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Okay. So now we've investigated the states are a well-defined momentum, which as you recall with these plane waves with a wave number.
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Pierpont Momentum upon HPR,
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we can go back to this problem of two split interference and ask ourselves why it is that quantum interference isn't observed
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in macroscopic objects and also assess what experimental setup will be required to see quantum interference with electrons.
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So we put some numbers into this experiment.
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So as you recall, we had some gun here that was symmetrically placed with respect to a couple of slits here in an obscuring screen screen,
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it fired out particles.
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Some of the particles got through the holes and they came to here and we said that the amplitude up here would be the sum of the quantum.
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The quantum amplitude to arrive at this point would be the sum of the amplitude to go via this route or to go for this route.
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These routes. Well, the the the distance from the gun to the two slits by set up is symmetrical is equal.
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So any difference in the amplitude that they come there is to do with the change in the amplitude when it goes along this route as
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against along this route so that D D plus b the distance from the upper slit to this place and D minus the distance to the lower slit.
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Then we can write a formula that D plus by Pythagoras theorem is going to be L squared plus x minus x squared square root,
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and correspondingly D minus is going to be the square root of l squared plus x plus s squared.
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And we make the reasonable conjecture.
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I mean, the right way to think about this is these particles.
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Let's imagine that we've got our gun here, has been tuned to emit particles with some well-defined energy.
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That means that as they go along here,
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the particles will have some reasonably well-defined momentum because they the energy will consist of that kinetic energy.
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So along here, we will have that, that the amplitude is going to be on the order of e to the i t upon h bar times x.
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So that's a reasonable model for what the how the amplitude varies with position.
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That's the new information that we bring to bear on this. So what will be the difference?
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So the probability to arrive at X, as you recalled, was equal to the amplitude to arrive by the top slot,
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plus the amplitude to arrive at the bottom slot mod squared.
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And the argument that this was a plus mod squared plus a minus mod squared plus twice the real part of a plus a minus.
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And these were about equal and well, these were the classical probabilities.
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So these were P plus plus, P minus.
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And then we have this quantum interference term, which we want to assess which.
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So so what is this quantum interference term?
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So the interference term. Is going to be.
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Mod plus mod minus neither.
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Not very interesting.
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And the crucial thing is that this one is going to be e to the eye p upon bar d plus and the other one's going to be e to the minus i pierpont bar.
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Sorry. Plus. Sorry.
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One of these needs to have a star in it, which means that one of these requires a minus sign.
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So if we're interested. Excuse me in the real in the real part of this and the real part so this thing
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can be written as cos Pierpont bar de plus de minus minus D plus etc. etc. etc.
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So this is looking like to the probability to go either way times through
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through each hole separately times the cosine of p over each bar d plus minus,
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d minus. Actually it's easier to do it the other way around it.
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Minus plus. So what is this difference of distances?
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This difference of distances from up there?
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Well, let's binomial expand this and because we can argue that el is going to be big experimentally compared to x, x will be mm.
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L will be a metre or so.
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So we can binomial expand this and say that this is l it's about half of l brackets one plus x minus x squared of a l squared plus a dot, dot, dot.
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And the other one's going to be about half L one plus x plus s.
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Squared over l squared plus dot, dot, dot. So when we tell you the difference of those two.
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So this is going to be two p plus minus this probability times the cosine of p upon bar of the
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difference is going to be x plus x squared over l minus x minus x squared over two l in fact.
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So when you take the difference of those two, you will be looking at two.
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X. S of l. Yes to excessive l is what that difference will be.
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So. So now let's put it now. We need to put in some numbers.
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All right. Suppose we take the energy, which is P squared over two.
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What do we want to do? We want to know. So what does this do?
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This gives us a probability of arrival as a function of X,
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which is going to consist of a of twice the sum of these two, which are going to be about equal.
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And so about twice this plus twice. This thing times this cosine.
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So it's going to be an oscillating commodity which will be doing this.
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And there's some characteristic distance between these, between these minima which so this will call this delta X say no, let's call it big x.
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That's what the so we call this big X the distance between the places where it's a
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minimum because the quantum interference term is cancelling the classical term.
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And this difference is what causes that.
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The argument is that cosine to become two pi so we can write the two pi is equal to p over four times two s of l times x.
