1 00:00:03,350 --> 00:00:08,900 So that's a summary of the important formulae we obtained yesterday. 2 00:00:13,680 --> 00:00:24,419 We'd, uh, Wednesday we had reduced hydrogen's problem to a one dimensional Hamiltonian one h 3 00:00:24,420 --> 00:00:28,440 sub l one for every particular value of the total angular momentum quantum number. 4 00:00:28,440 --> 00:00:37,820 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, 5 00:00:37,830 --> 00:00:43,740 in a certain amount of angular momentum. It constructed a state with the same energy and more angular momentum. 6 00:00:44,850 --> 00:00:51,120 And using that and the idea that the sequence of states of more angry momentum with the same energy had to stop. 7 00:00:51,840 --> 00:00:59,879 We concluded that the energy was given by a certain constant 13.6 V divided by end squared, 8 00:00:59,880 --> 00:01:05,130 where n is one more than the maximum angular momentum that you can afford at that energy. 9 00:01:06,030 --> 00:01:11,370 Put another way, the Anglo momentum quantum number is less than or equal to one minus. 10 00:01:11,370 --> 00:01:15,300 This number n controls the energy and is called the principle quantum number. 11 00:01:16,890 --> 00:01:22,950 So what we want to do now is move forward to get the energy eigen functions so as to get the. 12 00:01:26,990 --> 00:01:30,020 These are of interest from the perspective of hydrogen. 13 00:01:30,040 --> 00:01:36,450 If you want to do any detailed calculations, like how does hydrogen interact with the electromagnetic field? 14 00:01:36,980 --> 00:01:40,170 What happens if you scatter electrons of hydrogen and that kind of stuff? 15 00:01:40,190 --> 00:01:42,830 You'll need to know what these eigen functions, these wave functions are, 16 00:01:43,310 --> 00:01:47,740 but they're also the building blocks for atomic for studies of atomic structure generally. 17 00:01:47,780 --> 00:01:52,309 So there are a complete set of states which you can expand any state. 18 00:01:52,310 --> 00:01:57,260 For example, the stationary state of an oxygen atom. You can expand in these states. 19 00:01:58,100 --> 00:02:01,729 And that's what people do when they do atomic physics calculations. 20 00:02:01,730 --> 00:02:07,290 For the most part, what they do. So these these play a very big role in atomic physics. 21 00:02:07,510 --> 00:02:21,240 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, 22 00:02:21,960 --> 00:02:32,130 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 23 00:02:32,130 --> 00:02:41,010 function in question is this right the amplitude to find at X the the atom when it's in this state. 24 00:02:43,120 --> 00:02:53,310 The reduced particles, strictly speaking, when it's in this state. We know all about the dependence, the anger, the dependence of this. 25 00:02:53,340 --> 00:03:01,830 We know that this is takes the form of some function, which presumably depends on the energy and on the angular momentum. 26 00:03:03,750 --> 00:03:08,580 Times are our times. Why l am a feature in PHI? 27 00:03:08,610 --> 00:03:19,620 Because we know that these the spherical harmonics are the unique angular eigen functions of this operator with this eigenvalue. 28 00:03:22,270 --> 00:03:31,329 So this thing, which we know is such an eigen function, its angular dependence must be given by, strictly speaking, a sum. 29 00:03:31,330 --> 00:03:35,140 Potentially, it's a sum of these four different values of M and the same value of L, 30 00:03:35,500 --> 00:03:40,120 but it's sensible to look for them one by one, assuming a particular value. 31 00:03:40,390 --> 00:03:45,940 And so the angular dependence, you know, out of out of that statement up there is this. 32 00:03:45,940 --> 00:03:49,840 And all we're looking for right now is the radial dependence on the radio dependence. 33 00:03:50,230 --> 00:03:53,530 Uh, as I say, you have to expect it to depend on the energy. 34 00:03:54,010 --> 00:03:58,270 I mean, it has to depend on what's in here, which is the energy and the. 35 00:04:01,380 --> 00:04:06,150 The angle momentum, a quantum number. So that's what we're that's what we're looking for. 36 00:04:06,150 --> 00:04:10,290 And we're going to get it just as we got the wave functions for the harmonic oscillator 37 00:04:10,980 --> 00:04:18,090 stationary states by studying the equation that the the equation that this operator, 38 00:04:18,600 --> 00:04:21,809 your letter operator kills the thing off in the appropriate circumstances. 39 00:04:21,810 --> 00:04:33,360 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. 40 00:04:34,980 --> 00:04:43,350 The maximum value of this is N minus one. So we use a N minus one on E and minus one. 41 00:04:44,100 --> 00:04:50,310 We get nothing. It's the end of the line. So we look at this equation in the position representation. 42 00:04:51,810 --> 00:04:58,130 And so. We can forget about that a zero over two, because it's just a constant. 