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Okay. Good morning, everybody. Let's begin. So yesterday we looked at this very artificial problem of a particle trapped in a square potential.
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Well, and the motivation, what I build is the reason for discussing these problems was because you could these these potential worlds with
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sharp edges was that it enabled you to solve some simple problems which illustrated features of quantum mechanics,
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which are which are general, which would survive to other, more realistic potentials.
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So but the amount of physics you can do with a single square well is rather limited.
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There is some physics you can do with it, and I would urge you to play with this, to do things like supposing that your state,
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your your state is a linear combination of two adjacent stationary states so that there is some motion.
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See that the particle moves to and fro across the well at about the speed with about the period that you would expect.
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The energies in that square well are proportional to the square and squared of some integer,
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whereas the energies in the harmonic oscillating potential well go like.
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And that's to say when we took this simple case of the infinitely deep potential of the energy that was like and square like,
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and so it gives you a very different behaviour from the simple harmonic oscillator.
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And I think it's interesting to investigate that in the, to try and recover classical results and see how that dependent on the energy levels
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and then squared rather than on n manifests itself in the in the bottom line.
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But we won't take time to do that. Electricals is short.
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We're going to move on to this problem, which is more fun and will allow us to understand how a pneumonia maser works,
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which was which is a timekeeping device and was the first amazing lasing device.
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So what we do here is we imagine so first of all, we're going to solve an abstract problem and then I'll explain why this is relevant for for ammonia.
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We imagine we've got two potential wells which are divided by some barrier.
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Ideally, these potential wells would only be a finite depth.
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So the walls here would cruise off like this with V zero.
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But the computations are made much easier if we let these walls go off to infinity.
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So the potential goes off to infinity. If you go to left of minus x is minus B or right is of x is plus B, the potential is zero.
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Here the potential is V zero, some number there. So these basically two of these square wells put adjacent to each other with a finite.
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So there's only a finite barrier between the two and nothing essentially is changed as the
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computations are made a bit easier by making these go off to infinity left and right.
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So what are we going to do? We want to find, again, the stationary states, the states, a well-defined energy for this system,
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because they will have an interesting property which will lead to the ammonium.
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So again, the here is the origin X is zero.
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This potential well is and is symmetrical on reflection around the origin.
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So it's an even function of particles and even a function of X that guarantees that the parity operator competes with the potential energy operator,
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which guarantees that we can look for states of well-defined parity so the stationary
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states can have could be eigen functions of P so they have they have well-defined parity.
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The usual argument that computing operators mean we can find a complete set of mutual aid in states, and we're going to assume so.
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So there will be states which have even parity. There will be states that have odd parity.
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And the as yesterday the equation we have to solve is minus bar squared over to m d to you by the x squared plus v of x u is equal to EU.
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So we have to solve this equation and we're looking for even functions and we're looking for odd solutions.
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We are going to assume that we can find we're going to look for states which are low enough in energy that they are,
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as it were, bound by either this well or by this well.
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As I say, they're classically forbidden in this region. So we're going to look for states.
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Which have less energy with e being less than this barrier height so that classically the particle will be stuck in one
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well or stuck in the other well and could go between the two as we'll see quantum mechanically it will go between the two.
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So that being so the solutions. So what does that mean?
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That means that we've got do you buy the x squared? Well,
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do you buy the x squared is going to be to overage bar squared V nought minus e times u and this
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quantity is going to be positive when we're in the middle between x is when mod x is less than a,
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so this quantity here is going to be greater than zero.
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And that means that in this middle zone, the solutions to this equation are going to be things like E to the plus big x.
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So this means for four more x less than a,
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we're going to have you goes like E to the plus or minus big x is what we wrote down
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yesterday where big K is the square root of two and V0 minus E over bar squared.
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Okay. Or because we're looking for solutions of well-defined parity and these objects don't have well-defined parity,
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we can take linear combinations of those. Oh, and we can say or looking for well-defined parity, we can say u goes like kosh, k, x or shine big tags.
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So these are the solutions of well-defined parity to that very boring differential equation, given that this quantity here is positive.
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So those are the solutions. We know that our solutions will have that form in that middle section as drawn up there.
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This is this is X equals A, this is X equals minus C, and then in the allowed regions, B will be somewhere over here,
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right in the allowed regions either side of the of the central barrier,
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we know that we will have sinusoidal behaviour that will be trying to solve the equation.