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In other words, we have a formula for the distance between the minima, which is two pi to pi h bar, but h bar is h planck's constant over two pi.
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So that's going to be h over p h l sorry.
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Yes. Of S.
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Yeah. And we could also write this.
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I guess we we could also say that.
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Oh, no, it's not, Paula. So let's put some numbers in.
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Let's say that e the energy is.
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So in order to get a big value of X, we want to take a big value of L.
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Needless to say, we want to take a small value of P and of course a small value of S two on the bottom of.
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I've lost a two. Yeah, you're quite right. So, so let's take L to be a metre.
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Let's take S to be what got I think.
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What did I do. I think I think it's Yeah.
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A micron because you want to make it as small as you can.
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But if you make it much smaller than a micron, you'll find it difficult to make the hole using ordinary materials.
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And let's and we also in order to get a small value of P, we want to take a small value of the energy,
79
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but you can't take too small a value of the energy or your particles will be deflected by stray electromagnetic fields and stuff,
80
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and it'll be difficult to keep any kind of coherence.
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So let's take 100 ev say as a, as a sort of convenient low speed if you plug all this stuff into there.
82
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So that gives you what does it give you a 20th of a millimetre or something.
83
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I think it's is it 0.06 millimetres which is obviously perfectly, perfectly observable.
84
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Yeah. So this this such an interference experiment is, is, is possible but hard using electrons if you do the same thing with bullets.
85
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When we're not expecting anything to happen, what could we do?
86
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We would we would take the velocity for a gun is say 300 metres a second might be a bit faster these days.
87
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I'm not sure. But that's a classical that's, you know, faster than sound.
88
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So that's sort of a reasonable ballpark figure.
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Suppose we took L to be one kilometre a thousand yards, pretty reasonable shooting distance for a rifle.
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And if we took the mass to be ten grams, put it into the same formula and we discover that X is some ridiculous figure ten to the -29 metres.
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So it's obvious that you cannot observe this interference using anything like a bullet, any kind of macroscopic, any,
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any kind of macroscopic object, because it's going to be vastly bigger itself than the than the size of the interference pattern.
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Obviously, an absolutely basic requirement for this experiment to work is the physical size of
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your particle has to be smaller than the value of X that you derive out of this.
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So you haven't a hope of measuring this interference. So that's why classically we don't we are unaware of this interference term.
96
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But I would remind you that in the last lecture we recovered classical results,
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which which explain why cricket balls move as they do, why satellites and so on move as they do by interpreting.
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We calculated this.
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We obtained results which recovered classical physics by decomposing the amplitude to arrive into a sum of contributions from states of different,
100
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well-defined momentum. And these were all interfering with each other.
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And the classical physics came back as a result of quantum interference.
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So this quantum interference, on the one hand, is something which is very hard to observe with classical objects.
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On the other hand, our entire picture of the classical world, a classical world, is only recovered through quantum interference.
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It's not is it's not some esoteric corner of the subject, but it is hard to.
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It's hard to have it happened in a controlled way. Okay.
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So we. Yeah.
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We should just. So we've done, we've done the position representation in just one dimension.
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Everything has been a one dimensional motion motion long x.
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We obviously need to generalise the position representation to three dimensions because we live in a three dimensional world for whatever reason,
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and the generalisation is nice and trivial. We don't need to worry about it.
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We have we now have three position operators X, Y and Z, also known as X, Y.
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All right. And we have, of course, three momentum operators, three more operators, X, Y and Z, also known as P.
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And we have that every one of these operators commutes with the other one.
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So we have that exi comma sj.
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It's nothing. And every one of the momentum operators commutes P-I, comma, PGA.
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Equals nothing. So it is possible to simultaneously know your x, coordinate,
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your y coordinate and your z coordinate as a complete set of eigen function of can states.
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Of well-defined states. States. Where you know all those strings simultaneously.
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Or you can know all three components of momentum, but you can't know there's not a complete set of states for knowing and so on.
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And the only other interesting thing we have to have is XY commuted with PJ.
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He's h baa delta. So it is possible to know the exposition and the Y momentum, but it's not possible to know the exposition and the momentum.
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So most of these operators commute with well, each operator commutes with five of the of the sorry,
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four of the remaining five operators, but it does not commute with its own momentum.
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That's what each of these position operators. So that's the generalisation there.
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What else do we have to say? Well, we used to have a wave function of PSI being a function of scalar x.
