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Okay. Okay. Let's get going.
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So on Friday,
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we used the commutation relations that we had deduced for the angular momentum operators to find what the spectrum of J squared and any one of the JS,
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for example, Jay-Z could be.
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And now we can use this information to understand, to interpret the spectra,
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the infrared and the near infrared and terahertz spectra of diatomic molecules, which are which are very common right there in the room.
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Hero to an end to much of the mass of the is a barrels.
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The universe is contained in h two and a very important, from a practical point of view,
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a very important molecule for reasons I'll explain, turns out to be carbon monoxide.
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C0 So molecule like this consists for our purposes of two masses being the nuclei, for example, of the oxygen and the carbon.
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On a light springs we have two massive nuclei and a light spring formed by the electrons,
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and this is capable of being a simple harmonic oscillator in that it does this, but also at lower energies.
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It's capable just of rotating and of rimmed as a dumbbell and having rotational kinetic energy.
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So if you go to classical mechanics and ask yourself, what is the energy of this thing?
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The energy of this thing is given by the classical angular momentum around the X-axis squared over twice the moment of inertia around the x axis,
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plus the classical angular momentum squared around the Y axis over twice the moment of inertia around the y axis,
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plus the classical angular momentum squared over to I z.
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So this is reminiscent of this is a piece of classical physics that the energy of a chaotic energy moving particle is
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p squared over two in this is so this is classical mechanics of classical mechanics and those who did short options.
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Seven will I hope recognise this formula.
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So we have angular momentum x component here instead of x component of linear
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momentum divided by moment of inertia around the x axis and sort of divided by mass.
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So this is what the classical expression for the energy of a thing like this is.
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So what we say now, we conjecture that quantum mechanically we put this is a conjecture,
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we put a j x the classical thing goes to h bar times j x the operator.
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So we define these operators many people to find them to have dimensions of but we would define them so they would dimension less.
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This thing has the dimensions of h bar. So we make this transformation.
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We took, we take this classical expression and turn it into h by a times the quantum mechanical operator.
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ET cetera. And then we infer that the Hamiltonian should be what we guess should be h bar squared over to j x the operator squared over twice.
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Sorry over taking that out plus q y squared the operator over i y plus jay z squared over it.
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So these are operators now. So this is this you should think of this as a guess.
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It'll be confirmed by experiments to be to be shown soon.
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Now we say if for this diatomic molecule.
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We take. We take the X. We take the Z axis to be the symmetry axis of the molecule.
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That's the axis that runs through both nuclei. And then the moment of inertia around that axis,
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around that now z-axis is negligible because all the mass in this molecule is almost entirely contained in these nuclei, which are essentially point.
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They're very close to being point particles. And if you rotate around this axis, those nuclei have this provide very little moment of inertia.
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The electrons could provide a certain amount of moment of inertia.
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So it's possible for the electrons to have some. It's feasible for the electrons to have some momentum around this axis.
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And in some molecules they do. But in most molecules, they don't.
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So at the moment at the moment of inertia is, in any case, very small.
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So this implies that Z is very much less than I Y with an eye X, which is the moment of inertia about an axis that sticks out here,
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where the if you spin it around there, the, the two nuclei go around in a decent circle of radius, half the length of that bond say.
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And for a diatomic molecule by symmetry, x is going to equal a Y.
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So given that X is equal to a Y, it's natural to rewrite this Hamiltonian as H is equal to H bar squared over two brackets
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j squared which is g x squared plus two y squared plus jay z squared divided by x,
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which is equal to i y.
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So those first two I've combined together into here, I've now in here also got a jay z squared over x which I don't really want,
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so I simply get rid of this jay z squared brackets one over A-Z minus one over i y ix.
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So this is just a repackaging of that expression to group together the terms which have a common factor one over i x and now we.
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The beauty of this is that we know what the eigenvalues of this operator are and what the eigenvalues of this operator are.
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So we can immediately say what the eigenvalues of this operator are.
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So this this says immediately what the spectrum.