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So this is, this is far more x less than a if model X is bigger than A and less than B,
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we will be trying to solve D to you by the x squared plus nothing times plus
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v being nothing in that zone times u equals to m over bar squared e times u.
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So we're going to have solutions like sine k x plus some phase constant where k is going to be the square root of 2me on h bar squared.
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And this is. So if I put you in a if I if I put it in a phase and unknown phase, this is an unknown phase here,
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then this sinusoidal will represent any linear combination of sine and cosine and so will be a general solution to this.
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Oops. Sorry. No, that's correct.
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I, I want it to be negative. Right.
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Sorry. No, no. Is positive. Is supposed to be as positive. Yes.
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There's a minus sign. There's a minus sign here coming from the from the minus bar squared over to M.
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So that's why there's a minus sign here and that's why we have a sinusoidal behaviour.
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Right. So we know what the solutions look like in the middle,
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we know what the solutions look like on the sides and all we have to do now to fix
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everything up is solve is make sure the boundary conditions are satisfied where they join.
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So we have to. So we say that at X equals a, we have to make sure as yesterday that our wave function is continuous and has a continuous derivative.
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So we so the continuity of the wave function, let's specialise on the even parity case.
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Continuity of the wave function is going to say that kosh times K must equal B that's my constant that multiplies the sine times sine k plus phi.
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That's the continuity of u. And then I have to deal with continuity of the gradient of u as yesterday.
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So we have the k shine k is equal to k b cos k plus phi.
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Yeah. So we are not really very interested in B, it's the same sort of set up as yesterday.
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We're not very interested in the same B yesterday we had a similar constant.
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A We're very interested in in the value that K takes.
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And this thing here, Big K as yesterday can be expressed as a function of little K.
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So the name of the game is to get rid of B. So we, we divide this equation by this equation.
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So one over two leads us to the conclusion that put the hyperbolic cost of K is equal to K over k times.
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The. Sorry, sorry, sorry. Sorry. I'm doing which division.
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Sorry I put that the wrong way up so that becomes big K of a little K.
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Right, because I am, I am taking this equation and dividing it by this equation.
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So this K was on the bottom, as it were,
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and I brought it up to the top of the other side times the tangent of K plus five and
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now we have also no we haven't quite yet because there's one thing we haven't fixed,
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which is we, we need to find out about this five thing.
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What do we have to do with the Phi Phi has to be chosen so that the wave function in that right hand zone vanishes at x equals B.
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Remember when we had the infinitely steep sides? We concluded that the wave function vanished adjacent to the rise to infinite potential energy.
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That was the thing, the point we finished on yesterday.
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So what we can say is that at x equals B we require because the potential is about to go infinite, that you equals zero.
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So that means that sine k b plus phi is zero and kb fly is so that.
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What does that imply? It implies that k b plus phi is some integer number of PI's.
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So that tells us what phi is.
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It tells us that phi obviously is our pi minus B, so we put this information into here and we have our we have our equation.
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But while we're doing it, it's probably a good idea to express Big K in terms of Little K.
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So we have expressions for Big K and Little K on the board and let's work out exactly what we what we have.
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Actually, so big K is the square root of 2mv0 minus E on each our squared which we write as to m v nought a
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squared over a squared h bar squared minus 2me over h by a squared is in fact little k squared.
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And yesterday we identified this object as a dimensionless constant w.
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So this thing is the square root of w of a squared minus k squared.
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And so when I take this equation here and that result there in the right hand side and this result here to get rid of those big k's,
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we discovered that the hyper hyperbolic cotangent of I want big K,
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which is going to be the square root of w minus k squared is equal to on the right hand side.
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Big K of a little k which is going to be the square root of K.
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Sorry. Square root of what I'm talking about of big w over k a squared minus one times the size sorry.
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Eight times the tangent of of our pi minus k, b minus a.
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Now we have what this is. So W is the dimensionless constant that characterises the potential wells on each side as yesterday.
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K is the thing of interest to us because it controls the energy,
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it determines the energy of the stationary state and is the only unknown in this equation.
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Everything else is known. So what we have here is a pretty ghastly equation whose roots determine the values of K, which in turn determine the energy.
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And as yesterday, the way to solve this is graphically to plot the left hand side of the equation and
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the right hand side of the equation and separate curves and see where they intersect.
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And I live in hope that my computer has this has this here.
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So oh, the computer may have it here, but the it didn't come up.
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Do the systems perhaps go to sleep? Is anything showing, you know?