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We now it's trivial the argument of the wave.
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We we can now label a complete set of states by X,
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Y and Z so we can write that there's a we have states of well-defined position
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which are labelled by a vector now vector position x because there are three.
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This is an eigen state of the x operator. It's an elegant state of the Y operator.
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And then I can say to the Z operator, so we need three eigenvalues written inside here to describe what this is.
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It is that's the mathematical level of the physical level. This is the state of being at the location position vector x correspondingly,
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our wave functions become functions of X, Y and Z because they become these complex numbers.
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Right? That's still a complex number, this complex number, but it's a function of X and Y and Z, the locations of the particles.
135
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Similarly, we have states of well-defined momentum up.
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We have states. Yeah. Yeah. You. You p.
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Of X, which is x. P.
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So now we have. Here we have p, p, p, z.
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Because we have a state of well-defined momentum which is labelled with all three components of momentum.
140
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So we have this function of a complex of three components, right?
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This complex function of three variables X, Y and Z labelled by the momentum.
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This is just an identical notation. Whereas in single, when we were doing this in one dimension, we found that this was e to the.
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P over h bar times. X not vectors.
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Now that's a vector. That's a vector.
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And whereas on the bottom we used to have h bar to the one half, now we have h bar two,
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the three halves reflecting the fact that there are there's an X component to this, y component to this and Z component to this.
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So. So this the way this wave function of a state of well-defined momentum has now become a plain wave.
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Whose wave whose wave surfaces are normal to the vector p.
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And that's and that's what it is. It's easy to check that that stuff works.
150
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It's a it's a very straightforward generalisation of what we did before. And I think that's all we have to say.
151
00:17:40,610 --> 00:17:45,310
Oh, no, not quite. We also want to say what the momentum operator P looks like.
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So previously we had that x. P xp.
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Sorry. Not right. Talking about. Yeah. X p ci was minus h, but it was introduced by this formula here.
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DPD x of x ci. All right.
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That was what we did in one dimension. That generalises in three dimensions, very straightforwardly to x p psi.
156
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So. So that's become a vector.
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That's become a vector because we have to write down what it is for P and p z.
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This is really going to be a shorthand for three formulae and it's going to be
159
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minus H bar gradient operator on the function of three variables functional space.
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This one here. All right. The wave function. So this is a vector reflecting the fact that that's a vector.
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This is just a label which appears on both sides of the equation. That's what this this formula generates.
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Generalise this to that formula. I don't think we need to be detained about that any longer before we leave the position representation.
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It's good to do. And a useful result.
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Which falls into our laps now because of what we've already done call the variable theorem, which is a a theorem.
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It's a result in classical physics, which you may not have met, I don't know, but you in a way, should have met.
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Did you cover the variable theorem in classical mechanics anyway?
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No. Anyway, so it's there's nothing quantum mechanical about the variable theorem, but it has a classical counterpart,
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but it's going to fall into our laps because we've got this powerful machinery.
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So do you recall if we are in a stationary state, that is to say, a state in which the result of measuring energy is certain then.
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All expectation values for such a state are constants.
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That's why we call it a stationary state. It's going nowhere.
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So every expectation value for a stationary state for a state of well-defined energy is independent of time.
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So we want to exploit that result. So for a stationary state.
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Is this just recalling what we already had? It was it was a consequence of our first theorem for a stationary state.
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We have that DVD time or even deep DVD time of E Q E equals nought for all for all operators.
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Q It doesn't matter what observable you stuff in there, as long as the observable doesn't is defined in a way that is independent of time.
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So it's something like position, momentum, angular momentum, whatever it has,
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it has a vanishing rate of change or with respect to time, it's a constant.
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So we now apply this result to CU is equal to x dot p.
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So then we have that nought is equal to.
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T. Sorry.
182
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Subtle. Some moment of doubt. Yeah.
183
00:21:39,380 --> 00:21:46,580
Yeah. So I want to apply this to export p e and let's divide.
184
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No stinking H-bomb. That by Ehrenfest theorem is x dot p comma h.
185
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Whoops. So Aaron, First Theorem tells us that this rate of change which vanishes is equal to this here and now.
186
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Let's take suppose we're dealing with a particle which has.
187
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Which has kinetic energy and potential energy. So we'll take the Hamiltonian to be of that form, which is.
