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Shall we? Shall we begin then? So yesterday we looked at this this pair of wells that was separated by a barrier

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so that classically the particle couldn't get from one well to the other.

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And we found that the particle got from one well to the other and used that to make a model of

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an ammonia maser with a nitrogen atom passing through the barrier formed by the hydrogen atoms.

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So let's look at this phenomenon from another another perspective, from the perspective of scattering it, a scattering experiment.

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And we'll we'll come on to what this has to do with radioactivity.

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I hope at the end. So consider this set up.

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We have a we have a potential barrier here of height v nought usual square form for computational convenience.

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And we have some incoming stream of particles. We have a beam of particles here represented by the wave function plus each of the ICS.

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So this these are of course particles with well-defined momentum approaching the barrier.

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We expect to see some of these particles reflected classically.

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If the energy of these incoming particles were less than V nought, they would all be reflected.

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So we put in a reflected wave here which goes like e to the minus I kicks remember the time dependence of everything in quantum mechanics.

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These are in for a state of well-defined energy is e to the minus i.e. upon bar t so these things are

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going have sort of e to the minus omega t type dependence and therefore we have a minus sign here.

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You're looking at a wave which is travelling to the left.

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If you have a plus sign here, you're looking at wave which is travelling to the right and then we expect some of the particles to get through.

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So we put in a wave.

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In this portion we say the trial solution should look like C with some unknown constant C each of the six and then within the barrier.

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If because we can look for solutions at energies lower than V zero, we're going to have B plus each, the k x plus.

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The trial trial solution will be a combination of a of a of a of an exponential growth and an exponential decay.

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So this is a more complicated so so this is a bit different from what we've done before in two ways.

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One is that. Well, the main thing is that we are initial a problem inherently has a lack of left right symmetry right.

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We the the potential that we're discussing here has left right symmetry symmetrical around the origin, which I've got to say.

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But the origin is here in the middle of the well. This is minus A and this is a over here.

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So the potential is going to be an even function of X, same as ever.

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But our problem, our initial conditions, the physical situation we wish to discuss,

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has a built in asymmetry because the particles have to come in from one side or the other.

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Now we could. So that's. And that is computationally very inconvenient.

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It stops us using this nice trick, which doesn't it makes it difficult for users for us to use this nice trick of looking for solutions to the

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problem which have well-defined parity and thus discussing only what happens that this boundary condition.

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With this setup, we're going to have to discuss this boundary condition and this boundary condition,

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because you can see there are fundamental differences between what's happening on those two sides.

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So you can handle this problem using looking for solutions of well-defined parity, but it's slightly unnatural.

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And I think well, it's actually a very well, it's actually a very good way to go.

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But it's it's not such an obvious and intuitive way to go, even though it's computationally simpler.

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I think it's worthwhile just seeing what happens when you when you play the

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game straightforwardly and you'll see the algebra becomes quite unpleasant,

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which illustrates the benefits that we had before by assuming well-defined parity.

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All right. The other thing that's different here is that because we are considering particles which are free,

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you know that because the potential goes to zero outside this interval here, the particles.

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And we're going to consider particles with with positive energy, the particles are going to be able to push off to infinity.

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So we are not going to find discrete energy levels. We're going to be able to find solutions for any energy.

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Right? That's whereas previously we were we had a potential which forbade going to infinity and that made the energy levels discrete.

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So those are the those are the differences because we're we're dealing with a different physical situation and it has implications for the mass.

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Right. So what do we have to do? Well, we it's very boring. We have to impose continuity of the wave function.

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So the wave function here is this wave, plus this wave.

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And then it 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.

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So let's just quickly write that down. So we have a plus e to the minus i k which is the incoming wave evaluated at that barrier x equals minus AA,

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plus a minus e to them plus i k and that would better equal b plus E to the minus k minus sorry.

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Plus B minus E to the minus.

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No that one's got plus k. Right. Many double negatives here unfortunately.

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Oh, we forgot I forgot to say, of course,

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that we will have as ever that K is equal to the square root of two m times the energy of our squared because p squared over two m is the energy and

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p is h bar squared k squared three p is h mark and we will have the big k is equal to the squares of two and v zero minus the overage bar squared.

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So this is the continent, this is the condition for the wave function to be continuous at x equals minus say we

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require as yesterday that the gradient of the wave function is also continuous there.

