 W shall we begin then? Yesterday we looked at this pair of wells that were separated by a barrier so that classically the particle couldn't get from one well to the other. And we found that the particle got from one well to the other and used that to make a model of an ammonium asa with a nitrogen atom passing through the barrier formed by the hydrogen atoms. Ond dweud yw'n amser ymlaen i'r perthynau ar y gwaith iddyn nhw'n gweithio'r perthynau o gyllidio'r gwaith lleol. A dwi'n gofyn honno ar gael y brifysgol oedd rhaid i hoffi'r diwethaf. Adwg yma. Ond yma, dyma'r cyflawn yma. Mae yma'r bwysig fawr o'r ffairfynol yr hyn o hyn sy'n bynnag yma. Mae yma'r bwysig yn bwysig i'r bwysig i'r bwysig i'r bwysig i'r bwysig. ymbygwyd o hyd at y byddwyr'r cyfleu gwleid diwetheid i ymbygwyd. Ac rydych i'r gwleid eir eich rwynd am gyfyn fwyntio gweithio'r bario. Rydyn ni wedi'i gweithio'r cyfleuydd yma ymgylch, a yn gweithio'r cyfleu gwleid eich cyfleu'r cyfleu eich cyfleu'r cyfeithio'r cyfleu, byddai allwch yn cyfleu, ac le oedd yn cyfleu'r cyfleu'r cyfleu hwn. Mae wedi'u cyfleu ei yw gyffredd i ymgyrch ymwynt i'r cyfleu hwn. remember the time dependence of everything and quantum mechanics. For a state of well defined energy is e to the minus iE upon h bar t. So these things are going to have e to the minus i omega t type dependence and therefore if you have a minus sign here you are looking at a wave which is travelling to the left. If you have a plus sign here you are looking at a wave wrth gwrs honno. A dwi'n meddwl, sy'n gweithio allan o'r particle a'r ystod. Felly mor gwneud yn ystod yn y porfyn, sy'n ddod yn gwneud yn y trollwysion sy'n ffordd C, gyda'r ddod yn gwneud yn y trollwysion C, i'r I kx. A'r ystod yn y barion, er hynny byddwch gydag weithio sydd sy'n gwneud yn y trollwysion C ym Y, mae'n gwneud yn y trollwysion B ym Y, yn y trollwysion C ym Y, mae'n gwneud yn y trollwysion will be a combination of an exponential growth and an exponential decay. So this is a more complicated, so this is a bit different from what we've done before in two ways. One is that, well the main thing is that our initial problem inherently has a lack of left-right symmetry, right? The potential that we're discussing here has left-right symmetry. It's symmetrical around the origin, which I forgot to say, but the origin is here in the middle of the well. This is minus a and this is a over here. So the potential is going to be an even function of x, same as ever, but our problem, our initial conditions, the physical situation we wish to discuss, has a built-in asymmetry because the particles have to come in from one side or the other. So that's, and that is computationally very inconvenient. It stops us using this nice trick. It makes it difficult for us to use this nice trick of looking for solutions to the problem which have well-defined parity and thus discussing only what happens at this boundary condition. With this setup, we're going to have to discuss this boundary condition and this boundary condition because you can see there are fundamental differences between what's happening on those two sides. You can handle this problem using looking for solutions of well-defined parity, but it's slightly unnatural and I think, well, it's actually a very good way to go, but it's not such an obvious and intuitive way to go even though it's computationally simpler and I think it's worthwhile just seeing what happens when you play the game straightforwardly and you'll see the algebra becomes quite unpleasant, which illustrates the benefits that we had before by assuming well-defined parity. All right. The other thing that's different here is that because we are considering particles which are free, because the potential goes to zero outside this interval here, the particles, and we're going to consider particles with positive energy, the particles are going to be able to push off to infinity so we're not going to find discrete energy levels. We're going to be able to find solutions for any energy. Whereas previously, we had a potential which for that going to infinity and that made the energy levels discrete. So those are the differences because we're dealing with a different physical situation and it has implications for the maths. Right. So what do we have to do? Well, it's very boring. We have to impose continuity of the wave function. So the wave function here is this wave plus this wave and that has to be continuous at this boundary x equals a. So it has to give you the same numerical value as the sum of these two things here. So let's just quickly write that down. So we have a plus e to the minus i ka, which is the incoming wave evaluated at that barrier, x equals minus a plus a minus e to the plus i ka and that had better equal b plus e to the minus ka minus, sorry, plus b minus e to the minus. That one's got plus ka. Many double negatives here unfortunately. I forgot to say of course that we will have as ever that k is equal to the square root of 2m times the energy of h bar squared because p squared over 2m is the energy and p is h bar squared k squared, sorry, p is h bar k. And we will have that big k is equal to the square root of 2m v0 minus e over h bar squared. So this is the condition for the wave function. It could be continuous at x equals minus a. We require as yesterday that the gradient of the wave function is also continuous there. So we have to take the gradient of that function on the left and evaluate it at x equals minus a. And we find that ik common factor a plus e to the minus i ka minus a minus e to the i ka clens brackets is equal to big k common factor b plus e to the minus ka minus b minus e to the ka clens brackets. Then we have, so it's two equations. Now we have two more equations because we have to get everything on the right-hand boundary which is not now dealt with by symmetry as it was yesterday. So this is where life becomes, everything becomes difficult. So we have c e to the i ka is equal to b plus e to the ka plus b minus e to