188
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Pretty useful form. And stuff it in there and we're going to have that nought is equal to E!
189
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X dot p comma p squared over two m plus v close square brackets.
190
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So now we need to work out what this comitato is.
191
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And this is where a little bit of so this is where we get a bit of practice in using the three dimensional generalisation.
192
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We obviously have two things to work out. We've got a combination of X come up with P squared, so let's work that out.
193
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X come up with P squared.
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Now we write that in components. We write x x dot piece, x dot peters I say comma exactly comma squared.
195
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This x dot can be written as a sum over j equals 1 to 3 of x i p i sorry sj pj so that's just a way of writing that.
196
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And now I have a some peak. Well.
197
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P-square. Okay, so I'm something of a K as well.
198
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All right, p squared is p squared plus p y squared plus p said squared.
199
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Now we can work out this using our rules for a commentator.
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We had that rule that a comma B sorry, ap comma C was equal to a comma c.
201
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B plus a, B, C, P.
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We know that peak commutes with PGA that's been written down up there.
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So that commentator vanishes. That's this one here in some sense.
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Sorry, this this one here in some sense. And so what we're left with is so we have this double sum, we're going to have X Jay Peak, PJ.
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Sorry. That squared. Squared. Comma.
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No, no, no comma. So that's what we get. So.
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So this has to be commuted with that? That's what I've written down, I hope.
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And then there should in principle, be another term, this commuting with this.
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But that vanishes because commutes would peak for all for all jank.
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So we have to work out what this one is now and we can use the same rule.
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If we're being pedantic, we would say this is X on peak peak.
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So we would say that this is x j peak, peak plus peak.
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The commentator of x and peak. I'm using the same rule and that all has to be multiplied by p j.
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The same because. Because I'm not writing p squared is peak. Peak.
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But this is h bar. This is bar.
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So these two terms actually contribute the same thing. This becomes to h bar p p times, delta j k times, peak times.
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PJ And I'm sorry, I've lost track of the sum sign.
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Here we have a sum sine. With something over J and with something over K.
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Some have a J and you get nothing because of this delta j k unless j is equal to K, so this becomes PCP.
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K Some do have a k, but PCP k some of AK is the same thing as p squared.
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So this is 2ih bar p squared.
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That's what the commentator is of x dot p with p squared.
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Now let's let's right now, let's do the exact P commentator with V, which is itself a function of X, of course.
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These things ought to have hats, really. But one gets difficult to write down enough things.
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Well. What we want to do is is write this thoroughly in the position.
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Representation in the position representation x dot p is minus i h bar x dot gradient.
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Right? That's what this becomes in the position representation on v, which becomes a function of x just so this is in the position representation.
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So what does that mean. That means I ball minus age brackets x dot gradient.
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Uh, working on v minus the x dot gradient.
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And this is an operator statement. So it's waiting for you to put in the function of your choice of PSI on the right.
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Right. There's a virtual function there for it to work on.
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That's what that's the meaning of this vehicle gradient.
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And this exact gradient V doesn't mean extra gradient only V means of everything that is the right of it, including Europe PSI.
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So when you use this exact v on v times alone, you'll get a term and then you will still have to use the V on the up side.
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But the result of using the V on the up side will be killed by here and next v on up say.
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So what this is equal to is minus h bar x dot gradient of v.
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That's all it survives. This is the action of of the nebula, the gradient operator on the potential itself,
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the operation of the action of the gradient operator on the wave function that's virtually sitting here is cancelled by this contribution here.
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So we now have we we can put these results back into what we had up there.
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So what we had was nought is equal to yeah.
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Is equal to the sum of these of these commentators is equal to e x dot p comma p squared.
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E plus e x dot p v.
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That's just summarising where we stand. This we've discovered to be.
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This is to i h bar. This commentator turned out to be p squared.
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So it becomes the expectation. Oops, there should have been over two m on this, shouldn't it?
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Because it was the Hamiltonian p squared over to him. Yeah.
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This P came from the Hamiltonian where it was P squared over two m this v came from the Hamiltonian where it was just V.
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So we have over two M Uh, no, let's leave it alone of p squared over to m e plus we figured out that this one was a minus h bar.
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So we want to cancel what we.
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This is the expectation value of the kinetic energy. Clearly.