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So we have to take the gradient of that function on the left and evaluate it x equals

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minus j and we find that i k common factor a plus e to the minus i k a minus a minus.

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Each of the i k close brackets is equal to big k common factor b plus e to the minus k minus b minus each of the k close brackets.

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Then we have sets two equations. Now we have two more equations because we have to get everything hunky dory on the right hand boundary,

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which is not now dealt with by symmetry as it was yesterday.

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So this is the because this is where life becomes, everything becomes difficult.

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So we have c e to the i k is equal to B plus e to the k plus b minus E to the minus k

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and we have that i k over k I'll write it us of c e to the well maybe I should do that

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one like c e to the I k that's the gradient on the right side is equal to big k common

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factor open brackets B plus E to the k a minus B minus E to the minus k close bracket.

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And I live in hope. And so my anxiety that that that has been those equations have incorrectly stated.

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So what do we have to do? We now have four equations and five unknowns, I think.

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Right there are two ways to Bayes and a C, so we will not be able to get rid of all of them.

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We will be able to express in principle any one of a B of the APIs and C's in terms of the other one.

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And that physically corresponds to the point that the flux of incoming particles is controlled by a plus, and that's in your control.

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You can put in more particles or fewer particles,

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and that will obviously lead to more particles coming out or fewer particles coming out depending on the coming flux.

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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,

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as a multiple of plus, and we expect them to be linear and A-plus. So that's why we've got two few equations.

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We don't physically expect to be able to determine everything. So what we should do is what we should do is engage in an elimination exercise.

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A reasonable way to go is to take these two equations here.

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Divide this equation by this equation, say, and that will get rid of C and we'll give you a relationship between B plus and B minus.

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And then you can take that relationship between B plus and B minus and use it in these

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two equations to to express the right to get rid of B minus from these equations,

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these two right hand sides. So they both become simple multiples of B plus.

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And then you could divide these two equations one by another, the B plus, which will be a common factor on the right hand side.

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It will go away and you will be left with a relationship between a plus and a minus, a single relationship, an A-plus and a minus.

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So that will be the promised relationship that expresses the number of reflected particles is a multiple of the number of incident particles.

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So once you found what a minus is in in terms of a plus, you can go back to your original expression here,

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which had only B plus on the right hand side, A minus can be expressed as a function of a plus, the well-defined what B pluses and minuses.

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And they can all be they can all be determined. So let me not do all that algebra.

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That's the strategy. The execution, of course, is quite tedious and the scope for making errors is quite large.

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And in fact I find that there's a typo right there in Equation 540 in the book because when you when you

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do eliminate between these two equations here to find out the relationship between B minus and B plus,

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it should be that B minus is one minus I K over k over one plus i k over k e to the two big k a b plus.

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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.

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Now in in the doing the typesetting. Okay.

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So we have that relationship there. We stuff this back into the other places and we find that a minus is equal to AA plus.

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So.

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So I've described how we what we do, we take we take this B minus, use it to get rid of B minus from here and replace that with B plus in some factor.

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Then we divide these two equations and then we get this relationship I'm about to write down between a minus and D

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plus and it is a minus is a plus E to the minus to i k a q minus one over Q plus one where q is itself pretty yucky.

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It's kosh to k minus i.

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K on k. Hyperbolic shine of two.

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K all over.

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Kosh two k minus.

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Big K over eight k of shine.

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So the algebra is, as promised, all altogether more yet more messy than it was yesterday.

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Because we're not exploiting parity. We're not dealing with finding a right.

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So what do we want to know about this physically?

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What we want to know about this physically, I think, is what is the chance that the particle is reflected?

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What is the chance that the particle gets through?

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So classically everything will be reflected and the modulus of a minus would be the same as the modulus of a plus.

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Right. And you can see that that isn't looking very promising, because that would require that.

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Well, basically, the cube is simply enormous, right.

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If Cube were very large, then Q minus one would be the same as Q plus one and and everything would be reflected.

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But in reality, it's not all going to be reflected.

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Something is going to get through how to find see what we could you could take take this a minus expression from it,

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as I've described, obtain B plus from B plus, obtain B minus to put these back into the into this equation here, say and find.

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See, that's too much like hard work.

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It's easier to say that, look, there's going to be conservation of particles.