43 00:04:58,490 --> 00:05:09,200 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 00:05:09,200 --> 00:05:13,220 So that L plus one, I suppose is going to be an end. 45 00:05:16,620 --> 00:05:20,660 Overall. Plus he said, over. 46 00:05:21,960 --> 00:05:29,490 And a zero that's operating on you and and minus one must give you nothing. 47 00:05:31,940 --> 00:05:38,480 Now we we we know that PR is in the position representation. 48 00:05:38,510 --> 00:05:44,570 We figured it out. It was minus h bar with deep idea plus one overall. 49 00:05:45,320 --> 00:05:48,800 So we put that fact into this equation. 50 00:05:49,760 --> 00:05:54,530 We have the I and the minus sign make a one the bars cancel. 51 00:05:54,830 --> 00:06:01,090 So we're looking at D by the plus one overall minus this. 52 00:06:01,090 --> 00:06:10,790 So we're going to have minus N minus one overall plus Z over and a zero. 53 00:06:11,570 --> 00:06:14,780 You and n minus one is nothing. 54 00:06:16,410 --> 00:06:21,299 Now, this is a premiums equation, right? 55 00:06:21,300 --> 00:06:24,510 It's a first order linear differential equation. 56 00:06:24,510 --> 00:06:37,300 So it has an integrating factor. E to the integral of with respect to all of the set of this stuff, right. 57 00:06:37,760 --> 00:06:50,610 This just from prelims maths. This, of course on integrating just gives me an R on top. 58 00:06:50,850 --> 00:06:56,910 This gives me a logo. And so we're looking at E to the minus some multiple of log R. 59 00:06:57,240 --> 00:07:07,170 So that means that this is going to become R to the minus n minus one from the E to the minus n, 60 00:07:07,170 --> 00:07:15,660 minus one log r times the exponential that we get from the E to the z r over and a zero. 61 00:07:15,930 --> 00:07:20,969 So now we know pretty much what's going on because the original differential equation, 62 00:07:20,970 --> 00:07:31,710 remember, says that D by the R the integrating factor times u and minus one equals nought. 63 00:07:31,890 --> 00:07:40,770 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. 64 00:07:41,790 --> 00:07:46,919 So we have the U and and minus one is some constant which must be determined by 65 00:07:46,920 --> 00:07:55,650 the normalisation times R to the end minus one E to the minus Z and a zero. 66 00:07:56,370 --> 00:08:04,330 So that's wonderfully simple expression. So let's let's ask ourselves a few things. 67 00:08:06,440 --> 00:08:09,590 Let's have a look at the ground state of hydrogen. 68 00:08:13,270 --> 00:08:17,649 So this is the case and equals one. How do we know that? 69 00:08:17,650 --> 00:08:21,580 Because Elle has to be less than or equal to N minus one. 70 00:08:21,880 --> 00:08:26,920 L Is the numbers the possible candidates for Ellen nought, one, two, blah, etc. 71 00:08:27,130 --> 00:08:30,910 So when has to start at one and then go up to three, etc. 72 00:08:30,920 --> 00:08:34,420 So the ground state, the state with the least energy is n equals one. 73 00:08:35,650 --> 00:08:39,550 So what is the wave function? So you won. 74 00:08:40,210 --> 00:08:46,840 And of course there should be a zero. Here is going to be a constant E to the minus in hydrogens. 75 00:08:46,840 --> 00:08:51,730 That is one. So this is this is just going to be r over a zero. 76 00:08:51,880 --> 00:08:56,200 So the ground state wave function is this mere exponential, a beautifully simple result. 77 00:09:01,240 --> 00:09:06,700 What else was I going to say about this? Yeah. 78 00:09:07,870 --> 00:09:13,540 One interesting given that that beautiful exponential one thing you notice is that this thing is never zero. 79 00:09:13,850 --> 00:09:21,610 It it's the ground state wave function has nonzero modulus all the way to our equals infinity. 80 00:09:22,630 --> 00:09:27,970 Although the particle is classically forbidden to go beyond a certain radius. 81 00:09:27,970 --> 00:09:34,470 And in fact, so what this does was graph up here plots is the probability of finding the 82 00:09:34,480 --> 00:09:42,650 reduced particle at a radius are measured in units of a zero of a over of a z. 83 00:09:44,710 --> 00:09:50,490 So a radius bigger than this. And the classically forbidden region stops at that number two. 84 00:09:51,010 --> 00:09:57,580 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 00:09:58,210 --> 00:10:00,510 where the kinetic energy as well would be negative. Right. 86 00:10:00,520 --> 00:10:07,809 So if you go beyond R is to the potential energy is more than the total energy of the particle. 87 00:10:07,810 --> 00:10:13,990 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 00:10:14,950 --> 00:10:21,610 So so that's, I think, an entertaining result. It says that there's the p forbidden. 89 00:10:23,680 --> 00:10:29,230 Looks forbidden. It's about 24%. 90 00:10:35,210 --> 00:10:39,920 Let's get the normalisation of this thing sorted out so we can work out expectation. 91 00:10:39,920 --> 00:10:45,380 A few expectation values working out in the normalisation is is fundamentally very straightforward. 