 and welcome to module 5 of chemical kinetics and transition state theory. Today we are going to use the dynamics in phase space that we learnt in the last module and calculate what is the equilibrium density matrix. This is a very famous relation which is attributed to Boltzmann. We will use this equilibrium density matrix to calculate average quantities and then we will calculate the Maxwell Boltzmann relation. So, just a quick recap in the last module we looked at the density in phase space rho of q, p. This density is the density of finding the system at the given point in space which is q, p and again q, p represents the phase space. We also derived, not derived really, but verified the Hamilton's equations of motion given by these equations del qi over del t is del h over del pi and del pi over del t is minus del h over del qi, very symmetrically looking equation. So, we will move forward with this and what we want to find now is how does density matrix changes with time? I have told you how q and p changes with time, good. But as I have mentioned in the last module, we are really not interested in dynamics of q and p completely because it is a very, very large space. I am more interested in finding the dynamics of rho. So, we are going to assume one statement or the derivation can be slightly mathematical and complex. So, we are not going to cover that in this course. It is called the very famous Lieuvel's theorem, another very famous mathematician and he was also exploring the dynamics of our nature, the differential equations that govern our nature and he figured out one very, very strange and powerful theorem which is called the Lieuvel's theorem, d rho over dt is 0. I must here emphasize this is a full differential, I have not made a mistake. This is a full differential with respect to time, not partial differential. So, this is not partial differential. For those interested students who want to read a proof of this, I have provided a link here. This is one of the rather concise proofs. The proof, there can be more complex proofs coming from divergence theorems that you can also look at. But this one will give a proof in a rather simple language. So, if you are interested, you can look at this proof, you do not have to. First, let me tell you what this Lieuvel's theorem mean, what am I talking about? What the implication of this Lieuvel's theorem is, let me draw a one-dimensional phase space. Well, there is no one dimension, it is always 2D, q and p. So, I have a one-dimensional system which is only q and the corresponding momentum p. What this is saying is, let us look at some point in this phase space. And essentially, let us look at a little box here of length delta x and length delta p. And I have some density of particles in this box. And we are going to look at delta x and delta p being very small. Lieuvel is asking the question, fine, I have this very small box. How does this box changes with time? How does this box evolves with time? So, each particle in this box is having some dynamics. And I end up getting a new box that perhaps looks like this with some new delta q prime and with some new delta p prime. Lieuvel says that delta q into delta p is equal to delta q prime into delta p prime. So, the volume of this little box does not change with time. It can get reconfigured, its shape may change, but the volume will not. And in retrospect, it is not a very strange thing really, is it? What we are saying is that the number of particles is effectively conserved. We are starting with some particles. Well, they may expand or contract a little bit, but the net volume does not change, its shape may change. So, that might be intuitive or non-intuitive depending on you, but we do have a rather mathematical proof of this. And in this course, we are not going to write the proof, we are not going to cover the proof, but we will assume this statement to be true. And we are going to look at the consequence of this theory. So, let us analyze this d rho over dt. d rho of q comma p comma t, we are being more formal, over dt is given by, you remember your chain rule, it is given by 3 partial differentials, del rho over del t plus sum over i del rho over del q i del q i over del t plus sum over i del rho over del p i del p i over del t. So, here and here I have dropped the brackets just for notation is, but at all points, the rho is the same as rho of q comma p comma p. So, this is corrected for all rho. We are interested in rho equilibrium, that is what we are trying to find in this module and play around with it. So, rho equilibrium, in the last module we specified an important property of this, which is that rho del rho equilibrium over del t is 0. That is rho equilibrium does not explicitly depends on time, it depends only on q and p, that is the very meaning of equilibrium. So, I can get rid of this term, this whole thing is 0, value this theorem. So, I end up with only this equation for equilibrium, if rho is non-equilibrium, then del rho over del t may not be 0. Well, what about it? We get an equation, let us massage this equation a little bit more and what we get is, so this differential equation will hold true only for equilibrium density matrix. So, now we will put in the Hamilton's equations of motion for d q i over d t and del p i over del t. So, I will get this is equal to sum over i del rho equilibrium over del q i and del q i over del t, look back into your notes, this is equal to del h over del p i over del p i and del p i over del t, you can again look back into your notes, this is equal to minus del h over del q i this is equal to 0. Let me just take this equation and write it slightly cleaner, I will take the summation together of these two terms, I will take the negative outside and I get this equation. So, I have this equation that I have just copied here in this slide, what do I do with this equation? As it