's higher order of corrections too which were actually further refinement of the original WKB pharmacist. All right, so to get back to where it was just a moment ago, let's take this expression for psi and plug it into the Schrodinger equation. We have to evaluate the kinetic energy term of the second derivative of psi. So let's start taking derivatives. So psi prime you see is equal to the i over h bar times w prime multiplied times psi. First derivative. Taking the second derivative, psi double prime is equal to first I differentiate with w so I get i over h bar w double prime times psi. And then the difference in psi brings out another factor of i over h bar times w prime. So we get minus one over h by squared times w prime squared times psi. And it will allow me to do a bad thing which is to use my eraser here. I'll just factor it out and come into factor of psi in both these terms and write it like that. All right, so that's just the algebra. Now if I plug this psi double prime into the Schrodinger equation since you now see psi double prime as proportional to psi and the potential energy term is also and total energy are also both proportional to psi. And I make this substitution the size to cancel from all three terms. And so what's left over is this and it's minus h bar squared over 2m times this coefficient in the round brackets. I'll reverse the order of the terms minus one over h bar squared w prime squared plus i over h bar times w double prime. And then add the other two terms which is the potential energy v of x is equal to v. This is what we get. And let me now just clean this up, multiply through by minus h bar squared over 2m. And what we get then is to set one over 2m times w prime squared minus i h bar over 2m times w double prime plus v of x minus e, excuse me, is equal to v. Like this. And I'll put a box around this because it's a useful intermediate step to look at. This is the Schrodinger equation which has been transformed from psi into its order. In fact, there's no approximation at this point. It's exactly equivalent to the original Schrodinger equation. However, what I would do now is to substitute this expansion in for the w here. So this becomes one over 2m times w prime and becomes w zero prime plus h bar times w one prime plus dot dot dot quantity squared. That's what it says here. Then we get minus i h bar over 2m times w double prime which is the same as w zero double prime plus h bar w one double prime plus dot dot dot. Plus potential energy v of x equals e. So this is to take this series and to collect it order by order in powers of h bar. So h bar to the zero power which is a leading term. As you see, we just take the value zero prime here which is squared. So the first term is one over 2m times w zero prime squared. And then the next term is already order h bar so it doesn't contribute to h bar to the zero. And there's potential energy which don't depend on h bar so this becomes plus v of x is equal to e. And take back the soft little bit from the h bar to the zero so you can see the equation. So this is the lowest order equation to get. Now, at first order in h bar we're going to go to the first two orders of this. At first order in h bar there's a cross term here and I multiply these two terms together so I get 2w zero prime times h bar times w one prime divided by 2m. So this gives me one over m times w zero prime times w one prime. Over here, like, they do get a order h bar term from the w zero double prime so it becomes minus i into 2m times w zero double prime. And then v and e don't contribute to h quarter h bar so the whole thing is equal to zero. And this is the beginning of a hierarchy of equations. I'll stop there, but if you promote it, you can go on to the second order to get another equation. And observe the general structure of these equations. The first one is an equation for w zero alone. It doesn't involve any higher returns. And if you solve this for w zero then you can plug that into the second equation here and here for w one. And likewise you go to the next order so you solve this one for w one. Then you go to the next order you can plug w zero and w one to get an equation for w two. It's a hierarchy that goes on this way and you can solve it order by order. We're only going to be able to do the first two quarters. Now, let's take these equations here in terms of w zero and w one and let's put them back in terms of the original W Gaby-Hansatz. So with the first equation that's easy it becomes one over two m times s prime squared plus v of x equal to b. And for the second equation let me do some algebra on that. First of all we have to cancel what you see here. So the w zero prime becomes s prime. The w one prime since w one is minus i log a this becomes minus i times a prime divided by a. The second term is minus i times s double prime that whole thing is equal to because w zero is s. So it's minus i s w prime to s double prime equals zero. And you change the two minus sides to a plus and cancel the i's to make it go away. And this can be written now as a prime over a times s plus one half s double prime equals zero. And since that's sort of money out of the rules we carry out over here. And if I take this equation and let's multiply it times twice a squared which clears it up. Then the first term becomes two a a prime times s which should be s prime. And the second term becomes a squared s double prime equals zero. This in turn is an exact derivative