1 00:00:01,540 --> 00:00:05,632 To calculate the remaining metrics elements, in particular, the metrics 2 00:00:05,632 --> 00:00:09,687 element the first metrics element that we saw in the previous slide. 3 00:00:09,688 --> 00:00:14,599 So let me recall a distinction between the [inaudible] vector of psi used in the 4 00:00:14,599 --> 00:00:17,964 direct notations, and the wave function, psi of x. 5 00:00:17,965 --> 00:00:21,070 As it appears, let's say, in the Schrodinger equation. 6 00:00:21,071 --> 00:00:26,200 So the former really represents an abstract state of the system which is sort 7 00:00:26,200 --> 00:00:29,430 of unrelated to our choice of describing it. 8 00:00:29,430 --> 00:00:32,389 So, it sort of doesn't rely on the particular choice of the coordinates. 9 00:00:32,390 --> 00:00:36,635 Well psi of x is a specific what we call representation, which is much like 10 00:00:36,635 --> 00:00:40,083 coordinates of a vector. So, the relation between the two is the 11 00:00:40,083 --> 00:00:43,164 full length. So it's basically matrix elements of psi 12 00:00:43,164 --> 00:00:45,620 of x is a matrix element between x and psi. 13 00:00:45,620 --> 00:00:48,920 You know, and both are abstract vectors and some Hilbert space. 14 00:00:48,920 --> 00:00:52,888 Now a particular example of this wave function wave for example is this plane 15 00:00:52,888 --> 00:00:56,830 wave which just grabs the particle. Moving with a certain momentum. 16 00:00:56,830 --> 00:01:00,232 I don't want to discuss too much the choice of the coefficient at this stage, 17 00:01:00,232 --> 00:01:02,300 we're going to discuss it later in the course. 18 00:01:02,300 --> 00:01:07,490 In any case, we definitely have seen this function before and we know it describes a 19 00:01:07,490 --> 00:01:11,948 particle with a given momentum. Now the question that you're really asking 20 00:01:11,948 --> 00:01:16,096 now in relation with this matrix element It was a, what is, what is the wave 21 00:01:16,096 --> 00:01:19,890 function of a particle with the given coordinate, let's say x pri. 22 00:01:19,890 --> 00:01:24,182 So, in order to answer this question we need to remind ourselves to introduce for 23 00:01:24,182 --> 00:01:28,662 those of who[UNKNOWN] know the notion of the Dirac delta-function which basically 24 00:01:28,662 --> 00:01:32,913 is exactly this. Matrix elements we're after. 25 00:01:32,913 --> 00:01:38,380 So in order to introduce it, let me recall a very few, very simple facts from the 26 00:01:38,380 --> 00:01:42,295 usual linear algebra. So let's say if we have a finite 27 00:01:42,295 --> 00:01:48,703 dimensional vector space with some cartesian coordinate system and basis 28 00:01:48,703 --> 00:01:54,666 vectors, let's say e1 E2 well in principle it's every 3 or 4 so the Cartesian 29 00:01:54,666 --> 00:02:00,362 Coordinate system implies that we, the product, the dot product of two basis 30 00:02:00,362 --> 00:02:05,347 vectors, e sub i and e sub g is equal to zero if they are different. 31 00:02:05,348 --> 00:02:09,715 Which means they are orthogonal to each other or is equal to 1 if this is the same 32 00:02:09,715 --> 00:02:12,939 vector. Which means that the, normal trajectory is 33 00:02:12,939 --> 00:02:16,148 equal to 1. And, so the, symbol, which sort of 34 00:02:16,148 --> 00:02:21,692 summarizes, this, relation, is the chronicer delta symbol, delta IG, which is 35 00:02:21,692 --> 00:02:25,464 defined as this. Of course if we're going to calculate the 36 00:02:25,464 --> 00:02:30,958 sum, over all let's say i's, of delta IG, we're going to have the sum equal to one, 37 00:02:30,958 --> 00:02:36,452 just because all of them are equal all, all these terms are equal to zero apart 38 00:02:36,452 --> 00:02:41,315 from the one we, where I is equal to G. Now the difference between the sort of 39 00:02:41,315 --> 00:02:46,273 simplistic picture of the usual basis vectors and the usual linear vector space, 40 00:02:46,273 --> 00:02:50,561 and the basis vectors in quantum mechanics in, i-, i-, is that the basis we're 41 00:02:50,561 --> 00:02:55,580 dealing now is sort of continual base. So the basis vectors are really sort of 42 00:02:55,580 --> 00:02:59,960 some [unknown] vectors which are labeled by coordinates, and there are infinitely 43 00:02:59,960 --> 00:03:03,894 many coordinates and there's sort of a continuum space of this coordinates. 