turns out we can solve this equation and the solution is slightly hard. So, in this course, we are simply going to state you the solution. The solution is rho equilibrium is equal to sum n into e to the power of minus beta and h is a function of q comma. So, I will leave this as an assignment problem to you, this will be one of the assignment problems to verify equilibrium satisfies the above equation. So, you have to substitute rho equilibrium in this above equation, take its derivative with respect to q i and p i and show that this sum over i will come out to be 0. So far, what we have got is rho equilibrium is n into e to the power of minus beta h, where beta is a constant so far, I have not specified you what beta is. But this particular rho equilibrium might be reminding you of something very specific. In statistical mechanics, you must have seen the distribution from a different perspective for canonical ensemble, some constant into e to the power of minus beta e i at a particular e. In that statistical mechanics, beta was 1 over kT, where kB is the Boltzmann constant, you have to be as famous as Boltzmann to get a constant after you. Note that h is the equivalent of energy, it is a function, but that is what gives me the energy. So, we note these two equations and we will postulate that beta is 1 over kT for us, even in this equation. So, with this, I want to make a few comments on this distribution, this distribution is called the Boltzmann distribution. This distribution holds only at equilibrium, never ever make the mistake of assuming this distribution out of equilibrium, out of equilibrium I have no idea what this rho might be, that is for a totally different course. Another somewhat confusing point that might appear to you, this rho is a statistical answer, it is a highly averaged answer. What I mean by that? Suppose I take a small system, I take a box with a 1000 gas particles in it and I find the distribution of q and p at a given instant of time, that distribution might not follow this Boltzmann distribution. However, if you increase the size of this system, instead of using 1000, I use million, I use billion, I use 10 to the power of 23, I use 10 to the power of 100. So, as this number keeps on increasing, my distribution will tend to this Boltzmann distribution. Or if you do not want the large system, you may ask that is all good and fine, but my system is finite only. After all in this room that I am sitting in, I have a finitely many number of gas particles, will Boltzmann distribution not hold in it? It will if you average over time. So, for even a finite size of particles, if I take the distribution at different snapshots of time, I find it at some time t0, I find it again at time t1, I keep on doing it and I average them, I am going to get back this distribution. This is by the way called the ergodic hypothesis, just an extra bit of information. So, what we have got is this Boltzmann distribution in phase space. It is very famous, I have just written it out clearly here. Let us get into what is the normalization constant n now. So, one property that we had listed is integral over all phase space will always be 1. Well, this should also be true for equilibrium density matrix. If it is true for any rho, rho equilibrium is just a special case of a density matrix. So, this must also be true. So, I get integral dq integral dp n into e to the power of minus beta h, n is a number it is a constant. So, I take it out of the integral and I get n equal to 1 over integral dq integral dp e to the power of minus beta h of q, p. As it turns out this integral that you are looking at is a very important integral. This is called the partition function. Whole of statistical mechanics is dependent on this integral. Now, we can calculate any property we want in statistical mechanics using this partition function. So, at the end I get n equal to 1 over q or rho equilibrium is equal to e to the power of minus beta h q, p divided by q. So, now we will look at if we have this rho equilibrium. What do I do with it? Can I comment on something useful out of this rho equilibrium? And the answer is to calculate any useful property you always average over phase space. Again, I do not care about where each particle is. I also usually do not care about a property at a given point in space space. I want to calculate an average quantity with like temperature. So, once more a reminder of our 1D example have let us say q and p and I have some quantity some function defined over q and p. And I want to find the average of a. What am I going to do? Well, I will integrate over q. I will integrate over p over all possibilities. I will find the probability that I am at that q, p that is given by rho multiplied by a of q, p. So, that is our usual trick of finding averages. We find this property that I want to find average over at that given q, p. I find what is the probability that I am at that point. And then I just integrate over all phase space. So, in general for an n dimensions I will let me write 3 n dimensions. I get average of some quantity a where a is some function of q, p to be an integral over dq integral over dp rho of q, p a of q, p. So, I have calculated an average quantity now. And once more just to remind you integral over dq integral over dp is a shorthand for dq 1 minus infinity to plus infinity minus infinity to plus infinity of dq 2 till minus infinity to plus infinity of dq n minus infinity plus infinity of dp 1 minus infinity to plus infinity of dp n, it should be 3 n. So, I am integrating over every possible coordinate and momentum. And rather than writing this rather ugly and long many integrals, I denote it by integral over dq vector, integral over dp vector. So, let us take an example. Again let us revert back to one dimension and let us find what is average momentum. So, I have a property which is p. So, a is p now and