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No. Oh, good. Meanwhile, my computer is gone. Buried.
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It's. Here we go.
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Right. So that should show it's very faint from here.
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It's warming up. Right? But that should show the. The two sides of the equation, the right hand side of the equation.
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I'm sorry. No concern.
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This is not. Right. That was that problem.
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We need to go further down. That's not the right solution. Here we go. So here is the double potential.
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Well, and what we this is the each of these curves is the right hand side of the equation.
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It's coming down like this. And there's a there's a different one because we have this tangent so that so
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the right hand side of the equation contains a tangent which goes to zero.
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We can forget about the other pi as a matter of fact. Right, because the tangent of our pi plus an angle is the same as the tangent of the angle.
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So I could write this,
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I probably should write this as w over k squared minus one times the tangent of this angle here k b minus A and there's a minus sign out front here,
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which is that minus sign. Okay, so you can add another pi to a tangent and you don't affect its value.
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So this tangent keeps having zeros right?
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As K increases, this tangent hits pi, two pi, three pi, etc. The argument to this tangent hits pi to pi, three pi, etc. and the tangent vanishes.
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And those give you the points where these vertical lines crash down through the origin.
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These points, these points, these that's why there are many branches.
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The other thing that's happening here, so the dotted line there is sorry, there are two.
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Yeah. Focus for the moment. Only on the on the top curve because.
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Listings. So there are two curves. Where is this?
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I think this thing is needs a new battery.
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There are two curves running horizontally, almost horizontally, a dotted one, and the dashed one.
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The dotted one is the one we should look at, the one that curves upwards at the end.
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That is a plot of this cloth. All right.
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So when if W is fairly large and K is fairly small, then we're looking at the cloth of a large number, and that's always one.
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So this so this thing runs along at one.
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And then eventually when K becomes on the order of W, we're looking at the cloth of something smaller than a large number.
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And that becomes that then goes up towards infinity.
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That becomes a lot. That then goes up towards yeah.
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So that's what's happening there.
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And we add the into and the the magic values a k are given by the intersection of the dotted line and the full line's coming down.
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And you can see you get a series of you know, you get a series of integers.
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There's only a finite number of them because the when K becomes bigger than W, when K becomes bigger than W, the square root goes imaginary.
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And we no longer have any any solutions to the equation. Bad things happen here, too.
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So there are only a finite number of energy levels. If if the well has any finite value of w, if we would repeat all this.
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So that's so that's what we have. We have a finite number of of energy levels.
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This is for the even parity case. If we were looking for the odd parity solutions.
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So for odd parity, what would happen is that only the only thing that would happen is that this would become a shine and that will become a cosh.
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Everything else would stay the same so that the only thing that would happen as
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we followed through this logic would be that this cloth would become a fan.
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A hyperbolic tangent. All right. So this is even parity you can find through the algebra afterwards.
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But believe me, I think it's very plausible that all that happens is because you're swapping over a cosh and a shine.
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This becomes odd. We get fan of the square root.
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Equals minus the square root, etc. So the the vertically crushing downed lines remain the relevant lines.
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The right hand side hasn't changed, but the left hand side has changed.
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So it starts off as the fan of a the hyperbolic tangent of a large number I1 and
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then it becomes the tangent of a smaller number as K becomes comparable to W,
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so it turns downwards, not upwards because. Right. So, so what?
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You want to pump the point now the crucial physical point is the crucial point that has physical consequence is that when car is small.
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In other words, when and K is the measure of the energy. Right?
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The energy is proportional to K squared because a spark is the momentum of the particle as it rattles around inside as well.
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So the energy of the low energy solutions are associated with the poor.
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These things come in pairs.
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The energies come in pairs agreed because we have the even in odd parity, things intersect low k almost exactly in the same place.
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On that diagram you can't see the distinction as you get to higher values of k and the particle is only marginally
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bound has an energy which is comparable to V zero the the two the two curves for the right and the left.
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Science diverge and you can see you get different values of K so so what's the key thing is the solutions.
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They come in pairs because we had two potential wells.
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We get two adjacent solutions. In fact, if we had three potential wells, we would get three.
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If we have an infinite number, we would have an infinite number. And this is in solid state physics.
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This is this is crucial because in a crystal you have a, you know, ten to the 24,
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you have some vast number of potential wells, one for each atom, and you get a vast number of solutions all crowded together.
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Anyway, we just have two wells, we have a pair of solutions.