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Right. P squared over two m is the kinetic energy.
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That is the expectation value of it.
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So cancelling the ball, we can say that two times the expectation value of the kinetic energy is equal to this stuff.
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That's as far as we can go in general. But consider now very important cases have that V of X is proportional to model X to the alpha.
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So, for example, for a simple harmonic oscillator we're about to discuss, the potential energy goes like x squared alphas two.
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If we were dealing with a dealing with a Coulomb interaction, the potential and potential energy goes like one over radius.
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So it would be V of all is proportional to one overall.
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Well, this model X is R, so alpha would be minus one.
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So we can say that alpha equals two in simple harmonic motion.
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Alpha equals minus one is coulomb.
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There are, you know, you can think of other power laws which are relevant in this case.
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So then we ask ourselves, what is X dot?
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Gradient of. Evie.
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Well, that's going to be so sad. We'll say that this is equal to some constant A times, X to the alpha.
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What is this going to be? It's going to be alpha model X to the alpha minus one times X dot the gradient of model X and the gradient of Model X is.
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The gradient of Model X is. X the unit vector x, so it's the vector x over model x.
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So this is equal to. Sorry, this is a we have a X to the alpha minus one times x dot x over monarch's.
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So here this mod x is going to make this an alpha to the minus two.
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But from this x dot x we're going to get mod x squared.
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So this is going to be and I've lost sorry this was an alpha.
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That was also an A unfortunately. Yeah. Sorry. We need an A and an alpha.
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This is going to be alpha times a x to the alpha, which is alpha times V.
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So if V has a power lower dependence on distance from the origin, then x dot grad v is simply alpha times v.
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So when we put this result back into that formula, back into this statement,
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here we have that twice the k e expectation value is equal to alpha times the expectation value of the potential energy.
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So that's our Kepler formula. In the case of simple harmonic motion, alpha is two, and kinetic energy is equal potential energy.
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In the case of Coulomb interaction, where alpha is minus one, you have that the potential energy is minus twice the kinetic energy,
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which is to say that the particle has lost two units of energy, radiate in falling in from infinity into a bound orbit.
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It's lost two units of energy, one units being sent off to infinity and radiation or something, and one unit is used as kinetic energy of its orbit.
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So that's the this is a very theorem. So now we open a new chapter, as it were, by talking about harmonic motion.
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The harmonic oscillator is the single most important dynamical system in physics.
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Most of field theory, most of contents of quantum field theory,
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most of condensed matter physics is fiddling with more or less with harmonic oscillators, which which are which are decorated in some way.
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So the basic physics is that of the harmonic oscillator.
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And it's worth just taking a moment to understand why harmonic oscillators are all over the place, the universe, the physicists.
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A fundamental position of physicists. Well, physicists like to represent the universe is a collection of harmonic oscillators.
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And this is partly because physicists may be brighter than some other people, but they're still pretty stupid.
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We have a quite a small bag of tricks and harmonic oscillators is a trick that we we have.
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And it's an incredibly useful trick for this reason that if you plot force in some direction versus displacement from a point of equilibrium,
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you will get a curve which does something like this, the force.
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Vanishes at an equilibrium at the point of equilibrium of a system.
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The force on it obviously vanishes. So if you do a plot of force versus distance, you'll get a curve.
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Something like this passing through zero at the point of equilibrium, which I happen to put at the origin of X.
294
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But, you know, that's by construction, clearly.
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And the general idea is that most of the time you can destroy.
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That's meant to go through the through the origin. Most of the time, you can represent this to some, to a good approximation,
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you can say that F of X is about equal to x plus order of x squared or whatever,
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so to lowest order approximation, because f has to vanish to the point of equilibrium in the neighbourhood of the point of equilibrium.
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F is going to be proportional to x and if is if we neglect this, if this is small,
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then we have harmonic motion for displacements, these small displacements around here.
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So this is why harmonic oscillators are ubiquitous, a very credibly, an incredibly valuable model.
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We can apply. We can use to understand many, many systems because many systems for small displacements,
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almost all systems for small displacements look like a harmonic oscillator.
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Okay, so let's agree what the Hamiltonian of this thing should be.
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The Hamiltonian of our harmonic oscillator should be P squared over two M plus a half K X squared.
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Right. Because if the force is going like that, you integrated up, this becomes the potential energy.
