1 00:00:01,780 --> 00:00:05,956 In this video we are going to use the creation and angulation operator 2 00:00:05,956 --> 00:00:10,169 [UNKNOWN] introduced in the previous segment. 3 00:00:10,169 --> 00:00:14,400 To generate the energy spectrum of the harmonic oscillator. 4 00:00:14,400 --> 00:00:19,015 And this calculation as we discussed is going to involve in very important way 5 00:00:19,015 --> 00:00:23,650 calculating various commentators of the operators. 6 00:00:24,900 --> 00:00:29,508 So what we just proved in the previous video was that the harmonic oscillator 7 00:00:29,508 --> 00:00:34,480 Hamiltonian can be represented as so. As the sum of two terms. 8 00:00:34,480 --> 00:00:38,968 The first term is, a, dagger, a, where, a, dagger is this creation operator and, 9 00:00:38,968 --> 00:00:44,492 a, is this annihilation operator. And also there was a second term which is 10 00:00:44,492 --> 00:00:49,607 proportional to the commutator x of x and p position and momentum. 11 00:00:49,607 --> 00:00:53,627 And by the way this commutator of x and p is one of the so callled canonical 12 00:00:53,627 --> 00:00:57,648 commutation relations in quantum mechanics. 13 00:00:57,648 --> 00:01:02,340 And we actually should have discussed it in more details earlier in the course, 14 00:01:02,340 --> 00:01:07,224 but better later than never. So let me just first give you this result 15 00:01:07,224 --> 00:01:10,437 for this commutator. And this commutator in fact is 16 00:01:10,437 --> 00:01:14,784 proportional just to so-called c number, or in the, the quantum, early quantum 17 00:01:14,784 --> 00:01:19,238 mechanical jerk one. Or in other words to the identity 18 00:01:19,238 --> 00:01:21,884 operators. So in, in other words, this commutator 19 00:01:21,884 --> 00:01:27,768 effects and p, is not an operator at all. Its actually just a constant and to see 20 00:01:27,768 --> 00:01:33,815 this we can go to the position representation. 21 00:01:33,815 --> 00:01:41,553 And let's see in postition representation x times p acting on a wave function psi 22 00:01:41,553 --> 00:01:48,904 is equal to x. Which is just multiplication operator, 23 00:01:48,904 --> 00:01:54,960 followed by minus i, h bar d over dx, acting on psi. 24 00:01:54,960 --> 00:01:59,734 And so what we're going to have is minus i h bar psi x psi prime, so this is the 25 00:01:59,734 --> 00:02:05,794 result of the action of these two operators in this order. 26 00:02:05,794 --> 00:02:07,564 Now if we consider a different order of the operators. 27 00:02:07,564 --> 00:02:09,393 If we first put you know x is going to act first and then we're going to have 28 00:02:09,393 --> 00:02:11,800 the momentum so in this case we can write it as false. 29 00:02:11,800 --> 00:02:20,119 It can be be minus i h bar g over dx x times psi and in this case we have to use 30 00:02:20,119 --> 00:02:33,620 the chain rule to calculate the derivative of both terms in this product. 31 00:02:33,620 --> 00:02:40,460 So its going to involve therefore minus i h bar derivative for facts is just 1. 32 00:02:40,460 --> 00:02:46,824 Psi minus i h bar x, psi prime and now, if we if we put everything together and 33 00:02:46,824 --> 00:02:53,553 apply this commutator to an arbitrary function psi of x. 34 00:02:53,553 --> 00:02:59,573 So let me actually write it in red, so if we now have x commutator of x and p, sort 35 00:02:59,573 --> 00:03:07,070 of acting on the wave function psi. So what it means is that we want to act 36 00:03:07,070 --> 00:03:12,170 by the following operator xp minus px on psi and so we essentially have to 37 00:03:12,170 --> 00:03:19,976 subtract result from this result. And so this two guys are going to cancel 38 00:03:19,976 --> 00:03:24,926 each other out and the only thing which should going to remain is plus ih bar 39 00:03:24,926 --> 00:03:29,966 psi. So the action of the commutator position 40 00:03:29,966 --> 00:03:34,864 momentum on an arbitrary function psi x is equivalent to multiplying this 41 00:03:34,864 --> 00:03:40,575 function by i h bar. Which is exactly the result that I 42 00:03:40,575 --> 00:03:46,164 advertised well in the beginning and this is again perhaps most of you or many of 43 00:03:46,164 --> 00:03:52,300 you already knew it. So this is again one of the canonical 44 00:03:52,300 --> 00:03:56,550 commutation relations in quantum mechanics. 