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They come in pairs for we of of similar energy, very similar energy,
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similar e so one is even parity and this actually will be the lower energy solution.
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It has the smaller value of K, you can check from the diagram up there and one for the odd parity and this is the higher energy.
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So the the lowest energy.
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So for this system, for a particle trapped like that in a well, what do we have?
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We have a lowest state and just above it, which is that which is of even parity.
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So the ground state here we are we've got we've got a picture here. I think the ground state.
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So so the top diagram there sort of shows these wave functions.
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We have a sinusoidal on the right, the sinusoidal of the on the left, the ground state is the two upper four curves.
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It's up to sinusoidal with depression in the middle, where the barrier is in the particles classically forbidden.
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And then at an energy just a tiny bit higher, you have the you have a very similar pair of sinusoidal.
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They are in fact subtly different values of K. So they're very, very subtly different.
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But essentially the two a very good approximation.
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The odd parity stage is the same as the even parity state except reflected around the origin.
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So it becomes negative on this side. So we have two states, we have.
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So the wave functions are going to be what we'll call Amu plus of X, which is the even parity case and u minus of x,
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sorry, sorry, u e of x and you order x, which is the odd parity case.
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If we take linear combinations of these two states, we get two other states, ABC Plus,
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which will say is one over two of X, which is you even have x plus you order x and you end up saying minus of x.
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So if we take so ABC plus is the sum of those two way functions up there.
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So it essentially it vanishes on the left and as a non-zero value on the right.
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So this is the this is this is state of being on the right.
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And this is the state of being on the left, because if you take them away, they end up on the left and they subtract on the right.
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If we put a particle, if we actually put the particle into the right hand, well, we will set our system up in this state of PSI.
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Plus the state of PSI Plus is not a stationary state.
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It is not an energy eigen state because it's a linear combination of two eigen states of different energy.
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So. So if we drop it on the right, what our initial condition t equals nought.
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This is what our wave function looks like. What does our wave function look like?
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Generally of psi of x and t is going to be one over two times you even of x e to the
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minus i e even t over h bar the usual boring time evolution of a stationary state,
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plus you order x e to the minus, i.e. odd t over each bar, which we can write more conveniently.
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These are the minus i e even t of bar of a root to an even of x plus u of x e to the minus i e od minus even t h bar.
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And the crucial thing is that if we wait long enough, if we wait such a time that the argument of this exponential becomes equal to pi,
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we'll be looking at each of the i pi, which is minus one here, and our state will have evolved into some phase factor.
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Who cares? Times you even minus u odd which is the state of being on the left.
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So. So after a time the time required for this to become pi, which is t equals pi h bar over e of minus the even the particles on the left.
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And you can see that this will go on forever. You put the particle in on the right.
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It's not a stationary stage. After this time, it moves to the left.
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And then after twice this time, it will have come back to the right.
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It's going to oscillate between these two, between these two wells forever and ever, according to the theory.
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And the and crucially, the timescale is long.
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It takes a long time to get from the right to the left.
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If this energy difference is small and we've seen that this energy difference is small.
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Right. So this energy difference being small means that this timescale is long.
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Why is this and when is this energy difference? Small. It's small when the barrier between the two wells is high.
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So what we say is that the particle takes a certain time to tunnel through that barrier.
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So we say that the particle warps tunnels through the barrier.
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Well. And in a time which grows.
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It grows very rapidly as a fact with V0 with with w with the height of the barrier.
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So a high barrier or a wide barrier means it takes a long time to get through, but according to this analysis, it will eventually get through.
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Okay. So that's just that's just a toy problem. Now, let's see what the [INAUDIBLE] that has got to do with the real world by talking about ammonia.
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Somewhere here we have a picture of ammonia. So ammonia is a.
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We don't have a picture of Romania. No.
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Sorry. I've somehow lifted up. Okay. So what does ammonia consist of?
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It consists of three hydrogen atoms and a nitrogen.
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So these are ages and it consists of a nitrogen atom. And the three hydrogens form some kind of a triangle, roughly speaking.
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And the nitrogen atom sits. Well, this is the classical picture.
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We'll see. Quantum mechanical picture isn't like this, but this is the classical picture.
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We think of the hydrogen of the nitrogen atom as sitting either above the triangle
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formed by the hydrogen atoms or below the triangle formed by the hydrogen atoms.
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Now, this is a this is a complicated system. It's got four nuclei and ten electrons.