of the left-hand side. In fact it can be written as d d s of the quantity a squared times s prime equals to zero. And that's the most convenient way of writing this first order equation after we've converted back into these s and a variables. So over here on this box then is the version of the equations that I want with the function s and function a. Which are the two functions that have to do with the phase which are the w-p and the on socks. I really started writing the wrong formula this morning so I'm afraid I can cover this up. I'll write out these two equations once again. And here they are. So the first equation over h part of zero is this. It's one over two m s prime squared plus d of s equals e. In the order h part of the first equation is d d s d squared times s prime equals zero. Pretty simple equations actually. The first equations is actually a version of the so-called Hamilton-Dakota equation which has a long history in classical mechanics that goes way back before the invention of quantum mechanics in 1830s and 1840s. The second equation is sometimes called an amplitude transport equation and that's what I'll call it today. So we need to solve these two equations. Now let's we need to solve the let's solve the Hamilton-Dakota equation first. Obviously the way to do that is this is s prime dot s 12 s prime squared. The way to do this is to solve algebraically for s prime. And if we do, we find that s prime of x is equal to plus or minus just bring two m across and take the square root. Plus and minus the square root of twice the mass times the energy v minus v of x like this. Let's just take the plus sign of this for now and we'll worry about the minus later on. I'll remind you from the previous board that we figured out that s prime of x was the momentum which was the momentum determined by the local de Broglie wavelength. If we take that as this function, definition of this function of p of x, what we now see is that we have a little additional information now as we see what we have in formula for p of x of that de Broglie wavelength, local de Broglie wavelength of the wkd wave function. What does it mean, however? Let's get an interpretation of that. In order to do that let me take an example. Let's suppose we have a potential energy that looks like this. It's still a one dimensional problem. Let's suppose the potential is essentially zero for large negative x but that is the movement of the region of positive x that rises up like going up to a mountain and let's say it stays up forever so there's no possibility of entertaining through with this mountain and let's suppose there's an energy which looks like this. Now the energy that I want to talk about is the energy that appears in the Hamilton-Chicobbi equation and actually that energy is the same as the energy eigenvalue in the original Schrodinger equation. That's where that came from. So but in a picture of the potential it looks like this. Now, classically a particle coming in from negative x would originally be a free particle and it would be a natural zero. This would be an x term here. But as it comes in it starts to feel the potential and it slows down. It reaches, it comes to a standstill when the energy is at energy zero just right here. This is the class of turning point which I'll call XTG and then of course it reverses and comes back. If I take some x-value from a typical x-value here like this then you can see a particle going in the forward direction and one is going in the backward direction and the magnitude of the velocity of course is the same in both cases but the sign is changed exactly for it. Now, on the same scale allow me to make a phase speed a phase speed of classical phase space position momentum phase space. Let me bring this x turning point down here also to the lower graph The orbit in phase space of large negative x is a straight line in a constant momentum because the particle is going free so the momentum is constant and it just keeps moving forward in that. However, when it starts to get to where the potential is active it slows down so the momentum goes down like this and then goes right through momentum equals zero at the turning point and it turns around like I sketched this well and it comes back out like this so the orbit in phase space looks like this, it's kind of a leaf plane. This is really a scattering problem it's a one dimensional scattering problem of an impenetrable barrier. Likewise, if I take this x to the typical x scale here I bring this dotted line down like this and again here's some typical x scale I'll do what you see once again is the velocity of the particle passes this x point twice once going in the momentum is positive and the momentum is negative like this. Now, this curve in phase space what is the equation of this curve? The answer is it's just given by conservation of energy. If you write down the total energy of the system which is p squared over 2n plus v of x this is now the classical energy and instead of equal to the total energy which of course is conserved then you get an equation for a curve and the x p plane and it has to be the orbit the orbit is just a plot of a curve which you get by setting the classical Hamiltonian equal to the energy but in particular this allows us to solve for the momentum values that correspond to a given x position and that momentum just solves this equation for p and that momentum gives us p as a function of position as you see and it's