45 00:03:56,550 --> 00:04:00,906 Now to continue this algebraic exercise in calculation commutators let me now 46 00:04:00,906 --> 00:04:05,310 calculate a very useful commutator for the following. 47 00:04:05,310 --> 00:04:10,910 Namely the commutator of, a, and a dagger and this is another so called canonical 48 00:04:10,910 --> 00:04:15,327 commutation relation. Let me write the result first and then 49 00:04:15,327 --> 00:04:18,743 I'm going to prove it. So and we will see that the commutator 50 00:04:18,743 --> 00:04:22,560 of, a, and, a, dagger is going to be equal to 1. 51 00:04:22,560 --> 00:04:27,013 So to prove this we use the definition of, a, and a dagger from the previous 52 00:04:27,013 --> 00:04:33,240 video and this commutation relation of position and momentum. 53 00:04:33,240 --> 00:04:39,450 So if we calculate a, and, a, dagger, the commutator of, a, and a dagger so we will 54 00:04:39,450 --> 00:04:48,000 have basically two types of terms. So here we're going to have m omega 2 h 55 00:04:48,000 --> 00:04:53,670 bar x plus i p 2 m omega h bar and the second term in this commutator is going 56 00:04:53,670 --> 00:05:01,830 to be the same thing. But with a minus sign in front of the i. 57 00:05:07,340 --> 00:05:10,842 So here it is. And we'll see that we have commutator of 58 00:05:10,842 --> 00:05:13,700 x and x. We have commutation of p and p. 59 00:05:13,700 --> 00:05:16,835 But of course, a commutator of an operator with itself is equal to 0 60 00:05:16,835 --> 00:05:21,570 because of course, for example, x x minus x x is equal to 0, right? 61 00:05:21,570 --> 00:05:26,192 So any operator commutes with itself. So the only commutators that are going to 62 00:05:26,192 --> 00:05:32,430 be non-trivial are going to be these cross-commutators of x and p and p and x. 63 00:05:32,430 --> 00:05:37,190 And you can convince yourself that the commutator of x and b is of course equal 64 00:05:37,190 --> 00:05:43,331 to the minus of commutator of b and x. Because we simply switched the order in 65 00:05:43,331 --> 00:05:48,742 which the operators appear. So therefore we can write it so this 66 00:05:48,742 --> 00:05:53,660 result, so lets just go back to the blue fonts. 67 00:05:53,660 --> 00:05:58,285 So it's going to be a twice, let's say the commutator of this guy, and this guy. 68 00:05:58,285 --> 00:06:08,890 So m omega over 2 h bar, from this term, minus i, 2 m omega h bar from this guy. 69 00:06:08,890 --> 00:06:13,759 And here we're going to have x and p. But this is something we know, we just 70 00:06:13,759 --> 00:06:17,676 calculated this. So this is this commutation relation of 71 00:06:17,676 --> 00:06:21,360 position momentum and its equal to, i a h bar. 72 00:06:21,360 --> 00:06:25,440 And so indeed if we put everything together we will see that well m and 73 00:06:25,440 --> 00:06:30,617 omega cancel here. So h bar and h bar under this score would 74 00:06:30,617 --> 00:06:38,239 cancel here so i times minus i gives us 1, and 2 over 2 is equal to 1. 75 00:06:38,239 --> 00:06:44,224 So oh, no, it's equal to 1[UNKNOWN] . So and this eh, eh, you know completes a 76 00:06:44,224 --> 00:06:48,498 very important iteration. Namely the iteration of the Folic 77 00:06:48,498 --> 00:06:50,980 identity. And it's an operator identity. 