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And so it's a really complicated dynamical problem.
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But next next term, in the next course, you will discuss a thing called Li or C, a thing called the adiabatic approximation,
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which is what chemists use in order to understand the dynamics of complicated systems like this,
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it allows you to treat all these bonds between between the nuclei which are provided by the electrons as springs.
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So what we can do,
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what a what chemists routinely do do is they calculate the energy of this molecule in what's called a clement nuclear nuclei approximation.
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So they they put they say let the nuclei be, you know, let them be here and let's calculate the energy.
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And you can do that for the for the nitrogen atom,
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for the nitrogen nucleus being at every point along the line that passes through the central to the triangle.
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Right. So imagine doing that. That leads to a potential energy curve.
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If you like, for the nitrogen nucleus, which is going to be something it turns out sorry.
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It's not obvious that it's going to be what it turns out to be, something like this.
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So there are two worlds, the lowest potential energy where it's an appropriate distance, either above the plane or below the plane.
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There are two wells and this sort of a barrier B in the middle because it isn't
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comfortable being in the in the middle of the triangle formed by the form,
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by the hydrogens. So we have this we have this kind of potential energy curve.
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And now we know what the energy levels of this molecule are going to look like, because we know that the there are going to be even parity.
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The stationary states are going to divide this. It's obvious on geometrical grounds this is symmetrical that this well looks just like this.
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Well, because there's no fundamental difference seen above the triangle and below the triangle.
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So we know the solutions are going to come in pairs.
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There's going to be a of of even parity solutions and odd parity solutions just like in elsewhere.
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Well, so.
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We also know that if we if we start if we if we if we start with the nitrogen in this well, so we start with the nitrogen stay above the triangle.
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It's going on some characteristic time to move over here.
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And then it's going to move back there and it's going to oscillate between the above the triangle and below the triangle.
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So if we start it in, in either above the triangle or below the triangle, it's going to oscillate between the two.
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You both have done chemistry.
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Will know also that this nitrogen is going to be carrying a negative charge and these hydrogens some amount a positive charge.
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So what are we going to have? We're going to have an oscillating dipole.
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This molecule is going to have a dipole moment because the electronegativity of the nitrogen and it's
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going to be an oscillating if if we start it in this in this state of being above the triangle,
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it's going to oscillate to and fro. And what is not letting dipole electric dipole do it radiates so this thing is going to
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radiate and you will be able to if you know what this potential surface looks like,
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you'll be able to calculate the frequency which is going to radiate because the frequency which is going to
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radiate is going to be given basically by this formula here for the half period to go from there to there.
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Right? So twice this time is going to be the period of a complete oscillation and that's going to be the period of the of the radiation.
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Okay. So. So the molecule if started, it's a dipole, so started on top.
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If you can achieve that, it's going to be an oscillating dipole.
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And the experiments lead to the conclusion that the frequency new is about 150 megahertz.
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You can measure it. So it's a source of microwave radiation.
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From this 150 micro hooks. You can also work out what the change in the what the what this energy difference is.
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So e odd minus e turns out to be about ten to the minus four electron volts.
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So that's an energy difference, which is small compared to the thermal energies of molecules just here in the room,
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right at room temperature, the thermal energies of molecules.
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So K the Boltzmann constant times t the room temperature is something like 0.03 electron volts.
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So this is, this is much bigger than in minus E What does that mean from your statistical mechanics?
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Cause you either know now or will soon know that that means that we expect at room
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temperature there are essentially equal numbers of molecules in the even and the odd states.
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Right. Because the thermal energy the energy required to go from the ground state, this is the ground state.
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What is the energy of the ground state to the first excited state is less than the characteristic thermal energy knocking around in the room.
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So there will be large numbers of molecules in both of these, even in old states.
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And the idea behind an ammonia maser is that we is is to find a way of separating the molecules which are in the excited state, the old state,
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leading them into some kind of a resonant cavity where they will then because they're in the excited state,
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they will be, uh, they will decay into the ground state if you leave them.
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If they're left alone, they will decaying to the ground state, emitting a photon with a balance of the energy.
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Right. So they will radiate away at 150 megahertz. So that's that's that's the strategy behind a maser.
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So in a maser, you get a maser.
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You have to isolate the half the roughly a half of the molecules.
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In the first excited state. And the way you do this is you exploit the fact that.
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Well, let's just imagine putting these molecules into an electric field.
