Shall we? Shall we begin then? So yesterday we looked at this this pair of wells that was 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 ammonia maser with a nitrogen atom passing through the barrier formed by the hydrogen atoms.
So let's look at this phenomenon from another another perspective, from the perspective of scattering it, a scattering experiment.
And we'll we'll come on to what this has to do with radioactivity.
I hope at the end. So consider this set up.
We have a we have a potential barrier here of height v nought usual square form for computational convenience.
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.
So this these are of course particles with well-defined momentum approaching the barrier.
We expect to see some of these particles reflected classically.
If the energy of these incoming particles were less than V nought, they would all be reflected.
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.
These are in for a state of well-defined energy is e to the minus i.e. upon bar t so these things are
going have sort of e to the minus omega t type dependence and therefore we have a minus sign here.
You're looking at a wave which is travelling to the left.
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.
So we put in a wave.
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.
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.
The trial trial solution will be a combination of a of a of a of an exponential growth and an exponential decay.
So this is a more complicated so so this is a bit different from what we've done before in two ways.
One is that. Well, the main thing is that we are initial a problem inherently has a lack of left right symmetry right.
We the the potential that we're discussing here has left right symmetry symmetrical around the origin, which I've got 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.
Now we could. So that's. And that is computationally very inconvenient.
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
problem which have well-defined parity and thus discussing only what happens that 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.
So 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 well, it's actually a very good way to go.
But it's it's not such an obvious and intuitive way to go, even though it's computationally simpler.
I think it's worthwhile just seeing what happens when you 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,
you know that because the potential goes to zero outside this interval here, the particles.
And we're going to consider particles with with positive energy, the particles are going to be able to push off to infinity.
So we are not going to find discrete energy levels. We're going to be able to find solutions for any energy.
Right? That's whereas previously we were we had a potential which forbade going to infinity and that made the energy levels discrete.
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.
Right. So what do we have to do? Well, we 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 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.
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,
plus a minus e to them plus i k and that would better equal b plus E to the minus k minus sorry.
Plus B minus E to the minus.
No that one's got plus k. Right. Many double negatives here unfortunately.
Oh, we forgot I forgot to say, of course,
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
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.
So this is the continent, this is the condition for the wave function to be continuous at x equals minus say 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 x equals
minus j and we find that i k common factor a plus e to the minus i k a minus a minus.
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.
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,
which is not now dealt with by symmetry as it was yesterday.
So this is the because this is where life becomes, everything becomes difficult.
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
and we have that i k over k I'll write it us of c e to the well maybe I should do that
one like c e to the I k that's the gradient on the right side is equal to big k common
factor open brackets B plus E to the k a minus B minus E to the minus k close bracket.
And I live in hope. And so my anxiety that that that has been those equations have incorrectly stated.
So what do we have to do? We now have four equations and five unknowns, I think.
Right there are two ways to Bayes 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 a B of the APIs 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 coming 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 plus, and we expect them to be linear and A-plus. So that's why we've got two few equations.
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.
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 we'll 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 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, the B plus, which will be a common factor on the right hand side.
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.
So that will be the promised relationship that expresses the number of reflected particles is a multiple of the number of incident particles.
So once you found what a minus is in 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, the well-defined what B pluses and minuses.
And they can all be 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 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 one minus I K over k over one plus i k over k e to the two big 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.
Now in in the doing the typesetting. Okay.
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.
So.
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.
Then we divide these two equations and then we get this relationship I'm about to write down between a minus and D
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.
It's kosh to k minus i.
K on k. Hyperbolic shine of two.
K all over.
Kosh two k minus.
Big K over eight k of shine.
So the algebra is, as promised, all 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 cube is simply enormous, right.
If Cube were very large, then Q minus one would be the same as Q plus one and and everything would be reflected.
But in reality, it's not all going to be reflected.
Something is going to get through how to find see what 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, say and find.
See, 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 or 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 Bach over M,
you will get the flux of incoming particles and the flux of incoming particles has to equal to flux of the outgoing particles,
which is a minus square, the square of a minus, the density of outgoing particles.
Again times times h bach over M for the speed plus c mod squared.
Right. So conservation of particles implies this relationship between these amplitudes.
And of course, you can in principle check whether this relation algebraic relationship is satisfied by these equations,
by hard slog, because I've described how you can in fact find C we've already find out found a minus.
You could in fact in principle find C and check that it satisfied this equation.
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,
what we want to say is the following actually is the fraction the fraction of particles that get through.
Is obviously the ratio of the incoming flux and the LB Well, the ratio of the outgoing flux to the incoming flux.
So it's going to be this, this,
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,
namely H back over M between this quantity and the outgoing flux on the right is this is the
same as the console 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, 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,
but we've got a minus mod squared from that expression at the top and it 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 of the bottom.
