1 00:00:00,620 --> 00:00:03,664 Hi everybody. Up to this point in the course Charles 2 00:00:03,664 --> 00:00:08,200 and I have presented already a lot of material. 3 00:00:08,200 --> 00:00:12,961 Which included both material that can be found in traditional quantum mechanical 4 00:00:12,961 --> 00:00:17,584 text books and also some non standard exposure to quantum physics for example 5 00:00:17,584 --> 00:00:22,483 super conductivity path integrals some elements of quantum localisation, etc, 6 00:00:22,483 --> 00:00:27,982 etc. And I have to say, on my side, the 7 00:00:27,982 --> 00:00:32,880 preference has been for the latter. I want to tell you more, about things 8 00:00:32,880 --> 00:00:37,010 that we cannot find in traditional textbooks. 9 00:00:37,010 --> 00:00:40,786 But today is certainly going to, going to be an exception, because today I'm 10 00:00:40,786 --> 00:00:44,498 going to present a very traditional quantum mechanical problem that no 11 00:00:44,498 --> 00:00:48,594 reasonable quantum mechanical course can avoid, and this is the problem of a 12 00:00:48,594 --> 00:00:54,046 harmonic oscilllator. So the reason I cannot avoid this problem 13 00:00:54,046 --> 00:00:58,039 is two-fold. So first of all quantum harmonic 14 00:00:58,039 --> 00:01:02,126 oscillator, the problem of quantum harmonic oscillator is relevant to a 15 00:01:02,126 --> 00:01:06,615 variety of various experimental systems, ranging from cold atoms who are atoms 16 00:01:06,615 --> 00:01:10,769 that are trapped in a quadratic potential, to let's say molecules who's 17 00:01:10,769 --> 00:01:15,794 low energy oscillations are described by harmonic oscillator, to quantization of 18 00:01:15,794 --> 00:01:19,546 light, to quantization of oscillations increase those so-called 19 00:01:19,546 --> 00:01:24,520 [UNKNOWN]. 20 00:01:24,520 --> 00:01:26,590 We're going to actually discuss them soon. 21 00:01:26,590 --> 00:01:31,390 And many, many other things. And second, so the problem, on the 22 00:01:31,390 --> 00:01:36,010 theoretical side, the solution to the problem of harmonic oscillator, is very 23 00:01:36,010 --> 00:01:41,190 illuminating and useful, in that it gives you very useful tools, so-called creation 24 00:01:41,190 --> 00:01:45,530 and anti-elation operators that are useful in a variety of context across 25 00:01:45,530 --> 00:01:52,464 various fields of physics. And you're going to see this thing very 26 00:01:52,464 --> 00:01:56,260 often in other courses if you pursue. physics states. 27 00:01:57,400 --> 00:02:02,161 But before solving the quantum harmonic oscillator problem, let me remind you the 28 00:02:02,161 --> 00:02:07,360 story behind the classical harmonic oscillator. 29 00:02:07,360 --> 00:02:11,266 And here I have an example of the classical system which exhibits these 30 00:02:11,266 --> 00:02:15,440 small oscillations and near an equilibrium position. 31 00:02:15,440 --> 00:02:20,390 And these oscillations exactly are the harmonic oscillator motion. 32 00:02:20,390 --> 00:02:26,242 So, on the, mathematical side, what happens here, of course, is that, the, 33 00:02:26,242 --> 00:02:31,410 mass here experiences, the so-called Hook's force, which is proportional to 34 00:02:31,410 --> 00:02:40,040 the displacement x of the, spring relative to its equilibrium position. 35 00:02:40,040 --> 00:02:44,858 So let's say here if this is my x axis, and let's say we have an equilibrium 36 00:02:44,858 --> 00:02:49,470 position somewhere here so this is my zero. 37 00:02:49,470 --> 00:02:53,496 So this is x, and the force is proportional to the displacement, and in 38 00:02:53,496 --> 00:02:58,860 this case at this moment of time it's this Hook's force. 39 00:02:58,860 --> 00:03:02,010 X in this direction, in the negative x direction. 40 00:03:02,010 --> 00:03:06,040 Now, I can write this force, of course, as a minus gradient in this case it's 41 00:03:06,040 --> 00:03:10,256 just one-dimensional gradient, or just the derivative with respect to x, of some 42 00:03:10,256 --> 00:03:15,636 potential. And to produce a linear force, of course, 43 00:03:15,636 --> 00:03:19,495 I need to differentiate a quadratic potential. 