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So that's a summary of the important formulae we obtained yesterday.
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We'd, uh, Wednesday we had reduced hydrogen's problem to a one dimensional Hamiltonian one h
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sub l one for every particular value of the total angular momentum quantum number.
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L And we found that this thing at the top here was a ladder operator in the sense that it's out of a state of a certain amount of energy,
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in a certain amount of angular momentum. It constructed a state with the same energy and more angular momentum.
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And using that and the idea that the sequence of states of more angry momentum with the same energy had to stop.
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We concluded that the energy was given by a certain constant 13.6 V divided by end squared,
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where n is one more than the maximum angular momentum that you can afford at that energy.
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Put another way, the Anglo momentum quantum number is less than or equal to one minus.
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This number n controls the energy and is called the principle quantum number.
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So what we want to do now is move forward to get the energy eigen functions so as to get the.
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These are of interest from the perspective of hydrogen.
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If you want to do any detailed calculations, like how does hydrogen interact with the electromagnetic field?
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What happens if you scatter electrons of hydrogen and that kind of stuff?
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You'll need to know what these eigen functions, these wave functions are,
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but they're also the building blocks for atomic for studies of atomic structure generally.
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So there are a complete set of states which you can expand any state.
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For example, the stationary state of an oxygen atom. You can expand in these states.
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And that's what people do when they do atomic physics calculations.
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For the most part, what they do. So these these play a very big role in atomic physics.
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Okay. So we modify what they are. Now, from this from the fact that l squared on e l is equal to l l plus 1el from the fact that this thing,
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this state here, well defined in a stationary state, is an eigen function of the total angular momentum operator we know all about the so the wave
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function in question is this right the amplitude to find at X the the atom when it's in this state.
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The reduced particles, strictly speaking, when it's in this state. We know all about the dependence, the anger, the dependence of this.
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We know that this is takes the form of some function, which presumably depends on the energy and on the angular momentum.
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Times are our times. Why l am a feature in PHI?
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Because we know that these the spherical harmonics are the unique angular eigen functions of this operator with this eigenvalue.
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So this thing, which we know is such an eigen function, its angular dependence must be given by, strictly speaking, a sum.
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Potentially, it's a sum of these four different values of M and the same value of L,
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but it's sensible to look for them one by one, assuming a particular value.
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And so the angular dependence, you know, out of out of that statement up there is this.
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And all we're looking for right now is the radial dependence on the radio dependence.
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Uh, as I say, you have to expect it to depend on the energy.
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I mean, it has to depend on what's in here, which is the energy and the.
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The angle momentum, a quantum number. So that's what we're that's what we're looking for.
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And we're going to get it just as we got the wave functions for the harmonic oscillator
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stationary states by studying the equation that the the equation that this operator,
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your letter operator kills the thing off in the appropriate circumstances.
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So what we do is we look at the equation which says that a l well a l for the maximum value of l so we're not allowed to make.
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The maximum value of this is N minus one. So we use a N minus one on E and minus one.
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We get nothing. It's the end of the line. So we look at this equation in the position representation.
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And so. We can forget about that a zero over two, because it's just a constant.
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So what we want is we're going to be looking at IPR over each bar minus nine, putting L equal to and minus one.
44
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So that L plus one, I suppose is going to be an end.
45
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Overall. Plus he said, over.
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And a zero that's operating on you and and minus one must give you nothing.
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Now we we we know that PR is in the position representation.
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We figured it out. It was minus h bar with deep idea plus one overall.
49
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So we put that fact into this equation.
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We have the I and the minus sign make a one the bars cancel.
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So we're looking at D by the plus one overall minus this.
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So we're going to have minus N minus one overall plus Z over and a zero.
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You and n minus one is nothing.
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Now, this is a premiums equation, right?
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It's a first order linear differential equation.
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So it has an integrating factor. E to the integral of with respect to all of the set of this stuff, right.
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This just from prelims maths. This, of course on integrating just gives me an R on top.
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This gives me a logo. And so we're looking at E to the minus some multiple of log R.
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So that means that this is going to become R to the minus n minus one from the E to the minus n,
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minus one log r times the exponential that we get from the E to the z r over and a zero.
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So now we know pretty much what's going on because the original differential equation,
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remember, says that D by the R the integrating factor times u and minus one equals nought.
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In other words, this thing is equal to a constant. In other words, the wave function that we're seeking is is a constant over the integrating factor.
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So we have the U and and minus one is some constant which must be determined by
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the normalisation times R to the end minus one E to the minus Z and a zero.
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So that's wonderfully simple expression. So let's let's ask ourselves a few things.
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Let's have a look at the ground state of hydrogen.
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So this is the case and equals one. How do we know that?
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Because Elle has to be less than or equal to N minus one.
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L Is the numbers the possible candidates for Ellen nought, one, two, blah, etc.
71
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So when has to start at one and then go up to three, etc.
72
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So the ground state, the state with the least energy is n equals one.
73
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So what is the wave function? So you won.
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And of course there should be a zero. Here is going to be a constant E to the minus in hydrogens.
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That is one. So this is this is just going to be r over a zero.