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So we're going to put the molecules. On the strength of this electric field.
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I'm going to denote with Curly to distinguish it so that electric field is distinguished from energy, ordinary Roman.
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So we put these molecules into an electric field e and ask ourselves, how does that change?
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That's how is that since these are dipolar molecules, right?
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These molecules are electric which have our electric dipoles.
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We're expecting that that changes the energy of the molecules, putting them in a field and maybe we can get them separated by exploiting this fact.
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So you put them in an electric field and ask, So how does that change the way we're doing quantum mechanics?
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So changing energy means changing Hamiltonian. So we have to ask ourselves now, how does this change the Hamiltonian of these molecules?
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What term in the Hamiltonian is introduced by this electric field?
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And the way to do that is to think in terms of the Hamiltonian, right, the Hamiltonian in the basis.
315
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So we're going to let plus B.
316
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So it's going to be mathematically one of a route to the even parity the ground state,
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plus the odd parity now using a different notation, even parity plus all parity.
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So this is the ground state. This is the first excited state.
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This linear combination is we're going to call it plus and it'll be the state in which the
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molecule is definitely above the triangle and similarly minus is going to be one over route two.
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Even parity state minus all parity state is going to be below.
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So this is a low triangle. This is above this is exactly what we did over here with our PSI Plus and of PSI minus.
323
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We were working that with wave functions here. We're working with the underlying caps.
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The thing is, the dipole moment, if the dipole moment of these two states have opposite signs because this one has
325
00:41:03,020 --> 00:41:08,600
the negative charge at positive Z and this one has the negative charge negative Z.
326
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So they have opposite dipole moments and let P now be P, the dipole operator.
327
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It was the parity operator earlier on, but now let it be the dipole operator.
328
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Then these states are going to be eigen states of this dipole operator.
329
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They have well-defined dipoles. P plus is in fact going to be minus q s times plus.
330
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Well, well, what am I saying? I'm saying this is some charge.
331
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This is some distance. The product of a charge in the distance gives me units of units of dipole, electric, dipole moment.
332
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I've got a minus sign here because of because this state, I said, has a negative charge above positive z.
333
00:42:03,950 --> 00:42:12,800
So the dipole moment points in the opposite direction to the location of the, of the nitrogen atom.
334
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And therefore this one has a negative dipole moment and this is this eigenvalue, right?
335
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Is the dipole moment. This is the eigenvalue of this operator because this thing has a state is a state of well-defined dipole moment.
336
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And this is where the dipole moment appears.
337
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And basically this is just some number which has the dimensions of a charge times a distance, which the charge is going to be on.
338
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The order is going to be some fraction of electron charge.
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And this distance is going to be this characteristic size of the atom, a 10th of a nanometre.
340
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Similarly, P minus is going to be plus us minus.
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So this encapsulates the crucial point that these two states have opposite leave signs dipole moments.
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What is the energy? Of a dipole in a knee field.
343
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Well, the answer is that the energy is minus.
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This is just classical physics is minus the field times the dipole moment.
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So now we know how much the energy of this state and this state are changed by the electric field.
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It's it's given by by this here where we put in those kind of eigenvalues.
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So now what we want to do is write down the matrix that represents the Hamiltonian in this plus minus basis.
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So we want to write the we want to write the Hamiltonian is plus H plus matrix.
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This is a complex number, right? Plus H minus.
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Minus H plus. So these four complex numbers, we want to write down that matrix.
351
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And the Hamiltonian consists of the Hamiltonian of the undisturbed.
352
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Of the undisturbed nitrogen molecule.
353
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Plus the contribution here.
354
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This is plus the contribution here from the electric field.
355
00:44:33,790 --> 00:44:37,269
What do we have for the undisturbed? Let's.
356
00:44:37,270 --> 00:44:42,520
Let's calculate this separately. So I need to calculate what?
357
00:44:42,910 --> 00:44:46,780
What plus h plus is.
358
00:44:48,840 --> 00:44:58,510
For the isolated atom. Well, that is going to be one over two, even.
359
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Forbes Because these are linear combinations of the Asian states, even an odd even plus odd H even plus.
360
00:45:13,090 --> 00:45:20,680
All right, because this is this this is this.
361
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But Asian this is an Asian function of this operator if this is the isolated operator.
362
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So we get h on even produces even times even even is orthogonal to odd but meets up with this
363
00:45:37,390 --> 00:45:45,580
so we get an even and an H on odd produces e odd times odd which is orthogonal to this,
364
00:45:45,580 --> 00:45:47,410
but meets up with this to produce in the odd.
