1 00:00:01,230 --> 00:00:04,942 In the previous segment we saw that the Schrodinger equation for an infinite 2 00:00:04,942 --> 00:00:10,40 potential well without even writing down the Schrodinger equation itself. 3 00:00:10,40 --> 00:00:13,326 So of course it doesn't always happen this way, so for more complicated 4 00:00:13,326 --> 00:00:16,983 potentials we actually have to do some calculations, and then this segment I'm 5 00:00:16,983 --> 00:00:20,534 going to present an example of a general method actually that is used to solve the 6 00:00:20,534 --> 00:00:27,140 Schrodinger equation, and also consider the so-called finite potential well. 7 00:00:27,140 --> 00:00:31,892 So the, potential that we're actually going to study is, is illustrated here so 8 00:00:31,892 --> 00:00:36,446 here is, formal sort of mathematical expression for this potential and this is 9 00:00:36,446 --> 00:00:43,580 the, it's a illustration so, for, x. So this is one dimensional potential. 10 00:00:43,580 --> 00:00:47,530 For X is greater than A over 2 or smaller than minus A over 2. 11 00:00:47,530 --> 00:00:52,535 We have a finite wall, if you want, with a height of U naught and in between minus 12 00:00:52,535 --> 00:00:57,690 A over 2 and A over 2 we have a potential well. 13 00:00:57,690 --> 00:01:01,844 And so the question that we are going to be interested in is whether or not we 14 00:01:01,844 --> 00:01:06,199 have levels, energy levels, allowed states within the potential well, which 15 00:01:06,199 --> 00:01:10,554 implies that the energy of these states must be smaller than this maximum U 16 00:01:10,554 --> 00:01:17,88 naught but larger than 0. So these are positive energies here in 17 00:01:17,88 --> 00:01:21,434 this formulation. Now, clearly we have three regions that 18 00:01:21,434 --> 00:01:26,281 appear in this potential. So one region will be to the left of the 19 00:01:26,281 --> 00:01:30,545 point minus A over 2. The second region, which is the potential 20 00:01:30,545 --> 00:01:33,780 well itself, will be between minus A over 2 and A over 2. 21 00:01:33,780 --> 00:01:37,980 And the third region is X greater than A over 2 Two. 22 00:01:37,980 --> 00:01:41,214 And for each of the originals, we can write down the corresponding Schrodinger 23 00:01:41,214 --> 00:01:44,672 equation. Let's say, for the regions one and three, 24 00:01:44,672 --> 00:01:49,772 for absolute value of x greater than a over two we have the kinetic energy plus 25 00:01:49,772 --> 00:01:54,532 the potential energy u naught is equal to the energy that we're actually looking 26 00:01:54,532 --> 00:02:01,602 for e times psi of X. And for the region of the potential well 27 00:02:01,602 --> 00:02:05,698 we formally have just the free Schrodinger equation since the potential 28 00:02:05,698 --> 00:02:10,424 here is equal to 0. Now to solve this Schrodinger equation 29 00:02:10,424 --> 00:02:13,784 the main challenge in some sense of solving the Schrodinger equation is 30 00:02:13,784 --> 00:02:18,449 actually would be to match the solutions in different regions. 31 00:02:20,330 --> 00:02:25,840 And by match I mean the full link. We're going to demand that both the wave 32 00:02:25,840 --> 00:02:31,720 function itself and it's first derivative are continuous everywhere. 33 00:02:31,720 --> 00:02:36,420 In particular in these matching points minus a over 2 and a over 2. 34 00:02:36,420 --> 00:02:39,865 And so here I have formally represented the as so, so plus zero basically means 35 00:02:39,865 --> 00:02:44,560 to the right of a certain point. And minus zero means to the left of a 36 00:02:44,560 --> 00:02:48,160 certain point. And the reason one, our weight function 37 00:02:48,160 --> 00:02:51,760 or let's say its derivative to be continuous is because if we did not have 38 00:02:51,760 --> 00:02:56,292 this continuity. Let's say in this point if we let's say 39 00:02:56,292 --> 00:03:00,632 solve our Schrodinger equation three and origin two and department we don't bother 40 00:03:00,632 --> 00:03:04,848 to mention them, then, in general so our way function in this case is going to be 41 00:03:04,848 --> 00:03:12,382 something like this let's say. And there will be a jump, find a jump. 42 00:03:12,382 --> 00:03:16,30 In, well, in the wave function is derivative, and if we calculate the 43 