78 00:06:50,980 --> 00:06:53,785 It doesn't really matter what representation we use, what kind of 79 00:06:53,785 --> 00:06:58,291 methods we want to solve. We have proven that the harmonic 80 00:06:58,291 --> 00:07:02,791 oscillator Hamiltonian, so let me switch to red. 81 00:07:02,791 --> 00:07:08,285 The harmonic oscillator Hamiltonian that's written here can be represented as 82 00:07:08,285 --> 00:07:15,190 H omega, a, dagger a plus one half, so this is from this commutator. 83 00:07:15,190 --> 00:07:23,330 Where the, a and a dagger commute to 1. Or in other words you may see it so this 84 00:07:23,330 --> 00:07:27,740 annihilation and creation operators satisfy the economical commutation 85 00:07:27,740 --> 00:07:32,993 relations. So the second term in this expression for 86 00:07:32,993 --> 00:07:37,380 the Hamiltonian as we discussed is just proportional to a number. 87 00:07:37,380 --> 00:07:44,180 So it's not, it can not possible affect the solution of the ion value problem. 88 00:07:44,180 --> 00:07:49,540 Namely if we want to solve the equation h psi is equal e psi and if psi happens to 89 00:07:49,540 --> 00:07:56,606 be an ion function of, a dagger a. So it would indeed be ion function of the 90 00:07:56,606 --> 00:08:02,690 whole Hamiltonian. h with a slightly shifted energy because 91 00:08:02,690 --> 00:08:08,100 the identity operator just reproduces the function. 92 00:08:08,100 --> 00:08:13,326 So and therefore in order for us to find the spectrum of the operator h in order 93 00:08:13,326 --> 00:08:20,191 for us to solve this eigenvalue problem. And find the eigenvalue functions and 94 00:08:20,191 --> 00:08:25,082 eigenvalues, we have to focus on the eigenvalue problem for this operator, a 95 00:08:25,082 --> 00:08:29,668 dagger a. Now comes a very important part where 96 00:08:29,668 --> 00:08:35,119 we're going to generate the entire energy spectrum of the Hamiltonian. 97 00:08:35,119 --> 00:08:39,943 From just a single eigenstate of it, or more precisely, from an eigenstate of 98 00:08:39,943 --> 00:08:44,470 this guy a dagger a. And to generate the spectrum, we would 99 00:08:44,470 --> 00:08:48,726 need just two pieces of information. The fact that we can represent the 100 00:08:48,726 --> 00:08:53,279 Hamiltonian as so, and the commutation relations between the ca-, ca. 101 00:08:53,279 --> 00:08:56,509 Annihilation and creation operators that we just derived. 102 00:08:57,550 --> 00:09:01,975 So at this stage, let me assume that I know an eigenstate of this guy, of a 103 00:09:01,975 --> 00:09:06,544 dagger a. And let me call this eigenstate n, and 104 00:09:06,544 --> 00:09:11,480 the corresponding eigenvalue will also be n. 105 00:09:11,480 --> 00:09:15,960 So this would also of course imply that the state n is also a solution to the 106 00:09:15,960 --> 00:09:19,829 time. And depends I show in your equation for 107 00:09:19,829 --> 00:09:25,161 the harmonic oscillator Hamiltonian. Or in other words, it's an old Synagon 108 00:09:25,161 --> 00:09:30,354 state of the Hamiltonian with the energy h omega n plus 1 half. 109 00:09:30,354 --> 00:09:38,922 So this is the energy of this state. Now let us examine what happens with this 110 00:09:38,922 --> 00:09:45,929 state we assumed exists. Under the action of this creation and 111 00:09:45,929 --> 00:09:50,897 anihilation operators. And of course what we're interested in is 112 00:09:50,897 --> 00:09:57,060 whether or not construct new eigenstates out of an existing eigenstate. 113 00:09:57,060 --> 00:10:03,632 So first let me add on this n by the operator a dagger and lets see how this 114 00:10:03,632 --> 00:10:11,505 operator behaves with regards to this guy, a dagger a. 115 00:10:11,505 --> 00:10:18,417 So what we can do here, we can take advantage of the canonical commutation 116 00:10:18,417 --> 00:10:26,100 relations that we derived in the previous slide. 117 00:10:26,100 --> 00:10:35,688 And using this commutational relations we can arrive a dagger a, a dagger as a 1 118 00:10:35,688 --> 00:10:41,720 plus dagger a. So here, we are going to have so 119 00:10:41,720 --> 00:10:45,654 basically exchanged the order of these two operators. 