44 00:03:19,495 --> 00:03:23,338 So in this case is kx square over 2 and this is exactly the potential energy of 45 00:03:23,338 --> 00:03:27,364 the harmonic oscillator that eventually is going to appear in our Schrodinger 46 00:03:27,364 --> 00:03:32,195 Equation in the quantum version of this problem. 47 00:03:32,195 --> 00:03:37,907 Now the classical side though, what I have to do is I have to solve the second 48 00:03:37,907 --> 00:03:44,490 Newton law presented here. So, and the second Newton law, of course, 49 00:03:44,490 --> 00:03:48,390 is just mass times acceleration, the second derivative of x is equal to all 50 00:03:48,390 --> 00:03:52,120 the forces. In this case, there is just one force, in 51 00:03:52,120 --> 00:03:56,380 this example, it's if we ignore gravity, which is not going to modify too much. 52 00:03:56,380 --> 00:04:01,330 but in this case, it's going to be just one force which is the force due to the 53 00:04:01,330 --> 00:04:07,290 Hook's law. And if I write this equation, if I divide 54 00:04:07,290 --> 00:04:11,905 this equation by m and instead of writing the second derivative, I will write it in 55 00:04:11,905 --> 00:04:15,805 the compound form x with two dots on top So I can write it as x two dots, plus 56 00:04:15,805 --> 00:04:23,682 omega squared x, is equal to 0. Where omega, is equal to the square root 57 00:04:23,682 --> 00:04:29,025 of k over m. And the solution to this, differential 58 00:04:29,025 --> 00:04:36,325 equation is the sines, or cosines to some arbitrary phases, and it's up to us, 59 00:04:36,325 --> 00:04:43,835 which, solution to choose. So there will be, some amplitude here, 60 00:04:43,835 --> 00:04:47,950 amplitude here, so we can choose a sine or we can choose a cosine. 61 00:04:47,950 --> 00:04:51,085 And as long as we have two free parameters which will determine the 62 00:04:51,085 --> 00:04:55,018 initial conditions, the initial position and the initial velocity, these are all 63 00:04:55,018 --> 00:04:59,823 general solutions. And solve these classical problems so let 64 00:04:59,823 --> 00:05:04,350 me choose for example the sin solution and present it here. 65 00:05:05,729 --> 00:05:10,760 So this is this solution to this harmonic oscillator. 66 00:05:10,760 --> 00:05:15,316 Now to move further let me choose a particular intial condition which I can 67 00:05:15,316 --> 00:05:20,350 do which corresponds to x of t equals zero being zero. 68 00:05:20,350 --> 00:05:24,742 So let's say the initial moment of time the position of our harmonic oscillator 69 00:05:24,742 --> 00:05:28,829 was exactly at the equilibrium point just for simplicity and in this case this is 70 00:05:28,829 --> 00:05:34,916 my solution. Now if I want to find the velocity of the 71 00:05:34,916 --> 00:05:38,636 oscillator as a function of time, so all I have to do is differentiate the 72 00:05:38,636 --> 00:05:44,370 position, and in this case it's going to be just x naught. 73 00:05:44,370 --> 00:05:48,870 Some amplitude of oscillations times omega times cosign of omega t. 74 00:05:50,250 --> 00:05:53,672 So what I can do now, I can ride the energy of the system which is not a 75 00:05:53,672 --> 00:05:59,560 function of time as we shall see in accordance with the conversation law. 76 00:05:59,560 --> 00:06:04,135 Which would be a sum of the kinetic energy and m v squared over 2, and the 77 00:06:04,135 --> 00:06:10,290 potential energy v of x, which is equal to one half k x squared. 78 00:06:10,290 --> 00:06:13,746 k is the coefficient in the Hooke's law, or expressing it through the mass and 79 00:06:13,746 --> 00:06:19,510 the, and frequency, I can write it as so, one half, m omega squared, x squared. 80 00:06:19,510 --> 00:06:22,642 And, this is by the way, the form in which typically appears in the 81 00:06:22,642 --> 00:06:26,950 literature, in particular in quantum mechanical literature. 82 00:06:26,950 --> 00:06:31,650 So, now, indeed if we plug in this result into the energy. 