00:03:16,30 --> 00:03:19,934 second digit which is here, it uses the jump, we're going to have an infinite 44 00:03:19,934 --> 00:03:23,902 secondary digit which will be, which wouldn't be able to compensate in our 45 00:03:23,902 --> 00:03:29,930 Schrodinger equation. So there will be essentially a delta 46 00:03:29,930 --> 00:03:34,538 function coming up from this, contiu-, from this discontinuity, and, well, we 47 00:03:34,538 --> 00:03:39,173 don't have it in our problem. In our problem we don't have a delta 48 00:03:39,173 --> 00:03:41,969 problem. So, therefore, we must demand that the 49 00:03:41,969 --> 00:03:45,870 wave function is derivative so in our continuous everywhere. 50 00:03:45,870 --> 00:03:50,88 And it turns out that this along with these, with this condition would be would 51 00:03:50,88 --> 00:03:55,900 be enough to find the constraints which will determine the actual energy level. 52 00:03:55,900 --> 00:04:00,388 So now the latter constraint namely that the wave function should decay at 53 00:04:00,388 --> 00:04:05,610 infinity is essentially, physically a meaningful constraint. 54 00:04:05,610 --> 00:04:10,34 So if we were looking for a bound state of our particle, so we want the 55 00:04:10,34 --> 00:04:16,140 probability of finding this particle at infinity to be zero. 56 00:04:16,140 --> 00:04:19,920 And we want the probability to be sort of localized in the vicinity of the 57 00:04:19,920 --> 00:04:23,529 potential. And so this is where the second 58 00:04:23,529 --> 00:04:28,348 constraint is coming from. Now to simplify this equation Ev equals U 59 00:04:28,348 --> 00:04:33,974 naught is that, well we can rewrite it. It, let's say the first on we can 60 00:04:33,974 --> 00:04:41,242 introduce a new parameter gamma, which is, a 2m over x squared, U-not, minus, 61 00:04:41,242 --> 00:04:50,140 epsilon minus energy E, so let me actually called gamma squared, 62 00:04:50,140 --> 00:04:53,572 And, well, we can, we an just put the energy, in, in the left-hand side and 63 00:04:53,572 --> 00:04:59,82 multiply everything by minus. 2 m over h squared, and this would give 64 00:04:59,82 --> 00:05:05,400 rise to the full equation. Psy 2 prime minus gamma psi, gamma 65 00:05:05,400 --> 00:05:11,368 squared psi is equal to zero. And for the second equation we can 66 00:05:11,368 --> 00:05:17,38 introduce the parameter is the key, and, we're going to have this with psi 2k plus 67 00:05:17,38 --> 00:05:23,450 k squared cy equals to zero. So let me just rewrite in this slightly 68 00:05:23,450 --> 00:05:26,398 nicer way. So here are basically the two equations I 69 00:05:26,398 --> 00:05:30,986 just wrote and here are the parameters. Both these equations are actually very 70 00:05:30,986 --> 00:05:36,890 simple equations so, this one has a solution either power plus minus gamma x. 71 00:05:36,890 --> 00:05:40,320 You can just plug it in and see the well, the the inevitable of the exponential and 72 00:05:40,320 --> 00:05:43,505 the exponential is the exponential itself and well, clearly is going to satisfy 73 00:05:43,505 --> 00:05:48,122 this equation. And the second equation well which is the 74 00:05:48,122 --> 00:05:53,18 free Schrodinger equation, unsurprisingly gives us just plain waves as the solution 75 00:05:53,18 --> 00:05:57,530 either problem, plu-, plus minus i K times x. 76 00:05:57,530 --> 00:06:00,580 So and again, the challenge is going to be to match those guys in each of these 77 00:06:00,580 --> 00:06:05,186 points, minus a over 2 and a over 2. Now, it's not really necessary for the 78 00:06:05,186 --> 00:06:09,464 solution of the problem but, I would like to use the symmetry of the problem here 79 00:06:09,464 --> 00:06:13,990 and this would give me an excuse in some sense to introduce the symmetry concepts 80 00:06:13,990 --> 00:06:19,540 and quantum mechanics in a much Broader context. 