92 00:10:46,070 --> 00:10:56,990 What we require of course. So we require that the integral d cubed x over all space of the complete wave 93 00:10:56,990 --> 00:11:05,630 function which consists of u and n minus one of radius y l m should equal one. 94 00:11:06,500 --> 00:11:15,770 This thing we right of course, is the r r squared and then the integral sine theta dct defined to omega over the sphere. 95 00:11:16,070 --> 00:11:18,860 Oops. Sorry should mod squirted this right. 96 00:11:18,890 --> 00:11:28,550 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. 97 00:11:29,630 --> 00:11:40,160 So we're left staring at an equation which says that the r times r squared times times u this thing squared. 98 00:11:40,550 --> 00:11:43,730 Now what did we say that was? That was c squared. 99 00:11:43,740 --> 00:11:57,350 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. 100 00:11:59,440 --> 00:12:03,700 Z over and a zero. 101 00:12:05,380 --> 00:12:15,230 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, 102 00:12:15,230 --> 00:12:17,330 square it and integrate the R to get one. 103 00:12:17,330 --> 00:12:24,590 You have an additional factor, R-squared, because fundamentally this is the normalisation condition we wish to impose and the Y lens and normalised. 104 00:12:24,590 --> 00:12:35,360 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. 105 00:12:39,560 --> 00:12:43,700 Now this integral is is nice and easy. 106 00:12:45,560 --> 00:12:49,100 So this can be written a C squared. 107 00:12:50,990 --> 00:12:58,969 And what we need to do is to bring an integral like this under control is to declare that this is is ro. 108 00:12:58,970 --> 00:13:08,210 So we introduce a new variable, ROE, which is to Z to Z of and a zero. 109 00:13:08,510 --> 00:13:17,450 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, 110 00:13:17,720 --> 00:13:31,970 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. 111 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. 112 00:13:41,730 --> 00:13:46,980 Well, that's the definition o missing the E to the minus row. 113 00:13:49,220 --> 00:13:52,520 Now, this is a famous integral in mathematics. 114 00:13:53,570 --> 00:13:59,030 Which is it? So it's often called gamma of two N plus one. 115 00:13:59,030 --> 00:14:06,140 I believe Euler's responsible for that absurdity. But what it what it should be thought of as is two N factorial. 116 00:14:07,210 --> 00:14:14,290 So this integral if if is simply the this exponent up here factorial. 117 00:14:15,330 --> 00:14:23,729 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, 118 00:14:23,730 --> 00:14:25,140 Aha, that's two factorial. 119 00:14:25,410 --> 00:14:42,600 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, 120 00:14:42,690 --> 00:14:50,130 which enables me to write down the the relevant way function u and minus one of radius. 121 00:14:51,060 --> 00:14:54,210 I need this factor here and I'm going to write it as follows. 122 00:14:54,540 --> 00:15:00,840 I'm going to say this is two Z over and a zero to the three halves power. 123 00:15:01,260 --> 00:15:07,590 So I'm borrowing from that three halves power one over the square root of two and factorial. 124 00:15:08,660 --> 00:15:10,310 That's too. I have to be clear. 125 00:15:10,320 --> 00:15:19,460 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. 126 00:15:20,570 --> 00:15:31,810 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. 127 00:15:34,080 --> 00:15:38,730 Over to. That's the row is defined as to Z. 128 00:15:39,000 --> 00:15:44,340 R. Etc. So. So. The factors. 129 00:15:47,100 --> 00:15:52,979 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. 130 00:15:52,980 --> 00:15:57,270 Minus one. So what does this. 131 00:15:59,070 --> 00:16:04,830 Physically. What is this? We are looking at the states with the highest angular momentum for a given. 132 00:16:05,830 --> 00:16:13,090 For given energy. So these are the quantum mechanical analogues of circular orbits, not eccentric orbits, but circular orbits. 133 00:16:14,540 --> 00:16:17,060 So what do we expect qualitatively? 134 00:16:17,480 --> 00:16:27,080 Well, we expect classically, if it was a circular orbit, our probability would be a delta function of the radius of a circular orbit. 135 00:16:27,080 --> 00:16:36,320 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 00:16:36,830 --> 00:16:40,520 So we're expecting it to be sort of like this ish. 137 00:16:42,100 --> 00:16:45,050 So what's how is that how does that arise from this formula? 138 00:16:45,070 --> 00:16:55,630 Well, when R is nought, row nought, this is going to be zero and then it's going to shift itself off zero. 139 00:16:56,440 --> 00:17:07,390 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 00:17:07,690 --> 00:17:12,459 then this would rise ever so slowly from zero and we'd have it would hug the origin for a long time. 141 00:17:12,460 --> 00:17:17,320 It would then rise and then when it when it when Rho became on the order of one this 142 00:17:17,320 --> 00:17:21,370 exponential which previously been harmless being ease the mind is something small, 143 00:17:21,580 --> 00:17:26,320 would become a vicious cutting off thing. And that's how we get cut off on this side here. 144 00:17:26,800 --> 00:17:31,540 So here we're looking at row two, the end minus one growth. 