44 00:03:03,895 --> 00:03:08,625 But if we sort of use the analogy between, let's say e i, and x, and e g, and x 45 00:03:08,625 --> 00:03:13,739 prime, so this product is much like a [unknown] product of these two vectors. 46 00:03:13,740 --> 00:03:18,640 So using this analogy we can demand, therefore that if x is not equal to x 47 00:03:18,640 --> 00:03:23,740 prime, this measured [unknown] and therefore, this delta function, which is 48 00:03:23,740 --> 00:03:28,010 sort of a continuum version of this delta symbol, are equal to zero. 49 00:03:28,010 --> 00:03:32,906 And if x is equal to x prime and something else and so there's something else that is 50 00:03:32,906 --> 00:03:37,802 determined by a condition which is very similar to this condition but now instead 51 00:03:37,802 --> 00:03:42,122 of summing over a discreet set of vectors by i, we have to integrate over a 52 00:03:42,122 --> 00:03:46,280 continuum set of vectors labeled by x. And so these are all possible coordinates 53 00:03:46,280 --> 00:03:49,489 in one dimensional space going from minus infinity to plus infinity. 54 00:03:49,490 --> 00:03:53,896 And the corresponding the corresponding delta function, the integral of the 55 00:03:53,896 --> 00:03:56,375 corresponding delta function is equal to one. 56 00:03:56,376 --> 00:04:02,381 And so therefore, if we try to visualize this delta function, it's something which 57 00:04:02,381 --> 00:04:07,177 is equal to zero everywhere. So this is our delta of x, and this is x. 58 00:04:07,178 --> 00:04:12,550 So we have something which is equal to zero everywhere apart from just one single 59 00:04:12,550 --> 00:04:17,670 point where this function is so large, it has such a huge high peak that the 60 00:04:17,670 --> 00:04:22,320 integral collected from this single peak gives us a finite result. 61 00:04:22,320 --> 00:04:27,327 And well, this sort of implies that the value of this delta function at x equals 62 00:04:27,327 --> 00:04:32,406 x-prime is, is infinity. And these, conditions are the canonical 63 00:04:32,406 --> 00:04:36,903 conditions that determine the Dirac delta function. 64 00:04:36,903 --> 00:04:41,224 Now, this Dirac delta function has many, properties that can be discussed. 65 00:04:41,224 --> 00:04:46,338 So, I don't have time for that now but let me just mention one important property, 66 00:04:46,338 --> 00:04:50,392 which is essentially the Fourier transform or the delta function. 67 00:04:50,392 --> 00:04:54,182 So we, we, know that any reasonable function and, in this case, actually, 68 00:04:54,182 --> 00:04:58,987 unreasonable function too can be expanded in terms of these sort of, plain wave like 69 00:04:58,987 --> 00:05:01,832 functions. And this expansion is called the Fourier 70 00:05:01,832 --> 00:05:05,223 transform so if you the Fourier transform, the DIrac delta function. 71 00:05:05,223 --> 00:05:11,052 It will look like that so essentially an integral from minus infinity to plus 72 00:05:11,052 --> 00:05:16,252 infinity of dk over 2 pi either for ikx. So these are well, things that we need now 73 00:05:16,252 --> 00:05:18,974 to calculate the remaining matrix elements. 74 00:05:18,974 --> 00:05:21,274 So, one of them in some sense have already calculate. 75 00:05:21,275 --> 00:05:26,990 So let me just put everything together and clean up this space a little bit. 76 00:05:26,990 --> 00:05:32,760 So, one matrix element that we can immediately write down a by analogy with 77 00:05:32,760 --> 00:05:38,898 this relation we just arrived is the matrix element between the neighboring 78 00:05:38,898 --> 00:05:44,385 points x sub k and x sub k plus 1, which can be written just in one to one 79 00:05:44,385 --> 00:05:50,137 correspondence with this result. We just replaced x with xk plus 1 and x 80 00:05:50,137 --> 00:05:56,100 prime with x sub k and we get this delta function of the difference which we'll 81 00:05:56,100 --> 00:06:00,370 also call later on delta x. Now, in a very similar way. 82 00:06:00,370 --> 00:06:07,320 We can now calculate, also, the, matrix element of the, potential, v of x. 83 00:06:07,320 --> 00:06:11,120 So since v has the potential depends on the, on the coordinate. 84 00:06:11,120 --> 00:06:16,250 And this vector, this state vector is the eigenstate of the coordinate. 