So this can be written as one minus one Q minus one mod square over Q plus one mod square.
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's obviously on the top, it will have the existing top minus the bottom.
So when we take away the bottom from the top,
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.
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, cost to K minus K over K, etcetera, 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 code, the Cochise ad and the shine's irritating refused to cancel.
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.
So what we need to do now is take the mod square of these two numbers, ratio them and and take it from one.
So the fraction that gets through is going to be one minus that.
So the top of that is completely imaginary rights, pure imaginary.
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.
So that's, that's that's Q minus one mod squared.
As regards the top, the bottom we're not interested in because we're going to cancel with the other bottom.
And now we have to put underneath the mod square of this which will be for kosh squared to k, right.
Because this is the real part of it, this is the imaginary part of it.
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.
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 one is on.
The common denominator will be this this bottom plus that stuff there.
So this will be a four course squared to K and now we're going to have shine squared.
Let's write it in plus. Yeah, shine squared to K 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 squids of a K squids which will cancel because of that minus sign.
And what will not go away is the mixed term, the product of multiplying this on this,
which generates two and the product of multiplying this on this which generates another two.
So we will get four and will be with the minus sign because this minus sign will is there, you know,
when this comes up here, that minus sign will stick out and this minus sign will make the mixed term minus there.
So this is going to be times minus four and it's over the bottom as you see it,
over four squared to K plus K over K minus K over k squared shine squared to k and the top simplifies most beautifully
because kosh squared minus nine squared is one so the fours can be cancelled and this actually is nothing but one over.
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.
How much fun? So what do we learn from this?
What we learn from this is is most interestingly,
is what happens if we have a rather high barrier in the particles of very short of energy to get through.
All right. So so K is a measure of the deficits in energy that the particles.
Right. That they have 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,
then we're looking at the cost of a of a largish number and the shine of a largish number.
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.
All right. But we are interested in fact, in cost squared and shine squared.
So F is looking like one over E to the four.
Okay. So if K 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 is who? This machine goes to sleep as well as the trouble shouldn't go to sleep.
Give up. Is there anything there? I'll just draw it.
Is it. Is it sort of. Yeah. I know that we're saving the planet by having the machine turn itself off, but.
I can't see it. So what, you want to do some steps that sort of become typically what happens when.
KS Very large in detail.
You might want to know. The smaller the smaller k a.
Sorry. Right.
So these results offer a barrier which is so in these in these results, the barrier is not terribly high.
So. So we have V0. Sorry, we have an e is equal to 0.750.
No, no, no. Sorry. What if I don't? What I don't want to over.
Yeah, that is correct. Sorry.
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,
which is a measure of the width and the height of the parameter of the barrier.
So it's two v0 a squared over each bar.
This animal, right. 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.
Is the probability of getting through as a function of your energy of av0.
For barriers of different WS. So I think it's a 0.5 at the top there.
Yeah. So here's a relatively weak barrier which gives you a fairly small energy is a chance of getting through.
It's not a very fat barrier as the crucial thing. This is a fatter barrier.
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.
Okay. So if we can get these things to stay alive for later,
what's what's physically interesting about this or an interesting application of this is to radioactive decay.
So this is obviously a very simple minded, 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 two, three, eight uranium, which is the non fissile source of uranium, you have a number of alpha particles.
It is a simple minded picture. So what does the potential energy of an alpha particle.
So. So we kind of consider this to be so.
Two, three, eight. Uranium which decays to 2 to 3, four.
Thorium and an alpha particle with a half life of I think it's 6.4 giga years.
So it takes the age of the universe, typically for a uranium two, three eight made in some supernova to eject an alpha particle.
So what's happening here from this perspective? What's happening?
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.
So the alpha particle, when it's a long way from from when it's a decent distance,
more than ten to the -15 metres or so away from the thorium nucleus is repelled by the electrostatic repulsion.
So the potential energy curve has a sort of one overall type behaviour here.
If you get when it gets close enough to the thorium nucleus, the strong interaction,
it's able to exchange gluons and stuff with, with the, with the alpha particles.
Well with the nucleons inside there and it and it feels an attraction.
So there is a, well it looks a bit like this, except this is extremely narrow.
So the width of this right is in say ten to the -15 metres.
So typical nucleus size. So inside that uranium 238, did you mine 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 got ten to the -15 metres to cover and you're travelling at some speed comparable to the speed of light,
uh, that means that you, you cross this thing.
What, what does this give me ten to the -23. Sorry.
Yeah. You need about ten to the -20 3 seconds to cross.
So roughly ten 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 for on the order of 6.4 giga is so far for many giga years.
So for on the order of, shall we say, ten to the 17 seconds, which is a third of the age of the universe.
So it makes so it makes about ten to the 40 impacts on the barrier.
And then wonderful moment. It gets out on the 10th to the 40th attack, whatever.