81 00:06:19,540 --> 00:06:22,154 So namely I would like to point out the following fact. 82 00:06:22,154 --> 00:06:27,526 if, the Hamiltonian commutes with an operator a, this can be an operator of 83 00:06:27,526 --> 00:06:32,898 any physical quantity, now, then the solutions to the Schrodinger equation, 84 00:06:32,898 --> 00:06:37,243 that is an equation which is an Eigenvalue problem for, for the 85 00:06:37,243 --> 00:06:44,922 Hamiltonian. So then, the solutions can also be chosen 86 00:06:44,922 --> 00:06:49,992 to have definite value of 8, which means that the same solutions are going to be 87 00:06:49,992 --> 00:06:54,828 the Eigen states of the Eigenvalue problem for this operator E whatever it 88 00:06:54,828 --> 00:07:00,560 is. So in general if the 2 operators were not 89 00:07:00,560 --> 00:07:05,695 to commute, so I wouldn't necessarily be able, have been able to find As a set of 90 00:07:05,695 --> 00:07:12,761 way of function which would both A and E, the energy definite. 91 00:07:12,761 --> 00:07:18,298 So, but in the presence of a symmetry, you want this possibility exist. 92 00:07:18,298 --> 00:07:20,310 Now, what does that have to do with our problem? 93 00:07:20,310 --> 00:07:24,286 So in our problem, we have a sort of rather obvious Symmetry which is 94 00:07:24,286 --> 00:07:28,546 essential inversion symmetry of the potentials of a flip x if it x goes to 95 00:07:28,546 --> 00:07:33,286 minus x. So the potential is unchanged and so the 96 00:07:33,286 --> 00:07:37,688 kinetic energy also doesn't change so its a second derivative with respect to x, so 97 00:07:37,688 --> 00:07:41,718 therefore the Hamiltonian one can determine that Hamiltonian is indeed 98 00:07:41,718 --> 00:07:48,680 symmetric or invariant under this. Under this operation, and so for 99 00:07:48,680 --> 00:07:53,230 [INAUDIBLE] that forces this inversion is this operator i, which basically thinks 100 00:07:53,230 --> 00:07:57,715 adds to minuses. And this operature has two item values 101 00:07:57,715 --> 00:08:01,760 the barrier. Either plus or minus 1. 102 00:08:01,760 --> 00:08:05,918 So the functions feature high end function of this inversion operature 103 00:08:05,918 --> 00:08:09,750 either even or odd function of the coordinate. 104 00:08:09,750 --> 00:08:13,782 And so what, what we are saying in this, sort of, very complicated language is 105 00:08:13,782 --> 00:08:18,3 that we can find the Eigenstates of our problem the solutions to the Schrodinger 106 00:08:18,3 --> 00:08:23,650 equation which would be either, odd or even functions. 107 00:08:23,650 --> 00:08:26,90 That's all we are saying and, well, another statement that I would like to 108 00:08:26,90 --> 00:08:29,630 make is that if there is a symmetry in the problem, you'd better use it. 109 00:08:29,630 --> 00:08:33,446 So it always simplifies things. Never makes things more complicated, 110 00:08:33,446 --> 00:08:36,356 always simplifies things. And so that's what I'm going to do here 111 00:08:36,356 --> 00:08:39,64 as well. So in general if I were to ignore, this 112 00:08:39,64 --> 00:08:42,47 symmetry. So the solution, to the[UNKNOWN] 113 00:08:42,47 --> 00:08:45,752 equations we just discussed in this, region of the will would be arbitrary 114 00:08:45,752 --> 00:08:51,638 linear combination of two plane wave. Now I'm saying so see there is this 115 00:08:51,638 --> 00:08:55,408 symmetry in the problem. So I would like instead of just having 116 00:08:55,408 --> 00:08:59,56 this arbitrary efficiency c 1 and c 2, I would like to consider separately 117 00:08:59,56 --> 00:09:05,5 solutions with definite parity. Either even functions which is basically 118 00:09:05,5 --> 00:09:11,810 cosign with the c 1 equals to c 2 or a sign which is a node function. 119 00:09:11,810 --> 00:09:16,804 In which case c1 is equal to a complex conjugate. 120 00:09:16,804 --> 00:09:20,745 and these are purely measuring constants. So, in the following I am going to 121 00:09:20,745 --> 00:09:24,371 specifically consider to focus on this even solution just for simplicity but you 122 00:09:24,371 --> 00:09:27,556 can find the general analysis This in a number of books so this is the problem we 123 00:09:27,556 --> 00:09:32,909 are solving by the way. Is a absolutely classical problem, but 124 00:09:32,909 --> 00:09:37,319 classical not in the sense of classical physics, but in the sense that it appears 125 00:09:37,319 --> 00:09:43,400 in just about every course or every book on quantum physics. 