365
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So at the end of the day we get a half B even plus E odd, which is the average energy, let's call it E bar.
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Similarly, we have that.
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Let's have a look now at two plus H minus the off diagonal element that is going to be one over two even plus odd h even minus odd.
368
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So things are rather similar except we now get from here, here and here.
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Even from here, here and here we get minus e odd.
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So we get the we get a half of even minus E odd.
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00:46:45,850 --> 00:46:51,080
Which I think I have been calling we're going to call this minus a.
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So A is. So we going to it? This is the definition of AA.
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And we're going to put this minus sign in here, because we we know that odd the odd energy is the is bigger than the even energy.
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So this quantity here is a negative number. And put in that minus sign, we get this positive number.
375
00:47:07,150 --> 00:47:11,020
So this matrix up here, what's it looking like?
376
00:47:11,260 --> 00:47:16,960
It's looking like e bar from here.
377
00:47:20,510 --> 00:47:27,500
Sorry. A this is a permission matrix that what appears down here has to be the complex conjugate of that real number.
378
00:47:27,650 --> 00:47:33,220
In other words, A and we will find that this one is also E bar.
379
00:47:35,060 --> 00:47:43,960
That's for the isolated atom. Then we have to add on to it the contribution for the from this which which.
380
00:47:44,150 --> 00:47:47,390
But but this thing is an eigen function of P.
381
00:47:47,990 --> 00:47:56,120
So the only contributions down the diagonal, there are no off diagonal contributions from this additional piece to the upper to the Hamiltonian.
382
00:47:56,870 --> 00:48:02,779
And what I have is plus this is what I need is the dipole operator.
383
00:48:02,780 --> 00:48:08,780
So I have p on plus which produces minus Q s times plus.
384
00:48:09,800 --> 00:48:20,030
So this minus sign in the minus sign that I've just spoken cancel and we get plus Q K and here we get minus Q Curly.
385
00:48:20,280 --> 00:48:26,720
S So this is what the Hamiltonian looks like in this in this, in this basis.
386
00:48:28,670 --> 00:48:33,020
I'm running out of time. So let me just sketch how the how the thing goes.
387
00:48:34,190 --> 00:48:40,970
So what we have got is an explicit expression for the Hamiltonian in the presence of an electric field.
388
00:48:42,380 --> 00:48:45,620
What we would like to know is what the energy levels are, right.
389
00:48:46,120 --> 00:48:51,080
That results from this and how those energy levels vary as a function of the electric field.
390
00:48:51,950 --> 00:48:55,550
But that's a dead cinch. What you've got up there is a two by two matrix.
391
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Anybody can find the eigenvalues of that two by two matrix,
392
00:49:00,470 --> 00:49:04,340
and those will be the energies that the molecule has when you stuff it in this electric field.
393
00:49:07,820 --> 00:49:16,010
So we find the eigenvalues of this matrix and we plot how they depend on the E.
394
00:49:20,020 --> 00:49:36,280
So. What they actually are is the average energy plus or minus the square root, I believe of A squared minus Q.
395
00:49:38,960 --> 00:49:44,930
Just make sure I didn't make a mistake that I'm bringing. Yeah.
396
00:49:48,520 --> 00:49:53,050
So these are the possible energies. And if we plot this graphically, what do we see?
397
00:49:53,860 --> 00:49:58,510
What we see here, we have the strength of the electric field.
398
00:49:59,230 --> 00:50:02,860
We find that we have solutions that behave like this.
399
00:50:03,220 --> 00:50:12,220
Here we have EPA plus A, here we have E, bar minus A, and here we have the strength of the electric field.
400
00:50:12,370 --> 00:50:15,460
That's just this is just a graph of those two of those two things.
401
00:50:15,850 --> 00:50:19,030
You can easily check that. That's the case. Here, of course, is EPA.
402
00:50:19,750 --> 00:50:28,270
So this is the ground state. If you switch off the electric field, we have two states here and here separated by this small amount, which is two.
403
00:50:30,190 --> 00:50:37,180
As you switch on the electric fields, the energy of the ground state goes down, but the energy of the first excited state goes up.
404
00:50:39,160 --> 00:50:43,510
There's a lot of interest in the way in which this energy behaves here.
405
00:50:43,510 --> 00:50:45,510
It's behaving quadratic as a function.