I want to ask you the question what is average momentum at equilibrium. So, we will use the prescription that we had. So, in 1 b q, p at rho equilibrium of q, p is given by e to the power of minus beta h of q, p divided by partition function q. So, we will simplify this a little bit. My Hamiltonian is given by p square over 2 m which is the kinetic energy plus v of q. And q is integral over dq, integral over dp, this e to the power of minus beta h. So, the average over momentum as defined in the last slide is dq, dp, p the quantity I want into rho equilibrium. So, let us substitute these quantities that I have written. Rho equilibrium is e to the power of minus beta h divided by q. And I will substitute the q here in this equation. So, I get integral of dq, integral of dp e to the power of minus beta h divided by something that will look very similar. So, q is the number q I have taken out of these integrals and substitute q equal to integral dq, integral dp e to the power of minus beta h. In all of these places I have limits from minus infinity to plus infinity. Let me put all these integrals there minus infinity to plus infinity. So, how do we calculate this? How do we simplify this? So, let me just write down what I had got in the last slide dq minus infinity to plus infinity dp minus beta h into p. And now we are going to substitute our Hamiltonian here. Hamiltonian was p square over 2 m plus v of q into p by the same I am not going to I am going to stop writing integral limits. So, we will open this exponential and note that this is equal to. So, I have taken the part that is dependent only on q on the integral of q only and the part that is dependent on p only in the integral of p. And in the denominator I do the same this looks like a complex integral to solve. How do I do it? It is actually trivial to solve this is actually equal to 0. Why you ask? Well, you see this cancels with this. This is easy these look a bit nasty, but you note that the numerator this is this function that is there what you notice is that this is an odd function. So, f of minus p is equal to minus of f of p. And what do you know about integrating an odd function over all space? It is equal to 0. So, the numerator is 0 and so the whole thing becomes 0. But well this is perhaps you already something you expect do not you? The average momentum should be 0. It does it at thermal equilibrium all directions are equivalent to the particle has equal tendency of moving forward as going backward. It does not mean that every particle is at rest the average is 0. So, I have a question challenge for you can you estimate what the average kinetic energy is going to come out? So, let us assume I have two different chambers at same temperature T one filled with helium and other filled with argon. What do you think will be the average kinetic energy per atom for these two chambers? So, take a moment think about this answer very carefully go to the link that is provided here and provide your answer there for an immediate feedback. Do you think the helium is will be larger for helium it will be smaller or both will be equal and if you do not know be honest and you can write I do not know this is all anonymous. So, take a pause and answer this question. So, we will quickly look at how to solve this question. Well they have to be equal that is my claim we are going to prove it mathematically, but we already know they must come out equal. How? Because the average kinetic energy is essentially related to temperature and two boxes come connected to each other of whatever molecules are filled with in helium, neon, argon, nitrogen, hydrogen, atmosphere whatever at equilibrium the average kinetic energy must be equal. Because the temperatures are equal ok. So, we are going to prove this statement this intuition that I have. So, we are going to calculate the average kinetic energy p square over 2m now just like we had calculated the average p. So, now we will play the same tricks is well integral over dq integral over dp overall space that is your rho equilibrium into p square over 2m. So, just like we did in the for average momentum we will separate these and write the Hamiltonian and the denominator will be the partition function which looks very close to the numerator, but not exactly and we again do the separation of q and p. So, I get minus infinity of plus infinity dq to p square over 2m dq minus beta v of q integral over p d to the power of minus beta p square over 2m. So, the first thing that you notice at the potential term exactly cancels. So, it my gas might even be interacting it might be hydrogen it might be whatever gas you want even water vapor the potential term will exactly cancel. And so, we have to solve these nasty integrals that are here. So, to solve that I have provided you the integrals here. So, use those plug them in and calculate these. So, this is going to be another assignment to be able to solve these integrals using the provided integrals. And you have to show that this is equal to half kb so, 1 over 2 beta. So, this is another assignment problem. We will stop here what we have today looked at in this module is how equilibrium density matrix comes about from the fundamental equations of motion. We have skipped a few steps in the proof we do not derive the Louisville's theorem completely, but it is to motivate where does this partition where does this density matrix comes about. So, you do not need to know the details of the proof that we have not covered. But what you need to know is what is the origin of this equilibrium density matrix. We have used this equilibrium density matrix to calculate averaged quantities. We have looked at average momentum and average kinetic energy today. So, with that we will stop today. And in the next module we will look at one of the very famous distribution which is the distribution of speeds. Thank you very much.