126 00:09:43,400 --> 00:09:47,195 Now having determined the solution we're interested in, in region two and now have 127 00:09:47,195 --> 00:09:50,645 to To determine the solutions in regions one 128 00:09:50,645 --> 00:09:53,350 and three that we're going to be focusing on. 129 00:09:53,350 --> 00:09:57,112 And by the way, you, the symmetry of the problem, we can actually pick just one of 130 00:09:57,112 --> 00:10:00,707 these regions. So we don't really have to use, in this 131 00:10:00,707 --> 00:10:04,869 case, both matching points, we can do well just using one point. 132 00:10:04,869 --> 00:10:09,221 And as we discussed, a general solution to the Schrodinger equation, to this 133 00:10:09,221 --> 00:10:13,573 equation in the region three is given by a linear combination of these two 134 00:10:13,573 --> 00:10:18,265 exponentials, which are no longer oscillating functions, but are functions, 135 00:10:18,265 --> 00:10:23,433 that, are either, decaying, a rapidly decaying function, or a rapidly, growing 136 00:10:23,433 --> 00:10:30,58 function. And, so to, simplify to find the proper 137 00:10:30,58 --> 00:10:33,649 solution in the region three, let's say, for x is greater than a over 2 We have 138 00:10:33,649 --> 00:10:37,696 recalled, the constraint and infinity namely that the wave function must remain 139 00:10:37,696 --> 00:10:42,110 finite. Well actually I should correct. 140 00:10:42,110 --> 00:10:46,340 It should actually go to zero as we go to infinity. 141 00:10:46,340 --> 00:10:50,244 So otherwise if we don't impost this constraint, if we allow, in this case 142 00:10:50,244 --> 00:10:56,740 this, churum in the wave function, the probability of a particle to leak. 143 00:10:56,740 --> 00:10:59,707 in some sense to infinity would explode exponentially, which doesn't make any 144 00:10:59,707 --> 00:11:03,573 sense. So therefore we simply draw the B term in 145 00:11:03,573 --> 00:11:09,117 this in this wave function, and only we'll focus on the on this wave function 146 00:11:09,117 --> 00:11:14,940 in the region 3. So now we're in, in a position to 147 00:11:14,940 --> 00:11:20,250 actually match the solutions. We know the solution in each region. 148 00:11:20,250 --> 00:11:25,780 So here I do present a solution in all of the regions, including region one. 149 00:11:25,780 --> 00:11:28,366 So this guy's for region one. This guy's for region two, and this guy's 150 00:11:28,366 --> 00:11:31,505 for region three. But as I said, due to symmetry we can 151 00:11:31,505 --> 00:11:35,770 focus on only one matching point. And this is what I'm going to do. 152 00:11:35,770 --> 00:11:39,991 So now we can ride the wave function to left at this point, which is going to be 153 00:11:39,991 --> 00:11:44,279 c times cosign key a over two and, it must be equal to the way function to the 154 00:11:44,279 --> 00:11:52,36 right, which is this guy, which is a ethopro minus, gamma, a over two. 155 00:11:52,36 --> 00:11:56,836 So this is the first matching condition and the second matching condition is 156 00:11:56,836 --> 00:12:01,158 going to be the derivative of this function. 157 00:12:01,158 --> 00:12:05,733 So the derivative of cosine is minus sine, so were going to have minus C times 158 00:12:05,733 --> 00:12:11,760 k sine of k a over 2, to the left. Must be equal to minus A gamma e to the 159 00:12:11,760 --> 00:12:18,714 power minus gamma a over 2, to the right. And now what I'm going to do, I'm 160 00:12:18,714 --> 00:12:22,0 going to divide so let me call this equation, equation number two. 161 00:12:22,0 --> 00:12:24,790 And this is going to be equation, equation number one. 162 00:12:24,790 --> 00:12:30,570 And so if we divide equation number two by equation number one. 163 00:12:30,570 --> 00:12:35,70 Which we're allowed to do, what we're going to get in the in the left hand 164 00:12:35,70 --> 00:12:42,210 side, we're going to have k times sine or a cosine is a tangent of ka over 2. 165 00:12:42,210 --> 00:12:45,806 And in the right hand side, so the minuses will go away and the exponentials 166 00:12:45,806 --> 00:12:49,400 are going to go away, we're going to get Just gamma. 