the minus ka and we have that ik over k, I'll write it thus, of c e to the, well maybe I should do that one way, c e to the i ka, that's the gradient on the right side, is equal to big k common factor open brackets b plus e to the ka minus b minus e to the minus ka clens bracket. I live in hope and some anxiety that those equations have been correctly stated. So what do we have to do? We now have four equations and five unknowns I think, right? There are two a's, two b's and a c. So we will not be able to get rid of all of them. We will be able to express in principle any one of the a, b's and c's in terms of the other one and that physically corresponds to the point that the flux of incoming particles is controlled by a plus and that's in your control, you can put in more particles or fewer particles and that will obviously lead to more particles coming out or fewer particles coming out depending on the incoming flux. So the general idea is you expect to be able, the goal is to express any one of these things as a function of a plus, as a multiple of a plus and we expect them to be linear in a plus. So that's why we've got too few equations. We don't physically expect to be able to determine everything. So what we should do is engage in an elimination exercise. A reasonable way to go is to take these two equations here, divide this equation by this equation say and that will get rid of c and will give you a relationship between b plus and b minus and then you can take that relationship between b plus and b minus and use it in these two equations to express the right to get rid of b minus from these equations, these two right hand sides so they both become simple multiples of b plus and then you could divide these two equations one by another, b plus which will be a common factor on the right hand side will go away and you will be left with a relationship between a plus and a minus, a single relationship with a plus and a minus so that will be the promised relationship that expresses the number of reflected particles as a multiple of the number of incident particles. So once you've found what a minus is in terms of a plus you can go back to your original expression here which had only b plus on the right hand side a minus can be expressed as a function of a plus and you will be able to find what b plus is you will be able to find what b minus is and they can all be determined. So let me not do all that algebra, that's the strategy, the execution of course is quite tedious and the scope for making errors is quite large and in fact I find that there's a typo right there in equation 540 in the book because when you do eliminate between these two equations here to find out the relationship between b minus and b plus it should be that b minus is 1 minus i k over k over 1 plus i k over k e to the 2 b k a b plus so that differs from what's in the book partly by arrangement of this but more importantly by this having been left out that's got slipped out in the type setting. So we have that relationship there, we stuff this back into the other places and we find that a minus is equal to a plus so I've described how what we do we take this b minus, use it to get rid of b minus from here and replace that with b plus in some factor then we divide these two equations and then we get this relationship I'm about to write down between a minus and a plus and it is a minus is a plus e to the minus 2 i k a q minus 1 over q plus 1 where q is itself pretty yucky it's cosh 2 k a minus i k on k hyperbolic shine of 2 k a all over cosh 2 k a minus big k over i k of shine so the algebra is as promised altogether more yet more messy than it was yesterday because we're not exploiting parity we're not dealing with finding a right so what do we want to know about this physically what we want to know about this physically I think is what is the chance that the particle is reflected what is the chance that the particle gets through so classically everything will be reflected and the modulus of a minus would be the same as the modulus of a plus right and you can see that that isn't looking very promising because that would require that well basically the q was simply enormous right if q were very large then q minus 1 would be the same as q plus 1 and everything would be reflected but in reality it's not all going to be reflected something's going to get through how to find c well we could you could take take this a minus expression from it as I've described obtain b plus from b plus obtain b minus to put these back into the into this equation here saying find c that's too much like hard work it's easier to say that look there's going to be conservation of particles we've got a well defined theoretical apparatus apparatus here which is not going to which which conserves probability so the incoming particles the a plus are either are all going to go out at the end of the day either to the left or to the right so we can argue that a plus mod squared which is well that is the spatial density of incoming particles if you like if you multiply that by the speed of the incoming particles which is p over m so h bar k over m you will get the flux of incoming particles and the flux of incoming particles has to equal the flux of the outgoing particles which is a minus square the mod square of a minus the density of outgoing particles and times times h bar k over m for the speed plus c mod squared so conservation of particles implies this relationship between these amplitudes and of course you can in principle check whether this algebraic relationship is satisfied by these equations by hard slog because I've described how you can in fact find c we've already found a minus you could in principle find c and check that it satisfied this equation but we don't want to do all that algebra so the point is that what we want to say is the flux what we want to say is the following actually is the fraction of particles that get through is obviously the ratio of the incoming flux and the out