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We've got a well-defined theoretical apparatus, apparatus here, which is not going to which which conserves probability.

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So the incoming particles, the A plus or either are all going to go out at the end of the day, either to the left or to the right.

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So we can argue that a plus mod squared, which is well, that is the spatial density of incoming particles, if you like.

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If you multiply that by the speed of the incoming particles, which is P over M, so Bach over M,

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you will get the flux of incoming particles and the flux of incoming particles has to equal to flux of the outgoing particles,

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which is a minus square, the square of a minus, the density of outgoing particles.

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Again times times h bach over M for the speed plus c mod squared.

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Right. So conservation of particles implies this relationship between these amplitudes.

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And of course, you can in principle check whether this relation algebraic relationship is satisfied by these equations,

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by hard slog, because I've described how you can in fact find C we've already find out found a minus.

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You could in fact in principle find C and check that it satisfied this equation.

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But we don't want to do all that algebra. So so the point is that what we want to say is the, the, the, the flux of well,

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what we want to say is the following actually is the fraction the fraction of particles that get through.

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Is obviously the ratio of the incoming flux and the LB Well, the ratio of the outgoing flux to the incoming flux.

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So it's going to be this, this,

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this fraction that we want to call it F is going to be mod C squared over a plus squared because the contents of proportionality,

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namely H back over M between this quantity and the outgoing flux on the right is this is the

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same as the console proportionality between this constant and the incoming flux on the left.

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So the fraction of particles that get through will be given by this ratio here, which given that relationship.

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So in other words, see, let's, let's we can write that now is as a plus mod squared minus a minus mod squared of a plus mod squared,

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but we've got a minus mod squared from that expression at the top and it has a multiple of a plus mod squared.

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So we can write this as one minus Q minus one over Q plus one mod squared.

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Well the mod square of this ratio is the mod square, the ratio of the mod squares of the top of the bottom.

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So this can be written as one minus one Q minus one mod square over Q plus one mod square.

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So let's address ourselves to what these mod squares are. So what's Q minus one?

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Well, Q minus one is going to be well, it's obviously on the top, it will have the existing top minus the bottom.

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So when we take away the bottom from the top,

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the causes go away and we are left with I think over i k minus ik over k times shine to k and that will be over.

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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.

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It is the bottom that you see up there, cost to K minus K over K, etcetera, and Q plus one.

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The reason we won't care about the bottom is of course it will cancel when we take this ratio.

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So for Q plus one, we unfortunately find that the code, the Cochise ad and the shine's irritating refused to cancel.

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So this becomes two cosh, two K we're adding so we have minus I k over k plus k of ik only in a bracket shine to k and again that's over the bottom.

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So what we need to do now is take the mod square of these two numbers, ratio them and and take it from one.

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So the fraction that gets through is going to be one minus that.

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So the top of that is completely imaginary rights, pure imaginary.

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We should take out an AI from that bracket and then we will find we are staring at K over k plus k over k squared times shine squared to k.

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So that's, that's that's Q minus one mod squared.

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As regards the top, the bottom we're not interested in because we're going to cancel with the other bottom.

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And now we have to put underneath the mod square of this which will be for kosh squared to k, right.

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Because this is the real part of it, this is the imaginary part of it.

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We take out a factor awry and now we're staring at plus k over k minus k over k squared shine squared took nearly that.

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So now we put this all these two bits. It'll simplify if we put these two bits on a common denominator.

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So the top one is on.

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The common denominator will be this this bottom plus that stuff there.

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So this will be a four course squared to K and now we're going to have shine squared.

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Let's write it in plus. Yeah, shine squared to K brackets now brackets.

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What. We will have this bracket squared.

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Well we'll have this bracket squared sorry. Minus this bracket squared.

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And when we square these brackets we're going to get K squids of a K squids which will cancel because of that minus sign.

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And what will not go away is the mixed term, the product of multiplying this on this,

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which generates two and the product of multiplying this on this which generates another two.

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So we will get four and will be with the minus sign because this minus sign will is there, you know,

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when this comes up here, that minus sign will stick out and this minus sign will make the mixed term minus there.

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So this is going to be times minus four and it's over the bottom as you see it,

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over four squared to K plus K over K minus K over k squared shine squared to k and the top simplifies most beautifully

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because kosh squared minus nine squared is one so the fours can be cancelled and this actually is nothing but one over.

