1 00:00:00,530 --> 00:00:03,546 In the end of the previous video, I advertised to some degree, the main 2 00:00:03,546 --> 00:00:08,158 result that we are going to derive. Namely, that the electrons near the 3 00:00:08,158 --> 00:00:13,174 [INAUDIBLE] surface at low temperatures are going to form these balance states of 4 00:00:13,174 --> 00:00:20,30 2 electrons which are very condensed, and form this superconductor. 5 00:00:20,30 --> 00:00:24,318 And now, we're entering the technical part of the lecture, where we actually 6 00:00:24,318 --> 00:00:30,250 would want to get this calculation done and to see how it actually happens. 7 00:00:30,250 --> 00:00:33,814 And it connects somewhat to the calculations we were doing in the 8 00:00:33,814 --> 00:00:38,434 previous lecture, where we were studying particles in various potential So there, 9 00:00:38,434 --> 00:00:42,394 we were dealing with single particle quantum mechanics, so it was an 10 00:00:42,394 --> 00:00:46,684 extraordinarily imposed potential in which a single particle was moving, or 11 00:00:46,684 --> 00:00:53,748 which was localized there. Now, what we want to do, we want to study 12 00:00:53,748 --> 00:00:58,194 actually a two-particle problem with two particles here describing effectively the 13 00:00:58,194 --> 00:01:01,899 two electrons near the [UNKNOWN] service for the counting, in the counting of 14 00:01:01,899 --> 00:01:07,16 super conductivity. And it turns out that actually two 15 00:01:07,16 --> 00:01:10,656 particles problems often times, in quantum mechanics reviews the single 16 00:01:10,656 --> 00:01:14,785 particle problems. And in this segment I'm going to 17 00:01:14,785 --> 00:01:20,263 illustrate how exactly it happens. So here we're interested in analyzing 18 00:01:20,263 --> 00:01:25,319 this two-particle Schrodinger equation and to make things general for the sake 19 00:01:25,319 --> 00:01:29,863 of lectures later in the course, I'm considering here actually two different 20 00:01:29,863 --> 00:01:36,718 particles with masses m1 and m2. But in the next segment we'll work with a 21 00:01:36,718 --> 00:01:40,974 two-electron problem where one is actually equal to two. 22 00:01:40,974 --> 00:01:46,84 So so what we have here is essentially kinetic energy of two particles so this 23 00:01:46,84 --> 00:01:50,975 corresponds to b1 squared over 2m1 kinetic energy of the first particle plus 24 00:01:50,975 --> 00:01:58,30 b two square over 2m2 kinetic energy of the second particle. 25 00:01:58,30 --> 00:02:02,998 And the interaction between them which we expect to depend only on the distance 26 00:02:02,998 --> 00:02:08,243 between the particles. And now we have a wave function that 27 00:02:08,243 --> 00:02:13,94 depends on both coordinates of the particle r1 and r2 and well, the right 28 00:02:13,94 --> 00:02:18,720 hand side is the usual energy times of psi. 29 00:02:18,720 --> 00:02:23,869 So we're going to solve this equation. Now, the goal of this, video is to show 30 00:02:23,869 --> 00:02:28,117 that actually instead of solving this complicated, differential equation which 31 00:02:28,117 --> 00:02:33,500 depends two vectors r one and r two we can actually reuse it. 32 00:02:33,500 --> 00:02:37,468 To the good old single particle Schrodinger equations we really know how 33 00:02:37,468 --> 00:02:41,480 to solve. And to see how it happens lets do the 34 00:02:41,480 --> 00:02:46,512 following change of variables to go, we will go from the coordinates r1 and r2, 35 00:02:46,512 --> 00:02:51,618 the actual coordinates of each individual particles, to this capital R and 36 00:02:51,618 --> 00:02:58,856 uh,[UNKNOWN] are. So the former represents the center of 37 00:02:58,856 --> 00:03:03,680 mass of our particles, so it's defined