1 00:00:01,520 --> 00:00:05,302 So in this last video of this lecture, I'm going to derive the harmonic 2 00:00:05,302 --> 00:00:10,164 oscillator wave functions. But I will do so without even writing 3 00:00:10,164 --> 00:00:16,035 down the Schrodinger equation. we will rely entirely on the solution, 4 00:00:16,035 --> 00:00:20,725 that algebraic solution I presented that took advantage of the creation and 5 00:00:20,725 --> 00:00:25,629 annihilation operators. First, it turns out that they are very 6 00:00:25,629 --> 00:00:29,505 helpful in determining the explicit form of the oscillator wave functions of psi 7 00:00:29,505 --> 00:00:33,469 of x. So but before we write down these wave 8 00:00:33,469 --> 00:00:36,966 functions, we derive these wave functions. 9 00:00:36,966 --> 00:00:41,490 Let me summarize what we know so far, which is actually a lot. 10 00:00:41,490 --> 00:00:45,797 So first of all, we found a way to represent the harmonic oscillator 11 00:00:45,797 --> 00:00:50,542 Hamiltonian in this compact form, in terms of this creation annihilation 12 00:00:50,542 --> 00:00:56,182 operators here. And using the algebra or the commutation 13 00:00:56,182 --> 00:01:01,933 relations of this harmonic oscillator operators, we derived the energy spectrum 14 00:01:01,933 --> 00:01:06,548 of the oscillator, which has this form, h omega n plus 1 half, where n is equal to 15 00:01:06,548 --> 00:01:13,363 zero, one, two , etcetera. So and again, I reiterate that this 16 00:01:13,363 --> 00:01:17,651 spectrum is special in that the level spacing between the neighboring levels is 17 00:01:17,651 --> 00:01:24,100 always the same, so it doesn't change as we go higher up in energy. 18 00:01:24,100 --> 00:01:28,928 So another thing we derived, and here, I'm being a little more specific is that 19 00:01:28,928 --> 00:01:33,614 given in arbitrary Eigenstate of the Hamiltonian with the quantum number and, 20 00:01:33,614 --> 00:01:38,911 as here. So we can generate the entire spectrum by 21 00:01:38,911 --> 00:01:43,660 applying the creation operators and annihilation operators. 22 00:01:43,660 --> 00:01:48,283 The creation operators raise the energy of the system by each omega So we go from 23 00:01:48,283 --> 00:01:52,303 n to n plus 1, and in the case of normalized eigenfunctions there appears 24 00:01:52,303 --> 00:01:57,840 this coefficient. I haven't derived it in the previous 25 00:01:57,840 --> 00:02:01,920 video, but this is rather straightforward to do so using the same methods and it's 26 00:02:01,920 --> 00:02:07,690 done in many textbooks, so I'm not going to spend any time on that. 27 00:02:07,690 --> 00:02:12,382 And likewise, given the same eigenstate n, we can find the eigenstate first 28 00:02:12,382 --> 00:02:17,006 pointing to a lower energy with one less quantum of h omega by applying this 29 00:02:17,006 --> 00:02:24,000 annihilation operation . But what we don't know at this stage, we 30 00:02:24,000 --> 00:02:28,330 don't know an example of any wavefunction, explicit wavefunction. 31 00:02:28,330 --> 00:02:31,948 So, and they preferably would like to see a wave function within real space which 32 00:02:31,948 --> 00:02:35,566 is more intuitive than other sort of, let's say abstract space that have been 33 00:02:35,566 --> 00:02:43,304 operating with so far in this problem. So in order to find the in, the explicit 34 00:02:43,304 --> 00:02:47,651 form of an eigenfunction, let me actually focus on the ground state, that is the 35 00:02:47,651 --> 00:02:54,386 lowest energy state in our problem. And this eigenstate as we discussed in 36 00:02:54,386 --> 00:02:58,733 the previous video is characterized by the fact that the action of this A dagger 37 00:02:58,733 --> 00:03:04,272 a on it use simply 0. So it gives us, its an eigenstate with an 38 00:03:04,272 --> 00:03:10,980 eigenvalue 0 or the eigenenergy of the state is equal to[UNKNOWN] over 2. 39 00:03:10,980 --> 00:03:15,204 So in other words, so the annihilation operator acting on this state must 40 00:03:15,204 --> 00:03:19,956 produce 0, and we can you know, write this annihilation operator explicitly in 41 00:03:19,956 --> 00:03:24,837 real space. So we, we use the definition, if you 42 00:03:24,837 --> 00:03:28,863 want, of this a and so, and here, what I'm going to do, I'm going to just write 43 00:03:28,863 --> 00:03:35,626 down explicitly the form of the momentum operator and position operator. 