1 00:00:03,410 --> 00:00:09,220 Hello, bonjour, salamatsizbe, dobry den, 2 00:00:09,730 --> 00:00:16,390 welcome to the sixth tutorial of Statistical Mechanics: Algorithms and Computations, 3 00:00:16,390 --> 00:00:20,390 from the Physics Department of Ecole Normale Superieure. 4 00:00:20,390 --> 00:00:23,550 In this week's lecture and homework 5 00:00:23,550 --> 00:00:28,490 action was all about the Levy quantum path 6 00:00:28,490 --> 00:00:32,490 one-dimensional geometrical object 7 00:00:32,490 --> 00:00:34,450 that are easy to sample 8 00:00:34,450 --> 00:00:38,450 yet that carry a profound message. 9 00:00:39,010 --> 00:00:45,370 They lend a geometrical interpretation to the Heisenberg uncertainty principle: 10 00:00:46,180 --> 00:00:49,370 at high temperature and high energy 11 00:00:50,120 --> 00:00:57,260 a particle is at position x with probability rho(x, x, beta) 12 00:00:58,830 --> 00:01:02,680 the quantum path connecting x to x 13 00:01:02,680 --> 00:01:07,910 has no fluctuations and the particle is really located at x. 14 00:01:07,910 --> 00:01:11,540 Likewise, the density matrix 15 00:01:11,540 --> 00:01:11,910 has no off-diagonal part Likewise, the density matrix 16 00:01:11,910 --> 00:01:14,110 has no off-diagonal part 17 00:01:15,030 --> 00:01:18,110 In contrast, at low temperature 18 00:01:18,550 --> 00:01:22,110 the particle is uncertain in its position x 19 00:01:23,470 --> 00:01:26,110 as the quantum path 20 00:01:26,610 --> 00:01:32,390 fluctuates wildly from x to x, for beta being large 21 00:01:32,970 --> 00:01:39,270 This is also mirrored in the large off-diagonal part of the density matrix 22 00:01:39,920 --> 00:01:44,690 In this tutorial, we touch on the second pillar 23 00:01:44,690 --> 00:01:50,430 of quantum statistical mechanics, besides the Heisenberg uncertainty principle 24 00:01:50,430 --> 00:01:54,550 namely the indistinguishability of particles, 25 00:01:54,830 --> 00:01:57,480 in our case bosons. 26 00:01:58,740 --> 00:02:01,480 For our introduction, this week 27 00:02:01,480 --> 00:02:09,240 and next week we will concentrate on the case of non-interacting ideal bosons. 28 00:02:10,200 --> 00:02:13,240 This is not really a restriction 29 00:02:13,240 --> 00:02:17,240 because non-interacting bosons 30 00:02:17,240 --> 00:02:25,130 are the only system in all of physics that have a phase transition without interactions. 31 00:02:26,560 --> 00:02:31,320 In fact, a curious statistical interaction 32 00:02:31,320 --> 00:02:35,320 appears in a system of indistinguishable bosons 33 00:02:35,320 --> 00:02:39,710 and Michael and Alberto will trace it down 34 00:02:39,710 --> 00:02:41,710 in this week's tutorial. 35 00:02:43,430 --> 00:02:48,340 Unfortunately Vivien cannot be with us for a couple of weeks 36 00:02:48,340 --> 00:02:54,360 but he'll come back for our discussion of classical spin systems. 37 00:02:54,930 --> 00:02:59,900 We will put our ideal bosons into a harmonic trap 38 00:02:59,900 --> 00:03:03,900 and this same system has set the stage 39 00:03:03,900 --> 00:03:07,900 for ground-breaking experiments in atomic physics, 40 00:03:07,900 --> 00:03:13,030 where Bose-Einstein condensation has lead to a revolution 41 00:03:13,670 --> 00:03:17,930 in experimental glass cells just like this one. 42 00:03:19,030 --> 00:03:21,500 We will follow this revolution 43 00:03:21,500 --> 00:03:25,500 you will follow this ground-breaking experiment 44 00:03:25,500 --> 00:03:32,730 in your own path integral Monte Carlo calculation, using the path integral picture. 45 00:03:34,250 --> 00:03:38,170 but before doing this, Michael and Alberto 46 00:03:38,170 --> 00:03:45,400 will discuss Bose-Einstein condensation using wavefunctions and energy levels. 47 00:03:45,400 --> 00:03:50,470 This is necessary in order to understand what it is all about 48 00:03:50,990 --> 00:03:54,900 the Bose-Einstein condensation into the ground-state. 49 00:03:55,980 --> 00:03:59,250 They will then discuss all of this 50 00:03:59,250 --> 00:04:03,250 in a simple of a few trapped bosons. 