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Of H is e j m which is going to be h bar squared over two over two brackets.
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J j plus one. Because that remember,
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we showed the eigenvalues of this were j j plus one where j is allowed to be an
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integer or a half integer divided by x plus m squared one over i z minus one of i x.
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So we we have because we understand, because the Hamiltonian for this rotating system is simply a function of the angle meant by operators.
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And because we have already found out what the spectrum of the incremental operators is,
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we can immediately say what the spectrum of the Hamiltonian is, i.e. we now know what the allowed energy levels are of these rotating molecules.
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There's almost no further investment of effort. Now we know that.
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We know that M lies in the range minus j less than or equal to where less than or equal to J.
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That was one of the things we showed on Friday. So this number is sort of generically at the same order as this number.
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But this one over A-Z, I've explained is is a very much bigger number is it is very much smaller number than X.
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So this number is very much bigger than this number, which is the same as one over this number here.
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So the coefficient of m squared is huge. And that means that if you if m were anything other than zero, the energy would be enormous.
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So when you are dealing with when you're dealing with molecules in the room or in
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interstellar space or somewhere where they only have moderate amounts of energy,
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we can we so we can argue that this thing is going to be is going to be zero.
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So this this number here is very large.
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I mean, it really is huge because of that.
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What because that is so very small for a diatomic molecule, which means that this is always zero zero.
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Let's just say that these molecules have their angular momentum around an axis which is perpendicular to their symmetry axis.
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So that means that effectively. E the energy levels are simply e.j.
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H bar squared over two x times.
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J. J plus one. So those are the energy levels now.
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That's and that's that's all been totally generic for diatomic molecules.
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Let's not talk about carbon monoxide. Carbon monoxide has a much more interesting spectrum,
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much more easily observed electromagnetic spectrum because the carbon is slightly positive and the effect on the oxygen is slightly negative.
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I've lost this. Right.
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Oxygen has a great affinity for electrons. It borrows some of carbon's electrons and ends up being a little bit negative.
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This is a little bit positive. So a carbon monoxide molecule which is spinning and over end is a rotating dipole, electric dipole.
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So. So see, oh, this is plus.
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This is minus will be a rotating electric.
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Dipole. The same will not be true of a hydrogen molecule because obviously the two hydrogen atoms will have will have equal affinity for electrons.
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And so neither will be charged or an oxygen molecule or a nitrogen molecule.
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All of these molecules are not dipoles, so they can rotate without emitting electromagnetic radiation.
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But if CO is if a carbon monoxide molecule is rotating, it will emit or potentially absorb electromagnetic radiation at some frequency,
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which we would imagine would be the frequency at which it rotated and of range.
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Right. Classically, that would make sense. A rotating dipole should emit or absorb radiation at the frequency that it that it rolls over.
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Let's see whether that's true according to quantum mechanics. So.
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So it will emit or absorb.
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Am radiation. And the and get a frequency whether it's have the frequency.
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So the frequency is going to satisfy h new is equal to e j minus e j minus one.
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Sorry, another point. Photons carry one unit of angular momentum, it turns out.
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Right? Not more, not less.
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So a photon is can move a molecule which has a total increment from J to a state which has a total element of J minus one or j plus one.
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But it can't move it to J plus two or j plus three, or it can only move it by one unit in J because it's only got one use of angular momentum.
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So the angle and by conservation of the angular momentum of the photon plus the molecule.
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Right. So you can destroy the photon or create the photon, so you can add the angular momentum of the photon into the molecule,
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or you can dump into a photon one unit of angular momentum and hence make some step like this.
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But only adjacent JS can be moved to the key point by interaction with electromagnetic radiation.
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So we're going to have that h new the energy.
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The photon is equal to the difference in the energies of the molecule between the the excited state he started in the state he finished.
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And what's that going to be according to this formula here that's going to be h bar squared over
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to z i x sorry brackets j j plus one minus the same expression with j put equal to j minus one,
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so it becomes j minus one times j And I think you can see that this cleans up very nicely to simply two.