the square root of twice the mass times the energy minus v of x in the Hamiltonian well that agrees with the same p of x that came out of solving the Hamilton-Jacobbi equation from the WKB approximation so the local momentum connected with the local de Broglie wavelength from the WKB wave function is exactly the same as the momentum of a classical particle which is some moving moment together potential with two sides plus and minus so that's the interpretation of this function of p of x and you can see now right away that the momentum is going to be shorter where the momentum is higher where the velocity is higher and if we longer as the particle slows down I didn't exactly sketch that very well on this diagram but it would mean that as we rise up and the potentials of the velocity decreases this wavelength is going to get longer down here in this lambda now so that's the interpretation of this function p of x which is also s prime of s solution for s of x is easy it's just an integral with an upper limit of x of p of x prime d of x prime which to be a little more explicit is with an upper limit of x of the square root twice the mass energy minus p of x prime times of d of x prime of this what about the lower limit of this integral the lower limit of the integral is nothing but a constant of integration to help the Jacobi equation because you can set it equal to anything you want because it's a constant of integration we just call it x0 you can see that if you change x0 to something else x0 prime all it's going to do is just add a constant to s since s is the phase of the wave function let me write the wkv onslaughts down again so that you have here on this board if I change the phase s by a constant actually the phase is s over h but if I change s by a constant you can see an ounce of just changing the phase of the wave function so this lower limit acts as related to the phase convention for the solution and right now let's worry about phase conventions later so we can set this x0 to anything that does convenient anyway that's just the meaning of that lower limit there's also a geometrical meaning now if I take x0 let's say over here just to choose a value for it s of x is the integral of this p function between x0 and x which is to say this is the area here so the area here the area here under the classical curve is the same thing as this function s of x what I'm saying is the classical part moves along you can visualize that s of x is a little integrator with it and it integrates pdx and as it does so what it's accumulating is this function s of x which is divided by h bar is the phase of the quantum wave function so you can imagine that it's also creating oscillations so I'm not going down as it moves along it's a way of thinking about the solution now now I'd like to shift the tension now to the quantum interpretation but with this WKB Onsatz in particular to examine the probability density and probability current this is only a one-dimensional problem so the current is just a scalar and only a vector of one component but if we use the definitions we've talked about before the probability density is the square of psi and the probability current which I'll call j I'll remind you it's the real part and then you have the velocity operator which is minus i h bar over mass m acting on psi this is the definitions we get oh excuse me it's minus h bar on the m times the derivative pdx acting on psi this would have been a gradient in three dimensions but since this is one of the probabilities of the definitions we have now if I take the WKB Onsatz this is only an approximation of the number but in this approximation let's take this and plug it in here and see what we get when you can see right away the rub is just equal to the square of the ampers room that's assuming that s is real but s is real as you see up here in these pictures that I've stacked and so that's easy and so a squared is just the square of the wave function what about j well that's equal to the real part here I'll do it out for you, psi star is equal to the amplitude times either the minus i s over h bar and then we have minus i h bar over m which is a constant and then we have pdx applied to psi which is way up there there's two terms one comes to differentiating a and the other one from the exponent so one term is this is an a prime and the next term is a plus i times a i over h bar and this is all multiplied by times d by s over h bar and I close all of that so this is just a square of the algebra calculating j now this term i here times the amplitude a as you can see is purely imaginary and this term here times that second term that has an i and it's going to have a minus i times the plus i so this is purely real so the result is a prime term goes away because we're taking the real part and what's left over is is that this turns into an a squared which takes care of that a and that a there the phase here and here of two phases cancel out as a complex conjugate minus i times plus i cancel there's one over md h bar's cancel so what's left over is just s prime divided by m okay however we decided that s prime was the same thing as this momentum function p of x which we have an interpretation for up here and this is divided by m let's call p of x let's call it a velocity let's call it v of x so it's a logical definition here if p is a momentum of particle I can't reach it I'll just say v of x is v of x over m so this is just the velocity of the