well the ratio of the outgoing flux to the incoming flux so it's going to be this this this fraction that we want we'll call it f is going to be mod c squared over a plus squared because the constant of proportionality namely h bar k over m between this quantity and the outgoing flux on the right is the same as the constant of proportionality between this constant and the incoming flux on the left so the fraction of particles that get through will be given by this ratio here which given that relationship so in other words c let's we can write that now as a plus mod squared minus a minus mod squared mod squared over a plus mod squared but we've got a minus mod squared from that expression at the top and has a multiple of a plus mod squared so we can write this as one minus q minus one over q plus one mod squared well the mod square of this ratio is the mod square the ratio of the mod squares of the top and the bottom so this can be written as one minus one q minus one mod squared over q plus one mod squared so let's address ourselves to what these mod squares are so what's q minus one well q minus one is going to be well it'll obviously on the top it will have the existing top minus the bottom so when we take away the bottom from the top the coshies go away and we are left with I think k over ik minus ik over k time shine 2kA and that will be over I'll just call it the bottom because it's the we're not really going to take much interest in what this bottom is it is the bottom that you see up there cosh 2kA minus k over ik etc and q plus one the reason we won't care about the bottom is of course it will cancel when we take this ratio so for q plus one we unfortunately find that the coshies add and the shines irritatingly refuse to cancel so this becomes 2 cosh 2kA we're adding so we have minus ik over k plus k of ik shine 2kA and again that's over the bottom so what we need to do now is take the mod square of these two numbers ratio them and take it from one so the fraction that gets through is going to be one minus so the top of that is completely imaginary it's pure imaginary we should take out an i from that bracket and then we will find we're staring at k over k plus k over k squared times shine squared 2kA so that's q minus one mod squared as regards the top the bottom we're not interested in because it's going to cancel with the other bottom and now we have to put underneath the mod square of this which will be 4 cosh squared 2kA right because this is the real part of it this is the imaginary part of it we take out a factor i and now we're staring at plus k over k minus k over k squared shine squared 2kA nearly there so now we put this all these two bits it'll simplify if we put these two bits on a common denominator so the top when it's on a common denominator will be this bottom plus that stuff there so this will be 4 cosh squared 2kA and now we're going to have shine squared let's write it in plus yeah shine squared 2kA brackets now brackets what we will have this bracket squared well we'll have this bracket squared sorry minus this bracket squared and when we square these brackets we're going to get k squared over k squared which will cancel because of that minus sign and what will not go away is the mix term the product of multiplying this on this which generates 2 and the product of multiplying this on this which generates another 2 say we will get 4 and it will in fact be with a minus sign because this minus sign is there when this comes up here that minus sign will stick out and this minus sign will make the mix term minus there so this is going to be times minus 4 and it's over the bottom as you see it over 4 cosh squared 2kA plus k over k minus k over k squared shine squared 2kA and the top simplifies most beautifully because cosh squared minus shine squared is 1 so the 4s can be cancelled and this actually is nothing but 1 over so the fraction is 1 over cosh squared 2kA plus a quarter of k over k minus k over k squared shine squared 2kA not much fun so what do we learn from this what we learn from this is most interestingly is what happens if we have a rather a high barrier and the particles are very short of energy to get through so k is a measure of the deficit in energy that the particles by how much they don't have enough energy classically to get through the barrier if the barrier is very high and they don't have much energy we're working at the cosh of a large-ish number and the shine of a large-ish number and so what we can say is that for large kA we can say that cosh 2kA behaves pretty much like shine 2kA behaves like e to the 2kA but we're interested in fact in cosh squared and shine squared so f is looking like 1 over e to the 4kA so if kA is an appreciable number this probability of penetration is becoming small crucial result is that the probability of getting through there is decreasing exponentially fast in the height of the barrier so you don't need a very high barrier to make this quite a small effect and somewhere here we have so this machine goes to sleep doesn't it that's the trouble it shouldn't go to sleep give up is there anything there I'll just draw it is it sort of no doubt we're saving the planet by having the machine turn itself off I can't see it so you want to do some so that's sort of asymptotically what happens when kA is very large in detail you might want to know for smaller kA right so these results are for a barrier which is in these results the barrier is not terribly high so we have v0 sorry we have that e is equal to 0.7 v0 what have I done that is correct the height of the barrier is there's this parameter W which we talked about yesterday which is a measure of the width and the height of the barrier so it's 2m v0a2 over hbar this animal that's your dimensionless measure of the height and the width of the barrier in terms of the mass of the particle with no reference to the energy of the particle sorry that's not the case then what's being plotted here is the probability of getting