83 00:06:31,650 --> 00:06:35,750 So, the energy is going to be equal to mv squared over 2. 84 00:06:35,750 --> 00:06:41,062 So, we're going to have m over 2 x naught squared, omega squared, plus cosine 85 00:06:41,062 --> 00:06:47,635 squared of omega t. From this velocity term and potential 86 00:06:47,635 --> 00:06:53,290 energy plus m by two x squared, x not squared, omega squared from here, sine 87 00:06:53,290 --> 00:06:58,945 squared of omega t, and then we use the fact that cosine squared of an arbitrary 88 00:06:58,945 --> 00:07:08,610 angle plus and sine squared of the same angle is equal to one. 89 00:07:08,610 --> 00:07:15,770 And so we see that indeed the energy is simply this. 90 00:07:15,770 --> 00:07:20,906 So and the energy is conserved. Indeed in a, cor, in, in accordance with 91 00:07:20,906 --> 00:07:28,078 the energy conservation law. But of course, even though the energy E 92 00:07:28,078 --> 00:07:32,563 here, is not a function of time, the two terms in this equation, the kinetic and 93 00:07:32,563 --> 00:07:37,470 the potential energy are functions of time. 94 00:07:37,470 --> 00:07:42,670 So x is a function of time and the velocity is a function of time. 95 00:07:42,670 --> 00:07:47,924 but they work in unison with one another in such a way that their sum, the sum, is 96 00:07:47,924 --> 00:07:53,465 a constant. And if we plug this, so let's say the x. 97 00:07:53,465 --> 00:07:57,497 X axis here is going to be m omega times the coordinate, and the y coordinate here 98 00:07:57,497 --> 00:08:03,400 is going to be the momentum. So we see that this trajectory basically 99 00:08:03,400 --> 00:08:08,670 in the face space describes the motion of a harmonic oscillator. 100 00:08:08,670 --> 00:08:13,450 It's a motion on a circle. and the radius of the circle is 101 00:08:13,450 --> 00:08:18,900 determined by the energy of our harmonic oscillator that we. 102 00:08:18,900 --> 00:08:24,460 And this motion is a phase portrait of this classical system. 103 00:08:24,460 --> 00:08:28,035 And, by the way, I will just mention in passing that there is a relation between 104 00:08:28,035 --> 00:08:31,610 the phase portrait of the classical system and the corresponding spectrum of 105 00:08:31,610 --> 00:08:35,888 quantum systems. But that's the only thing I'm going to 106 00:08:35,888 --> 00:08:40,691 say here in this regard. So now let us move closer to the main 107 00:08:40,691 --> 00:08:48,120 subject of today's lecture and make the oscillator quantum. 108 00:08:48,120 --> 00:08:53,000 So the motivations to do so are many, as I said. 109 00:08:53,000 --> 00:08:57,760 So, for example, we can think about instead of having a classical mass and a 110 00:08:57,760 --> 00:09:02,450 spring, we can talk about, let's say a two-atom molecule and the oscillations of 111 00:09:02,450 --> 00:09:09,100 this molecule relative to the equilibrium position of the atoms. 112 00:09:09,100 --> 00:09:13,510 So at low energies these oscillations are going to be described by a quantum 113 00:09:13,510 --> 00:09:17,694 harmonnic oscillator. So of course there are many more examples 114 00:09:17,694 --> 00:09:20,590 as I mentioned in the beginnning of the lecture. 115 00:09:20,590 --> 00:09:25,804 Let me just mention another one, so for instance Talking about the behavior of a 116 00:09:25,804 --> 00:09:30,860 quantum particle in a rather arbitrary potential v of x so lets say somethign 117 00:09:30,860 --> 00:09:36,752 like this. So if we are interested in low energy 118 00:09:36,752 --> 00:09:41,580 behavior the particle somewhere near the minimum of the potential, we can always 119 00:09:41,580 --> 00:09:46,124 Expand the potential view effects near the minimum, so it's going to be v of x 120 00:09:46,124 --> 00:09:54,450 naught, so this point plus v prime of x naught x minus x naught. 121 00:09:54,450 --> 00:10:00,820 And this is zero because this is the minimum, plus the second derivative over 122 00:10:00,820 --> 00:10:06,840 2x minus x not squared. And model of this overall constant, which 123 00:10:06,840 --> 00:10:13,830 is, just shifts the minimum of energy and this shift of the equilibrium position. 