 So, good afternoon everybody. Welcome back. It is time to start with the second part of the first lecture by Deepak Da. Okay. So, let us start. So, in the first part of this today's lecture, I had discussed the classical mechanical tongs gas and the gas with nearest neighbor coupling which is called the Taghashi gas. And now we would like to discuss what happens to this problem if you introduce quantum mechanics. So, I call it quantum mechanical tongs gas. So, the Hamiltonian is equal to this kinetic energy V of x. So, the key point which is the difference from the previous problem is that now we cannot separate away the momenta. Momenta and positions are coupled by quantum mechanics by the uncertainty principle or you know you cannot I cannot just do this problem. I have to put p square. Okay. So, we will try to do this. We will take the same old system in which you have a total system of length L and you will have n particles which are looking like this of width sigma. And now each of these is a quantum mechanical particle. And we want to study this system. I guess in the beginning we just want to find the eigenstates of energy. Diagonalize the Hamiltonian. Okay. Once we have diagonalized the Hamiltonian, we will sum up over all the energy states that will give the partition function. And then if you have energy left, you can try to calculate the correlation functions of this system. Okay. So, this problem is actually quite hard. And let us discuss n equal to 1 case. That is actually rather easy and I can do this one here. So, there is a single particle and it moves in a potential. The potential is provided by its interaction with the walls. So, actually it can only move from sigma by 2 to L minus sigma by 2. Okay. So, there is a wave function psi of x minus d square h cross is set equal to 1 m psi equal to e psi and e psi of x. And x lies between sigma by 2 and L minus sigma by 2. And psi of x equal to 0 at boundaries of this interval. So, that is well studied and well known. All the answers are known. It is given in sheaf. The nth energy level has energy n squared h cross squared by 2 m. I think there is a pi squared. And then there is a L minus sigma squared. Okay. And the corresponding wave function equal to sine. We will also call it k n pi x minus sigma by 2 normalization k n squared h cross squared by 2 m equal to n squared h cross squared equal to n squared. Okay. Pi. Pi should be absorbed in the k n. I think this looks good. Okay. So, one particle problem has been solved. That is rather straight forward. And now we will try to do the n equal to 2 k's which as you will see is already quite complicated. Okay. So, n equal to 2. What is my state space? The same as what we discussed last time. It looks like this. The forbidden region x 1. This is x 2. This is allowed. Then there is all these are forbidden outside. Okay. Again, there is a wave function psi of x 1 x 2. And it satisfies minus t squared by dx 1 squared 2 square by dx 2 square 2 m sorry e times psi x 1 x 2. And the wave function has to vanish inside these regions where it is forbidden. And so it should also vanish at the boundaries of the regions. Okay. Bigger psi 1 x 2 equal to 0 at boundaries. So, can I solve this problem? So, the first thing to notice is that the solution here will always be degenerate because the wave function can be here or here. But I can certainly construct a wave function which is psi 1 of x 1 x 2 which is nonzero in top triangle and 0 in the lower triangle. And of course, psi 2 can be done with the way you know opposite. These two are linearly clearly linearly independent solutions because they do not even have any overlap. Okay. So, a general solution can be a linear combination of these ones. I should solve the problem in one triangle. Then I am kind of happy. Okay. So, give or take a little bit the triangle looks like this. And I want to solve the del squared phi Laplace's equation in this triangle with the wave boundary condition 0 at the boundary. Okay. Now, it turns out that this is sort of a generally a hard problem. There are a lot of people who spend a lot of time studying the Billiard's problem which is you take some funny boundary shape and you discuss the solutions of the Laplace's equation on with that boundary. Okay. All the different Eigen functions. And they show very complicated and difficult behavior. We are not worried about the most general boundary condition. But with this rather stupid boundary condition of a right triangle. For the sake of my discussion just now it is useful to consider m 1 m 2 here. m 1 m 2 are two masses of two different particles. Right. In classical statistical mechanics m 1 m 2 did not show up in the tungsten gas problem because they came in the part of the momentum integral. And it was you know it was a multiplicative factor which was thrown out in the configurational integral m 1 m 2 were not there. In the quantum mechanical problem m 1 m 2 have to be there because they determine a coefficient between these two. You cannot get rid of them if they are not equal that problem is not the same as if they are equal. You can scale m 2 you can scale x 2 to make m 1 m 2 equal. But then the triangle becomes a right angle triangle but not an isosceles triangle. Now it turns out that this problem when the triangle is unequal in size cannot be solved in closed form easily. Even this simple problem has not been solved in closed form. Of course there are solutions they exist you give numerically you can determine them. But there is no simple expression for E n as a function of n and the corresponding wave function psi n. They cannot be written in terms of are not explicitly known. So what is meant by explicitly known these things this is this is an important point. We have defined the problem it is well defined psi n is the nth eigen function of this particular shape. So that is the definition it