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J So we end up with h bar times j over.
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I think somebody must have a have a call.
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What does that tell us about new. It tells that new.
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Maybe we should get a subscript J to indicate that it's the frequency which will be emitted as it goes from a state j to J minus one.
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We can cancel one of the H bars. Well, this is an H on an H bar.
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Right? So this becomes equal to H bar J over to pi x.
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So that's the according to quantum mechanics. That's the frequency of emitted light.
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Can we reconcile this with with our classical picture?
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So how fast was the molecule? Was the molecule going around?
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So classically. We can say that the angular momentum j x is equal to o x times omega.
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This is the angular frequency. Of of the tumbling of the rotation.
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So this is a piece of classical mechanics. Again, the relationship between angular momentum and frequency.
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This is like the relationship between momentum. So this this mirrors the the classical statement that X is equal to m times x dot right.
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Omega is x dot rate of change of angle. I mean, it's the analogue of x dot.
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This is the rate of change of position. That's the rate of change of angle.
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Well, I've already said that moment of inertia plays a role, a bit like mass, and this momentum plays the role of.
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Right, so this that's that's where this thing comes from. It's just a piece of classical physics.
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So when you so when you are in a state so what do we think that omega should be?
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Omega should be on the order of well, it should be equal to j x over i x that's classical doing it quantum mechanically.
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This is going to be h bar times the square root of j j plus one.
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This is this is so this is a horrific sort of thing. We're saying that it's supposing it's angular momentum is along the x axis, right?
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Then J X squared is going to be on the order of h bar squared j j plus 1jx squared is going to be on the order of j squared,
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which will be this divide through by i x.
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And we want to relate this to the frequency we had up there, the factor of two pi.
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I dead wrong. Oh, it's because this is this is the angular frequency.
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That's the actual frequency. Right? So this is equal to two pi frequency of this is this was angular frequency.
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This would be frequency of rotation. So we have the frequency of rotation.
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We expect to be h bar over two pi x times the square root of j j plus one.
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So this is going to be. So what do we what can we say?
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This is slightly bigger. This is greater than just a bit.
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Just if J is a larger number, it's only a smidgen bigger than H bar j over two pi by x so the in its upper
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state when it had angular momentum total increment m j j plus one h bar squared,
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the rate at which it was tumbling was a bit larger.
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This is the frequency of emitted light. Sorry, this is new j from above.
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So in its upper state, its rate of tumbling on this classical picture will be slightly larger than the frequency at which it emits the light.
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And you can check that in its lowest state. We would have a minus sign here, a minus one.
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So we're just a bit below the frequency in which the light. So it.
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So the light is in fact emitted just the average of the expected rotation,
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tumbling rates at the upper and lower levels, which makes it, I think, perfectly good physical sense.
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So so in lower state. That's when j well.
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E J Minus one tumbling rate.
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It's just a bit lower. Oh, no.
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This thing is going to sleep again. It's not my computer that's gone to sleep.
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It's the wretched lecture room that's gone to sleep.
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It's too irritating. We must get them to stop doing this.
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We do want to show you something today. Is it coming back?
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Okay. Right. Yeah.
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So this is this is the actual experimental spectrum of carbon monoxide.
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And so this is this is in gigahertz.
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So this is a terahertz two, terahertz three terahertz along there.
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And you you have a line here.
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There's a line being drawn at the frequency, measured associated with the transitions from j equals one to j equals nothing.
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Then you have so you get photons out with this frequency.
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You get photons out at this frequency, which is almost but not quite twice the previous frequency.
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According to this mathematics that we've done so far, the next frequency up should be exactly twice.
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This frequency here should be exactly twice that frequency. Then you can't tell the difference actually on this plot and so on and so forth.
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So you get each line here is associate is telling you the measured frequency of a line, a spectral line from of from these carbon monoxide molecules.
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And you can see that they do form a regular grid just like this says.
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So this is the transitions from from one to nothing.
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This is the transitions from 1 to 2. Sorry, from 2 to 1.