85 00:06:16,250 --> 00:06:21,985 Means that v of x, when acting on the x sub key just gives us. 86 00:06:21,986 --> 00:06:27,386 The number v of x of k which can be factored out and the remaining matrix 87 00:06:27,386 --> 00:06:31,877 element we just calculate it so the result is simply this. 88 00:06:31,878 --> 00:06:38,676 So a slightly less trivial matrix element appears from the kinetic energy where we 89 00:06:38,676 --> 00:06:44,505 have to calculate the matrix element of the p squared over 2m and So the x of k is 90 00:06:44,505 --> 00:06:50,256 not and eigenfunction of momentum p. So in order to calculate it, do we can use 91 00:06:50,256 --> 00:06:55,500 again the resolution of the identity that we discussed before, but now formulated 92 00:06:55,500 --> 00:06:58,543 for the momentum rather than for coordinates. 93 00:06:58,544 --> 00:07:03,234 So this guy is equal to 1. And we can insert this resolution of the 94 00:07:03,234 --> 00:07:08,527 identity in between this operator p squared over 2m and x sub k, and if we do 95 00:07:08,527 --> 00:07:12,424 so we get this integral. So here are placed the dummy variable of 96 00:07:12,424 --> 00:07:15,959 integration. Instead of writing it as p here, I just 97 00:07:15,959 --> 00:07:20,977 write it as p sub k, and, but it's the same resolution of the identity. 98 00:07:20,978 --> 00:07:25,958 And so here now, so the, momentum operator, this kinetic energy, does act on 99 00:07:25,958 --> 00:07:30,064 its eigenstate. So therefore this becomes simply P 100 00:07:30,064 --> 00:07:34,817 sub-case squared over 2m, which is a number that can be factored out. 101 00:07:34,818 --> 00:07:39,726 And there are many matrix elements this guy and this guy has to be compared, has 102 00:07:39,726 --> 00:07:43,350 to be compared with this expression with the plain wave. 103 00:07:43,350 --> 00:07:47,619 So these are simply plain waves. And so if we put everything together for 104 00:07:47,619 --> 00:07:51,714 this room, we get the following results with the matrix element of the kinetic 105 00:07:51,714 --> 00:07:55,619 energy. So which involves an integral o over a 106 00:07:55,619 --> 00:07:58,820 momentum. In this case p of k is the momentum. 107 00:07:58,820 --> 00:08:04,780 So in order to write all measuring supplements in a similar fashion we can 108 00:08:04,781 --> 00:08:11,742 use this now this relation for the delta function which with the fourier transform 109 00:08:11,742 --> 00:08:16,682 put the delta function and just replace key with a p over e bar. 110 00:08:16,682 --> 00:08:22,870 So if we do so, we get essentially the same integral but now, integrate it over a 111 00:08:22,870 --> 00:08:28,977 momentum and the exponential here will, will i over each bar, p sub p times x. 112 00:08:28,978 --> 00:08:33,889 So but it's essentially the same the same expression for the delta function and we 113 00:08:33,889 --> 00:08:38,719 can use it here to write the matrix element in terms of this in terms of this 114 00:08:38,719 --> 00:08:43,030 integral. So you see that all three metrics elements 115 00:08:43,030 --> 00:08:48,022 that appear in the in our previous calculation can be written in a very 116 00:08:48,022 --> 00:08:51,922 similar fashion. Now, the good news is that we have reached 117 00:08:51,922 --> 00:08:57,214 our final slide and we are now in the position to put everything together and 118 00:08:57,214 --> 00:09:03,444 well, get to the main result of Feynman. We have done in the previous slide, we 119 00:09:03,444 --> 00:09:09,905 have calculated the, the matrix elements that we need to, to describe the 120 00:09:09,905 --> 00:09:13,917 propagator and so these are these matrix elements. 121 00:09:13,918 --> 00:09:16,890 So we write them sort of in the uniform fashion. 122 00:09:16,890 --> 00:09:21,312 So the, there is this one coming from his term and there is a kinetic and potential 123 00:09:21,312 --> 00:09:25,746 energy from the cumulitonian. Now why, let's recall why do we even 124 00:09:25,746 --> 00:09:29,162 bother you know to think about this [unknown] element. 125 00:09:29,162 --> 00:09:35,098 So, where it comes from is really the this sort of picture of the propagators. 