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So the ground state wave function is this mere exponential, a beautifully simple result.
77
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What else was I going to say about this? Yeah.
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One interesting given that that beautiful exponential one thing you notice is that this thing is never zero.
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It it's the ground state wave function has nonzero modulus all the way to our equals infinity.
80
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Although the particle is classically forbidden to go beyond a certain radius.
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And in fact, so what this does was graph up here plots is the probability of finding the
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reduced particle at a radius are measured in units of a zero of a over of a z.
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So a radius bigger than this. And the classically forbidden region stops at that number two.
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And it turns out there's a 20 there's a 24% probability that you'll find the reduced particle in the region that's classically forbidden,
85
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where the kinetic energy as well would be negative. Right.
86
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So if you go beyond R is to the potential energy is more than the total energy of the particle.
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So there's less than nothing left the kinetic energy and there's a very significant probability of finding the particle that far out.
88
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So so that's, I think, an entertaining result. It says that there's the p forbidden.
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Looks forbidden. It's about 24%.
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Let's get the normalisation of this thing sorted out so we can work out expectation.
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A few expectation values working out in the normalisation is is fundamentally very straightforward.
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What we require of course. So we require that the integral d cubed x over all space of the complete wave
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function which consists of u and n minus one of radius y l m should equal one.
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This thing we right of course, is the r r squared and then the integral sine theta dct defined to omega over the sphere.
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Oops. Sorry should mod squirted this right.
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That's our requirement when we integrate of angles over the sphere where we're integrating while m mod squared over the sphere and that comes to one.
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So we're left staring at an equation which says that the r times r squared times times u this thing squared.
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Now what did we say that was? That was c squared.
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Well, we don't need to moduli make it a modulus. We can declare it to be real times r to the two and minus one e to the minus two.
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Z over and a zero.
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So this should be one. So the only thing to notice here is that there's an additional fact ask where you don't just take the the radial wave function,
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square it and integrate the R to get one.
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You have an additional factor, R-squared, because fundamentally this is the normalisation condition we wish to impose and the Y lens and normalised.
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So integrating every sphere we get ones we've used here that the integral de to omega y l m squared is one that probably normalised.
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Now this integral is is nice and easy.
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So this can be written a C squared.
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And what we need to do is to bring an integral like this under control is to declare that this is is ro.
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So we introduce a new variable, ROE, which is to Z to Z of and a zero.
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And I want to make these. So what I have here is an R to the to N, and I want to make all of those R's into rows,
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which means I have to multiply by a load of factors and a zero of A to Z I've got or to the two N from these two and another one from there.
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00:13:31,970 --> 00:13:39,500
So it's two N plus one and then I can turn all these into rows d row, row two, the two in.
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Well, that's the definition o missing the E to the minus row.
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Now, this is a famous integral in mathematics.
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Which is it? So it's often called gamma of two N plus one.
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I believe Euler's responsible for that absurdity. But what it what it should be thought of as is two N factorial.
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So this integral if if is simply the this exponent up here factorial.
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Is the proof and, and you want to be able to recognise that because one often encounters that pattern and you want to just be able to say,
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Aha, that's two factorial.
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So this tells me what C is, C is equal to two Z over and a zero to the end plus a half, one over the square root of two and factorial,
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which enables me to write down the the relevant way function u and minus one of radius.
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I need this factor here and I'm going to write it as follows.
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I'm going to say this is two Z over and a zero to the three halves power.
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So I'm borrowing from that three halves power one over the square root of two and factorial.
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That's too. I have to be clear.
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So I need to bracket that to make sure it's the whole to end the gets factorial ized and then for the rest these other factors.
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So. The rest of the factors in here can be put together with those odds to make this row to the end minus one each of the minus row.
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Over to. That's the row is defined as to Z.
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R. Etc. So. So. The factors.
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Left over from here. Just what we need to bring make that which was an artsy and minus one into a rose at the end.
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Minus one. So what does this.
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Physically. What is this? We are looking at the states with the highest angular momentum for a given.
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For given energy. So these are the quantum mechanical analogues of circular orbits, not eccentric orbits, but circular orbits.
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So what do we expect qualitatively?
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Well, we expect classically, if it was a circular orbit, our probability would be a delta function of the radius of a circular orbit.
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And we know in quantum mechanics everything's a bit blurry because steep gradients of the wave function are associated with large kinetic energies.
136
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So we're expecting it to be sort of like this ish.
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So what's how is that how does that arise from this formula?
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Well, when R is nought, row nought, this is going to be zero and then it's going to shift itself off zero.
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Slower and slower, the bigger N is. So if end is ten to the 30 or whatever it would be for a classical classical particle,
140
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then this would rise ever so slowly from zero and we'd have it would hug the origin for a long time.
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It would then rise and then when it when it when Rho became on the order of one this
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exponential which previously been harmless being ease the mind is something small,
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would become a vicious cutting off thing. And that's how we get cut off on this side here.
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So here we're looking at row two, the end minus one growth.
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And here we're looking at E to the minus row over to.
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Well, if this is the probability, then we need to multiply by we need to square up.