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This is 3 to 2 and so on and so forth. And as we go up here, this is I don't know, can't count 10 to 9 or something, right?
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There's a there are a couple of missing lines in here where people where I wasn't able to find what the mean.
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Somebody hasn't published a measurement or something. The other interesting thing to note,
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so so that's that even spacing is confirming this frequency new J is proportional to J business and that's the interpretation.
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There's another interesting thing to notice here, which is that as you go along here,
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the the black lines get slightly to the left of the dotted lines,
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and the dotted lines are exactly at multiples of the frequency of this lowest transition.
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J equals one to nothing. So what's happening is that the measured frequencies almost conform to this, to this rule up here, but not quite.
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They, they, they turn out to be slightly smaller than the numbers new j given up there.
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And the physical interpretation of that is interesting it's that okay so new measured minus new j is slightly negative if you like.
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And that's because new measured is equal to these what you would think classically is equal to edge bar over two pi j over i x,
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which itself is a function of j right?
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As these molecules you make this molecule show you firing in circular polarised photons and making this molecule spin faster and faster and over end.
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Obviously the centrifugal force will stretch that spring out a bit.
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The spring is stiff, but not infinitely stiff. So when you're spinning it faster and faster, the centrifugal force pulls the spring longer,
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increases the distance between the nuclei, and in that way increases the moment of inertia.
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So this moment of inertia that's appearing on the bottom of this formula should really be itself a function of j, which increases.
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So some of this increase here is is sohc is cancelled by a slight increase in the bottom.
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And that's why that's the interpretation, anyway, of these spectral lines falling behind the measured numbers,
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which are the black lines falling behind the perfectly evenly spaced grid of dotted lines.
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And one of the so one of the problems on the problem set is to use this phenomenon to estimate the stretching and how stiff this spring is,
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because you can calculate using classical physics, you can calculate what the what this force is,
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the centrifugal force, you know, how fast the molecules are going around.
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So you can calculate what V squared overall is, which is the force pulling the spring from the change in the moment of inertia.
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You can I mean, from this change in the spectral line frequency, you can estimate this change in a moment of inertia.
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So you have a given force, a given displacement, so you can work out the spring constant and then you can check.
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Then you make a prediction for what the frequency is at which the carbon and
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the oxygen would oscillate to n and towards each other using the spring rate,
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which appears in a different part of the spectrum. And so that's what that problem is about.
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So I think that's all. There are other nice things you can do with carbon monoxide molecules,
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but I think we'll leave it at that because we have an important additional item to put on the agenda, which is orbital incremental.
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Maybe we should put it over here. Okay.
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So classically we know what angular momentum is for a particle classically.
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Well, let's be careful. So the world, the earth has angular momentum about the centre of the sun.
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For two reasons. One is that it's moving around the sun once every year, and that contains that motion contains a great deal of angular momentum.
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And the other is it's spinning on its own axis, which accounts for a slightly smaller amount of angular momentum.
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But its langland mentum and orbital dynamics are very much involved with the
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interchange of angular momentum between orbital motion and spin motion and so on.
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It's the same quantity, it's exactly the same on atomic scales or whatever.
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And as I've said, electrons and protons and neutrons are all gyros that like the Earth,
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they spin on their own axis, but of course they also move around.
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And in moving around they have angular momentum. So we in the end of last term,
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we introduce these angular momentum operators by considering by very general considerations to do with
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what happened when we generated the operators that would generate new states that were like the set,
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which gave us a system just like our old system, except rotated through some angle.
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But and that apparently has no connection with angular momentum as classically conceived or so.
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And in classical physics, we, we have a thing orbital angular momentum, which is going to be given by L is equal to X across P.
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So this is classical physics. This is all this is classical orbital.
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So this describes the angular momentum of the earth about the centre of the sun by virtue of the motion of the earth around the centre of the sun.
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Where P is the momentum of the sun in a frame of reference in which the sun is stationary?
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Right. So we can define an operator. So we define.