126 00:09:35,098 --> 00:09:41,762 So remember we want to find the propagator from an initial point xi or 0 to the final 127 00:09:41,762 --> 00:09:47,402 point xf and a time t. And so what we did, we split the time 128 00:09:47,402 --> 00:09:54,134 interval from 0 to t into n a very, very small time intervals delta t and in doing 129 00:09:54,134 --> 00:10:00,764 so we have represented our initial propagature as a product essentially of 130 00:10:00,764 --> 00:10:07,596 this small propagature. Going let's say from this point to x1, 131 00:10:07,596 --> 00:10:13,110 from x1 to x2 etc., etc. And so basically to calculate this, this 132 00:10:13,111 --> 00:10:18,659 many matrix elements you know to calculate the 0 without the delta t, we needed to 133 00:10:18,659 --> 00:10:21,730 calculate this guy and so we have the answer. 134 00:10:21,730 --> 00:10:24,769 So this is the expression. And now we can just use it. 135 00:10:24,769 --> 00:10:27,368 So notice the good thing about this expression. 136 00:10:27,368 --> 00:10:30,789 Well, it looks a little bit cumbersome, but the good thing about this expression 137 00:10:30,789 --> 00:10:32,867 is that it doesn't have any operators any more. 138 00:10:32,868 --> 00:10:34,760 The operators are gone. There are no hats here. 139 00:10:34,760 --> 00:10:37,955 There are only numbers, the regular numbers we're used to. 140 00:10:37,955 --> 00:10:42,856 So another thing we should, we can notice is that this time interval, delta t, is 141 00:10:42,856 --> 00:10:47,072 very small, because we try and keep our Interval from 0 to 10 to n pieces and n 142 00:10:47,072 --> 00:10:50,270 goes to infinity, so this guy is really, really small. 143 00:10:50,270 --> 00:10:53,750 At this point what we can recall is the[UNKNOWN] Taylor expansion for an 144 00:10:53,750 --> 00:10:56,979 exponential of some small number. So if we have a small number, let's say 145 00:10:56,979 --> 00:11:01,120 epsilon much smaller that 1. So the exponential can be written as 1 146 00:11:01,120 --> 00:11:04,190 plus epsilon, we can neglect the other terms. 147 00:11:04,190 --> 00:11:09,562 So what Feynman does at this, at this stage, he notices that, well this term 148 00:11:09,562 --> 00:11:13,019 here is a small, much like, the epsilon here. 149 00:11:13,020 --> 00:11:16,260 And he uses this, Taylor Expansion sort of backwards. 150 00:11:16,260 --> 00:11:18,365 So instead of writing the exponential as a liner. 151 00:11:18,365 --> 00:11:21,910 Well, 1 plus a linear term. He write this 1 minus a linear term, is an 152 00:11:21,910 --> 00:11:26,233 exponential. So, and, This is the, result that he has. 153 00:11:26,233 --> 00:11:31,200 So the reason he does that is because now the integral that we get is very 154 00:11:31,200 --> 00:11:33,790 convenient. Because this is really the Gaussian 155 00:11:33,790 --> 00:11:37,542 integral that we know how to calculate so you see we have this, we have an integral 156 00:11:37,542 --> 00:11:41,462 over momentum, we have a programmatic term and a linear term and this is something we 157 00:11:41,462 --> 00:11:44,310 have seen before. So if we do the calculation of the 158 00:11:44,310 --> 00:11:46,700 integral so what we'll going to get is this. 159 00:11:46,700 --> 00:11:53,840 So just, you know, the result of exponential m over 2 delta x squared over 160 00:11:53,840 --> 00:11:57,818 delta t squared. But now let's look again at this picture 161 00:11:57,818 --> 00:12:01,738 which shows the trajectory. So you see these are these trajectories 162 00:12:01,738 --> 00:12:06,277 that, various trajectories that go from the initial point to the final point. 163 00:12:06,278 --> 00:12:10,624 And so here, this term sort of corresponds to a single segment, let's say this guy. 164 00:12:10,625 --> 00:12:17,128 Where a delta t is this small dimension. And delta x is really the change in the 165 00:12:17,128 --> 00:12:20,236 coordinate. Well, basically, you may think about the 166 00:12:20,236 --> 00:12:23,520 distance that the particle propagates in this time delta t. 167 00:12:23,520 --> 00:12:28,004 So delta x or delta t simply, can interpret as simply the velocity of the 168 00:12:28,004 --> 00:12:31,629 particle. So therefore, if we write it, so we, we 169 00:12:31,629 --> 00:12:37,219 can see that this matrix element is nothing but either the power I over H, MV 170 00:12:37,219 --> 00:12:43,239 