44 00:03:35,626 --> 00:03:39,992 So in real speeds the position operator is really simple, it's just the 45 00:03:39,992 --> 00:03:44,654 multiplication operator while the momentum operator as we know is just 46 00:03:44,654 --> 00:03:50,966 minus ih bar d over dx. So therefore you know, here, I will have 47 00:03:50,966 --> 00:03:55,450 simply, I will just simply remove the measuring constant. 48 00:03:55,450 --> 00:04:01,150 So all in all the equations, so I should have written perhaps here, si not, 49 00:04:01,150 --> 00:04:06,175 because I'm specifically talking about the ground state of the harmonic 50 00:04:06,175 --> 00:04:12,036 oscillator. So explicitly, the equation for this psi 51 00:04:12,036 --> 00:04:15,250 naught, therefore, becomes the full length. 52 00:04:15,250 --> 00:04:25,078 So we'll have this square root of m omega over 2 h bar x plus h over 2m omega h bar 53 00:04:25,078 --> 00:04:34,936 in the denominator d over dx. So, which is coming from this momentum 54 00:04:34,936 --> 00:04:41,130 operator and this acts on psi naught, and this whole thing must be equal to 0. 55 00:04:41,130 --> 00:04:45,770 So here, what I'm going to do, I'm going to well simplify this thing a 56 00:04:45,770 --> 00:04:50,570 little bit, so I'm going to put the square root here on top, I'm going to get 57 00:04:50,570 --> 00:04:57,121 rid off the factor two. And so, what I see here is that I can 58 00:04:57,121 --> 00:05:02,651 introduce a new parameter. Let me call it x naught, which is the 59 00:05:02,651 --> 00:05:09,010 square root of h divided by m omega and it appears both here and here. 60 00:05:09,010 --> 00:05:14,626 And so, this, so you can check that the physical dimension of this square root of 61 00:05:14,626 --> 00:05:21,378 h over m omega is the is a length scale. So it, so it has a physical dimension of 62 00:05:21,378 --> 00:05:24,239 length. So this is, in some sense, a natural 63 00:05:24,239 --> 00:05:28,350 length scale for the harmonic oscillator problem. 64 00:05:28,350 --> 00:05:32,900 And so, it is convenient, therefore, for me to rewrite my equation as above in the 65 00:05:32,900 --> 00:05:38,424 full length form. I'll write it as x over x naught plus x 66 00:05:38,424 --> 00:05:44,633 naught times d over dx. And this whole this thing x on the wave 67 00:05:44,633 --> 00:05:49,520 function [UNKNOWN] and we want to produce 0. 68 00:05:49,520 --> 00:05:57,720 So now let me just rewrite this equation in a slightly different form. 69 00:05:57,720 --> 00:06:01,791 So I'm going to write d over dx is psi prime, so this is going to be psi prime 70 00:06:01,791 --> 00:06:06,861 and in with this guy in the left-hand sign. 71 00:06:06,861 --> 00:06:11,404 So I'm going to have minus x over x naught squared times psi and so I just 72 00:06:11,404 --> 00:06:16,794 want to solve this differential equation, and you, you can check you can solve it 73 00:06:16,794 --> 00:06:22,770 formally. Or you can just convince yourself that 74 00:06:22,770 --> 00:06:28,090 the solution to this equation is an exponential, it's a Gaussian basically, 75 00:06:28,090 --> 00:06:34,778 minus x squared over 2x naught squared. So and this is let me just put zero, 76 00:06:34,778 --> 00:06:38,483 zeros in, in here which correspond again to the ground state and this is pretty 77 00:06:38,483 --> 00:06:42,732 much the result. So c here is a coefficient which is 78 00:06:42,732 --> 00:06:46,670 determined by the normalization of this wavefunction. 79 00:06:46,670 --> 00:06:50,891 So again, this wavefunction describes a particle in the ground state of the 80 00:06:50,891 --> 00:06:55,715 harmonic oscillator, and so therefore, on the, you know, total probability for me 81 00:06:55,715 --> 00:07:01,660 of finding a particle somewhere in space must be equal to 1. 82 00:07:01,660 --> 00:07:05,442 So therefore, this integral for minus infinity to plus infinity of sine r 83 00:07:05,442 --> 00:07:09,360 squared. So don't bother here with the complex 84 00:07:09,360 --> 00:07:16,060 size because everything is real so far. So this must be equal to 1 and this gives 85 00:07:16,060 --> 00:07:20,696 me the value of the way of the coefficient that multiplies this ground 86 00:07:20,696 --> 00:07:26,743 state wave function. So I can just, let me just write it down, 87 00:07:26,743 --> 00:07:31,720 so the c is equal to m omega over 2 I'm sorry, pi, that is pi h bar to the power 88 00:07:31,720 --> 00:07:39,420 1 4th. So this is the normalization coefficient. 