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Let's go down here. So quantum in quantum mechanics, by analogy with this,
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it's of it's natural to define an operator l hat which is equal to one over h bar x hat cross P hat.
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That's right. That out in components to make sure we know what we're doing.
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That's one over bar the some epsilon ijkxj6.
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Oh, sorry. No. Which way around do I want to do this?
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Doesn't actually matter how I do do it, but I for consistency, I should try and keep the same as we have to.
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Yeah. So that's consistent with what was on the book. So this is some develop over J and K.
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So things to notice. One is we put a one over H bar in here to make this thing dimensionless.
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It's a close call whether you should make it dimensionless or not. Most people probably don't have it dimensionless.
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I think on balance, you're better off having dimension as we introduce the angle momentum operators in such a way that they were dimensionless,
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didn't have an undesirable, and they didn't have an edge bar which made our formulae simpler.
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So for consistency we need to make these dimensionless, right? So this has dimensions of this and therefore this ratio is dimensionless.
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Does this make sense? Do we have to worry about the order in which we write down X and P in classical physics we clearly don't.
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In quantum mechanics, you would think you would have to worry about the order.
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Does it matter whether this is XP or X?
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Well, it doesn't matter, because in this sum, the only terms which occur when the subscript on the x is different from the subscript on on the P,
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because this epsilon symbol, remember, vanishes if any two of its subscripts are the same.
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So for example,
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we have an x is equal to is equal to one over h bar the sum over j and k of epsilon i j f excellent x j k x j p k so j is sums goes over x,
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y and z, but when it's x you get nothing here.
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So there are only two classes to consider when it's when it's Y and when it's Z,
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and when it's why we don't need to consider the possibilities that K are either X or Y, because this will vanish.
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If it's if they're either if K is either X or y. So this is one over x bar of x, sorry, x y.
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Which is why has p z minus minus z had p y.
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So that's how it'll work out. And these two operators commute.
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These two operators commute. So it doesn't matter about the order. The order is not important.
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Alex is also going to be is going to be a mission operator.
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Because if we take the dagger of this, right, the dagger of this equation, we will have the dagger of what?
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Well, let's do it. Alex Dagger is going to be one over a bar.
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The rule is when you take the commissioner joint, you have to reverse the order of the operator.
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So it's going to be p dagger z y dagger minus p y dagger, z dagger.
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Well, these things are rule operators, but but each of these things is its own dagger because the momenta and coordinates,
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the permission operators and the order in which you write them down, we've already agreed, is unimportant.
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So this is, in fact equal to Alex. So it's a commission operator.
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So we expect it to be associated with an observable. And the observable is obviously going to be orbital angle momentum that down the next thing.
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Oh, yes. One more thing we introduce by analogy with angular momentum, we introduce a new operator,
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L squared, which by definition is l squared plus l y squared plus L.Z. squared.
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Again, it will be emission because this it's Alex's emission.
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So L x squared is emission and so on. Right? So this is another emission operator.
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The next thing to do is to work out some commutation relations.
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Now that we define these operators, find what computation relations we have and let's do l i x l well,
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mind if I leave off these hats now with thoroughly stuck into quantum mechanical operators?
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So. So watch this. COMMENTATOR Well, we should write in what this is.
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This is epsilon i j k x j p k as it will be summed over j and k.
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So this is this commentator on x l So this is the commentator of an operator with a product.
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So in principle, there are two terms. There's is this thing stands idly by wells that competes with that.
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And then there's this thing stands idly by in the back where this computes with this.
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But obviously the X is all commute, so forget that. So we only have to consider this.
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Commentator So this is equal to the sum epsilon i j oops j k of x j standing idly by while we do the
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commentator peak x l But this is minus h bar delta k l so we have minus I sorry I'm missing here.
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One on h bar am I not because l I is one upon bar of this product.
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So we have a one on h bar here. Then this generates a minus h bar.
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The bars cancel the minus. I sticks around in this pain times epsilon i j k x j delta k l.
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This is still summed over j and k.
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When we sum over K, this becomes an L, so this becomes a minus.