is well defined. Why is that not good enough for everybody? No some people are much happier if psi n can be written as an elliptic function of x 1 times some other function of x 2 and some complicated integral thereof. So that is called an explicit representation of psi. But just saying that psi is the nth eigen function of this shape is that not an explicit representation of psi. I think it is equally good in fact it is not known that you can write this for general shape even as as simple as shape as a square. It turns out that if there is these two sides a and b are rational multiples of each other then in some lucky cases you can determine the solution. But some of the solution is not all of them you can write down an eigen function which will actually work. We will write down some of them and but they do not exhaust all the eigen functions of the system. So that is a problem. But the case where a equal to b is a lucky case and there we can get actually all the eigen functions exactly. So that is what we do next sorry there is a explicitly known for all n for all a by b. So we take the case a equal to b and then I construct a solution. So the answer comes through some trick which of course is hard to realize but it is easy to teach like I mean if I were to discover it on my own it will be very hard I may not get it for 5 years. But if I tell you now then in 5 minutes you know the answer. So the trick is the following we want to solve the problem in this triangle we do not do that we solve it in the full square which is obtained by reflecting the triangle above its diagonal. And then choose the functions which are odd under reflection in this all the functions of psi which I can write down there are finite set of eigen functions which actually have the symmetry that they are odd under this thing ok. So then I work with them that is the answer. So a equal to b the solution is psi nn is equal to sin m pi x1 over a sin n pi x2 over a this is the solution on the full square. But I just make an anti-symmetric combination n pi x1 over a sin m pi x2 over a and I can normalize it ok. If I look at just the first term this is an eigen function of del squared x1 this is an eigen function of del squared x2 the product is an eigen function of del squared x1 plus x2 the eigen value is m squared plus n squared up to multiplicative constants the same thing is true for this one with the same eigen value ok. So if I make this combination then of course the linear combination also is an eigen function with the same eigen value. But this one has the additional property that if x1 equal to x2 it has to vanish. So on this line sorry I just flipped my directions on the line x1 equal to x to the wave function vanishes ok. Then I can go back to my original problem by introducing the shifts you know by sigma and so on that is not a problem that I will not write down the solutions here. Is that clear? I have once I have solved the problem on this equilateral sorry right triangle isocellist right triangle then I solved the problem for two particles with equal masses ok. So that is the solution. But this is odd under symmetry. So I will write here for x1 x2 belonging to top triangle and the corresponding eigen value is emn is equal to n squared plus n squared times some constant. I can write down these constants but I am deliberately not writing them because it is not necessary for understanding what is going on. If you get written notes of these lectures which I will actually not write. But I could have written then all the factors of h cross and m and pi will be there. You can provide them yourself ok. So we are discussing the fact that this solution psi 1 as written here here and psi 2 are both of them valid quantum mechanical solutions. If you start with the wave function which is nonzero here some wave function and evolve it in time using Schrodinger's equation. It will remain in that triangle it cannot escape out of there and the solution in the second will remain 0 ok. So that is just saying that you can exchange. So there are these sectors x1 less than x2 is a dynamical sector in which the particle stays. If you put it initially there it remains there throughout its life and if you choose the other one then the system will stay in that sector throughout and these two are not connected to each other. So if you want you can ignore one of them. If you want I can say that oh the particles are identical then I cannot say which one is one which one is two indistinguishable particles. Then I can say by convention in this case it is correct that psi x1 x2 will be equal to psi x2 x1 we may choose. So under exchange we can since in this case the two sectors do not communicate with each other. I am allowed to do whatever I like and in particular I can choose them to be psi x1 x2 equal to psi x2 x1 then the wave function is symmetric under exchange of particles and these particles are called Bosch ones vice versa. If the particles are Bosch particles then the wave function has to be symmetric. All the other hand I can choose psi x1 x2 equal to minus sign and then I will say these particles are fermions. And the particles perhaps in this case do not care and they will remain the same whether you call them Bosch ones or fermions because according to the interaction given they actually cannot exchange places. So what happens when they exchange places you know will the wave function change sign or not change sign is a known question that is an important point to realize. A more detailed way of saying this is that in one dimension the Bosch ones and fermions are the same thing there is no distinction between them if they are hardcore. So now we come to the case n equal to 3 and luckily that is rather easy. So 4 and less than x2. So now it is actually fairly straightforward. The equation I will write will be d squared x1 squared minus d squared x2 squared