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So the fraction is one over cost squared to k plus a quarter of k of a k minus k of a k squared shines squared to care.

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How much fun? So what do we learn from this?

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What we learn from this is is most interestingly,

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is what happens if we have a rather high barrier in the particles of very short of energy to get through.

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All right. So so K is a measure of the deficits in energy that the particles.

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Right. That they have by how much they don't have enough energy classically to get through the barrier.

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If the barrier is very high and they don't have much energy,

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then we're looking at the cost of a of a largish number and the shine of a largish number.

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And so what we can say is that for large K, we can say that kosh two K behaves pretty much like shine to K behaves like E to the two K.

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All right. But we are interested in fact, in cost squared and shine squared.

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So F is looking like one over E to the four.

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Okay. So if K is an appreciable number, this probability of penetration is becoming small.

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Crucial result is that the probability of getting through there is decreasing exponentially fast in the height of the barrier.

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So you don't need a very high barrier to make this quite a small effect.

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And somewhere here we have. So this is who? This machine goes to sleep as well as the trouble shouldn't go to sleep.

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Give up. Is there anything there? I'll just draw it.

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Is it. Is it sort of. Yeah. I know that we're saving the planet by having the machine turn itself off, but.

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I can't see it. So what, you want to do some steps that sort of become typically what happens when.

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KS Very large in detail.

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You might want to know. The smaller the smaller k a.

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Sorry. Right.

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So these results offer a barrier which is so in these in these results, the barrier is not terribly high.

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So. So we have V0. Sorry, we have an e is equal to 0.750.

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No, no, no. Sorry. What if I don't? What I don't want to over.

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Yeah, that is correct. Sorry.

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Yeah. The height of the height of the barrier is sorry there's this parameter w it wasn't that which we talked about yesterday,

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which is a measure of the width and the height of the parameter of the barrier.

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So it's two v0 a squared over each bar.

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This animal, right. That's your dimensionless measure of the height and the width of the barrier in

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terms of the mass of the particle with no reference to the energy of the particle.

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Sorry, that's not the case then. What's being plotted here is.

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Is the probability of getting through as a function of your energy of av0.

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For barriers of different WS. So I think it's a 0.5 at the top there.

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Yeah. So here's a relatively weak barrier which gives you a fairly small energy is a chance of getting through.

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It's not a very fat barrier as the crucial thing. This is a fatter barrier.

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This is a fatter barrier. And so you can see how, as a function of the energy, your chance of getting through rises in detail.

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Okay. So if we can get these things to stay alive for later,

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what's what's physically interesting about this or an interesting application of this is to radioactive decay.

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So this is obviously a very simple minded, very simple minded model that we have so far.

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But the general idea, for example, is this.

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So what we should say is that inside two, three, eight uranium, which is the non fissile source of uranium, you have a number of alpha particles.

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It is a simple minded picture. So what does the potential energy of an alpha particle.

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So. So we kind of consider this to be so.

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Two, three, eight. Uranium which decays to 2 to 3, four.

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Thorium and an alpha particle with a half life of I think it's 6.4 giga years.

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So it takes the age of the universe, typically for a uranium two, three eight made in some supernova to eject an alpha particle.

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So what's happening here from this perspective? What's happening?

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So what we should do is we should think about this alpha particle in this uranium two, three, four nucleus as a kind of dynamical system.

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So the alpha particle, when it's a long way from from when it's a decent distance,

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more than ten to the -15 metres or so away from the thorium nucleus is repelled by the electrostatic repulsion.

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So the potential energy curve has a sort of one overall type behaviour here.

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If you get when it gets close enough to the thorium nucleus, the strong interaction,

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it's able to exchange gluons and stuff with, with the, with the alpha particles.

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Well with the nucleons inside there and it and it feels an attraction.

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So there is a, well it looks a bit like this, except this is extremely narrow.

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So the width of this right is in say ten to the -15 metres.

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So typical nucleus size. So inside that uranium 238, did you mine in Australia or something?

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There's some alpha particle moving around in here with a large velocity,

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a sort of relativistic velocity motion inside nuclei is kind of relativistic, so it bangs to and fro across here, right?

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If you're moving, if you got ten to the -15 metres to cover and you're travelling at some speed comparable to the speed of light,

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uh, that means that you, you cross this thing.