120 00:10:45,654 --> 00:10:49,734 So we are going to have instead of a dagger a, a dagger we are going to have a 121 00:10:49,734 --> 00:10:57,307 dagger, 1 plus a dagger a, acting on n. But, so this is just a number, and this 122 00:10:57,307 --> 00:11:04,544 guy, we already know how it acts on n. So by assumption, this is an eigenstate 123 00:11:04,544 --> 00:11:09,436 of a dagger a. So therefore, we can write it as a dagger 124 00:11:09,436 --> 00:11:14,309 n plus one half. I'm sorry plus one and since this is 125 00:11:14,309 --> 00:11:19,574 operator and this is just a number we can just exchange the order and so we have n 126 00:11:19,574 --> 00:11:26,538 plus one a dagger n. Which is exactly the same function as in 127 00:11:26,538 --> 00:11:32,132 the left hand side. So what we have proven Is that if n is a 128 00:11:32,132 --> 00:11:38,180 nagging function of the operator a dagger a. 129 00:11:38,180 --> 00:11:42,124 And consequently if it's a nagging function of the operator h, the 130 00:11:42,124 --> 00:11:47,120 Hamiltonian, then a dagger is also a nagging function. 131 00:11:47,120 --> 00:11:52,716 But with a nagging value n plus 1. So it raises the energy of the 132 00:11:52,716 --> 00:11:59,000 oscillator, by one energy quantum, if you want, of h omega. 133 00:11:59,000 --> 00:12:04,870 In the very similar way, what we can prove is the following. 134 00:12:04,870 --> 00:12:10,718 So we can prove that if n is indeed an eigenfunction of the Hamiltonian then the 135 00:12:10,718 --> 00:12:18,985 action of the annihilation operator on n. So this function is also an eigenfunction 136 00:12:18,985 --> 00:12:23,444 of the[UNKNOWN] . but it is an eigenfunction with a smaller 137 00:12:23,444 --> 00:12:28,141 energy, which sort, sort of justifies the terminology of this being an annihilation 138 00:12:28,141 --> 00:12:32,166 operator. So, while if we were to repeat this 139 00:12:32,166 --> 00:12:36,902 exercise that we just did for a dagger. And if we were to apply a dagger a on 140 00:12:36,902 --> 00:12:42,689 this function, we would, we would see, and I'll leave it for you as an exercise. 141 00:12:42,689 --> 00:12:45,900 That this guy is also an eigenstate of a dagger a. 142 00:12:45,900 --> 00:12:49,275 But now with a smaller eigenvalue, n minus 1. 143 00:12:49,275 --> 00:12:56,145 And here I will have the same function. So by knowing an arbitrary eigen function 144 00:12:56,145 --> 00:13:00,669 of the Hamiltonian, I can construct an infinite number. 145 00:13:00,669 --> 00:13:04,893 What seems like an infinite number of other eigenfunctions, with different 146 00:13:04,893 --> 00:13:09,063 eigenvalues. And by applying this creation and an, 147 00:13:09,063 --> 00:13:13,884 annihilation operator. And so we can also construct the spectrum 148 00:13:13,884 --> 00:13:17,922 of the Hamiltonian. But we haven't yet answered the question 149 00:13:17,922 --> 00:13:21,497 of what n could be. But with ultimate thinking we can 150 00:13:21,497 --> 00:13:27,066 convince ourselves that n here must be an integer going from 0 up to infinity. 151 00:13:27,066 --> 00:13:31,595 0,1,2,3 etcetera. And, a way to see this is essentially by 152 00:13:31,595 --> 00:13:36,131 uh,science check, if you want, of going to the physical problem we're trying to 153 00:13:36,131 --> 00:13:41,019 solve. So remember that, what we're doing do 154 00:13:41,019 --> 00:13:45,504 here, we're trying to find the energy levels in these harmonic oscillator 155 00:13:45,504 --> 00:13:49,878 potential. Which is simply the probably potential 156 00:13:49,878 --> 00:13:54,420 proportional to x squared. So this is via fax proportional to x 157 00:13:54,420 --> 00:13:58,625 squared. Now if we assume that we know a and 158 00:13:58,625 --> 00:14:05,557 certain a I can function with an energy. E sub n, as so and this is, let's say 159 00:14:05,557 --> 00:14:12,487 this corresponds to the state n. Then part, by applying the creation 160 00:14:12,487 --> 00:14:17,856 operator a dagger. We can construct a series of higher 161 00:14:17,856 --> 00:14:23,330 energy states. So n plus 1, n plus 2 n plus 3 etcetera. 