minus d squared x3 squared sorry minus psi x1 x2 x3 xn equal to e times psi x1 x2 xn. The phase space configuration space will be a generalized pyramid you know you will have directions x1 x2 x3 x4 each of them is positive and there is a condition x1 plus x2 plus x3 ok. So what we did here with triangle in the 3d it looks like this you know there is an x1 x2 x3 and there is a cut which gives you a pyramid. Non-trivial to think of this in the three dimensional space but I encourage you to try to do it then you will see that this is the configuration space and then there are other copies of this which you can put x2 less than x1 will give you the same figure but rotated and put in space you know and if you put 6 of them together I get you get the whole cube. The cube can be decomposed into 6 such figures and we solve for one of them and put the wave function 0 at the boundary and the way to do it is to put psi of x1 x2 xn equal to determinant of sine given x1 sine k2 x2 sine kn xn with kixi pi kl equal to ni pi. Put all ni distinct otherwise you would not get a non-trivial wave function and that is like the fermion condition ok. So that all the ni should be distinct and then this provides the solution. And if you want Boson wave function psi for me is equal to free particle wave function. Sorry, yeah it is a square matrix n by n. Sir what? Ah it is sin k 1 x 2 sorry I beg your pardon I wrote it wrong this is k 1 x 1 k 1 x n thank you k 2 x 1 sin k 2 x 2 this is the so called slater determinant sin k 2 x n like that. So, sin k i x j would be the generic term yes that is correct you are right. So, in going from here to here the change from x to delta was made ok. And so psi bose is equal to mod value of psi for me. So, if you go to a different sector the Boson wave function will be symmetrical ok. So, now, so what have we learnt with this exercise? We can determine all the Eigen functions of the quantum mechanical tongs case with n particles if the masses are equal we just did it. Now, what is the partition function? Partition function is the same as the partition function of an ideal firming gas which you have studied in your BSE because the energy levels are the same. Each allowed energy level of an ideal firming gas gives you an energy level for this system. So, did I work out the partition function of an ideal firming gas? No, did not. Usually what people do is they say that we will work in the grand canonical ensemble and calculate the grand partition function that is easier to do. And then if you really want to do it you know you take the nth coefficient by doing some contour integral, but it is not done or it again the final answer cannot always be written in closed form for the partition function. For the grand partition function there is a nicer simple product form representation ok, but that is not my concern now. So, this issue about different masses behaving differently than equal masses is sort of an interesting question. And there have been lot of studies in literature you take a one dimensional line put particles with different masses m 1 m 2 m randomly selected you know you can take two masses, but each mass can take two possible values sorry you can take particles where each particle has one of the two possible masses and you make a line and then give them some initial velocity and let them collide and see what happens. Turns out all kinds of crazy things happen and people keep on studying this for long time and they cannot figure out. All the masses are equal then the behavior is kind of very simple. When two collisions occur the velocities are exchanged and so that problem can be studied or understood. When the masses are unequal that does not work and so all this stuff will not work if the masses are unequal ok. So, I do not know what happens to the tombs gas with unequal masses. Somebody might guess that no, no, no all these equivalences to free fermions etcetera are going to be valid still, but that is the hope it does not seem to explicitly work out ok. So, I only indicate the fact that a that different masses makes the problem much more complicated in quantum mechanics and secondly the connection between the equilibrium stat mac and these dynamical simulations in which particles collide and see what happens at long times is also an issue because you know whether equilibrium is reached or not reached becomes a question. If you have funny boundary conditions like this part is at a higher temperature then this part then all kinds of extra stuff happens which I will not discuss, but yes yeah no say it is correct I do not know what symmetry is broken yeah the masses are not equal. So, the symmetry between two particles is broken yeah ok. So, you know I do not know what is the consequence of which one which comes first the symmetry breaking comes first what the name you invent a symmetry when it is there when it is not there then you first imagine some symmetry and then it is broken. When I stated the problem to begin with there was no symmetry then it is not there no. So, I think it displays a mindset by saying some symmetry is broken because it was not there ok. So, now I still have some time and so what should I do in the remaining time. What we will do is to calculate some correlation functions of the tongs gas yes yes yes yeah no. So, the problem was well defined there will be the original problem was a equal to b you know that was how the problem was posed, but the coefficient d squared by dx 1 squared has a different coefficient than d squared by dx 2 squared which you can rescale to convert to a problem in which the sides are unequal length, but the Laplace operator is you know with d squared by dx 1 squared plus d squared by dx 2 squared with equal coefficients and I am saying that that problem cannot be solved completely in closed form already even that stupid simple problem with two particles right