124 00:10:13,830 --> 00:10:15,840 And this shift of the equilibrium position. 125 00:10:15,840 --> 00:10:21,860 So the remaining part of this potential is really the quadratic potential that we 126 00:10:21,860 --> 00:10:28,697 have in the harmonic oscillator problem. So what I'm saying here is that you can 127 00:10:28,697 --> 00:10:32,784 always almost always approximate a reasonable potential, well near the 128 00:10:32,784 --> 00:10:39,550 minimum of the well with the harmonic oscillator if you have a small function. 129 00:10:39,550 --> 00:10:43,775 Of course if you have square quantum well we can not do that but in most cases you 130 00:10:43,775 --> 00:10:47,935 have a smooth potential and in this case we can just replace it at low energies 131 00:10:47,935 --> 00:10:55,002 with this harmonic oscillator. Okay, now let me raise this stuff and 132 00:10:55,002 --> 00:11:01,366 move further so how to formally go from the classical harmonic oscillator in this 133 00:11:01,366 --> 00:11:07,570 but does show the quantum harmonic oscillator. 134 00:11:07,570 --> 00:11:13,140 So, to do so we have to pretty much do the following mathematical procedure. 135 00:11:14,260 --> 00:11:17,300 We can just put a hat here. And that's it. 136 00:11:17,300 --> 00:11:20,770 So that's how me made the oscillator quantum. 137 00:11:20,770 --> 00:11:24,676 So well seriously what we have to do, we have to replace the energy of the 138 00:11:24,676 --> 00:11:28,387 oscillator. So the energy we derived in the previous 139 00:11:28,387 --> 00:11:31,400 side for the classical harmonic oscillator. 140 00:11:31,400 --> 00:11:35,621 So now we replace the energy of the oscillator with the Hamiltonian which is 141 00:11:35,621 --> 00:11:41,115 an operator the momentum which is just a variable in the classical description, 142 00:11:41,115 --> 00:11:48,210 now we replace it with the operator that acts on the wave function. 143 00:11:48,210 --> 00:11:51,612 And in principle we can put a hat also on top of the cord metal, though with the 144 00:11:51,612 --> 00:11:54,798 action of the cord metal on a wave function in position space, is pretty 145 00:11:54,798 --> 00:12:00,030 simple action. It's just a multiplication operator, but 146 00:12:00,030 --> 00:12:04,320 this pretty much what what it means on the mathematical side to make, to go from 147 00:12:04,320 --> 00:12:08,808 classical description with no operators involved to a quantum description with 148 00:12:08,808 --> 00:12:14,955 operator involved. So this is the mapping, and the problem 149 00:12:14,955 --> 00:12:21,370 that we want to solve now is to find solutions to the Eigenvalue problem. 150 00:12:21,370 --> 00:12:25,150 So of course its a non its a time independent problem so there is no time 151 00:12:25,150 --> 00:12:30,782 involved and what we want, we want to find a solution to this equation. 152 00:12:30,782 --> 00:12:35,823 So when solving this Schrodinger equation and in particular when thinking about the 153 00:12:35,823 --> 00:12:40,890 allowed energy levels. So what we should keep in mind about this 154 00:12:40,890 --> 00:12:45,718 harmonic oscillator potential is one property which makes it very different 155 00:12:45,718 --> 00:12:52,875 from what we have seen before. Namely that this potential grows. 156 00:12:52,875 --> 00:12:58,530 The potential energy grows indefinitely to infinity as x become very large or 157 00:12:58,530 --> 00:13:04,070 very negative. So there is no room here for plane 158 00:13:04,070 --> 00:13:08,870 wave-like solutions propagating to infinity, and the only thing we may have 159 00:13:08,870 --> 00:13:14,870 here, are discreet states, corresponding to particles localized in the vicinity of 160 00:13:14,870 --> 00:13:21,092 this x equals 0. And to determine this these energy 161 00:13:21,092 --> 00:13:25,636 levels, these harmonic oscillator spectrum Is exactly the main problem 162 00:13:25,636 --> 00:13:31,431 we're going to be focusing on in the next couple of videos.