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00:17:39,880 --> 00:17:47,860
Right. And this is an exponential decline. So so precisely what it looks like is given in the next diagram.
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So the top picture there shows just the first three.
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So the pure exponential is end equals zero.
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The ground state, then the one that rises steeply at the origin and falls off after an early peak is n equals two and n equals three.
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Is, is, is the next one. And what you can see is that the characteristic radius is moving outwards.
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Quickly. So let's calculate some let's calculate some expectation values, because that's now easy to do.
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00:18:27,390 --> 00:18:32,400
Let's work out the expectation value of the radius, right? So if we want to make a connection back to classical physics,
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we should be thinking about expectation values because classical physics is the physics of expectation values.
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So this is easy to work out. It's going to be the integral d r r squared times r times you and minus one.
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Squared, right? That's what it should be. And now that we've got this normalisation, everything sorted, we can evaluate this.
157
00:19:04,040 --> 00:19:15,140
So we're going to have. Yeah.
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00:19:15,150 --> 00:19:24,780
Well, actually. Let's just go back to which is the best way to do this.
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All right. Let's turn this all into Rose. Let's turn these all into Rose now.
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So this we've already got more or less as a function of Roe. So what we need to do is to.
161
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Is to deal with these other ones. So there are four powers of all there are, and I need to turn those into rose,
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which means I need I need an and a nought over two z race to the fourth power.
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Then I need to write down this thing mod squared which is which is c squared which is two z over and a zero raise to the uh,
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so that's two and +11 over two and factorial as the rest of that is the rest of C squared, then we need the integral D row row cubed.
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So these three and then here we have row and I need to square that.
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So it's going to be n to n minus two.
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Easy. Linus Rowe. So what's this going to be?
168
00:20:52,380 --> 00:20:58,560
This is going to be two and this will be row two, the two and plus one.
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Each of the minus row. So this integral is going to be two and plus one factorial.
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And on the bottom I've got two n factorial. So this on the top and that on the bottom gives me simply a2n plus one.
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Everything else cancels in the editorial. And here something has gone wrong in that I've got far too many powers.
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And what have I done wrong? What are we doing wrong? Uh.
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Sorry. I got confused. Which one I was doing. Excuse me? I was using this formula here, which meant that the powers that I needed here.
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Right. I was using this formula for you.
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These require this. These powers. Now I'm using this formula.
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So it's two the three halves power to the three because I've squared it up.
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Exactly. So we end up with these three of these cancel.
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So at the end of the day, I'm going to have an eight nought over two Z.
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Just one of them. And we're going to have what we said was this was two and plus one.
180
00:22:10,440 --> 00:22:19,880
In other words, we're going to have four. If I put that two inside there, we're going to have n n plus a half of a nought over Z.
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So the expectation value of the radius is going sort of like N squared and it's going like the scale radius we defined for hydrogen divided by Z,
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which tells you that if you increase the nuclear charge, the size of the orbit's going to shrink like one over the nuclear charge.
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So the interesting fact here is that the expectation value of all is sort of like going like an
184
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squared and that's exactly what we expect because e remember goes like minus one over and squared.
185
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So therefore it's going like minus one over expectation value of R.
186
00:23:01,360 --> 00:23:06,620
But where? But we have a particle moving in a Coulomb field, so the potential energy goes like one overall.
187
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And from the visual theorem, we're expecting the potential energy to be minus twice the kinetic energy.
188
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So the total energy should be sort of a half, minus a half of the potential energy.
189
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So this is exactly what we're expecting. So that's that's a recovery of sort of classical ish stuff.
190
00:23:27,760 --> 00:23:37,750
Interesting fact here is because this grows like this, it means the volume occupied by the atom is going like.
191
00:23:38,530 --> 00:23:45,190
Which obviously goes like the expectation value of all cubed, which goes like end of the six power is growing very rapidly within.
192
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So states so so this so this grows very rapidly.
193
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This means that states in which you excite the electron to a large value of n cannot be seen.
194
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You won't be able to observe these, to measure these unless you're in an incredibly high vacuum.
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So. So. E.g. if ns a hundred the volume is going to be ten to the 12 times a regular atomic volume.
196
00:24:29,560 --> 00:24:36,760
And in interstellar space you can see hydrogen atoms transitioning from any is 100 to
197
00:24:36,760 --> 00:24:41,830
an equals 99 and stuff like that rate by making measurements of radio wavelengths,
198
00:24:42,790 --> 00:24:47,290
centimetre wavelengths. But you can't do that kind of thing laboratory because you can't get high enough vacuum.
199
00:24:50,980 --> 00:24:59,020
So in the brochure on earth we are restricted to relatively small values of n n less than ten typically.
200
00:25:00,100 --> 00:25:06,610
Right. What else can we say? What about? What about? It's interesting to work at the expectation value of R squared.
201
00:25:07,240 --> 00:25:10,840
It's essentially identical performance to what we've just done.
202
00:25:10,870 --> 00:25:14,490
I mean, all we have is an extra R and that integral at the top there.