triangle what is the wave function I cannot do it in closed form ok. So, now so incidentally once I have said that the partition function is the same as that of free ideal gas I can take the thermodynamic limit I can see how the specific heat varies at low temperature for ideal Fermi gas in one dimension anybody remembers the answer linearly yeah yeah yeah. So, in all these systems the Fermi systems there is a Fermi liquid theory and this stuff will satisfy the Fermi liquid theory assumptions and everything will vary as predicted calculated by Fermi ok alright so let me again define the problem. So, we have the system of size l there are n particles of size sigma and what I like to calculate is the mutual correlation between these positions of these particles. So, there is this stuff which is called the two point correlation function rho 2 r 1 r 2 equal to r 1 x 1 x 2 x 1 dx 2 is equal to probability of finding particle t equal between x 1 and x 1 plus dx 1 and 1 particle between maybe r 1 r 2 was better x and y because we have used x 1 x 2 as the coordinates of the particles and these are not the coordinates of the particle x plus dx and y and y plus dy. So, I take this line I pick a point x and pick a point y and make a tiny window around this of with dx dy and I guess you know what is the probability that the center of one of the particles will lie in this window and another here and I do not care where other particles are ok. The key point is that I am only asking what the probability of finding some one particle here and one particle here and I do not care what happens to the rest ok. So, these I hope to compute, but without calculating it can I sort of guess what this function does rho 2 of x y will be equal to rho 2 of x minus y away from boundaries or large n ok. So, I take the limit of system size becoming very large and the density remains finite then I ask this question then the answer only depends on the distance between x and y it does not depend on the precise position of where I take x and y because we will have translational invariance in the system. Maybe I have I have not established I have not shown this, but is this convincing enough for all of you. This argument that the two point correlation function will only depend on the distance between the points. So, long as the points are away from the boundary is a good behaviour condition which we expect will hold unless something awful happens and in my system that awful thing is not happening it is a good system. So, this will hold. So, this is a function of single variable and what is that can I plot this function rho 2 as a function of x minus y mod space yeah ok space. Well, let us write x minus y. So, at 0 there is a little bit of a problem because same particle may come twice no. So, let us not worry about 0, but what is the probability that very close to 0 here what is rho 2 when x minus y is very tiny let us say sigma by 10 0 precisely all the way up to distance equal to sigma this is going to be 0 and of course, the same thing on this side then it will have some non-zero value it will do something when x minus y is very large. We expect this to become equal to rho squared which is the mean you know each these particles will be independent and it tends to rho squared which is the mean density square again I have not proved this one. But is this obvious or you do not need a proof you will prove it yourself going after going home or some such thing or we do not need a proof we believe it anyway ok. So, very good known. So, now I want to sort of be able to guess the general structure of this without doing the calculation and then maybe one will actually do the calculation later. So, it turns out we will show this eventually, but you know even if you do not follow the details of the algebra later this point should be clear we will find that rho has a very interesting structure it is 0 up to here then it shoots up to a very large ish value and then it kind of goes down like this and then shoots up and goes down like this and then it goes down maybe and tends to an asymptotic value where why does this bump occur precisely at sigma because you know if you have put n particles. So, the mean density is rho, but you have removed particles cannot be here. So, the n minus 1 particles have to be in the rest. So, the some bumps have to occur, but suppose there is a particle here then again in very close to it other than particle cannot occur and. So, there is a dip and. So, this argument very qualitative works even in 3 dimensional heart spheres. So, the correlation function in 3D will have roughly this behavior not be exactly this, but of this sort which is that there is a region where it is 0 then there is a peak which is called the first shell and then there is a dip and there is a second shell and there is a dip and there is a third shell, fourth shell and after four shells you cannot see much because it becomes the this function is different from 0, but not very different from. So, the correlation function is different from rho square, but not very different it kind of asymptotes to a constant value. Now, this phenomenology can be qualitatively understood in as many words as I said just now, but the advantage of the Tung's guess is that we can actually calculate this function rho 2 exactly in this problem which you cannot do in most other problems in Stake Mac ok. Even for the 2D Ising model rho 2 cannot be calculated not in this way because you know Ising model can only be solved at h equal to 0 and we need we need to work with fine variable density and stuff like this. So, it will be an exaggeration, but let me make it. It will be only a finite number of cases are known where you can actually calculate the correlation function exactly this rho 2 and so, this is one of them. I am sure you can come up to me and say there is another one, yeah there are some