167 00:12:49,400 --> 00:12:53,740 So here I have the same equation written a exclusively in a nice way. 168 00:12:53,740 --> 00:12:57,400 And also I recall the definitions of gamma and k, that I made in a previous 169 00:12:57,400 --> 00:13:00,617 slide. So k itself is the square root of 2mE 170 00:13:00,617 --> 00:13:03,602 over h squared. And the energy is really what we are 171 00:13:03,602 --> 00:13:06,352 looking for. And energy appears both in the left hand 172 00:13:06,352 --> 00:13:10,50 side and in the right hand side, in a rather non-linear way. 173 00:13:10,50 --> 00:13:14,80 And as a matter of fact there is no way, we cannot solve this equation 174 00:13:14,80 --> 00:13:18,744 analytically. So this is, no linear algebraic equation 175 00:13:18,744 --> 00:13:24,200 and it doesn't have a formal sort of closed solution, that is four. 176 00:13:24,200 --> 00:13:27,350 So which by the way is quite amazing that the problem is simple as this finite 177 00:13:27,350 --> 00:13:30,438 potential. Well in one dimensional quantum mechanics 178 00:13:30,438 --> 00:13:33,462 which is just about the simplest problem you can think of It cannot really be 179 00:13:33,462 --> 00:13:36,726 solved, so this gives you something, it tells you something about the complexity 180 00:13:36,726 --> 00:13:41,500 now, of the technical complexity of the Schrodinger equation. 181 00:13:41,500 --> 00:13:45,770 Now to precede further towards a bit of numerical analysis or asymptotic 182 00:13:45,770 --> 00:13:49,627 analysis. It is always a good idea, in such cases, 183 00:13:49,627 --> 00:13:53,958 to get rid of the physical, dimensional quantities such as energy or this 184 00:13:53,958 --> 00:13:58,431 potential height which have physical dimension and move to dimensionless 185 00:13:58,431 --> 00:14:03,231 parameters. And so here the nature of the 186 00:14:03,231 --> 00:14:07,787 dimensionless parameters is going to be Well, the argument of this tangent which 187 00:14:07,787 --> 00:14:13,630 is EE over 2. And the parameter introduced here. 188 00:14:13,630 --> 00:14:19,545 Which is basically to be able to write our equation in a nice and dimensionless 189 00:14:19,545 --> 00:14:22,865 way. The reason why dealing with 190 00:14:22,865 --> 00:14:27,610 dimension-less quantities is much more convenient than with. 191 00:14:27,610 --> 00:14:31,858 quantities in case it has non-trivial physical dimensions is because, for the 192 00:14:31,858 --> 00:14:35,830 former, we can talk about them being large or small. 193 00:14:35,830 --> 00:14:40,360 So we can see that something is small or large as comparing it with one. 194 00:14:40,360 --> 00:14:44,86 While for a dimensional quantity it doesn't really make sense to talk about 195 00:14:44,86 --> 00:14:47,812 large or small unless we Specify what we actually mean by that for instance is one 196 00:14:47,812 --> 00:14:52,247 meter large or small. Well does we don't know unless we compare 197 00:14:52,247 --> 00:14:55,82 with the lets say size of an atom and which case it's clearly large or with the 198 00:14:55,82 --> 00:14:59,700 size with the size of the universe, in which case one meter is really small. 199 00:14:59,700 --> 00:15:05,706 So here one, once we get these parameter si it allows us to define basically a 200 00:15:05,706 --> 00:15:10,290 deep. And shallow potential so we will call the 201 00:15:10,290 --> 00:15:14,385 potential deep if this guide, the, of this squad squared is much larger than 202 00:15:14,385 --> 00:15:18,480 one and the potential is going to be called shallow if it's much smaller than 203 00:15:18,480 --> 00:15:23,773 one. And by the way, in the, full length two 204 00:15:23,773 --> 00:15:28,603 video segments we're going to be focusing on, shallow potentials which are well 205 00:15:28,603 --> 00:15:33,888 mottled by delta function. So, naively you would say, you would 206 00:15:33,888 --> 00:15:37,320 think, that delta function is something which is actually very deep because it 207 00:15:37,320 --> 00:15:40,908 goes to infinity if you, approach, x equals zero, let's say if we're talking 208 00:15:40,908 --> 00:15:46,586 about potential u of x, which is let's say minus u, delta of x. 209 00:15:46,586 --> 00:15:52,46 But, well, delta of x and potential delta of x is really, it means, to represent, a 210 00:15:52,46 --> 00:15:57,156 Potential with the very with the very small radius and the radius in order for 211 00:15:57,156 --> 00:16:01,986 the integral the delta function be equal to 1 so the radius of the potential, the 212 00:16:01,986 --> 00:16:11,590 size of the potential should scale with the depth of the potential as so. 213 00:16:11,590 --> 00:16:18,0 So as a goes to zero, well you not is sort of proportional to y and 1 