203
00:25:14,500 --> 00:25:18,610
Right. So what are we going to have if we come down here, we will have an N,
204
00:25:19,060 --> 00:25:27,790
a nought over two Z raised to the fifth power this time because we're going to have an extra power of oh before the use starts.
205
00:25:28,240 --> 00:25:35,350
Then we will have two Z over and a nought to the third power coming from the you.
206
00:25:36,130 --> 00:25:41,890
Then we will have to n plus one sorry to n factorial coming from the c and then
207
00:25:41,890 --> 00:25:46,020
we will have to do the integral d row and we'll have an extra power of row.
208
00:25:46,030 --> 00:25:52,660
So it'll be row two the fourth times that row to the to n minus two.
209
00:25:52,690 --> 00:25:56,830
So we'll end up with row two, the two N plus two, each of the minus row.
210
00:25:56,860 --> 00:26:03,310
In other words, this is going to be two and plus two factorial.
211
00:26:05,530 --> 00:26:11,800
Right? So now we're taking two N plus two factorial, two N plus one factorial dividing by two N factorial.
212
00:26:12,040 --> 00:26:14,850
So this is and we're going to get an extra power here.
213
00:26:14,860 --> 00:26:25,350
So this is going to be n squared coming from here because because this fifth power we reduce to the second power by when we multiply this one.
214
00:26:25,360 --> 00:26:38,110
So we left with n squared a zero over two Z squared and then we will have two N plus two, two N plus one.
215
00:26:39,220 --> 00:26:46,390
It's interesting to express that as a multiple of as a multiple of of the expectation value of R,
216
00:26:46,660 --> 00:26:52,030
which we've already derived as being an N plus a half a0 of a Z.
217
00:26:52,030 --> 00:26:57,430
So A0 of Z is essentially so expectation value of R.
218
00:26:57,790 --> 00:27:03,400
So this is going to be this these twos I can take out, there are two twos in here.
219
00:27:03,730 --> 00:27:19,090
I take them out and use them to clean that up. So this is going to be n squared, n plus one n plus a half of a zero over z squared,
220
00:27:19,360 --> 00:27:29,620
which itself is the expectation value of R squared over N squared and plus a half squared.
221
00:27:31,360 --> 00:27:42,040
So we can cancel many things and we find that that's n plus one over n plus a half of the expectation value of R squared.
222
00:27:43,060 --> 00:27:51,460
So what does that mean? That means that the uncertainty. Well, so, so what's the what's the arms variation.
223
00:27:55,660 --> 00:27:58,280
And Oh, now you think this thing would go to zero, right?
224
00:27:58,280 --> 00:28:04,160
Because what we're doing is we're looking at the quantum mechanical analogue of a circular orbit.
225
00:28:04,580 --> 00:28:12,180
Circular orbit. The particle does not move in and out. So we would expect that this arms variation in all went to zero,
226
00:28:12,200 --> 00:28:17,150
as in went to two large values and we would have thought we would recover classical physics.
227
00:28:17,540 --> 00:28:27,950
We'll see that that's not the case because what is this are mass variation where it's all squared expectation value minus R expectation squared.
228
00:28:28,490 --> 00:28:34,069
Take the square root of that. Okay, so here is the expectation value hit.
229
00:28:34,070 --> 00:28:43,640
Let me write it down again. R squared expectation. So what I want to do is from this I want to take R squared and then take the square root.
230
00:28:43,970 --> 00:28:55,010
So this is equal to the square root of n plus one over n plus a half minus one expectation value of R.
231
00:28:57,500 --> 00:29:07,490
So you can easily see that this is going to come to something like the square root of a half over n plus a half of the expectation value of all.
232
00:29:09,490 --> 00:29:15,340
So what's happening is that the arms variation in the radius is becoming small
233
00:29:15,340 --> 00:29:18,909
with respect to the radius relative to the expectation value of the radius.
234
00:29:18,910 --> 00:29:22,930
But. But jolly slowly. Right. That's on the order of expectation.
235
00:29:22,930 --> 00:29:26,530
Value of are divided by root root n.
236
00:29:28,030 --> 00:29:31,720
So it's becoming small relative to the radius itself, but only slowly.
237
00:29:32,200 --> 00:29:39,460
But it's absolutely large, right? This because this thing is growing like an squared.
238
00:29:40,540 --> 00:29:48,609
This is looking like ends of the three halves power. And I think you can just about see that in those pictures up there that as you as you well,
239
00:29:48,610 --> 00:29:52,870
I've only shown the first three, but you can't see the peak becoming narrow.
240
00:29:52,870 --> 00:30:03,910
It doesn't become narrow. So that's a remarkable result. Okay.
241
00:30:03,920 --> 00:30:09,020
So, so those those are the wave functions for the essentially circular orbits.
242
00:30:09,410 --> 00:30:16,130
What about the non-sequitur orbits? As we see, they're not very circular, but, you know, that's the best we can do.
243
00:30:21,370 --> 00:30:25,239
So how do we expect to get these way functions for non circular orbits?
244
00:30:25,240 --> 00:30:34,330
Well, in the case of the simple harmonic oscillator, we found the ground state wave function by solving a on ground state wave function equals zero.