89 00:07:39,420 --> 00:07:42,344 But notice that we have found the ground state wavefunction without actually 90 00:07:42,344 --> 00:07:46,042 solving the Schrodinger equation. We actually solved a much simpler 91 00:07:46,042 --> 00:07:49,995 equation, this first order differential equation just by by requiring that our 92 00:07:49,995 --> 00:07:53,712 wave function is in eigenstate with an eigenvalue of 0 of this annihilation 93 00:07:53,712 --> 00:07:59,080 operator. Okay, so this is great. 94 00:07:59,080 --> 00:08:02,162 So we can now work with this wavefunction if we want, but what about the higher 95 00:08:02,162 --> 00:08:05,748 energy state? So what we have just done was we 96 00:08:05,748 --> 00:08:09,934 determined this guy, the lowest energy state. 97 00:08:09,934 --> 00:08:13,556 So here, what I'm showing, let me just write down. 98 00:08:13,556 --> 00:08:18,029 So this is first of all the potential of this [UNKNOWN] x, so this is x, and these 99 00:08:18,029 --> 00:08:25,450 these levels here correspond to the energy levels of the harmonic oscillator. 100 00:08:25,450 --> 00:08:30,940 And the lowest energy has the wave function as so, which is usual Gaussian. 101 00:08:30,940 --> 00:08:34,570 So this is a typical Gaussian and high energy states have a more complicated 102 00:08:34,570 --> 00:08:38,445 sort of texture, so how do we get this eigenstate? 103 00:08:39,470 --> 00:08:44,974 Well, so here, we can take advantage again of the creation operators now, and 104 00:08:44,974 --> 00:08:50,048 so for, as we discussed in the first slide in this video, so the relation 105 00:08:50,048 --> 00:08:57,830 between the lowest, low energy states and high energy states. 106 00:08:57,830 --> 00:09:02,412 So let's say we can go from n to n plus 1, we can simply apply the creation 107 00:09:02,412 --> 00:09:08,210 operator divided by n plus 1, square root of n plus 1. 108 00:09:08,210 --> 00:09:12,935 So if we apply it to our eigen state n, we're going to get oops, there is no 109 00:09:12,935 --> 00:09:17,960 square root here, it's just n plus 1, a normalized state which corresponds to 110 00:09:17,960 --> 00:09:23,827 high energy. And so what we can do, we can apply these 111 00:09:23,827 --> 00:09:28,852 operators, let's say to aroused state, so we can apply this guide as many times as 112 00:09:28,852 --> 00:09:32,977 we want to the ground state, for instance, we can generate this 113 00:09:32,977 --> 00:09:39,616 wavefunction. So we are just applying this derivative, 114 00:09:39,616 --> 00:09:44,507 essentially to the Gaussian and this will produce a certain coefficients in, in 115 00:09:44,507 --> 00:09:50,273 front of the Gaussian. And this well, not coefficients better I 116 00:09:50,273 --> 00:09:55,379 say polynomials which are functions of x, and these guys, these polynomials they 117 00:09:55,379 --> 00:10:01,980 are called Hermite polynomials which are special functions. 118 00:10:01,980 --> 00:10:06,180 And you can Google them or you can look them up in Wikipedia and in many books. 119 00:10:06,180 --> 00:10:10,404 So this sort of, this solution using the explicit form of the wavefunctions and 120 00:10:10,404 --> 00:10:14,436 the explicit form of the Hermite polynomials appears in pretty much any 121 00:10:14,436 --> 00:10:20,004 book on quantum physics. And, I'm not going to spend time on 122 00:10:20,004 --> 00:10:23,985 discussing these details. I will simply notice that, well, you can 123 00:10:23,985 --> 00:10:27,681 actually solve the Schrodinger equation, starting from the spectrum up to the 124 00:10:27,681 --> 00:10:31,097 ground state and up to the, the excited states using only the creation and 125 00:10:31,097 --> 00:10:36,928 annihilation operator. So there is very little in, insight that 126 00:10:36,928 --> 00:10:39,808 is actually needed from Schrodinger equation which is in fact quite 127 00:10:39,808 --> 00:10:43,433 remarkable. So in your homework, you will need to 128 00:10:43,433 --> 00:10:47,398 play around a little bit with this equation, annihilation operators and with 129 00:10:47,398 --> 00:10:54,368 this ground-state wavefunction. And I hope that you found this solution 130 00:10:54,368 --> 00:10:59,672 illuminating and of course, I encourage you to look into the literature to see 131 00:10:59,672 --> 00:11:06,381 more about this very important problem of quantum physics.