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And this is of course the kinetic energy. We're familiar with that already. It's better, though, to write this in a different way,
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to anticipate results that are to come and to write this is p squared plus omega x squared over to M.
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So, uh, and of course omega squared is k over m so it's easy defining omega squared to be okay over him.
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It's easy to write this formula like that, and that's how I want to write it.
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So you want to reproduce this formula? Just think about dimensional analysis.
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We want to have piece going over to him because it's the kinetic energy. We're always saying that.
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And here I want something that's proportional to x squared and has the dimensions of momentum and obviously omega x has dimensions of speed.
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So M omega X's dimensions momentum. So that's why, you know, that enables you to recover that quickly from this.
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And that's the way to go for practical purposes. So that's our Hamiltonian and we're trying, of course, to solve.
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So these stationary states are the key to understanding dynamics because they have this
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trivial time evolution and by by decomposing any initial condition into with some of,
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of stationary states, into a linear superposition of stationary states,
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then evolving the stationary states, we find out how any arbitrary initial condition evolves in time.
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So that's why we want these stationary states. I've said that before and I'll say that again.
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So we want to find states of well-defined energy.
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This is the problem we want to solve. And this is a completely generic situation in physics.
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First of all, you think about your physical system on the grounds of physics, you write down the Hamiltonian.
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Then the next thing you do is you find the down stationary states, because once you've got those, you can do anything you want, pretty much.
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So that's what we're trying to solve. The the the way to do this is the proper way to, to find these states.
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So we need to find the energies that that are possible and we need to find the corresponding states.
327
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And the way to do this is to introduce some new operators.
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Let's introduce a which is m omega x plus i p over the square root of two m omega or h bar omega.
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Why do I write that down? Well, basically because I know where I'm going.
330
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But just to give you some sense of direction, the general idea here is that we want to factories, that that's the general idea.
331
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We want the factories, the Hamiltonian into.
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So it's a quadratic expression. It seems kind of reasonable to factories.
333
00:40:07,280 --> 00:40:13,690
It if these were if these weren't operators because these are operators, sorry.
334
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In future I'm not going to even attempt to put hats on operators.
335
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Right. These are operators.
336
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Despite the absence of hats, it's just too difficult to remember to put the hats on and takes too much time and grown ups never do.
337
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But these are operators now, if they but if they weren't operators this in its complex conjugate would factories that.
338
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So that's the drift.
339
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Okay let's write down it's it's a bit complex this is this of course is an operator and it's not an observable it's not a mission operator.
340
00:40:43,820 --> 00:40:52,219
It's what is its dagger? A dagger is not joint is this thing dagger, which is itself because X is a mission operation.
341
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It's own dagger. So it's M omega x plus this thing dagger p is its own dagger.
342
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But I has the the dagger of I the joint of eyes minus II.
343
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So this is minus I p over.
344
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Of course, this on the bottom is a real number. So it's own it's its own complex conjugate.
345
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So here we have two operators and the general idea is they're going to factories h or
346
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they almost do whatever because the plan and this is called in annihilation operator.
347
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And this is a correctional operator and. Well, the reason they have these names will emerge.
348
00:41:33,310 --> 00:41:36,310
But it is that if you use this on a state,
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this operator increases the excitation of our harmonic oscillator and this oscillator reduces the expectation of our harmonic oscillator.
350
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And since in quantum field theory, particles are excitations of the vacuum.
351
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This thing creates a particle because it creates an excitation, which is a particle,
352
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and this thing destroys a particle because it destroys the excitation.
353
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So what we next do is work out what a a dagger A is, because the idea was that this product would be more or less the Hamiltonian.
354
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So what exactly is it? Let's get this right. M omega x minus IP and omega x plus i p over to m bar omega.
355
00:42:24,670 --> 00:42:27,700
Now, when we write this out, we have the obvious terms.
356
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We have p squared and we have m omega x squared.
357
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So let's write those down. That's P squared plus m omega x squared all over two bar into m omega bar, whatever.
358
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And. And then we have some additional terms which would class that would cancel in classical algebra but don't now because we have an X,
359
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we have an M Omega X times IP and here we have an m omega x on the so we have an I minus IP times name and we are x.
360
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So the additional term is an m omega x.
361
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Am Omega I x comma p and it's a gain of a two and h bar omega.