particle that goes with x position really one of the two velocities because there's plus and minus but it's part of the momentum of the directions let's take this to be the positive velocity and so to summarize then we get j, rho is equal to a squared and we get j is equal to a squared times the velocity v which is the function of x like this okay so in the wk of the approximation this is the quantum density and current at least we have to satisfy a continuity equation this is the continuity equation which we talked about the last hour I believe except in one dimension instead of a divergence it's just going to be working on the derivative so well it's going to be partial derivative partial of j with respect to x equals zero so this is the one dimensional continuity equation however nothing here depends on time everything's, rho for example is a transition that's because of the original shorting equation they're looking at is the time-independent Schrodinger equation so we're really interested in stating our solutions so this term goes away and so the continuity equation just turns into derivative of j in fact let us call it an ordinary derivative since the original time derivative anymore and so the continuity equation reduces just dj v axis equal to zero however j is a squared times v so this becomes d dx of a squared times v of x is equal to zero and apart from the factor of 1 over m you'll see somewhere here yes up here that's the amplitude transport equation so the meaning of the amplitude transport equation is conservation of probability it's the continuity equation for that given that fact so now we have an interpretation of both of these equations and also for the function we'll have an interpretation of s in terms of the classical mechanics we'll have an interpretation of a let me address that question next let's say a classical interpretation it works like this to be specific let's take the case of an oscillator suppose I have an oscillator like this and in face space this is the orbit so the particle is going around in circles like this this topological circle for a harmonic oscillator this is an ellipse but for other oscillators not linear oscillators will be before we look like an egg or something but topologically it's a circle now let's take some small interval dx like this by the way let's let capital T here be the oscillator a lot of time it takes for the classical part of it to go around let's define let's define a I'll call it a classical probability density which is a function of x and let's define this way we'll say the classical probability at x times dx which is a small interval this of course is the probability of finding the particle in this small interval let's declare that this is equal to the amount of time dt which the particle spends in this small interval divided by the total period it's just the fraction of time the particle spends in that small interval then if we do this this clearly is normalized to 1 around the oscillator once the light thing to change however it's a question of the interpretation of dt the total time the particle spends in this interval is a dt from one above so this dt but it stands for the sum of those two small time intervals it might be preferable to let dt stand for the time but it takes the particle to cross this small interval on just one of these two crossings and if we do that we're going to factor a two here to double back with that new interpretation of dt this becomes the classical expression now we can then solve for the max explicitly it becomes 2 divided by the total period times dt divided by dx which is equal to 2 over the period times 1 over the velocity now why is that interesting it's because it agrees with the solution of the amplitude transport equation here it is using amplitude transport equation really can't really compute any equation so the dx of this thing is equal to 0 it says that a squared times the velocity of u of x is equal to a constant over a squared is a quantum probability instead of plugging us in we see the quantum probability loaded back to seem to be a constant fighting line with the velocity of u of x so it agrees with this classical calculation this quantum probability density in this semi classical interpretation is the fraction of time which a particle spends well times dx is the fraction of time a particle spends on that number of dx here it's explicitly normalized of unity for the case of an oscillator if we had a scattering problem which is the kind of thing I have in the company board here I don't even need to draw it again where the orbit goes out to infinity then in that case it's not possible to normalize the classical probability distribution because the time the period of t goes to infinity but it's still possible to talk about the classical density which is just now not normalize we can just in this case we can just say the classical density is equal to some constant divided by the velocity v of x and this is in fact precisely what we get from the solution of the amplitude transport equation so this thing gives an interpretation of the classical interpretation of the amplitude a in terms of particle density as the quantum row you see classical row and quantum row are the same the consequence of this result is that when the velocity gets low the density gets high in the case of this oscillator here if I plot again this now has two training points let me plot row classical as a function of x for this