squared over two, so basically the kinetic energy minus the potential energy times 171 00:12:43,239 --> 00:12:47,740 Delta T. So and this is something which has a name 172 00:12:47,740 --> 00:12:55,090 in classical physics, so maybe some of you know it, and this Name is called the 173 00:12:55,090 --> 00:12:59,557 Lagrangian. Let me call it L sub k. 174 00:12:59,558 --> 00:13:04,595 And so in these notations we can write the result for the matrix element of U of 175 00:13:04,595 --> 00:13:07,714 delta T is simply E to the power of I over H bar. 176 00:13:07,715 --> 00:13:15,114 L sub k times delta t. So using this result we can now rewrite 177 00:13:15,114 --> 00:13:19,910 our main object this propagature as follows. 178 00:13:19,910 --> 00:13:24,932 So we're going to have a bunch on integrals over x1, this point, this point, 179 00:13:24,932 --> 00:13:27,938 this point. Uih, and of this exponentials of this 180 00:13:27,938 --> 00:13:31,066 Lagrangians. First point it was at the first segment, 181 00:13:31,067 --> 00:13:36,316 second segment thir segment etcetera. So notice that eseentially these integrals 182 00:13:36,316 --> 00:13:40,537 over X1, X2 etcetera sort of correspond to moving aroudn these poitns tehse 183 00:13:40,537 --> 00:13:44,030 intermediate points between Xi and Xf. The endpoints. 184 00:13:44,030 --> 00:13:47,690 Are completely fixed. But the intermediates intermediate points 185 00:13:47,690 --> 00:13:51,046 are free. And these integrals essentially symbolize 186 00:13:51,046 --> 00:13:54,378 this freedom. So now the last step here in the 187 00:13:54,378 --> 00:14:00,328 continuation is you have to recognize that, so here the product of exponential, 188 00:14:00,328 --> 00:14:05,149 it can be written of the exponential of the sum, i over h bar. 189 00:14:05,150 --> 00:14:12,696 L sub k delta t. And as delta t the interval, time interval 190 00:14:12,696 --> 00:14:19,640 goes to zero, we can simply write it as e to the power i over h bar, an integral 191 00:14:19,640 --> 00:14:24,658 well from 0 to t l d t. And the integral of the Lagrangien over 192 00:14:24,658 --> 00:14:28,119 time is called is known as classical action. 193 00:14:28,120 --> 00:14:34,295 So what we got is essentially the main result that the propogator can be written 194 00:14:34,295 --> 00:14:39,398 as an integral over all possible trajectories, all possible intermediate 195 00:14:39,398 --> 00:14:44,760 coordinates of the weighted with this exponential of the classic election. 196 00:14:44,760 --> 00:14:47,758 So this is the result that we sort of advertised in the very beginning. 197 00:14:47,758 --> 00:14:53,372 And Feynman introduced essentially a symbol, a notation, instead of writing a 198 00:14:53,372 --> 00:14:59,662 bunch of these Intermediate step, so if you can introduce a single symbol which is 199 00:14:59,662 --> 00:15:04,608 this capital curly d of x of t which is called the path integral. 200 00:15:04,608 --> 00:15:08,689 So, it's just a symbol. But it turns out that it's actually a very 201 00:15:08,689 --> 00:15:11,500 convenient symbol. So now our derivation is over so this is 202 00:15:11,500 --> 00:15:14,416 again, this is the main result, so maybe let me just write it. 203 00:15:14,416 --> 00:15:19,542 Sort of underline it again, so this is our main result. 204 00:15:19,542 --> 00:15:24,414 And so, I know it's been a very long derivation, but as we'll see in the next 205 00:15:24,414 --> 00:15:29,202 lecture, so this result is extremely useful in that, it will allow us to get, 206 00:15:29,202 --> 00:15:34,750 well from now on, without extraordinary calculations, extremely profound and 207 00:15:34,750 --> 00:15:39,045 interesting physical facts. So we're going to discuss the 208 00:15:39,045 --> 00:15:43,961 correspondence between classical and quantum mechanics. 209 00:15:43,962 --> 00:15:46,998 We're going to discuss so called quantum localization, we're going to discuss a 210 00:15:46,998 --> 00:15:49,070 very interesting process of quantum dissipation. 211 00:15:49,070 --> 00:15:52,937 So there are a lot of cool things that can be studied now with this object. 212 00:15:52,938 --> 00:15:57,098 But well, it did take us a while to derive it, so hope some of you got the main 213 00:15:57,098 --> 00:15:58,693 points of this derivation.