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I summed over j of epsilon i.
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J l. X j.
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And we can get rid of this unattractive minus sign by swapping the order of these two.
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Right. So we can write this as plus i some over j of epsilon i l j excel.
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So let me just write in the left hand side again so we can see appreciate the pattern that we're getting.
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The commentator l i with excel because i times epsilon i l these two letters being those two letters.
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Sorry. This was some of j j some of j.
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So between here and here, I've merely reversed the order of the subscripts on the epsilon picking up of of dealing with the minus sign.
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Now, this is exactly the same as a result we already had.
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So we need to recall at this point that j i comma x l is equal to i sum of a j epsilon i.
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L j x j.
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So the commentator of this orbital angle mentioned operator with this position operator is the
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same as the coming iteration of this total angular momentum operator with the position operator.
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Similarly, just the same calculation imprecisely precisely analogous calculation implies that l i comma p l we just sit down and calculate this in
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exactly this way we will find it's i some over j of epsilon i l j p j which mirrors and there's an analogous relationship between j and P.
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The next thing to calculate is. So what's l i comma l j.
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We've introduced a family of three operators.
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We should investigate what the commutation relations are between any two of them so we know what the answer is.
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In the case of the angle momentum operators g j j is i epsilon i j k j k.
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So this is setting us up for expecting what the answer is here.
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Okay. But in order not to get so so it's an exercise that I would encourage you to do to work this out just as it stands.
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But I'm not going to do a simple calculation, just one component, so I'm going to do so.
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There's a question mark associated with this. It's a good exercise to do that, but it's slightly complicated to do it in the general case.
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So let's do something slightly simpler. Let's work out what x, comma, l y is.
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So what do we do? What we do is re replace one of these, shall we say, this one by its expansion in terms of X and P.
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So this is going to be one over bar of l x, comma, l y.
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So what is l y are well l y must be I think z p x minus x p z.
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All right. So this product divided by age bar is that if I'm if I've not got my signs wrong and we know
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now how Alex commits with this and how Alex commits with that so we can work it all out.
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So this is going to be one on edge bar open,
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a big bracket of X committed with Z with standing idly by plus Z standing idly by Alex commuting with x minus l x commuting with x and p,
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z standing idly by minus x standing idly by while Alex works on p z.
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The easy terms. Here are this and this.
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Because they're what? Zero, right?
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Because this would be epsilon x ei, epsilon x x k that vanishes similarly here.
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So these two are nice and equal to zero.
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And these we have to work out using horrible cyclical things.
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So this is going to be minus.
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So Alex, communities with a component of X is going to produce high times.
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The third component which will be Y. And I think we probably get a minus sign.
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So I think this will be one over bar brackets and there'll be an I sorry.
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And I then I think we'll have a well maybe I should give it a minus.
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I times Y from here, times X and this was zero.
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This was zero. This is the other interesting thing. One, this is the other interesting one.
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It'll be the same thing this computed with this will be minus i p y.
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So we have two minuses. So we'll have plus i x p y.
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So this is equal to i overage bar of x y minus y he x which this h bar and this stuff together make l z so it's i l z.
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So in summary, what we what we this again mirrors a result we saw with the total agreement of operators.
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Just so this is one component of again.
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Now write down the answer for that calculation up there without proving it.
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But you can see that it's going to happen. L i comma l j is I summed over k of epsilon i.
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J k. L k. So let me repeat a result we've already got l i x j is equal to i epsilon i j k x k so l
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computed with any components of a vector produces i times the third component of the vector,
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and that rule even works for l itself, which is itself a vector.
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All right. So this is we regard this as analogous to this relationship here because we also had sorry,
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we also had that l i comma PJ is equal to i epsilon i j k p.
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Okay. The point is that what goes in here for l to work on can be any component, any vector, a component of any vector.
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And then you always get out times the other component of the vector you put in here.
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So here also we get out times the other components of the vector we put in here, which in this case will itself and this is mirroring precisely.
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So these results exactly the same. If you replace all those L's with JS, everything remains true.