362
00:43:31,010 --> 00:43:43,660
Right. So this is the Hamiltonian overreach for Omega and this is an H bar which and the I's make a minus one with this.
363
00:43:43,670 --> 00:43:48,910
So this is going to be minus a half and everything else will cancel.
364
00:43:51,120 --> 00:43:54,560
Right. Because we'll we've got an Omega here.
365
00:43:54,570 --> 00:44:01,649
We're going to get an H bar from there. So the rest cancels. So I should have I should have explained.
366
00:44:01,650 --> 00:44:07,170
Sorry, what is the factories, this and this on the bottom, this normalising factor on the bottom is put in.
367
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It's not really essential, but it's very convenient and it's put in in order to make this dimensionless.
368
00:44:14,100 --> 00:44:19,770
So just to check that that's true. H Bar has dimensions of position, times momentum.
369
00:44:20,730 --> 00:44:23,910
Right. So it has the dimensions of position, times momentum.
370
00:44:24,210 --> 00:44:33,570
So what we have here is m x sorry, m omega x which we've agreed has dimensions of momentum, times p which has dimensions of momentum.
371
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Then we take the square root.
372
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So this on the bottom has dimensions of the whole square root of momentum and therefore cancel the dimensions of what's on the top.
373
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So it's dimensionless. That's the purpose of the that's the purpose of the horrible square root.
374
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So we find that this product, which is dimensionless, is equal to the Hamiltonian divided by H Bar Omega,
375
00:44:56,760 --> 00:45:01,440
which has the dimensions of energy because H Bar also has the dimensions of energy,
376
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times, time, omega, of course, has dimensions of one over time it's a free is the frequency of the oscillator.
377
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So this is has dimensions of energy minus a half, which is obviously dimensionless.
378
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So we have indeed almost fact arised.
379
00:45:15,270 --> 00:45:24,720
We have the statement now that H can be written as H Bar Omega, which carries the dimensions time z dagger a plus a half.
380
00:45:25,800 --> 00:45:36,720
We've almost factorisation just this that there. The next thing we want to do is calculate the commentator a dagger, a comma, a yes.
381
00:45:36,720 --> 00:45:40,950
We just got time to do this, a dagger of these two operations.
382
00:45:41,910 --> 00:45:52,110
So of course we will have a one over two H Bar Omega as a factor on the bottom because each of these A's brings in its own square root.
383
00:45:52,830 --> 00:46:10,500
And then we will have the commentator of m omega x minus i p on m omega x plus i p now.
384
00:46:11,400 --> 00:46:16,170
We and we have this breaks down into four commentators in principal.
385
00:46:16,440 --> 00:46:23,639
There's the commentator of this with this and the commentator of this with this the commentator of this
386
00:46:23,640 --> 00:46:30,000
with this obviously vanishes because excuse with itself and the commentator of this with this is.
387
00:46:36,030 --> 00:46:44,579
So we're going to have an m omega AI times X comma p, that's the computation of this with this.
388
00:46:44,580 --> 00:46:50,520
And now we have to deal with these with these terms. This produces a non-negotiable commentator.
389
00:46:50,520 --> 00:46:57,270
With that, we're going to have minus M omega ai times p, comma x.
390
00:47:00,050 --> 00:47:02,960
And then we'll have the commentator repeat with self, which will vanish.
391
00:47:05,780 --> 00:47:13,070
If I swap those two over, then clearly I change the sign in front and then this becomes a plus x comma p.
392
00:47:13,250 --> 00:47:14,690
It becomes this thing all over.
393
00:47:14,960 --> 00:47:27,980
So that cancels this and this whole caboodle is going to equal i x comma p over h bar that because we're going to get a two.
394
00:47:27,980 --> 00:47:30,920
These two terms are going to add together to make us a two, which cancels with this.
395
00:47:31,220 --> 00:47:36,020
And the moment is clearly go x, come a p is itself equal to h bar.
396
00:47:36,200 --> 00:47:41,270
So the i's make a minus one, the bars cancel and this is equal to minus one.
397
00:47:41,930 --> 00:47:48,710
So these two operators have non vanishing comitato actually equal to minus one.
398
00:47:51,780 --> 00:47:56,190
Yeah. Well, we seem to still have time to to nail this this problem, I think.
399
00:47:56,580 --> 00:48:01,770
So let us suppose we have got a state of a stationary state.