more, but this is one of the few cases which can be done and it can be done in 15 minutes. So, so that is an extra advantage ok. So, I will calculate the correlation function for this rho 2 in the next few minutes 15 yes. So, there is the mean spacing between particles. So, they actually correspond only to sigma, but the healing length corresponds to rho and the rate at which these things die down like this or like this they correspond to the mean density ok. So, I want to calculate this function probability there is a particle between a plus d r even. So, how is this different from this? I have shifted the origin that is one thing I have done and this is a conditional probability. So, there is it is divides by rho when r goes to infinity this function will go to rho while the other function went to rho squared ok is this point clear ok. So, now equal to what this is what we want to determine the key to the solution is the fact that the variables delta i are nearly uncorrelated variables and in the thermodynamic limit they become uncorrelated variables ok. Had another way in the constant pressure ensemble which is we introduced last time they are independent variables. So, I will study this problem in the constant pressure ensemble where life is easier. So, in constant pressure ensemble delta i are independent i i d ok. Now, what is the distribution of delta i? Because they are independent variables in our problem they it was just I forgot the answer somebody can help me with this what is the distribution of delta i? It is a so probability of delta i I guess looks like this it is it is shifted exponential it is 0 up to something and after that there is an exponential minus beta p delta i which is in the distribution function. So, it is e to the power minus beta p delta i minus sigma that is the function and there is some value here and there is some normalization I will not do ok. So, so this quantity is equal to well suppose there is in our problem there is one particle here and I am asking what the probability that the second particle is here then the answer is 0 you cannot have it. So, I guess my r should be bigger than sigma otherwise the answer is 0 suppose r is bigger than sigma, but it is less than 2 sigma then they cannot be any other particle between them then delta i is equal to r that is the probably it must be there is nothing else that can happen ok. So, this probability is equal to probability that delta 1 equal to r the full answer good, but this answer is valid only if r lies between sigma into sigma suppose r is a little bit bigger bigger than 2 sigma then it is possible that delta 1 equal to r, but it is possible delta 1 is not bigger than r n, but there is one more particle in there ok. So, I should add to this plus probability that delta 1 plus delta 2 equal to r minus sigma probability that delta 1 plus delta 2 plus delta 3 is equal to r minus 2 sigma and terms like that ok. So, the first one is very simple that was this result which we drew it is a something like this. What is this thing delta 1 and delta 2 both are exponentially distributed distributions what is the distribution of delta 1 plus delta 2. So, that is easy that is if x and y have distribution which is p of x is equal to e to the power minus x then p of x plus y is equal to x e to the power minus x is x plus y equal to alpha is equal to alpha e to the power minus alpha because you take two exponentials and you have to integrate along the diagonal that gives you one factor x, sorry x and y this is the function is e to the power minus x minus y and the positive quadrant then I want to determine the probability that x plus y equal to a given number then I have to integrate along this line, but along this line this function is a constant. So, the length of the line comes which gives you a factor proportional to the value this is x plus y equal to s. So, there is an extra s which comes here and that gives me the term. So, this looks like s e to the power minus s and this term by the same argument now there are three terms, but they have to add up to r minus 2 sigma then I have to do this in three dimensions. Same thing, but now the cut is two dimensional is proportional to s square. So, the answer will be s square e to the power minus s give or takes a little bit. So, now, let me write it in go read detail just to show that I was not cheating this is equal to exponential minus r minus sigma theta of sigma minus r this ensures that this term does not contribute when sigma r is less than sigma then I will write the second term plus, but I will write r minus 2 sigma minus r sorry I beg your pardon it is r minus sigma not sigma r minus sigma this is r minus 2 sigma and this term is actually s minus r minus 2 sigma whole time e to the power minus r minus 2 sigma theta p theta sorry bigger yeah bigger this part is visible they just so it was too small this is the last equation I will write. So, e to the power minus r minus let me write r minus sigma theta of r minus sigma and let me call it p tilde which is beta p plus theta of r minus 2 sigma then there is an r minus 2 sigma e to the power minus p tilde of r minus 2 sigma oh I did not write it much small bigger r minus 2 sigma exponential minus p tilde of r minus 2 sigma plus theta of r minus 3 sigma r minus 3 sigma squared exponential of the same thing minus p tilde r minus 3 sigma and so on yeah this is seen looks right it is actually still wrong and there are some factors which will come here to properly normalize this stuff I can do that you know it is not a problem I can put in all the algebra, but the key point was to make you realize that this problem can be done tracking those factors is straight forward and my time is up. So, I do not want to do it here if I concentrated on fixing all the factors in all these formulas then the key point would not have been conveyed ok. So, I will stop here.