51 00:04:08,840 --> 00:04:14,440 Before putting many particles into a three-dimensional trap, let us first consider 52 00:04:14,440 --> 00:04:15,850 a single particle. 53 00:04:15,850 --> 00:04:22,450 This means you enumerate all single particles states with a given energy E. 54 00:04:23,030 --> 00:04:27,630 In three dimensions, the harmonic trap has three spring constants 55 00:04:27,630 --> 00:04:33,270 one in x, one in y and one in z direction, as shown in the picture here. 56 00:04:34,560 --> 00:04:36,210 Just like last week 57 00:04:36,210 --> 00:04:41,820 we set all the spring constants omega_x, omega_y, omega_z = 1, 58 00:04:41,820 --> 00:04:47,850 so that the energies E_x, E_y, E_z 59 00:04:47,850 --> 00:04:50,990 are equal to 0, 1, 2, 3, 4, .. 60 00:04:51,600 --> 00:04:55,620 For simplicity, we set the ground-level energy equal to zero 61 00:04:55,620 --> 00:04:57,660 and label all the states 62 00:04:57,660 --> 00:05:06,010 by the energy E_x, E_y, E_z, so that they have the total energy E = E_x + E_y + E_z 63 00:05:07,390 --> 00:05:10,010 Let us now write a little program 64 00:05:10,010 --> 00:05:12,340 naive_single_particle.py 65 00:05:12,340 --> 00:05:18,820 to enumerate all these states. Output of the program is shown here 66 00:05:27,730 --> 00:05:31,200 For example here we have state 12 67 00:05:31,200 --> 00:05:34,460 it is in the ground state in x 68 00:05:34,460 --> 00:05:36,910 in the second excited states in y 69 00:05:36,910 --> 00:05:39,920 and in the first excited state in z. 70 00:05:41,140 --> 00:05:44,130 There are many states, but notice that 71 00:05:44,130 --> 00:05:46,990 each tuple E_x, E_y, E_z 72 00:05:47,250 --> 00:05:49,600 corresponds to a unique 73 00:05:49,600 --> 00:05:53,600 single-particle quantum state of energy E 74 00:05:53,600 --> 00:05:56,790 E = E_x + E_y + E_z 75 00:05:57,370 --> 00:06:02,360 Let's have a look at them, there's one ground-state of energy zero 76 00:06:02,360 --> 00:06:05,000 there are three states of energy one 77 00:06:05,000 --> 00:06:08,440 there are six states of energy two 78 00:06:08,440 --> 00:06:10,890 ten states of energy three 79 00:06:10,890 --> 00:06:15,980 fifteen states of energy four, and so on.. 80 00:06:16,650 --> 00:06:18,130 Do you see a pattern? 81 00:06:19,260 --> 00:06:23,300 So let us now compute the number of choices 82 00:06:23,300 --> 00:06:27,300 E_x, E_y, E_z that give an energy three. 83 00:06:27,300 --> 00:06:32,160 For a harmonic trap, the number of states with a given energy E 84 00:06:32,160 --> 00:06:36,160 can be calculated explicitly, as shown here. 85 00:06:36,160 --> 00:06:41,250 This expression means that for each choice of E_x 86 00:06:41,250 --> 00:06:47,660 we can choose E_y as an integer between 0 and (E - E_x) 87 00:06:49,230 --> 00:06:51,660 given E_x and E_y, 88 00:06:51,660 --> 00:06:55,660 then the remainder E_z is fixed. 89 00:06:55,660 --> 00:06:59,000 This means for each choice of E_x, 90 00:06:59,000 --> 00:07:04,250 we have (E - E_x + 1) choices for E_y 91 00:07:05,660 --> 00:07:08,250 This makes that we have 92 00:07:08,250 --> 00:07:12,250 (E + 1) + (E) + (E - 1) + .. 93 00:07:12,250 --> 00:07:17,020 .. + 1 choices 94 00:07:17,020 --> 00:07:24,120 In total, this makes (E + 2) (E + 1) / 2 choices. 95 00:07:25,240 --> 00:07:28,850 This corresponds to one choice for E=0 96 00:07:28,850 --> 00:07:36,390 three choices for E=1, six choices for E=2 and so on 97 00:07:38,360 --> 00:07:41,780 But let use now use a more systematic method 98 00:07:41,780 --> 00:07:45,780 to compute the number of states with an energy E 99 00:07:45,780 --> 00:07:50,060 Alberto will soon use it in a more involved context 100 00:07:50,060 --> 00:07:54,750 The method consists in writing the number of states 101 00:07:54,750 --> 00:07:58,750 as a free sum over E_x, E_y and E_z 102 00:07:58,750 --> 00:08:03,280 with a condition that we implement by means of Kronecker delta functions 103 00:08:03,280 --> 00:08:08,310 you simply sum up the states, as is shown in this equation here 104 00:08:08,310 --> 00:08:12,310 where the Kronecker delta is defined as shown here 105 00:08:13,500 --> 00:08:16,310 Because of the Kronecker delta function 106 00:08:16,310 --> 00:08:19,790 only combinations of E_x, E_y, E_z 107 00:08:19,790 --> 00:08:24,230 that sum up to E contribute to the number of states. 