245
00:30:34,330 --> 00:30:38,960
And that's essentially what we've just done. And then we found the excited state wave functions by,
246
00:30:38,980 --> 00:30:45,580
by taking that wave function we first found and multiplying it by a dagger an appropriate number of times.
247
00:30:45,580 --> 00:30:49,059
And every time we multiply by a dagger, we got a more complicated wave function, right?
248
00:30:49,060 --> 00:30:56,380
So the ground state wave function, the harmonic oscillator was a Gaussian, the ground state wave function here was a well.
249
00:30:58,010 --> 00:31:03,890
It's a slightly more complicated problem, this more complicated problem because we have all these different values of the angular momentum.
250
00:31:04,130 --> 00:31:15,200
So here the starting point is R to the L is R to the N minus one times an exponential as in the
251
00:31:15,200 --> 00:31:18,980
same sense that our starting point in the case of a harmonic oscillator was just a Gaussian.
252
00:31:19,370 --> 00:31:26,630
But that's but that's the strategy. And what we would hope is that a l dagger does the business right.
253
00:31:27,140 --> 00:31:37,940
A l increased our angular momentum at fixed energy and drove us up against the equation that we solved to find the circular orbit
254
00:31:37,940 --> 00:31:44,030
wave function and l dagger we would hope would move us from the circular wave function back down to more eccentric orbits.
255
00:31:47,510 --> 00:31:52,670
But this has to be done in a slightly but in a slightly subtle in a slightly subtle matter manner.
256
00:31:54,700 --> 00:31:58,840
Okay. So let's look at this for at a formula that we have here somewhere.
257
00:31:59,260 --> 00:32:02,889
Let's look at this formula here. Ale, comma, ale.
258
00:32:02,890 --> 00:32:13,840
Dagger is equal to the difference of h's. So let me write that down with l reduce by 1al minus one comma l minus one dagger.
259
00:32:14,830 --> 00:32:21,250
This is just a relabelling operation, right? Is equal to can I remember which way up it is?
260
00:32:21,250 --> 00:32:33,280
No h0 squared of amu is zero squared mu over h bar squared of h l minus h l minus one.
261
00:32:38,720 --> 00:32:41,900
Now. Let me take the dagger. Sorry.
262
00:32:41,900 --> 00:32:50,330
Let me commute this entire equation. Both sides of it, with respect to a L minus one dagger.
263
00:32:50,450 --> 00:32:52,670
You'll see why we're doing this when we. When we've done it.
264
00:32:53,060 --> 00:33:07,040
So we're going to say that this is a L minus one comma, a L minus one dagger comma, a L minus one dagger.
265
00:33:07,050 --> 00:33:12,380
So that's the left side of the equation, commuted with a L minus one dagger.
266
00:33:12,950 --> 00:33:16,220
And that's going to be boring, constant.
267
00:33:19,860 --> 00:33:32,900
Times h l comma a l minus one dagger, which is what I want, minus h l minus one comma a l minus one.
268
00:33:32,910 --> 00:33:38,910
Oops. Minus one dagger. Why am I doing this?
269
00:33:38,940 --> 00:33:44,519
I'm doing this because I want to calculate this, which we haven't so far calculated.
270
00:33:44,520 --> 00:33:49,470
We could calculate by going back to first principles and stuff, but working out these commentators is quite wearisome.
271
00:33:50,010 --> 00:33:53,310
So it's best this is this is a reasonably slick route.
272
00:33:53,550 --> 00:33:59,220
But what I want to do is calculate the commutation of this with with h sub l and
273
00:33:59,220 --> 00:34:06,370
what I know at the moment is only the commentator of this with h l minus one K.
274
00:34:06,720 --> 00:34:15,470
So I'm going to rearrange this equation now because this is my target as h l comma, a l minus one dagger.
275
00:34:16,110 --> 00:34:19,560
That's what I want to find the value of, because it'll turn out to be the key.
276
00:34:21,030 --> 00:34:29,660
Is equal to. Each password over a zero squared.
277
00:34:29,890 --> 00:34:35,570
Mu open a big bracket. Then let's, let's write this out.
278
00:34:37,760 --> 00:34:40,459
Turning this into its products. I'm going to expand this in a commentator.
279
00:34:40,460 --> 00:34:45,850
As a general rule, I hate to expand commentators, but you'll see in a moment that it's an expedient thing to do.
280
00:34:45,860 --> 00:34:55,010
So this is going to be expanded and it's going to be a L minus one times, a L minus one dagger.
281
00:34:56,210 --> 00:35:00,860
Still, the outer commentator in place so computed with a L minus one dagger.
282
00:35:01,850 --> 00:35:04,999
Close that bracket minus is that room?
283
00:35:05,000 --> 00:35:17,660
Yes, minus the commentator a l minus one dagger a L minus one comma a L minus one close brackets dagger.
284
00:35:17,970 --> 00:35:21,770
Sorry, that one's dagger writes this one here. So all I've done is taken the contents.
285
00:35:21,770 --> 00:35:31,549
I've taken this in a commentator and expanded this into its two bits and then I should bring this onto this side of the equation.