through as a function of your energy over v0 for barriers of different Ws so I think it's a 0.5 at the top there so here's a relatively weak barrier which gives you a fairly small energy as a chance of getting through in other words it's not a very fat barrier which is a crucial thing this is a fatter barrier so you can see how as a function of the energy your chance of getting through rises in detail see if we can get these things to stay alive for later what's physically interesting about this or an interesting application of this is the radioactive decay so this is obviously a very simple minded model that we have so far but the general idea for example is this so what we should say is that inside 238 uranium which is the non-fysile sort of uranium you have a number of alpha particles this is a simple minded picture so what does the potential energy of an alpha particle so we kind of consider this to be so 238 uranium which decays to 234 thorium and an alpha particle with a half life of I think it's 6.4 gigae years so it takes the age of the universe typically for a uranium 238 made in some supernova to eject an alpha particle so what's happening here from this perspective so what we should do is we should think about this alpha particle and this uranium 234 nucleus as a kind of dynamical system so the alpha particle when it's a long way when it's a decent distance more than 10 to the minus 15 meters or so away from the thorium nucleus is repelled by the electrostatic repulsion so the potential energy curve has a sort of 1 over R type behaviour here when it gets close enough to the thorium nucleus the strong interaction it's able to exchange neurons and stuff with the alpha particles well with the nucleons inside there and it feels an attraction so there is a well that looks a bit like this except this is extremely narrow so the width of this is say 10 to the minus 15 meters typical nucleus size so inside that uranium 238 did you mind in Australia or something there's some alpha particle moving around in here with a large velocity a sort of relativistic velocity motion inside nuclei is kind of relativistic so it bangs to and fro across here right if you're moving if you've got 10 to the minus 15 meters to cover and you're travelling at some speed comparable to the speed of light that means that you cross this thing what does this give me 10 to the minus 23 sorry you need about 10 to the minus 23 seconds to cross so roughly 10 to the 23 times a second this alpha particle bangs to and fro to and fro to and fro this will be the classical picture and it needs to do this so it does this on the order of 6.4 gigae years for many gigae years so for on the order of shall we say 10 to the 17 seconds which is a third of the age of the universe so it makes about 10 to the 40 impacts on the barrier and then wonderful moment it gets out on the 10 to the 40th attack whatever it slips through here and goes off to infinity so this astonishing phenomenon of systems with incredibly small dynamical times the smallest dynamical times in the typical physical world doing something on a timescale which is the age of the universe it is the most astonishing phenomenon but how does it happen? it happens through this exponential decay the height and width of this barrier are substantial but that e to the 4 is that e to the 4 times the height and width of the barrier amplifies this so much that your chance of getting out turns out to be only one in 10 to the 40 so that a neutron that got trapped in there and a supernova before the sun was born pops out today so we should now so that's the end of games with square potential wells I hope you get the idea that it's a rather artificial it's a scheme for finding solutions to the time-independent Schrodinger equation which can illustrate interesting physical phenomena although the potentials themselves are very artificial and we should now just ask ourselves what of the results that we've obtained would be spoiled would change if the changes in potential weren't abrupt in the real world they're not going to be just step potentials we've used step potentials as a computational convenience in the real world they're going to have to extend over some distance and it's important to understand which of these results would survive and which would be spoiled by taking a more realistic potential and I've focused on problems where stuff would survive and tried to neglect problems or haven't spoken about problems which would be seriously damaged but you can be misled so in particular if we would do a calculation precisely analogous to this for particles encountering a square potential well all this calculation could be pushed through with a minor modification that in here we would have B plus E to the I kx and B minus E to the minus I big kx so if we had particles moving in here from infinity with an energy greater than zero the particles when they got here would speed up and slow down when they got here and stuff and classically all the particles would pass through if you solve this problem using this apparatus here what you're going to find is that some of the particles are reflected from this barrier well some of the particles are reflected sorry from the whole setup I don't want to say which barrier they're reflected from because there are two barriers they can be reflected from and the results are a superposition of those and some particles get through and if you do this calculation you are learning something which will be profoundly changed if you're more realistic and say well my real potential well is going to have somewhat slopey boundaries and the issue is how steep does something have to be for this to be a decent guide the good news is that the results to that kind of calculation are not going to be profoundly affected by the steepness they'll be somewhat affected but not enormously affected so long as we stick we would be misled if we put particles in at sufficient energy that they were classically able to get over the