108 00:08:25,650 --> 00:08:30,880 We can represent the Kronecker delta function by this integral 109 00:08:31,600 --> 00:08:37,120 We can then enter this integral representation of the Kronecker delta function into the above sum 110 00:08:37,120 --> 00:08:41,120 we then exchange sums and integrals 111 00:08:41,120 --> 00:08:45,120 then we see that the sums have become independent 112 00:08:45,720 --> 00:08:47,370 as is shown here 113 00:08:47,930 --> 00:08:53,520 These are geometric sums that can be evaluated explicitly, as you see here 114 00:08:55,220 --> 00:08:58,560 Finally, we can evaluate the remaining integral 115 00:08:58,560 --> 00:09:03,930 using brute-force Riemann integration, as Alberto will demonstrate in a few minutes. 116 00:09:04,730 --> 00:09:08,590 This numerical integration is all that we will need here 117 00:09:08,590 --> 00:09:12,590 but those of you who feel at home in the complex plane 118 00:09:12,590 --> 00:09:18,290 know of course that this integral can also be evaluated exactly 119 00:09:18,290 --> 00:09:22,290 The substitution exp(i lambda) = z 120 00:09:22,290 --> 00:09:27,420 leads to a complex contour integral that can be evaluated using the residue theorem. 121 00:09:27,420 --> 00:09:33,590 and of course the result is (E + 2) (E + 1) /2 . 122 00:09:33,590 --> 00:09:37,590 Now, let us do two things 123 00:09:37,590 --> 00:09:41,590 first: in our list of single particle states 124 00:09:42,090 --> 00:09:47,340 (shown right here) let us retain only the 35 states of lowest energies 125 00:09:47,340 --> 00:09:51,340 these are states that have energy smaller or equal to four. 126 00:09:52,620 --> 00:09:56,210 We now have a list of 35 states 127 00:09:56,210 --> 00:09:59,160 and we can still look at them here on the screen 128 00:10:00,010 --> 00:10:04,740 second thing: we now put five bosons into these states 129 00:10:06,220 --> 00:10:09,870 Remember that the partition function of the physical system 130 00:10:09,870 --> 00:10:12,990 is equal to the sum over all states 131 00:10:12,990 --> 00:10:15,680 of the probability of each state 132 00:10:15,680 --> 00:10:19,160 and this probability is given by the Boltzmann weight 133 00:10:19,160 --> 00:10:22,050 exp(-beta E) 134 00:10:23,450 --> 00:10:26,300 the partition function of our five bosons system 135 00:10:26,300 --> 00:10:30,300 is then put together from all the different ways we have 136 00:10:30,300 --> 00:10:34,300 to put the five bosons into the single particle states 137 00:10:34,830 --> 00:10:40,160 For example, we might put our red particle into this state here 138 00:10:40,160 --> 00:10:44,160 the blue particle into this state here 139 00:10:44,160 --> 00:10:48,160 the yellow particle into this state here 140 00:10:48,160 --> 00:10:54,800 the green particle into the state here, and finally the pink particle into this state here. 141 00:10:55,980 --> 00:11:00,350 The sum of these possibilities is the partition function 142 00:11:00,350 --> 00:11:02,420 something we can measure 143 00:11:02,710 --> 00:11:08,090 While we should be careful not to forget any of the five-particles states 144 00:11:08,090 --> 00:11:10,690 in our partition function 145 00:11:10,690 --> 00:11:13,800 we are not allowed to overcount them either 146 00:11:14,400 --> 00:11:18,620 The problem arises because bosons are identical particles 147 00:11:18,620 --> 00:11:25,830 they don't just look the same, they are indistinguishable. There's no way to tell them apart. 