oscillator it looks like this it actually goes to infinity as the velocity goes to zero you will find classically this is still normalizable the intertural area of the curves is far right but nevertheless there is divergence in the classical classical density here's another way of thinking about this classical density as a measure that we populate this orbit with classical particles in such a way that individual particles are separated by the same amount of the last time you see each particle is following an orbit it's the same orbit they're just displaced they're just in different places so they're all moving around the orbit at the same time when I say that this place by a certain amount of fixed amount of time there's some delta t here and the idea is that after delta t each particle moves into the position of the next particle if you think of delta t getting smaller and smaller then the particles are all moving but they're moving in such a way that the density of particles that they create is constant in time so one can say that it's an ensemble of classical particles whose density is stationary in time that's what's occurring in the solution of the eigenvalue equation stationary state problem in quantum mechanics the particles pile up here at turning points because you see there's no pile up in the orbit and phase space but they pile up in the density and next space because we're getting projected down like this and there's slope of this curve coming down that's the reason for that okay so most of what I've been saying here has to do with classical interpretations and solutions of these two equations to help the Jacobian amplitude transport equation it's actually easy to solve the equations it's just a matter of knowing what they are now so we have these two equations still here so to summarize the solution then we've got s prime of x is equal to p of x is equal to the square root of 2 nanotimes e minus p of x so s of x is the integral of that of the limit x of p of x prime of dx like this with some lower limit and as far as the amplitude transport equation is concerned it says that a squared is whatever s prime is whatever the momentum so a of x is equal to 1 over the square root of the momentum p of x times the constant here like this and that's the solution for the amplitude transport equation if I factor a mass out of the square root of mass out of the constant it inverts this momentum into a velocity it takes impact these classical densities that I'm talking about here this constant that appears here is just the constant generation of the amplitude transport equation and you can see that it also was just related to the normalization of the quantum wave function in which you can sort of lost it then the wk beyond spots is the amplitude times the phase and so obviously a constant multiplying a is just an overall multiplicative constant inside so we can likewise as I said x0 is related to the phase of the wave function we can worry about normalization and phase later so these two constants we can set as something which is convenient of a later point all right Delvin this means that if I go back to this case of the scattering problem where the potential rises like this here's the energy B here's the turning point sketch it again here's the turning point here's x here's the phase phase orbit coming down like this all right so this is the solution for a and s here there's a couple of things that I need to a couple of minor points I need to pay attention to one is that in solving the amplitude transport equation there were really two solutions for s' having to do with a plus or minus on the square root and I just took one of the two solutions from writing these equations down but the other one that's in there is just the opposite side so s is equal to plus or minus the integral of the effect of the two solutions the result is we actually get two solutions of the WKD equation so if I write psi of x out explicitly there's an amplitude for one of them is one over the square root of p of x I'll set just one there for this constant and this is all the y times e to the i over h bar s of x but there's another solution which is one over the square root of p of x times e to the minus i over h bar s of x there's really two linearly independent solutions this of course is what we'd expect because the Schrodinger equation is the second order and has two linearly independent solutions so the general solution is going to be a linear combination of these two now for one of these solutions the function s is increasing as I move to the right because it's the area under the curve so the phase is increasing as we move to the right and that corresponds to a wave just traveling in the forward direction just like in a positive rental whereas the second term s is the phase is decreasing as x increases which means there's a wave here which is moving in the other direction and so the linear combination is a linear combination of forward and backward going waves which corresponds to the two particles which are crossing into an exposition like this if you think of this in terms of one particle you don't see it because you see a positive velocity at one point and then a negative velocity if you think about this curve as being populated by a whole ensemble of classical particles and then there's always one particle above and one particle below where their phases are increasing in opposite directions here's something I mentioned a moment