over a. 214 00:16:18,0 --> 00:16:21,720 But here we see that the definition of the shallow potential is u naught a 215 00:16:21,720 --> 00:16:25,626 squared times m over a squared must be much more than one which means that the 216 00:16:25,626 --> 00:16:29,718 Delta potential is actually closer to shallow potential which is an important 217 00:16:29,718 --> 00:16:35,610 observation and will take advantage of later on. 218 00:16:35,610 --> 00:16:39,764 But in its stage let me just say that we have these two special cases and even 219 00:16:39,764 --> 00:16:44,655 though we cannot solve exactly this equation, for arbitrary side, we can do 220 00:16:44,655 --> 00:16:52,10 two separate analysis of this equation in such two special cases. 221 00:16:52,10 --> 00:16:56,276 And here, I present this analysis. So this is again the equation, the same 222 00:16:56,276 --> 00:17:00,497 equation, so the red tangent of x is plauded here, so basically applauded the 223 00:17:00,497 --> 00:17:06,700 functions on the left of this equation and on the right of this equation. 224 00:17:06,700 --> 00:17:12,590 And you see by the way that when x becomes larger than psi. 225 00:17:12,590 --> 00:17:16,790 So this right hand side becomes purely an imaginary constant which implies that 226 00:17:16,790 --> 00:17:20,630 there is definitely no solution to this equation as the tangent here does not 227 00:17:20,630 --> 00:17:27,250 become measured. So so it makes sense only to plot the 228 00:17:27,250 --> 00:17:33,990 well the this function for a axis smaller than psi. 229 00:17:33,990 --> 00:17:38,61 So here I have an example of deep potential where I say a set psi to 220 230 00:17:38,61 --> 00:17:42,408 and so I have all these red, red curves corresponding to the tangent and this 231 00:17:42,408 --> 00:17:47,169 blue curve corresponds to the right hand side and where, when, wherever I see it 232 00:17:47,169 --> 00:17:56,428 crossing between these two guys, so let's see, here, here or here or here, etc. 233 00:17:56,428 --> 00:17:59,849 So, I have levels. In this case, I can count the number of 234 00:17:59,849 --> 00:18:03,988 levels that are allowed in this quanta. Well, one, two, three, four, five, six, 235 00:18:03,988 --> 00:18:08,566 seven levels are allowed. Well for the symmetric wave functions if 236 00:18:08,566 --> 00:18:12,196 size equal to 20. Now lets consider relatively shallow 237 00:18:12,196 --> 00:18:16,510 potential that this on the other hand, so lets say with si equals 1. 238 00:18:16,510 --> 00:18:20,719 In this case I only have to plot this functions for x as more than 1 and there 239 00:18:20,719 --> 00:18:26,94 is just 1 crossing. Between the blue curve and the red curve, 240 00:18:26,94 --> 00:18:31,450 which means that there is just one single level in the shallow potential well. 241 00:18:31,450 --> 00:18:34,852 Which by the way is a very important result that we're going to confirm on the 242 00:18:34,852 --> 00:18:37,930 case of a one dimensional delta potential, and we're going to see how 243 00:18:37,930 --> 00:18:41,913 this result is modified in higher dimension. 244 00:18:43,130 --> 00:18:46,670 Now, the last comment I'm going to make is that in principal I could have 245 00:18:46,670 --> 00:18:50,390 continued this analysis of this self-consistency equation and there's a 246 00:18:50,390 --> 00:18:54,170 lot to do to determine the actual numerical value of the energy levels to 247 00:18:54,170 --> 00:19:01,940 see what happens with the anti-symmetric levels with the old wave functions. 248 00:19:01,940 --> 00:19:05,905 And I'm not going to do this right now in the lecture because admittedly, it's 249 00:19:05,905 --> 00:19:10,53 probably not the most exciting thing in quantum mechanics so I will encourage you 250 00:19:10,53 --> 00:19:14,689 to take a look at various text books and other lectures online, for example these 251 00:19:14,689 --> 00:19:22,764 set of lectures by professor Michael Fowler at University of Virginia. 252 00:19:22,764 --> 00:19:27,769 and also you will see a few problems going up on this solution in your 253 00:19:27,769 --> 00:19:32,440 homework. So you will have a chance to look into 254 00:19:32,440 --> 00:19:36,360 this problem in more detail and understand the solution at a deeper 255 00:19:36,360 --> 00:19:38,416 level.