148 00:11:26,380 --> 00:11:30,420 This three states here are different for particles 149 00:11:30,420 --> 00:11:34,020 that are distinguishable, for example by their color, 150 00:11:34,020 --> 00:11:37,230 but for bosons they are one and the same. 151 00:11:40,310 --> 00:11:44,980 In our partition function, we should count only one of them 152 00:11:45,940 --> 00:11:51,260 this is one of the most profound insights in all of physics 153 00:11:51,790 --> 00:11:55,900 it is due to Bose (in 1923, for photons) 154 00:11:55,900 --> 00:11:59,900 and Einstein (in 1924, for massive bosons) 155 00:11:59,900 --> 00:12:06,530 We can translate this great achievement of Bose and Einstein into a simple algorithm 156 00:12:06,530 --> 00:12:11,340 in order to avoid overcounting many-particle states 157 00:12:12,160 --> 00:12:17,560 we count only states that satisfies that the single-particle state of particle 0 158 00:12:17,560 --> 00:12:21,560 is smaller or equal to the state of particle one 159 00:12:21,560 --> 00:12:24,930 is smaller or equal to the state of particle two 160 00:12:24,930 --> 00:12:28,390 is smaller or equal to the state of particle three 161 00:12:28,390 --> 00:12:32,390 is smaller or equal to the state of particle four. 162 00:12:33,240 --> 00:12:35,400 Out of our three states 163 00:12:35,820 --> 00:12:41,620 we thus pick the last one. The other two have to be dropped from our partition function 164 00:12:42,220 --> 00:12:46,180 Alberto will now leave this graphical discussion 165 00:12:46,180 --> 00:12:50,180 and put our five bosons model with the 35 states 166 00:12:50,180 --> 00:12:51,700 onto the computer. 167 00:12:53,170 --> 00:12:56,080 But before doing so, please take a moment or two 168 00:12:56,080 --> 00:13:00,080 to go over our discussion of single-particle states 169 00:13:00,080 --> 00:13:03,580 and five-particle states, and to download and to run 170 00:13:03,580 --> 00:13:07,580 our nice little program: naive_single_particle.py 171 00:13:07,840 --> 00:13:13,750 that helped us to create these nice states that we have been discussing all along. 172 00:13:18,140 --> 00:13:21,610 The ordering trick that was introduced by Michael 173 00:13:21,610 --> 00:13:25,860 allows us to compute the bosonic partition function 174 00:13:25,860 --> 00:13:28,420 with distinguishable particles 175 00:13:28,420 --> 00:13:30,250 Remember 176 00:13:31,110 --> 00:13:37,010 you put the particle 0 in a state between 0 and 34 177 00:13:37,780 --> 00:13:39,760 then you put the particle 1 178 00:13:39,760 --> 00:13:45,050 in a state between the state of the particle 0 and 34 179 00:13:45,660 --> 00:13:50,770 then you put the particle 2 between the state of the particle 1 180 00:13:50,770 --> 00:13:54,770 and 34, and so on.. 181 00:13:54,770 --> 00:13:58,770 This five-particles state has an energy 182 00:13:58,770 --> 00:14:01,720 E given by this formula 183 00:14:01,720 --> 00:14:06,790 and a statistical weight exp(-beta E) 184 00:14:07,260 --> 00:14:12,580 this allows us to write the bosonic partition function as follows 185 00:14:14,100 --> 00:14:18,460 But maybe you prefer to see this as an algorithm 186 00:14:21,230 --> 00:14:23,830 Note the curious line 187 00:14:25,010 --> 00:14:27,830 the multiplication of a list 188 00:14:27,830 --> 00:14:31,830 with an integer factor k simply concatenates 189 00:14:32,040 --> 00:14:34,010 the list k times 190 00:14:34,620 --> 00:14:40,080 So that [1] * 3 is simply [1, 1, 1]. 191 00:14:41,160 --> 00:14:44,080 Here we compute the partition function 192 00:14:45,630 --> 00:14:48,770 the multiple loops are really naive 193 00:14:48,770 --> 00:14:52,770 but we can compute the number of states 194 00:14:52,770 --> 00:14:55,420 which contribute to the partition function 195 00:14:55,420 --> 00:15:00,500 It is a large number: 575757 196 00:15:00,500 --> 00:15:04,500 but it is 90 times smaller 197 00:15:04,500 --> 00:15:08,500 than the number of states for distinguishable particles 198 00:15:08,500 --> 00:15:11,050 which is 35^5 199 00:15:12,030 --> 00:15:15,050 can we compute this number? Of course 200 00:15:15,050 --> 00:15:20,470 Let's look again at the bosonic states, and erase all information 201 00:15:21,430 --> 00:15:23,770 We have 35 boxes 202 00:15:23,770 --> 00:15:27,770 which means 34 inner walls 203 00:15:28,760 --> 00:15:32,910 we can now put the 5 bosons inside the boxes 204 00:15:33,790 --> 00:15:37,320 the number of states correspond 205 00:15:37,320 --> 00:15:45,220 to counting how we can arrange 34 + 5 walls and bosons 206 00:15:45,220 --> 00:15:51,900 taking into account that walls as well as bosons are indistinguishable. 