145 00:17:31,960 --> 00:17:35,450 And here we're looking at E to the minus row over to. 146 00:17:35,770 --> 00:17:39,879 Well, if this is the probability, then we need to multiply by we need to square up. 147 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. 148 00:17:48,760 --> 00:17:54,340 So the top picture there shows just the first three. 149 00:17:54,640 --> 00:17:57,640 So the pure exponential is end equals zero. 150 00:17:57,640 --> 00:18:11,920 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. 151 00:18:12,220 --> 00:18:19,600 Is, is, is the next one. And what you can see is that the characteristic radius is moving outwards. 152 00:18:21,120 --> 00:18:27,380 Quickly. So let's calculate some let's calculate some expectation values, because that's now easy to do. 153 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, 154 00:18:32,610 --> 00:18:37,200 we should be thinking about expectation values because classical physics is the physics of expectation values. 155 00:18:38,670 --> 00:18:54,780 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. 156 00:18:56,030 --> 00:19:03,710 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. 158 00:19:15,150 --> 00:19:24,780 Well, actually. Let's just go back to which is the best way to do this. 159 00:19:26,760 --> 00:19:29,910 All right. Let's turn this all into Rose. Let's turn these all into Rose now. 160 00:19:30,600 --> 00:19:36,300 So this we've already got more or less as a function of Roe. So what we need to do is to. 161 00:19:39,260 --> 00:19:45,530 Is to deal with these other ones. So there are four powers of all there are, and I need to turn those into rose, 162 00:19:45,530 --> 00:19:54,920 which means I need I need an and a nought over two z race to the fourth power. 163 00:19:55,850 --> 00:20:11,040 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, 164 00:20:11,630 --> 00:20:32,540 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. 165 00:20:33,020 --> 00:20:40,249 So these three and then here we have row and I need to square that. 166 00:20:40,250 --> 00:20:43,490 So it's going to be n to n minus two. 167 00:20:45,970 --> 00:20:52,370 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. 169 00:20:59,220 --> 00:21:06,360 Each of the minus row. So this integral is going to be two and plus one factorial. 170 00:21:07,230 --> 00:21:14,210 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. 171 00:21:14,220 --> 00:21:24,210 Everything else cancels in the editorial. And here something has gone wrong in that I've got far too many powers. 172 00:21:24,360 --> 00:21:27,670 And what have I done wrong? What are we doing wrong? Uh. 173 00:21:30,390 --> 00:21:37,170 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. 174 00:21:37,850 --> 00:21:42,090 Right. I was using this formula for you. 175 00:21:42,510 --> 00:21:47,069 These require this. These powers. Now I'm using this formula. 176 00:21:47,070 --> 00:21:53,520 So it's two the three halves power to the three because I've squared it up. 177 00:21:53,550 --> 00:21:57,930 Exactly. So we end up with these three of these cancel. 178 00:21:58,170 --> 00:22:02,220 So at the end of the day, I'm going to have an eight nought over two Z. 179 00:22:03,150 --> 00:22:09,270 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. 181 00:22:19,890 --> 00:22:31,140 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, 182 00:22:31,140 --> 00:22:37,650 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. 183 00:22:39,060 --> 00:22:45,900 So the interesting fact here is that the expectation value of all is sort of like going like an 184 00:22:45,900 --> 00:22:53,070 squared and that's exactly what we expect because e remember goes like minus one over and squared. 185 00:22:53,940 --> 00:22:59,010 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 00:23:07,450 --> 00:23:13,479 And from the visual theorem, we're expecting the potential energy to be minus twice the kinetic energy. 188 00:23:13,480 --> 00:23:18,070 So the total energy should be sort of a half, minus a half of the potential energy. 189 00:23:18,580 --> 00:23:25,840 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 00:23:47,410 --> 00:23:55,210 So states so so this so this grows very rapidly. 193 00:24:01,830 --> 00:24:08,310 This means that states in which you excite the electron to a large value of n cannot be seen. 194 00:24:08,700 --> 00:24:13,380 You won't be able to observe these, to measure these unless you're in an incredibly high vacuum. 195 00:24:13,590 --> 00:24:24,630 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.