as so and the second one is the distance 38 00:03:03,680 --> 00:03:08,422 between the particles. So the relative distance between 39 00:03:08,422 --> 00:03:13,290 particles are one and are two. Now what we're going to do, we're 40 00:03:13,290 --> 00:03:18,810 going to rewrite this equation in terms of this capital R and letter case R. 41 00:03:18,810 --> 00:03:23,155 So in order to do this, we well, first we notice that the, of course the, potential 42 00:03:23,155 --> 00:03:26,730 energy depends only on the distance, basically about the instruction, we've 43 00:03:26,730 --> 00:03:31,713 chosen that, this way. And the only non trivial part, we have to 44 00:03:31,713 --> 00:03:35,674 analyze is this kinetic energy. The derivatives. 45 00:03:35,674 --> 00:03:39,800 This Laplacian which appear in this brackets. 46 00:03:39,800 --> 00:03:41,918 And so we're going to work with this guys now. 47 00:03:41,918 --> 00:03:46,166 So to calculate this part of kinitec energy, we can focus for the sake of 48 00:03:46,166 --> 00:03:50,962 simplicity on just one component of this Laplacian. 49 00:03:50,962 --> 00:03:54,922 So Laplacian is a sum of second derivatives to the respect of, with 50 00:03:54,922 --> 00:03:59,602 respect to each other directions in our problem x, y, z so let's just focus this 51 00:03:59,602 --> 00:04:04,138 one x. And in this case the capital is X is 52 00:04:04,138 --> 00:04:08,170 going to be the center of mass x component and this is going to be a 53 00:04:08,170 --> 00:04:15,359 relative coordinate in the X direction. Now in order to change the variables in 54 00:04:15,359 --> 00:04:21,272 the Laplacian and the derivatives we can. go from the derivative with respect to X 55 00:04:21,272 --> 00:04:26,450 one to a derivative with respect to the capital X and the lower case x. 56 00:04:26,450 --> 00:04:30,110 And this transformation can be done by rewriting it soon. 57 00:04:30,110 --> 00:04:34,214 So we basically add in some sense derivative in the numerator and 58 00:04:34,214 --> 00:04:38,470 denominator and so this derivative[UNKNOWN] capital X over one is 59 00:04:38,470 --> 00:04:43,486 calculated from here And derivative of d lowercase x over x1 is calculated from 60 00:04:43,486 --> 00:04:49,157 here. And the first one is, as you can see, 61 00:04:49,157 --> 00:04:54,54 uses m1 over the total mass and the second one is give us basically minus 62 00:04:54,54 --> 00:04:57,712 sine. So this is the result. 63 00:04:57,712 --> 00:05:03,96 of this derivative when expressed through capital s and small s. 64 00:05:03,96 --> 00:05:06,876 We can do the same thing of course for x two and this is the result with the only 65 00:05:06,876 --> 00:05:11,470 difference here being m two and here being the plus sign. 66 00:05:11,470 --> 00:05:15,340 And that's, pretty much it. So we actually calculate the quantity 67 00:05:15,340 --> 00:05:21,850 that we're after, written here. by, essentially squaring this result. 68 00:05:21,850 --> 00:05:25,510 So this would give us the x particles[UNKNOWN] for the first particle 69 00:05:25,510 --> 00:05:29,770 and this will, this guy squared will give us the, x component that applies into the 70 00:05:29,770 --> 00:05:35,710 second particle. So therefore this guy goes here and you 71 00:05:35,710 --> 00:05:40,497 know, this result goes here. And so you can see if we square this 72 00:05:40,497 --> 00:05:43,687 operation differ it in all derivatives we're going to get now the second 73 00:05:43,687 --> 00:05:47,317 derivative for this vector center of mass, the second derivative with respect 74 00:05:47,317 --> 00:05:50,837 to the center of coordinate the, to the relative coordinate but also we're 75 00:05:50,837 --> 00:05:54,577 going to get the cross terms and actually the coordinates were chosen in such a way 76 00:05:54,577 --> 00:05:58,152 that the cross terms in these 2 brackets cancel each other out because of the 77 00:05:58,152 --> 00:06:07,695 minus sign here and the plus sign here. So well you can verify this, of course, 78 00:06:07,695 --> 00:06:11,471 just by writing it down, or just by staring it, staring at it for a minute or 79 00:06:11,471 --> 00:06:15,760 so, but it, it's going to be pretty clear. 