207 00:15:53,600 --> 00:15:58,140 The number 575757 208 00:15:58,140 --> 00:16:06,640 is then 39 ! / 5 ! / 34 ! 209 00:16:07,750 --> 00:16:10,910 We can easily extend 210 00:16:10,910 --> 00:16:16,330 the program naive_bosons_trap.py 211 00:16:16,630 --> 00:16:22,090 counting the number of particles that are in the ground state 212 00:16:22,900 --> 00:16:27,250 In this state we have zero particles in the ground state 213 00:16:27,420 --> 00:16:31,500 here we have three particles in the ground state 214 00:16:31,500 --> 00:16:35,500 and here we have one particle in the ground state 215 00:16:36,610 --> 00:16:44,000 we can implement this in naive_boson_trap.py 216 00:16:44,000 --> 00:16:49,710 using the count of zeros in the list state 217 00:16:50,120 --> 00:16:55,070 We can compute the average number N0 218 00:16:55,070 --> 00:16:57,780 and the condensate fraction 219 00:16:58,600 --> 00:17:05,470 considerably larger numbers of particles and states can be achieved if we introduce 220 00:17:05,470 --> 00:17:11,290 the occupation number associated to a single particle state 221 00:17:12,040 --> 00:17:15,930 In our scheme of 35 states 222 00:17:16,220 --> 00:17:25,320 instead of drawing colored particles, we just indicate that the state 0 is occupied by n0 particles 223 00:17:25,320 --> 00:17:30,950 the state 1 is occupied by n1 bosons and so on 224 00:17:32,020 --> 00:17:39,140 So for example this state can be written with these occupation numbers 225 00:17:39,140 --> 00:17:49,460 and the energy associated to any state can be written as follows 226 00:17:50,610 --> 00:17:55,260 the partition function of the system writes as follows 227 00:17:57,130 --> 00:18:04,890 This expression is equivalent to what we have computed in naive_boson_trap.py 228 00:18:04,890 --> 00:18:08,890 but it is even more complicated 229 00:18:08,890 --> 00:18:14,040 instead of the 575757 states 230 00:18:14,040 --> 00:18:19,460 here we have 6^35 terms 231 00:18:20,020 --> 00:18:25,260 However now we have a bunch of sums and a delta constraint 232 00:18:25,260 --> 00:18:29,260 does this ring a bell with you? 233 00:18:29,260 --> 00:18:33,550 This is exactly what was considered by Michael 10 minutes ago 234 00:18:34,060 --> 00:18:41,260 So we can use his providential representation of the Kronecker delta function 235 00:18:42,850 --> 00:18:48,820 Note: the sums are geometrical sums, so that we can compute them 236 00:18:49,760 --> 00:19:00,120 and of course each sum depends only on the energy of the state and not any more on the occupation number 237 00:19:01,950 --> 00:19:07,410 we have to think two minutes to understand if we have to consider 238 00:19:07,410 --> 00:19:11,410 the finite geometrical sum or the infinite geometrical sum 239 00:19:11,880 --> 00:19:16,970 and this is also done in the fact-sheet which is associated to this tutorial 240 00:19:19,000 --> 00:19:30,020 In any case, we can now arrive to an expression of the partition function which is a simple one-dimensional integral over the variable lambda. 241 00:19:31,280 --> 00:19:38,140 This expression for the partition function allows us to compute all thermodynamic quantities 242 00:19:38,630 --> 00:19:44,540 as the mean energy or the condensate fraction, which is shown here 243 00:19:45,650 --> 00:19:51,430 we can compute this integral numerically, using a simple discretization 244 00:19:51,430 --> 00:19:57,410 and moreover the cut-off of the energy E_max which was fixed to 4 245 00:19:57,410 --> 00:20:04,110 can now be increased as much as we like, and we can also increase the number of particles. 