286
00:35:31,550 --> 00:35:44,330
Step one step to replace this h with the corresponding expression in terms of a dagger, a dagger, a right.
287
00:35:44,330 --> 00:35:50,810
So a dagger it says up there is, is, is a multiple of l plus a constant.
288
00:35:51,620 --> 00:35:55,489
I'm interested in h l inside a comitato so I don't need to care about that constant
289
00:35:55,490 --> 00:36:04,700
because it will commute with everybody and produce the vanishing comitato. So for my purposes I need to replace h l by a l dagger a l times.
290
00:36:04,700 --> 00:36:07,880
That constant. That constant is already present and correct.
291
00:36:08,360 --> 00:36:15,440
So this term here becomes minus a l.
292
00:36:15,770 --> 00:36:24,620
Gosh, which way round is it? It's dagger. On the left is a L minus one dagger a L minus one.
293
00:36:26,040 --> 00:36:32,280
So that's this to get together with this. That's this insider commentator.
294
00:36:33,310 --> 00:36:36,970
Comma. Now this a dagger.
295
00:36:37,690 --> 00:36:41,560
L minus one close commentator. Close break bracket.
296
00:36:42,550 --> 00:36:46,510
Now, we should find that two of these terms. Cancel.
297
00:36:51,520 --> 00:36:56,260
Ow, ow, ow, ow. They don't know this.
298
00:36:56,590 --> 00:37:01,900
And this is a minor catastrophe. Oh, because I brought it on to the other side of the question.
299
00:37:01,930 --> 00:37:06,520
Thank you. Yep. People. So. So I brought this across here.
300
00:37:08,170 --> 00:37:13,430
No, but. I thought I though, she says, then arrives.
301
00:37:13,610 --> 00:37:16,730
It arrives with a plus sign. It rises. A plus sign, of course, is brilliant.
302
00:37:17,240 --> 00:37:25,160
So we can kill to these two, fight each other, and leave a blank piece of board.
303
00:37:26,600 --> 00:37:30,930
So. So what do we need to do now?
304
00:37:30,950 --> 00:37:34,279
What we need to do is concretely evaluate this. COMMENTATOR Right.
305
00:37:34,280 --> 00:37:38,210
Because so. So these two kill each other. And we're left looking at this.
306
00:37:38,600 --> 00:37:43,160
This is the usual computation. Rubbish. So what we have is.
307
00:37:43,250 --> 00:37:55,879
So I'll write it down explicitly because it is a bit of a mess. A zero squared mu, a l minus one comma, a l minus one dagger.
308
00:37:55,880 --> 00:38:00,020
Commentator. Right. He works on this one while he stands idly by.
309
00:38:02,920 --> 00:38:07,090
And the other term vanishes because it's because everybody commits with themselves.
310
00:38:07,390 --> 00:38:09,730
So we don't need the big bracket because that's the end of the discussion.
311
00:38:10,030 --> 00:38:15,280
And all we have to do now is plug in what this is, because it turned out to be the difference of the two.
312
00:38:15,310 --> 00:38:31,140
Right. It's the difference of the two. Hamiltonians. So this is is this is h bar squared over a nought squared mu of h l minus h l minus one.
313
00:38:31,150 --> 00:38:38,050
I have to do a little translation because his l minus one times a l minus one.
314
00:38:38,380 --> 00:38:49,510
Okay. So now with that expression, we can, we can go to business because we can go to I'm sorry, should have gone away from the content out the front.
315
00:38:49,510 --> 00:38:52,660
Should have gone away. Exactly. It was what I needed to.
316
00:38:52,690 --> 00:38:55,780
Yes. There was that constant the other way up there. Thank you very much.
317
00:38:55,960 --> 00:39:03,280
Get rid of that. So we have an unbelievably simple result after slightly scary computation.
318
00:39:05,820 --> 00:39:08,070
So what do we do with that?
319
00:39:08,080 --> 00:39:26,790
What we do with that, of course, is we go and say that a of a l minus one dagger e l is equal to a l minus one dagger h l e l right.
320
00:39:26,790 --> 00:39:37,290
So h l l is e e l multiply both sides of the equation by this baby, and we have what we've got on the board.
321
00:39:37,560 --> 00:39:45,870
Now, of course, we're going to swap these two over. So we right this is h l a l minus one dagger plus.
322
00:39:45,870 --> 00:39:55,380
So we're not allowed to write that down. So we put in what should sort it out, which is a commentator, a l minus one dagger comma h l close brackets.
323
00:39:55,650 --> 00:39:58,830
Brackets, brackets, e l.
324
00:40:01,650 --> 00:40:08,059
This is what we've just laboriously worked out and found is is h l minus h l minus one.
325
00:40:08,060 --> 00:40:14,600
Right. So this is going to give me an h l minus h l.
326
00:40:15,790 --> 00:40:27,940
Minus one. Time's a crucially times h l minus one subscript times a l minus one dagger.
327
00:40:27,950 --> 00:40:35,000
That's that's. That's right. So that this expression goes in where that commentator is, we can see.
328
00:40:39,290 --> 00:40:45,109
Oh. The commentators the other way around.