80 00:06:15,760 --> 00:06:18,444 So in the, if you put everything together from these two brackets for a 81 00:06:18,444 --> 00:06:21,795 straightforward calculation, this is what we're going to get. 82 00:06:21,795 --> 00:06:25,689 So this is essentially a result for this, for the X component of this quantity we 83 00:06:25,689 --> 00:06:30,295 are calculating. And so a Laplacian in terms of X1 and X2 84 00:06:30,295 --> 00:06:35,650 can now be written essentially as a combination of 2 Laplacians with a Yeah, 85 00:06:35,650 --> 00:06:41,580 there should be, of course, a plus sign here. 86 00:06:41,580 --> 00:06:46,821 There's no minus sign so it was a typo. And so the first Laplacian with respect 87 00:06:46,821 --> 00:06:51,330 to the center of mass involves the total mass in the denominator. 88 00:06:51,330 --> 00:06:55,70 And the second applies within respect to the relative coordinate this guy, which 89 00:06:55,70 --> 00:07:00,302 actually is called the reduced mass. So now the way we can write it as m1 m2 90 00:07:00,302 --> 00:07:05,739 divided by m1 plus m2, and this is called reduced mass. 91 00:07:07,610 --> 00:07:11,940 So, now we're in the position to actually put everything together. 92 00:07:11,940 --> 00:07:17,125 So what we have proven, as a matter of fact, is that minus h squared over 2 m 1 93 00:07:17,125 --> 00:07:22,480 Leplacian with respect to 1, the kinetic energy, the first particle minus h 94 00:07:22,480 --> 00:07:32,240 squared over 2 m 2, leplacian of 2 the kinetic energy of the second particle. 95 00:07:32,240 --> 00:07:40,692 Can be written as minus h squared over 2 m 1 plus m 2. 96 00:07:40,692 --> 00:07:46,695 the glassian with respect to the sensor of mass minus h squared over 2 mew the 97 00:07:46,695 --> 00:07:52,400 reduced mass that I introduced in the previous. 98 00:07:52,400 --> 00:07:52,913 Slide, La Plassian with respect to the relative coordinate. 99 00:07:52,913 --> 00:08:01,418 If I stick it into the Schongerd equation, I will get therefore, the full 100 00:08:01,418 --> 00:08:09,923 length, I will get minus h squared over 2, this La Plassian of r and 1 plus m2 101 00:08:09,923 --> 00:08:21,507 plus Laplacian lower case r. Over mu, plus the, potential energy and 102 00:08:21,507 --> 00:08:26,577 the interaction energy between the two particles, which depends only on this 103 00:08:26,577 --> 00:08:32,110 relative coordinate, acting on my wave function. 104 00:08:32,110 --> 00:08:34,900 And the right-hand side is just the energy as usual. 105 00:08:34,900 --> 00:08:38,786 So what I see from this equation and that, is that there is just one part 106 00:08:38,786 --> 00:08:44,820 Which depends on the relative coordinate and which involves interaction. 107 00:08:44,820 --> 00:08:50,515 And this part represents essentially a free Shrodinger equation for a combined 108 00:08:50,515 --> 00:08:56,920 particle with the mass M1 plus M2. It essentially describes a free motion. 109 00:08:56,920 --> 00:09:01,31 Of 2 particles as a whole. So for instance, if they were to form a 110 00:09:01,31 --> 00:09:04,346 bounce state this exactly what we're going to be discussing in the last video 111 00:09:04,346 --> 00:09:08,958 in this lecture. So this term essentially describes the 112 00:09:08,958 --> 00:09:14,940 motion of this bounce state this sort of molecule like state as a whole. 