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What, what does this give me ten to the -23. Sorry.

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Yeah. You need about ten to the -20 3 seconds to cross.

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So roughly ten to the 23 times a second. This alpha particle bangs to and fro, to and fro, to and fro.

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This will be the classical picture and it needs to do this.

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So it does this for on the order of 6.4 giga is so far for many giga years.

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So for on the order of, shall we say, ten to the 17 seconds, which is a third of the age of the universe.

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So it makes so it makes about ten to the 40 impacts on the barrier.

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And then wonderful moment. It gets out on the 10th to the 40th attack, whatever.

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It slips through here. It goes off to infinity.

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So this astonishing phenomenon of of a systems with incredibly small dynamical times, the smallest dynamical times,

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you know, in the in the typical physical world, doing something on a time scale, which is the age of the universe.

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It is the most astonishing phenomenon. But how does it happen? It happens through this exponential decay.

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The height and width of this barrier are substantial,

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but that each of the four is that each of the four times the height and width of the barrier

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amplifies this so much that your chance of getting out turns out to be only one in ten to the 40.

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So that of a neutron that got trapped in there in a supernova before the sun was born pops out today.

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So we should now. So. So that's.

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That's the end of games with square potential. Well,

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so I hope you get the idea that it's a it's a rather artificial it's a it's a scheme for finding solutions

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to the to the time independent Schrodinger equation which can illustrate interesting physical phenomena.

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Although it's the potentials themselves are very artificial.

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And we should now just ask ourselves what of the results that we've obtained would be spoilt?

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What would change if the potential if the changes in potential weren't abrupt?

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Right. And in the real world they're not going to be just step potentials.

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We've used step potentials as a computational convenience in the real well, they're going to have to extend over some distance.

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And one wants to understand it's important to understand which of these results would

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survive and which would would be spoilt by by taking a more realistic potential.

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And I focussed on problems where stuff would survive and,

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and tried to neglect problems or haven't spoken about problems which would be seriously damaged, but you can be misled.

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So in particular, if you if we would do an a calculation precisely analogous to this for particles encountering a square potential.

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Well we could all this calculation could be pushed through with the minor modification that in here we would have B plus E to the I,

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k, x and B minus E to the minus like big k x.

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Right. We would have to. So if we had particles moving in here from infinity with an energy greater than zero,

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they the particles when they got here would speed up and slow down when they got here and stuff.

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And classically, all the particles would pass through.

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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.

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Well, some of the particles reflected, sorry, from the whole set up. I don't want to say which barrier that are reflected from,

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because there are two barriers they can be reflect from and the results are a superposition of those and some particles get through.

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And if you do this calculation, you were learning something which will be profoundly changed.

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If you are more realistic and say, well, my real potential well of course,

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has he's going to have somewhat sloppy, you know, somewhat sloppy boundaries.

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And the issue is how steep does something have to be for this to be a decent guide?

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The good news is that the the results to that kind of calculation are not going to be profoundly affected if by the by the statements,

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they'll be somewhat affected, but not enormously affected.

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So long as we stick, we would be misled if we put particles in it, sufficient energy that they were classically able to get over the top.

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But if we stick to particles which are classically forbidden in here, we're not going to be enormously deceived by taking sharp boundaries.

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How do we do this? Well, what you need to do is numerically solve the time.

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It solved the wave equation to solve the time independent Schrodinger equation for some kind

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of a for some kind of a potential change which can be made either steep or less steep.

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So if you take that, the potential as a function of X is equal to some constant brackets, times nought if model if X is less than minus say.

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And in this zone here is something like one minus, uh, sorry, one plus sign pi x over a that.

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For Model X less than A and you take it to a one down here, if not x,

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if x is greater than I hope I've done that the way I should have done that, then you will.

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So this is this is just a simple functional form that describes a curve that looks like this.

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Right? It goes from the note here. It's precisely v nought when you're more than a way and it's precisely zero if you're to the left of minus

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A and it moves smoothly and continuously with a continuous gradient from one thing to the other thing.

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And by changing a, you can make this steeper or less steep.

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And it's very straightforward. I urge you to, to try it on your laptop to solve the to solve the time independent Schrodinger equation.

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Numerically, there's a problem describing how to do it. I think it's certainly in the book, possibly in a problem set.

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And what do you find when you do it? You get this kind of curve here.