329
00:40:45,110 --> 00:40:50,470
Yes, the commentators are up there is h a not we want h so this becomes a minus.
330
00:40:50,480 --> 00:40:51,080
Exactly.
331
00:40:51,320 --> 00:41:05,840
So the undesired terms which are this on this cancels with this on this and we are left with minus minus is plus h l minus 1al minus one dagger e.
332
00:41:06,620 --> 00:41:20,110
L. Which is the equation that we require, because it says that this object is an eigen function of this operator with the same old icon value.
333
00:41:20,150 --> 00:41:33,680
Right. So this is establishes that e l minus one is actually equal to a is some multiple of to be discussed a l minus one dagger.
334
00:41:34,550 --> 00:41:40,420
E. L. And that enables this by mere differentiation,
335
00:41:40,420 --> 00:41:44,620
by just using more and more of these to work away from the circular orbit wave
336
00:41:44,620 --> 00:41:48,880
function down to the wave function associated with no angular momentum at all.
337
00:41:49,370 --> 00:41:56,830
Let's just begin to see how that how that pans out. So let's ask ourselves about.
338
00:41:57,220 --> 00:42:11,620
So you kn and minus two you should get by using this a l dagger stuff on on you kn minus one which we already know what it is.
339
00:42:11,950 --> 00:42:16,270
So that's proportional to I'm not going to chase down these proportionality constants now.
340
00:42:17,980 --> 00:42:21,010
Any time it's proportional to.
341
00:42:23,490 --> 00:42:35,290
Okay. Now we have to think what to put in here. This is the angular momentum with which you arrive and I'm interested in arriving with and minus two.
342
00:42:35,560 --> 00:42:43,900
So this is going to be in minus two dagger operating on on the wave function associated with with where we were before,
343
00:42:44,200 --> 00:42:48,970
which was U and N minus one of radius.
344
00:42:49,480 --> 00:42:52,480
Let's write that out in the position where it's more or less in the position representation.
345
00:42:52,480 --> 00:42:59,379
Let's be more concrete about it. So let's find our expression for a l we we're not interested in the constants in
346
00:42:59,380 --> 00:43:02,170
front because we're just we're going to normalise this when we're all sorted.
347
00:43:02,530 --> 00:43:07,270
So I want the I want the emission adjoin to if I want the dagger of that thing at the top there.
348
00:43:08,080 --> 00:43:14,080
So this is equal to minus I pr on h bar.
349
00:43:17,420 --> 00:43:25,700
Minus. Now, what's l? We put al equal to and minus two.
350
00:43:28,560 --> 00:43:31,830
No, no, no, no. Uh, yes, I, uh.
351
00:43:32,370 --> 00:43:42,030
This is. This is l it's been put equal to n minus two, and I'm supposed to have a L plus one there, so I'm supposed to have one more than this.
352
00:43:42,720 --> 00:43:48,120
So that's n minus one overall plus z over.
353
00:43:48,120 --> 00:43:52,490
Is it n minus one. Yep.
354
00:43:53,020 --> 00:43:59,080
A zero. Close the bracket.
355
00:43:59,440 --> 00:44:10,840
What's that working on? It's working on this which is ah to the end, minus one e to the minus, uh, z ease of the minus Z.
356
00:44:13,820 --> 00:44:16,830
Over and. A zero.
357
00:44:20,280 --> 00:44:28,740
If I did that right. And so now we have to we have to gain to replace PR with minus body or plus one overall.
358
00:44:29,310 --> 00:44:33,120
So we we're going to have too many minus signs.
359
00:44:33,120 --> 00:44:44,940
Now we'll get a minus D by the R minus one overall coming from here minus.
360
00:44:46,500 --> 00:45:00,930
And minus one overall plus z over N minus one a zero plus brackets O to the n minus one E to the minus set of and a zero.
361
00:45:03,730 --> 00:45:09,580
So we could write this, taking out an overall minus sign to make life a bit less negative.
362
00:45:10,900 --> 00:45:18,400
This is an overall minus Z over N minus one, a zero or N minus one.
363
00:45:24,100 --> 00:45:27,790
Okay. So what's going to happen?
364
00:45:28,190 --> 00:45:32,800
One thing is I want to say what's going to happen? This is this is a this is a particular value.
365
00:45:33,240 --> 00:45:38,830
This is this is a. This is a specimen of the L dagger sort of thing.
366
00:45:38,830 --> 00:45:42,040
Right. If we change and L, we'll be changing this number here.
367
00:45:42,430 --> 00:45:47,980
But the main idea is what's crucial is that this is going to contain every one of these is going to contain a derivative operator.
368
00:45:48,220 --> 00:45:54,000
When the derivative operator meets this, it will produce a term which goes like R to the end minus two.
369
00:45:54,010 --> 00:46:00,130
It will produce a term which has less R dependence. Also, there is this something over our term does the same thing.
370
00:46:00,490 --> 00:46:04,360
So we get an amount of R to the end minus two.
371
00:46:04,840 --> 00:46:11,170
But we also get from this or from differentiating the exponential, we get an amount of out of the end plus one.