246 00:20:04,110 --> 00:20:09,420 For N=5 and E_max=4 247 00:20:09,420 --> 00:20:16,250 we recover exactly the same results as in naive_boson_trap.py 248 00:20:17,890 --> 00:20:22,980 Here we show the plot of the condensate fraction 249 00:20:22,980 --> 00:20:26,610 as a function of the rescaled temperature 250 00:20:26,610 --> 00:20:31,000 for a system of N bosons in a harmonic trap 251 00:20:32,280 --> 00:20:38,750 Here we see that all particles are in the ground states at very low temperatures 252 00:20:38,750 --> 00:20:42,750 this is a simple consequence of Boltzmann statistics 253 00:20:43,090 --> 00:20:47,470 At zero temperature all the particles populate the ground state 254 00:20:48,370 --> 00:20:51,980 the Bose-Einstein condensation is something else 255 00:20:51,980 --> 00:20:56,440 it means that a finite fraction of the system 256 00:20:56,440 --> 00:21:00,440 is in the ground-state for temperatures which are 257 00:21:00,440 --> 00:21:09,070 much higher than the gap between the gap between the ground-state and the first excited state, which is one, in our system. 258 00:21:11,680 --> 00:21:24,260 You see here that this fraction goes to zero for a rescaled temperature T / N^(1/3) = 1 259 00:21:24,260 --> 00:21:29,720 this means that the critical temperature for the Bose-Einstein condensation 260 00:21:29,720 --> 00:21:34,530 grows like N^(1/3) for a harmonic trap. 261 00:21:35,110 --> 00:21:41,010 This is a much higher temperature than 1, for large system 262 00:21:42,230 --> 00:21:48,250 Clearly this observation can be confirmed by explicit calculation 263 00:21:48,250 --> 00:21:54,590 so you can really believe us when I say that the critical temperature 264 00:21:54,590 --> 00:21:58,590 grows like N^(1/3) for a harmonic trap 265 00:21:59,160 --> 00:22:03,920 Finally take a moment to download, run and modify 266 00:22:03,920 --> 00:22:07,920 naive_boson_trap.py 267 00:22:07,920 --> 00:22:13,450 that studied the bosonic statistics and the Bose-Einstein condensation 268 00:22:13,450 --> 00:22:17,940 for a problem of five bosons, that was really too small 269 00:22:18,450 --> 00:22:21,220 but we came up with some idea 270 00:22:21,220 --> 00:22:27,120 actually a one-dimensional integration, that allowed us to go much further. 271 00:22:29,640 --> 00:22:34,800 In conclusion, in this tutorial 6 of Statistical Mechanics: 272 00:22:34,800 --> 00:22:42,180 Algorithms and Computations, we made our first step in the studying of quantum indiscernability 273 00:22:43,090 --> 00:22:49,850 the second pillar of quantum statistical mechanics, next to the Heisenberg uncertainty principle 274 00:22:51,350 --> 00:22:58,840 Five bosons in a trap allowed us to shed light on the very essence of quantum statistics 275 00:23:00,340 --> 00:23:07,990 The same model also allowed to understand something about Bose-Einstein condensation 276 00:23:08,930 --> 00:23:12,580 when all of a sudden a finite fraction 277 00:23:12,580 --> 00:23:16,790 of particles populate the single-particle ground state 278 00:23:16,790 --> 00:23:24,230 In a trap this happens at higher and higher temperature as we increase the particle number 279 00:23:26,050 --> 00:23:32,190 Next week we will take up the study of bosons and Bose-Einstein condensation 280 00:23:32,190 --> 00:23:37,390 in the language of density matrices and path integrals 281 00:23:37,390 --> 00:23:43,730 and you will produce your own path integral Monte Carlo calculation 282 00:23:43,730 --> 00:23:55,330 to produce Bose-Einstein condensation in a situation very similar to the experimental one that takes place in this glass cell. 283 00:23:56,010 --> 00:24:01,980 So in the meantime have fun with homework session 6 284 00:24:01,980 --> 00:24:10,120 and see you again next week in session 7 of Statistical Mechanics: Algorithms and Computations 285 00:24:10,120 --> 00:24:14,370 for a study of Bose-Einstein condensation.