It slips through here. It goes off to infinity.
So this astonishing phenomenon of of a systems with incredibly small dynamical times, the smallest dynamical times,
you know, in the in the typical physical world, doing something on a time scale, 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 each of the four is that each of the four 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 ten to the 40.
So that of a neutron that got trapped in there in a supernova before the sun was born pops out today.
So we should now. So. So that's.
That's the end of games with square potential. Well,
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
to the to the time independent Schrodinger equation which can illustrate interesting physical phenomena.
Although it's the potentials themselves are very artificial.
And we should now just ask ourselves what of the results that we've obtained would be spoilt?
What would change if the potential if the changes in potential weren't abrupt?
Right. And 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 well, they're going to have to extend over some distance.
And one wants to understand it's important to understand which of these results would
survive and which would would be spoilt by by taking a more realistic potential.
And I focussed on problems where stuff would survive and,
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 you if we would do an a calculation precisely analogous to this for particles encountering a square potential.
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,
k, x and B minus E to the minus like big k x.
Right. We would have to. So if we had particles moving in here from infinity with an energy greater than zero,
they 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 reflected, sorry, from the whole set up. I don't want to say which barrier that are reflected from,
because there are two barriers they can be reflect from and the results are a superposition of those and some particles get through.
And if you do this calculation, you were learning something which will be profoundly changed.
If you are more realistic and say, well, my real potential well of course,
has he's going to have somewhat sloppy, you know, somewhat sloppy 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 the results to that kind of calculation are not going to be profoundly affected if by the by the statements,
they'll be somewhat affected, but not enormously affected.
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.
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.
It solved the wave equation to solve the time independent Schrodinger equation for some kind
of a 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 model if X is less than minus say.
And in this zone here is something like one minus, uh, sorry, one plus sign pi x over a that.
For Model X less than A and you take it to a one down here, if not x,
if x is greater than I hope I've done that the way I should have done that, then you will.
So this is this is just a simple functional form that describes a curve that looks like this.
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
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, to try it on your laptop to solve the to solve the time independent Schrodinger equation.
Numerically, there's a problem describing how to do it. I think it's 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 K.
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.
So all of these solutions are for energy equals point is 0.7 V zero,
which in the square with the if we have an abrupt you know sudden change in the
in the potential gives us this probability of roughly 0.1 of being reflected.
So this is the property of reflection. I say something different, this is the probability of reflection.
And the square one gives you the shot, one gives you this.
The numerics reproduce this if you take a and a is now this not the width of a well,
but the width of the transition well to a really is the width of the transition.
If is less than one, then the numerics reproduce the analytic solution.
But if K is bigger than one, you see there's a very look at this is a logarithmic scale, right?
This is this is a probability of 0.1.01.001.
So the probability of reflection drops like a stone as K becomes bigger than one.
So the 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 between two zones within within which the particle is classically allowed.
Right. So the step between classically allowed regions.
Is misleading. It exaggerates reflection if K is greater than on the order of one that is to.
So what does that tell me? K is two pi over lambda.
So that tells me that a if a the transmission width is greater than two pi over the debris wavelength.
Right. So the transition really has to be quite abrupt in terms of this natural, 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 ten to the minus nine energy over one EVA to the half metres.
So the de broglie wavelength,
this quantity for an electron and I mentioned electrons obviously because they're things that we do far around laboratories.
People used to follow them around their homes even when they had cathode ray tubes.
So it's a it's a it's a typical kind of particle you want to understand about.
Then the debris wavelength is is a nanometre or so times the energy in electron volts.
Look, that's a minus a half, isn't it? Because the the higher the energy, the shorter the debris wavelength.
So if you're if you're constructing a a step potential,
typically you 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,
the 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 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 5.1 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, you so you can make if you if you are using atoms to make the barrier,
you know you're propagating an electron through some kind of solid state material.
You can probably 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 one EV.
But that's extremely challenging. In practice, your energies will typically be higher than one.
So these results are going to be basically misleading. What you see here is, is return of common sense and rationality.
If you if you roll a piece of chalk off the edge of this table, it will, of course, fall.
It won't be reflected. It's not gonna be reflected by the lower potential onset of lower potential.
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,
for example, the reflection of chance is going to be in fact, very small because this is not going to be abrupt.
It's going to be like this tiny bit easy, and then everything is basically going to get through.
So what happens? What actually happens is that when you when you have a slow change, a gradual change in the potential,
is that the wavelength as the as the as the electron or the particle comes along, it comes this region of lower potential energy.
We would say it speeds up. The numerics will show you is that the wavelength of the wave is getting shorter.
So the momentum is getting larger because as P is H mark.
Yes, it's speeding up and it's just it the there's no reflected wave.
So the whole thing just just moves into a new regime with a short wavelength of 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.