372
00:46:13,510 --> 00:46:20,080
And minus one. Sorry. So in other words and of course, the exponential will live on in the way that exponentials do and differentiated.
373
00:46:21,550 --> 00:46:24,820
So what does the wave function look like? You and.
374
00:46:24,830 --> 00:46:28,450
And minus two is now a linear function of radius.
375
00:46:28,450 --> 00:46:32,070
It's going to be A plus B oh.
376
00:46:32,980 --> 00:46:37,060
And actually that will be negative. Let me write it. I mean, concrete and it's going to be negative.
377
00:46:39,250 --> 00:46:43,970
Uh, times are to the end, minus two.
378
00:46:43,990 --> 00:46:48,970
So I've taken this out e to the minus said R over in a zero.
379
00:46:49,420 --> 00:46:58,450
So since this is a linear function of R and moreover B and a a positive, we get one node, right?
380
00:46:58,450 --> 00:47:09,999
So that is to say, you have some particular radius vanishes when we go to get you and minus two and sorry,
381
00:47:10,000 --> 00:47:18,549
minus three and we'll use another of these differential operators and we will find that this is some it'll be sort of a primed minus,
382
00:47:18,550 --> 00:47:22,420
B prime, double plus C prime to R squared.
383
00:47:22,880 --> 00:47:26,740
It will be a quadratic expression E to the minus stuff.
384
00:47:26,980 --> 00:47:33,340
And we will have two nodes because it'll turn out this quadratic has real roots and so on and so forth.
385
00:47:33,340 --> 00:47:40,030
Right? So every time we, we use an a dagger, we get one more note.
386
00:47:40,690 --> 00:47:41,919
Why? Why is that?
387
00:47:41,920 --> 00:47:50,170
Physically, what we're doing is we're taking kinetic energy out of the out of the tangential motion, remember, and stuffing it into the radial motion.
388
00:47:51,250 --> 00:47:57,940
And kinetic energy motion in quantum mechanics is associated with oscillating wave function.
389
00:47:57,950 --> 00:48:03,340
So we're getting more and more wavelengths radially and therefore more and more nodes radially.
390
00:48:05,920 --> 00:48:10,330
And here we are. And here is a picture of what a few of these things look like.
391
00:48:10,900 --> 00:48:16,390
So, oops. These are these are the wave function, the radial wave functions.
392
00:48:16,440 --> 00:48:24,910
Well, this is a picture. Sorry, this is a picture of the meridional plain. So this is radius here and this is z direction here.
393
00:48:25,010 --> 00:48:29,440
Okay, so we're discussing the wave function in question is fitting a volume.
394
00:48:31,780 --> 00:48:38,140
And it begins. Right. So so that's kind of how you should think about it.
395
00:48:38,530 --> 00:48:45,100
This is the circular orbit case. Blackness means high probability of finding the electron or the reduced particle.
396
00:48:45,580 --> 00:48:48,670
Whiteness means not much. So this is the circular orbit.
397
00:48:48,670 --> 00:48:53,440
It's zero on the axis. Because, you know, we saw it with zero on the axis before.
398
00:48:53,440 --> 00:48:56,739
It's reasonable. It's got a lot of tangential motion.
399
00:48:56,740 --> 00:48:57,880
It can't get to the axis.
400
00:48:58,180 --> 00:49:06,820
But then otherwise the the amplitude of finding it rises quite quickly, peaks and then very slowly falls away as you go off to infinity.
401
00:49:07,930 --> 00:49:20,790
But it's rather boring, right? It's sort of everywhere. If we are using an A dagger and a N minus two dagger on this, we get this wave function.
402
00:49:20,800 --> 00:49:26,820
This is for the case. N equals three. By the way, we get this situation where we have one, we have one radius.
403
00:49:26,830 --> 00:49:37,209
This is the node. So it rises at the origin, it peaks sooner, so it reaches a high peak at a smaller radius than this,
404
00:49:37,210 --> 00:49:40,060
which is associated on a plunging a more plunging orbit. Right.
405
00:49:40,060 --> 00:49:45,370
It has a pair of centre, but then quantum mechanically it has this node where you won't find it.
406
00:49:45,370 --> 00:49:49,420
You have zero probability of finding it around that circle and then you have a property of finding it further out.
407
00:49:49,780 --> 00:49:56,350
Use another of these, a dagger things and you'll come over here where you have two nodes here you see them.
408
00:49:57,340 --> 00:50:01,840
And now in this case, because N is three, you've run out of energy.
409
00:50:02,350 --> 00:50:05,590
In this case, there's no orbital angular momentum. There's no tangential motion.
410
00:50:05,800 --> 00:50:09,010
This is what a plunging orbit looks like in quantum mechanics.
411
00:50:09,280 --> 00:50:13,600
Right? That particle in classical physics, we just diving at the nucleus,
412
00:50:14,650 --> 00:50:19,360
slipping around it and coming back out again and in an arbitrary, elongated ellipse.
413
00:50:21,510 --> 00:50:30,540
So it doesn't look at all like that. Well, that is clearly the moment to stop with that review of the hideaway functions.