162 00:14:23,330 --> 00:14:27,755 And there is no problem with that. But also by applying a, we can lower the 163 00:14:27,755 --> 00:14:29,932 energy. And we can do so seemingly indefinitely. 164 00:14:29,932 --> 00:14:31,272 But, clearly we have, this procedure has to stop somewhere, because we cannot go 165 00:14:31,272 --> 00:14:36,975 below 0. So in other words, we cannot go below the 166 00:14:36,975 --> 00:14:47,363 minimum of the potential, which in this case is 0. 167 00:14:48,490 --> 00:14:53,182 And this also implies that therefore, this procedure of generating new eigen 168 00:14:53,182 --> 00:14:59,098 functions, using this annihilation. op, operator, must terminate, and so 169 00:14:59,098 --> 00:15:03,666 there must exist, therefore, basically, the ground state. 170 00:15:03,666 --> 00:15:10,440 For which the action of the annihilation operator simply is equal to 0. 171 00:15:10,440 --> 00:15:13,402 Okay? Or in other words so this guy the vacuum 172 00:15:13,402 --> 00:15:20,175 state is the eigenstate of this guy a dagger a with the eigenvalue being 0. 173 00:15:20,175 --> 00:15:25,199 Or using this the result. So the energy of ground state therefore 174 00:15:25,199 --> 00:15:30,464 is equal to h omega over two which is the so called vaccum energy of the harmonic 175 00:15:30,464 --> 00:15:35,706 facilator. But this is really important and this 176 00:15:35,706 --> 00:15:41,257 tells us that this is really the property of the ground state. 177 00:15:41,257 --> 00:15:44,139 Okay? And this is by the way the simplest way 178 00:15:44,139 --> 00:15:49,208 to find an analytical expression for the ground state. 179 00:15:49,208 --> 00:15:54,990 We simply have to find a function that is annihilated by this operation. 180 00:15:54,990 --> 00:16:00,190 And finding this function is really the only remaining step in completing the 181 00:16:00,190 --> 00:16:04,617 iteration. Because here we essentially have done 182 00:16:04,617 --> 00:16:10,879 everything which needs to be done. We found a series of energy levels so we 183 00:16:10,879 --> 00:16:15,736 just write it again. So E sub n, the energy levels of the 184 00:16:15,736 --> 00:16:20,904 harmonic oscillators is equal to h omega n plus 1 half where n is equal to 0,1, 2, 185 00:16:20,904 --> 00:16:26,759 et cetera. And this was constructed under the 186 00:16:26,759 --> 00:16:32,302 assumption that eigen state exists. But we haven't yet shown that it is 187 00:16:32,302 --> 00:16:35,845 indeed so. We haven't yet, constructed explicitly, 188 00:16:35,845 --> 00:16:39,536 the eigenstate. And this is really the last piece of the 189 00:16:39,536 --> 00:16:44,560 puzzle, that we will put together in, in the next video. 190 00:16:44,560 --> 00:16:50,800 Where we also will summarize all the findings about the harmonic oscillator. 191 00:16:50,800 --> 00:16:54,220 And go a little closer to physical reality from this abstract algebraic 192 00:16:54,220 --> 00:16:58,536 derivation. Which, however, I hope shows the power of 193 00:16:58,536 --> 00:17:02,872 this algebraic method. So here, again, we are solving initially, 194 00:17:02,872 --> 00:17:07,210 what looks like a differential equation. But we get the results, and the 195 00:17:07,210 --> 00:17:11,850 quantization of the energies, without even calculating any derivatives. 196 00:17:11,850 --> 00:17:15,510 So all based on this commutation relations. 197 00:17:15,510 --> 00:17:19,800 And well I find it very elegant, and, I hope some of you also enjoy this 198 00:17:19,800 --> 00:17:23,450 derivation, but there is more to do.