400
00:48:07,590 --> 00:48:12,900
Let us now apply the operator a dagger to both sides of this equation.
401
00:48:13,080 --> 00:48:16,440
Right then, this is just an eigenvalue. It's only a number.
402
00:48:16,440 --> 00:48:19,470
So I can then write E a dagger.
403
00:48:20,310 --> 00:48:23,940
E is equal to a dagger.
404
00:48:24,660 --> 00:48:28,260
H e that's obvious.
405
00:48:29,820 --> 00:48:37,410
I would like to swap these over so I jolly well do. I say this is equal to AJ Dagger plus a dagger.
406
00:48:37,980 --> 00:48:50,660
Commentator H. So this commentator puts in what I'm supposed to have and takes away what I'm not supposed to have but have previously written down.
407
00:48:53,390 --> 00:49:00,620
But we know what age is in terms we have that age is equal to there it is age bar, omega eight Dagger.
408
00:49:00,740 --> 00:49:08,450
So let's use that. So this is H, a dagger plus commentator of dagger.
409
00:49:09,170 --> 00:49:14,810
And H turns out to be a dagger. A plus a half.
410
00:49:15,800 --> 00:49:19,520
Close brackets. H bar omega to carry the dimensions.
411
00:49:20,420 --> 00:49:24,590
Close that, close that and stick in our E that we first thought of.
412
00:49:27,130 --> 00:49:30,460
So all I have done is replace H by an expression we already derived.
413
00:49:32,080 --> 00:49:38,110
Yeah. Now I have to take the commentator of a dagger with this.
414
00:49:38,230 --> 00:49:44,740
And with this. The commentator of Dagen with a half clearly vanishes because a half is just a number, not an operator.
415
00:49:45,790 --> 00:49:49,089
The commentator of a dagger with itself then vanishes.
416
00:49:49,090 --> 00:49:54,610
So. So when we do the commentator with this product, there should in principle be two terms, but only one of them survives.
417
00:49:55,060 --> 00:50:00,540
And that term is. That term is this sticks.
418
00:50:01,390 --> 00:50:04,870
This stands idly by while the dagger works on that.
419
00:50:08,980 --> 00:50:12,310
And then I have an omega. Omega.
420
00:50:12,910 --> 00:50:16,240
Close brackets. Close brackets. He.
421
00:50:18,530 --> 00:50:22,640
But we just worked this thing out and found that it's minus one, right?
422
00:50:22,820 --> 00:50:26,840
A dagger turned out to be minus one. So this is equal to.
423
00:50:29,340 --> 00:50:40,110
H a dagger e minus H bar omega a dagger e.
424
00:50:41,070 --> 00:50:46,440
Just to remind you, remind us what we had on the left. What we had on the left was e a dagger.
425
00:50:46,450 --> 00:50:49,590
E is just a restatement of what's been at the top.
426
00:50:50,460 --> 00:51:02,880
So we take this a dagger E and we obviously join it on to that degree and we discover that h on a dagger e is equal to well, h working on this.
427
00:51:03,240 --> 00:51:15,210
The cap that you get by using a dagger on E is equal to E plus h bar omega of dagger working on E, what does this tell us?
428
00:51:15,630 --> 00:51:26,070
It tells us that we have out of a state which had energy e we have constructed a state by multiplying by a dagger which has energy e plus h by omega.
429
00:51:26,730 --> 00:51:42,180
So this means that a dagger E is equal to a constant normalising constant not discussed times e plus h bar omega, a new stationary state.
430
00:51:50,350 --> 00:51:55,360
This is an incredibly powerful result because it immediately follows that we have states,
431
00:51:55,480 --> 00:52:03,340
if we can find a state e we can immediately generate E plus h bar omega by using this a dagger based.
432
00:52:03,670 --> 00:52:11,890
And also if we use a dagger on this, it follows we're going to get E plus two H Bar Omega and we're going to get another.
433
00:52:12,040 --> 00:52:17,710
If we use a dagger on this, we're going to get E plus three H Bar Omega.
434
00:52:18,100 --> 00:52:27,040
So we're going to get a whole infinite series of states of ever increasing energy simply by applying a dagger again and again and again.
435
00:52:27,640 --> 00:52:34,540
So what remains is to find what e what number E is, and that we will do first thing tomorrow.