ago the WKB theory expresses an approximation expresses the solution of the quantum problem in classical terms but the physical interpretation of quantum mechanics of course is very different from classical mechanics and in particular quantum mechanics is statistical fundamentally and classical mechanics is not so how does this very different physical interpretation come about in this approximation it's one of the questions to ask when we start this business and we can already see part of the answer to that is that the statistical aspects of the quantum solution is contained in this classical ensemble of particles that are orbiting around and around this orbit or in the case of an oscillator there's an infinite number of coming in scattering going back out again it's not a single classical particle but a whole ensemble of them they all however have the same energy because they all have the same orbit alright in any case these two solutions one is moving to the right and one is moving to the left so let me give this a coefficient of a call CR and this one a coefficient of a call C L where R and L just stand for right and left going waves and if we have this then this is the solution this is the WKB solution that begins alright now there's another point that I've been admitting up to this point which is that these solutions that I've worked out here actually only apply the classically allowed region in this diagram here for example the classically allowed region C A R let's call it is the region which is the left of the turning point this is where the actual particles are moving but there's also a classical combination that's called a CFR which is over to the other side there are some classical particles there may not be any classical particles over there but there are solutions that the Palomkin to Covinae have to do transport equation the classically allowed region the energy is greater than the potential energy that's what the definition of that whereas the classical convenient region is one where the energy is less than the potential energy that means that in the classically convenient region you would say the kinetic energy is negative which would mean the momentum is imaginary that doesn't have a meaning in classical mechanics but it does have meaning in solving these quantity equations and so in the classical convenient region what's going to happen is this quantity here is negative so this little box here let me just say that in this box that I've got here applied in the classically allowed region and let's write down some things that apply in the classical convenient region let's just do it this way let's define this let's write p of x p of x now has to be purely imaginary let's write it as i times the square root of 2m times p of x minus e this is just a way of defining one of the two the two with solutions plus and minus for the function p of x and let's define an integral on k of x which is an integral with some of the limit and it would be the absolute value of p of x prime p of x prime which is the same thing as the integral with an upper limit x it's a lower limit of the square root of twice the s times the potential of p of x minus e of x prime minus e of x prime okay and if we do this then s of x which is is equal to excuse me, pi of s of x is equal to minus k of x so over here in the classical so this solution I wrote down in the book this board is for the classical in the classical convenient region we can say this that sine of x is going to be a linear combination of a term which is one of the square root of x the absolute value of the moment can make it real and then we get e to the k of x over h bar and then another solution is one of the square root of the absolute value of p of x times e to the minus k of x over h bar in the classical convenient region there again are two many independent solutions k is increasing with x and so the term that has e to the k of x over h bar is a term which is diverging exponentially as you move in the positive x direction so let's call this c g growing the minus sign is a solution which is decaying exponentially as you move to the right so let's call this damping solution c v growing in damping solutions as you move to the right and the result of this is that we have now the two solutions here and the classical and the classical and the regions now to proceed with this let's take the case as I sketched it here exactly as I already sketched in which the classical and the left of the turning point in the classical convenient region is to the right we can also call these regions and let's do it this way in the classical area of the the lower limit of integration let's make a turning point which I'll call the XTP let me call it XR but R is the center of the right means the right turning point let's call it XR as a right turning point let's integrate S as I mentioned any lower limit can be chosen it's equivalent to a phase convention for the way function so let's just make it the turning point it's the obvious place to choose anyway then that gives us a let's take the lower limit here to be XR that means both K and capital S manage exactly that and so those are definite conventions for that alright so now what we have then is definite functions here with unknown coefficients actually to four definite functions and four unknown coefficients when applied in two different regions the Schrodinger equation is second order in time in space excuse me, second order