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So this is the reflection probability as a function of K.

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So so that's right. And this is for this is was what I did for an energy E which was equal to 0.7 V zero.

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So all of these solutions are for energy equals point is 0.7 V zero,

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which in the square with the if we have an abrupt you know sudden change in the

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in the potential gives us this probability of roughly 0.1 of being reflected.

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So this is the property of reflection. I say something different, this is the probability of reflection.

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And the square one gives you the shot, one gives you this.

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The numerics reproduce this if you take a and a is now this not the width of a well,

305
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but the width of the transition well to a really is the width of the transition.

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If is less than one, then the numerics reproduce the analytic solution.

307
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But if K is bigger than one, you see there's a very look at this is a logarithmic scale, right?

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This is this is a probability of 0.1.01.001.

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So the probability of reflection drops like a stone as K becomes bigger than one.

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So the the abrupt transition is going to be profoundly misleading when unless the transition.

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So the step in this case where we have what's crucial here is that we have a transition

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from between between two zones within within which the particle is classically allowed.

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Right. So the step between classically allowed regions.

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Is misleading. It exaggerates reflection if K is greater than on the order of one that is to.

315
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So what does that tell me? K is two pi over lambda.

316
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So that tells me that a if a the transmission width is greater than two pi over the debris wavelength.

317
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Right. So the transition really has to be quite abrupt in terms of this natural, natural sense of scale.

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If you ask, so what's the debris wavelength for an electron?

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The answer is that it is on the order of 1.2 times ten to the minus nine energy over one EVA to the half metres.

320
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So the de broglie wavelength,

321
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this quantity for an electron and I mentioned electrons obviously because they're things that we do far around laboratories.

322
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People used to follow them around their homes even when they had cathode ray tubes.

323
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So it's a it's a it's a typical kind of particle you want to understand about.

324
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Then the debris wavelength is is a nanometre or so times the energy in electron volts.

325
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Look, that's a minus a half, isn't it? Because the the higher the energy, the shorter the debris wavelength.

326
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So if you're if you're constructing a a step potential,

327
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typically you you are going to be doing it by having some kind of doing some kind of solid state physics so

328
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that those sheets of glass provide pretty much a step change in they provide a change in the refractive index,

329
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which affects photons. Right. So photons hitting the window have a chance of being reflected,

330
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the chance of being transmitted basically as if it were being bounced off a step potential.

331
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Why? Because the photons have wavelengths.

332
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Those photons that we're we're bouncing off the windows have wavelengths of 500 nanometres or something and atoms.

333
00:41:46,790 --> 00:41:53,300
So the size of an atom is, of course, on the order of 5.1 nanometres.

334
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So it's easy using atoms to make, to make changes that occur over a few atoms.

335
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Therefore over on the order of a nanometre, you so you can make if you if you are using atoms to make the barrier,

336
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you know you're propagating an electron through some kind of solid state material.

337
00:42:12,650 --> 00:42:19,430
You can probably you can probably make a step change which has a you can change the

338
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effective potential of the electron experiences within on the order of a nanometre.

339
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So you may be able to get useful results out of this provided your energies are lower than one EV.

340
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But that's extremely challenging. In practice, your energies will typically be higher than one.

341
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So these results are going to be basically misleading. What you see here is, is return of common sense and rationality.

342
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If you if you roll a piece of chalk off the edge of this table, it will, of course, fall.

343
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It won't be reflected. It's not gonna be reflected by the lower potential onset of lower potential.

344
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And that's what's what the numerics are saying here, that unless you have a that in practice when something encounters a drop in potential,

345
00:43:10,670 --> 00:43:16,340
for example, the reflection of chance is going to be in fact, very small because this is not going to be abrupt.

346
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It's going to be like this tiny bit easy, and then everything is basically going to get through.

347
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So what happens? What actually happens is that when you when you have a slow change, a gradual change in the potential,

348
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is that the wavelength as the as the as the electron or the particle comes along, it comes this region of lower potential energy.

349
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We would say it speeds up. The numerics will show you is that the wavelength of the wave is getting shorter.

350
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So the momentum is getting larger because as P is H mark.

351
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Yes, it's speeding up and it's just it the there's no reflected wave.

352
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So the whole thing just just moves into a new regime with a short wavelength of everything changing continuously.

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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.

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And on Monday we can start on.