top but if we stick to particles which are classically forbidden in here we're not going to be enormously deceived by taking sharp boundaries how do we do this well what you need to do is numerically solve the time solve the wave equation solve the time independent Schrodinger equation for some kind of a potential change which can be made either steep or less steep so if you take that the potential as a function of x is equal to some constant brackets times nought if mod x is less than minus a and in this zone here is something like one minus sorry one plus sine pi x over a that's for mod x less than a and you take it to a one down here if mod x is greater than a I hope I've done that the way I should have done that then you will so this is just a simple functional form that describes a curve that looks like this it goes from v0 here it's precisely v0 when you're more than a away and it's precisely 0 if you're to the left of minus a and it moves smoothly and continuously with a continuous gradient from one thing to the other thing and by changing a you can make this steeper or less steep and it's very straightforward I urge you to try it on your laptop to solve the time independent Schrodinger equation numerically there's a problem describing how to do it I think certainly in the book possibly in a problem set and what do you find when you do it you get this kind of curve here so this is the reflection probability as a function of ka so this is what I did for an energy E which was equal to 0.7 v0 so all of these solutions are for energy E equals 0.7 v0 which in the square if we have an abrupt sudden change in the potential gives us this probability of roughly 0.1 of being reflected sorry is that this is the probability of reflection this is the probability of reflection and the square one gives you the sharp one gives you this the numerics reproduce this if you take ka and a is now this not the width of a well but the width of the transition well 2a really is the width of the transition if ka is less than 1 then the numerics reproduce the analytic solution but if ka is bigger than 1 you see there's a very look at this as a logarithmic scale this is a probability of 0.1 0.01 so the probability of reflection drops like a stone as ka becomes bigger than 1 so the abrupt transition is going to be profoundly misleading when unless the transition so the step in this case where we have what's crucial here is that we have a transition from between two zones within which the particle is classically allowed so the step between classically allowed regions is misleading it exaggerates reflection if ka is greater than on the order of 1 that is to say what does that tell me ka is 2pi of a lambda so that tells me that a if a the transition width is greater than 2pi over the debris wavelength right so the transition really has to be quite abrupt in terms of this natural sense of scale if you ask so what's the debris wavelength for an electron the answer is that it is on the order of 1.2 times 10 to the minus 9 energy over 1ev to the half meters so the debris wavelength for an electron I mentioned electrons obviously because they're things that we do fire around our laboratories people used to fire them around their homes even when they had cathode ray tubes so it's a typical kind of particle you want to understand about then the debris wavelength is a nanometre or so times the energy in electron volts that's a minus a half isn't it because the higher the energy the shorter the debris wavelength so if you are constructing a a step potential typically you are going to be doing it by having some kind of doing some kind of solid state physics so that those sheets of glass provide pretty much a step change in they provide a change in the refractive index which affects photons right so photons hitting the window have a chance of being reflected a chance of being transmitted basically as if it were being bounced off a step potential why because the photons have wavelengths those photons that we're bouncing off the windows have wavelengths of 500 nanometres or something and atoms so the size of an atom is of course on the order of 0.1 nanometres so it's easy using atoms to make to make changes that occur over a few atoms therefore over on the order of a nanometre so you can make if you are using atoms to make the barrier you're propagating your electrons through some kind of solid state material you can probably make a step change which has a you can change the effective potential of the electron experiences within on the order of a nanometre so you may be able to get useful results out of this provided your energies are lower than 1EV but that's extremely challenging in practice your energies will typically be higher than 1EV so these results are going to be basically misleading what do you see here is a return of common sense and rationality if you roll a piece of chalk you know this table it will of course fall it won't be reflected it's not going to be reflected by the lower potential and that's what the numerics are saying here that unless you have a that in practice when something encounters a a drop in potential for example the reflection chance is going to be in fact very small because this is not going to be abrupt like this a tiny bit easy and then everything is basically going to get through so what happens is that when you have a slow change gradual change in the potential is that the wavelength as the electron or the particle comes along it comes to this region of lower potential energy we would say it speeds up the numerics will show you that the wavelength of the wave is getting shorter so the momentum is getting larger because p is h bar k yes it's speeding up and it's just there's no reflected wave so the whole thing just just moves into a new regime with a shorter wavelength everything changing continuously well I think that's pretty much all I want to say so we'll finish there and that's the end of step potentials and on Monday we can start on