in space so it has two linearly independent solutions and so the general solution of the Schrodinger equation can only involve two arbitrary constants and not four so it has to be a connection between these four constants to the right and the left and the growing and the dabbing constants in different regions of space so what we need is some way of connecting together the solution on the left side with the solution on the right side these are called connection rules now it's not possible to extend the class that allowed solution right up to the turning point and the class that forbidden region right up to the turning point and then that is the matching of there in order to get the connection rules connecting to coefficients the reason for that is that neither one of these solutions is valid in an immediate neighborhood of the turning point in fact it can be shown that there is some small region around the turning point where both of these solutions are invalid why is that so? they're invalid there because the momentum lost in the momentum are going to zero at the turning point but remember that the Devois white point is 2 pi h r divided by the momentum p so this actually goes to infinity at the turning point and the result is the condition of Lydian WPB which is that the wavelength should be small compared to the scale of the potential breaks down so if you look at this more carefully you can actually estimate what size of the region around here the plunger is not valid I won't go into that but let's just say because of this small region around the turning point where neither one of these solutions is valid it's not possible to find the connection coefficients by just bringing them both up to the turning point it would be nice if you could so instead of what do we do what we need is to connect with yet a third solution on top of those two a third solution which is valid in an immediate neighborhood of the turning point this is a stuff like this if I sketch my potential again the events let's take this case here here's our energy here's the turning point like here like this I'm calling XR now let's approximate the potential of my straight line in the neighborhood of the turning point so let's start here V of X is approximately equal to V of X at the turning point XR what's X minus XR times V prime of XR and if we plug that in in the Schrodinger equation then it becomes minus H bar squared over 2M times psi whole prime plus this thing V of X so it's V of XR and it's copied at X minus XR times V prime of XR times psi equals E psi V of XR however is equal to is equal to the energy E because that's the definition of the turning point and the energy is equal to the potential so this term V of XR cancels the E term on the other side and the result is the beginning equation which now just has this linear term this of course V prime of XR is a constant here just has this linear term on my side the right hand side is 0 this equation can be cleaned up and put into a standard form if we do a change in variables that's why X equals XR plus the constant A times Z so that Z is a new variable A is a constant here C is a new variable which has been advantaged exactly at the turning point and it will be negative on the left side and it will be positive on the right side and otherwise it's just X scaled by a certain factor A so we do a linear transformation in this result of the equation now we choose the constant A in order to clean up the equation all the constants that are in here and if you do this you'll find that the nicest place for A is this it's h bar square whereas here over 2M times V prime elevated at the turning point to the 1 third power and if you make that choice then the joining equation becomes this it becomes Z squared psi VZ squared minus Z times psi and it's been cleaned up and all the physical constants disappear this is one of the standard equations of mathematical physics you can read about it in books like The Member of Itz and Stegen and it has two solutions psi and Z what is called the airy function A, I, Z and then it's called the airy function B, I, Z and the general solution is linear combination of these two look in the books on special functions and get all the properties of the airy and airy functions but in order to understand what you're reading from a physical standpoint it helps to have a physical model to understand the results the way that I derive this is the airy equation is by using an approximation of the potential of a straight line approximation of course it's possible the potential really is a straight line and not just a continuity point could be like this for example if we had a particle in the gravitational field then B of Z let B of X that's called it would be equal to Mgx or the strictly linear in the position and so the airy and airy functions are the approximate solutions that can be approximated in the neighborhood of a continuity point of an arbitrary potential but there are also the exact solutions for a particle in a uniform gravitational field or for that matter a charged particle in a uniform electric field which is the same story because it's linear and x so thinking about the physics of a particle the quantum mechanics of a particle in a uniform electric or gravitational field helps us to understand the mathematical properties of the airy and airy functions and that's where I'll pick up the next slide because that I I I I I I I I I I I I I I I I I I I I I I I I I I I