406
00:50:45,520 --> 00:50:52,540
This is a sort of parabola as a function of V and eventually it becomes a straight line as a function of V, both top and bottom.
407
00:50:53,170 --> 00:50:59,739
And what that is telling you is that in this regime here, the even state well, so sorry, the the ground state,
408
00:50:59,740 --> 00:51:03,549
which is the even state has no dipole moment intrinsically because the particle is
409
00:51:03,550 --> 00:51:07,240
equally likely to be above the nitrogen is equally likely to be above or below the plane.
410
00:51:08,230 --> 00:51:15,980
So in the ground state there is no dipole moment. You switch on the electric field and you make a bias so that the, the,
411
00:51:16,570 --> 00:51:24,100
the upper fields say the upper the upper position gets to have a lower energy in the lower position because the electric field is pushing it that way.
412
00:51:24,790 --> 00:51:29,200
So the nitrogen spends starts to spend a bit more time above than below.
413
00:51:29,530 --> 00:51:33,430
And now the molecule acquires a dipole moment, right?
414
00:51:34,090 --> 00:51:37,090
Because it's not spending, the nitrogen is not spend equal time above and below.
415
00:51:37,780 --> 00:51:46,480
So it requires a dipole moment. The the magnitude of this dipole moment is proportional to the energy and therefore the energy that you get,
416
00:51:46,750 --> 00:51:52,180
the change in the energy which is equal as it sets up there e is minus electric field times.
417
00:51:52,180 --> 00:51:55,330
The dipole moment becomes proportional to dipole moment squared.
418
00:51:55,930 --> 00:52:02,590
So that's why we have a quadratic dependence like this. Once you're in this regime, you have a fairly strong electric field.
419
00:52:02,980 --> 00:52:13,270
You've made such a strong bias, you've made being up so energetically more favourable than being down the nitrogen, spending all its time up.
420
00:52:13,810 --> 00:52:18,610
The dipole moment is now independent of the strength of the electric field,
421
00:52:19,240 --> 00:52:24,470
because it's because because increasing electric field doesn't cause causes to spend any more time up than it than it.
422
00:52:24,910 --> 00:52:30,070
It's already spending all this time up so that the energy becomes proportional to this.
423
00:52:30,520 --> 00:52:34,030
How do you make your maser work? You make your maser work?
424
00:52:34,090 --> 00:52:36,640
Well, the fact that the energy of one of these goes down,
425
00:52:36,670 --> 00:52:45,760
the energy the other one goes up means that as the field increases, that if you have an inhomogeneous field,
426
00:52:45,760 --> 00:52:55,450
if you pass your beam of NH three molecules through a region in which the electric field is varying in strengths,
427
00:52:55,450 --> 00:52:59,080
if I, if I put so I have some sort of capacitor plates will have plates.
428
00:52:59,260 --> 00:53:01,900
I have a pointy thing, right, a needle in a cup.
429
00:53:02,230 --> 00:53:07,719
Then the electric field goes like the field lines go like this and we have a high field up here and a low field down there.
430
00:53:07,720 --> 00:53:16,180
So I have a region of inhomogeneous field. The ones in this state are going to be bent, sorry, they can be bent upwards.
431
00:53:16,330 --> 00:53:23,920
They like the field, it lowers their energy. So they, they, they move into the region of high fields, these ones move into the region of low field.
432
00:53:24,460 --> 00:53:27,970
So this is the ground and this is the excited.
433
00:53:29,380 --> 00:53:34,900
And then you can put these ones into your resonant cavity and here it's safe.
434
00:53:36,900 --> 00:53:43,740
So there's an example of of how with with a very simple minded model,
435
00:53:44,100 --> 00:53:51,780
you can you can explore some quite interesting physics and some physics which is very, very inherently quantum mechanical.
436
00:53:51,780 --> 00:53:56,190
This this energy levels coming in a couple of in pairs.
437
00:53:57,360 --> 00:54:03,989
This because we have even an odd parity states this ability of a type of a molecule
438
00:54:03,990 --> 00:54:08,879
which you would think was inherently dipolar to to in the ground state have
439
00:54:08,880 --> 00:54:14,780
no dipole moment because the nitrogen can be simultaneously above and below of
440
00:54:15,270 --> 00:54:18,719
all of these features of very quantum mechanical and with the simple model,
441
00:54:18,720 --> 00:54:20,850
we are able to explore them. Okay.