113 00:09:14,940 --> 00:09:19,540 And so to sort of extract this overall motion. 114 00:09:19,540 --> 00:09:24,714 We can, just look for a solution. This psi tilda of capital R and, lower 115 00:09:24,714 --> 00:09:29,4 case r can be written as a solution to this, sort of free Schrodinger equation, 116 00:09:29,4 --> 00:09:35,50 which is simply a plane wave as we know. It's e to the power i. 117 00:09:35,50 --> 00:09:38,230 Some momentum of this bound state or whatever it is. 118 00:09:38,230 --> 00:09:43,105 That's capital r divided by h bar, times, well, the wave function, which depends on 119 00:09:43,105 --> 00:09:48,170 the, relative coordinate that we actually want to find. 120 00:09:48,170 --> 00:09:53,60 And, the energy of this, motion from here is going to be just this capital p. 121 00:09:53,60 --> 00:09:59,126 squared divided by m one plus m two, twice and one plus m two. 122 00:09:59,126 --> 00:10:03,502 So, which is just a parameter. It doesn't appear anywhere else in the 123 00:10:03,502 --> 00:10:07,590 problem and we can just move it, to the, right hand side subtract the total energy 124 00:10:07,590 --> 00:10:12,380 from the total energy of this. Energy. 125 00:10:12,380 --> 00:10:15,996 That's it. So the resulting, the final result that 126 00:10:15,996 --> 00:10:20,316 we're going to get, which doesn't lot include this theorem, is going to look, 127 00:10:20,316 --> 00:10:24,204 if we just write it here, is going to look as so, so it's going to be minus 128 00:10:24,204 --> 00:10:31,80 eight squared over two[UNKNOWN], so this is a reduced mass. 129 00:10:31,80 --> 00:10:39,760 So this plus which I can write as D two over D R Squared plus v of r side of r. 130 00:10:39,760 --> 00:10:46,224 And the left hand side is going to be this e minus p squared over 2 and 1 plus 131 00:10:46,224 --> 00:10:51,300 m2. So let me just write it as e prime psi of 132 00:10:51,300 --> 00:10:55,207 r. Okay and this as you can see looks like 133 00:10:55,207 --> 00:11:00,780 Single particle Schrodinger equation. So, it depends only on, the wave function 134 00:11:00,780 --> 00:11:04,360 depends only on one coordinate. Everything else is exactly the same. 135 00:11:04,360 --> 00:11:08,248 The only difference is here I'm dealing with some reduced mass. 136 00:11:08,248 --> 00:11:11,896 So, let's see in the context of the problem of two electrons if they're 137 00:11:11,896 --> 00:11:19,0 actually going to start here. So m1 and so e, m1 is equal to m2. 138 00:11:19,0 --> 00:11:25,700 So if m1 is equal to m2. So then fo course, the total mass is 139 00:11:25,700 --> 00:11:30,940 equal to well, 2m. 2, 2m. 140 00:11:30,940 --> 00:11:39,380 And the reduced mass of this new Is equal to m squared over 2m, or just m over 2. 141 00:11:39,380 --> 00:11:43,382 So therefore, in our electron problem, this guy is going to be just the electron 142 00:11:43,382 --> 00:11:48,208 mass. So and just to summarize what we have 143 00:11:48,208 --> 00:11:54,284 proven By this sort of brute force straight forward calculation is that in 144 00:11:54,284 --> 00:12:01,144 this simple two particle problem, in two particle quantum mechanics, we can always 145 00:12:01,144 --> 00:12:09,150 get treat of one sort of unnecessary quantum. 146 00:12:09,150 --> 00:12:12,990 It describes a motion of our two particles as a whole And we can reduce 147 00:12:12,990 --> 00:12:16,766 our problem to the standard single-particle quantum mechanics which 148 00:12:16,766 --> 00:12:20,798 makes things much easier because we can now use the result and some of the 149 00:12:20,798 --> 00:12:25,406 conclusions of single-particle quantum mechanics that we saw in the previous 150 00:12:25,406 --> 00:12:29,673 lecture.