 Right, we will today discuss the answers to this quiz that we had last week and what I would like to do is to also make comments. Ideally I should have given you the answer sheet so you could check against it but I will read out the questions and then you will be able to recollect what the question was about and we will try to give justifications for the various choices. So the first question was in 2 of all was that as follows, consider a quantum mechanical system with Hamiltonian H0, its normalized density operator in the canonical ensemble is given by e to the minus beta H0 over the trace of the same thing. So you are given this row equilibrium is e to the minus beta H0 divided by the trace of and the question asked is what does this tend to as t tends to 0 from above of course as you go to absolute 0. The statement is that it tends to the unit operator. Is that true or false? Tends to the unit operator as t tends to 0. What happens to the state at t tends to 0? That means beta tends to infinity, this thing goes to infinity here, right? So you cannot just put beta going to infinity because then this goes to 0. But then this denominator has to be taken into account. So if you took this system to have a discrete set of energy levels e0, e1, e2 etc with corresponding Eigen functions, so if you set that H0 on phi n equal to en on phi n, if you did this then these states, normalized states form a complete set, an orthogonal set, an orthonormal set of Eigen vectors. N now labels all the possible states, stationary states of the system, all the state, all the Eigen states of the Hamiltonian. There could be degeneracy in which case n will be more than one label. It will have more than one quantum number and we are just schematically writing everything down as with an n here. So this trace is over all the states, diagonal elements over all the states clearly. Now in this representation of these phi n's, it is clear that this operator itself has a representation which is summation over n e to the minus beta en times the portion that projects onto the ground, onto the state n. So this is phi n, phi n divided by summation over n e to the minus beta en. This is over the states n. And out here when you take phi n, phi n, it is normalized to unity and you get this here. So the question is what happens to this? Now we can say what happens to this as beta tends to infinity. Clearly if all the en's are positive, then the numerator is also 0 and the denominator is also 0. But if e is 0 is finite, we do not care if it is positive or negative. We really do not care where you start measuring the spectrum from. Wherever is a ground state, it should not be minus infinity. Wherever you have this ground state, you can pull out e to the minus beta e 0 as a common factor and then whatever is inside all the other exponentials will be the difference of e 0 minus en's and they will all go to 0 as beta tends to infinity. So this thing will tend as beta tends to infinity to just the contribution from the lowest energy level, the ground state which is equal to of course e to the minus beta e 0, phi 0, phi 0 divided by e to the minus beta e 0. Because all the other terms will have decaying exponentials will all go to 0 and this cancels out. So it tends not to the unit operator but to the projector of the ground state as it should because the system goes to its ground state at absolute 0 of temperature, right. The probability of it being in any higher state is 0 essentially and it is with probability 1 in the ground state. Therefore the density operator is not any longer, there is no uncertainty, thermal uncertainty here, it is just the ground state operator, projector, that is it. Which is not the unit operator, which is not the unit operator because recall that it is this quantity, this is equal to the unit operator. At t equal to infinity, you can ask what happens if the temperature becomes infinite, then all these Boltzmann factors e to the minus beta en's essentially all of them go to 1, everything becomes equally probable and then the system will end up with this density operator. You see, if these, if these are the energy levels of the system, they do not have to be equally spaced and this is the ground state e 0, what statistical mechanics is telling you is that once you have a certain finite temperature then there is an average energy and there is a variance, a scatter about it, okay. On the other hand, if the temperature becomes infinite then k t which is the thermal noise, the quantum of thermal noise becomes much larger than the gap between energy levels, so these gaps are irrelevant completely and the system will just freely move around from state to state and therefore there is no bias at all in this case and then every state is equally probable essentially in the limit. But as the temperature is lowered, the bias comes more and more strongly towards the ground state till at t equal to 0, everything sits in the ground state. So that is the fundamental property of equilibrium, statistical mechanics and the answer therefore is false, it is just the projector of the ground state, okay. The next statement is given that the x component of the velocity of a particle of mass M in a fluid satisfies a generalized Langevin equation with a memory kernel which is a decreasing function of its argument and A t of t is a stationary noise, all you are given is that this is a stationary noise, you are not given anything more. So the statement is m b dot of t plus m integral minus infinity to t d t prime, some memory kernel gamma of t minus t prime, v of t prime is equal to on the right hand side, random force A t of t and this is stationary. So we are told that this quantity here has 0 mean at any time and A t of t t naught, A t of t naught plus t equal to function of t, the origin does not matter, that is all you are given and you are also told that this quantity gamma is a decreasing function, positive decreasing function of its argument for all positive values of its argument, right. So that is all you are told, nothing more. And then the statement is v of t is a Markov process and its density function, conditional density pdf satisfies a Fokker-Planck equation, these two are false. It is false because this incredible memory sitting here, this is an integral differential equation, it is not a Planck-Jama equation at all, the usual type of Planck-Jama equation. If you insist on writing this as a differential equation, you will have derivatives of all orders in the time variable. So there is a lot of memory in this business and it is not a Markov process. So there is no chance that the probability density function pdf satisfies the Fokker-Planck equation, none whatsoever, it is not a Markov process. But there is actually an interesting theorem, it says that if this is stationary and Gaussian, which I have not said in this problem, but if this is stationary as well as Gaussian, then it turns out that the density function pdf p of vt v0, this is the conditional density, turns out in that case when this is also a stationary Gaussian process to obey an equation which looks very much like the Fokker-Planck equation, but it is not a genuine Fokker-Planck equation. In particular, in particular you have to recognise that if it is not a Markov process, then this 2 point conditional density alone does not determine the rest of the process. You have got all the joint densities on top of it. So the fact that this satisfies a certain Fokker-Planck like equation might help in calculating this quantity and therefore the autocorrelation, but it does not help as far as understanding the nature of this stochastic process. That is much more intricate than this. Well then one of the questions says that this is inconsistent for eta to be delta correlated. And we can see right away we will see in a minute that it is not, this cannot be delta correlated. So the question is, suppose I tell you this is a stationary process, it is not delta correlated, it has got some finite correlation time. What is it you can say about the output process here? Well for that you have got to take recourse to these fluctuation dissipation here which we will talk about in some detail. So you can make some statements about the autocorrelation of V of t, compute it, but in general to find out what is this process, what is its probability density function exactly, that is not doable from this as it stands. You need to know all the properties of eta of t and case by case you have to analyse it which is not a trivial task. So the statement is, it is false, the statement is false. Next thing says, the assumption that the autocorrelation of eta of t is proportional to a delta function leads to an inconsistency. Is that true or false? That is true. It definitely does as follows so we can see what the problem is. You see all you have to do is to ask what is the second fluctuation dissipation theorem in this case. Recall that the original Panjama equation which had this m v dot plus m gamma v, just a local function with a constant gamma, the ordinary Panjama equation, this was set equal to square root of gamma times eta of t. This was a Gaussian white noise with 0 mean and a delta function autocorrelation. Then consistency required that 2 m gamma k Boltzmann t had to be equal to capital gamma. That was the second fluctuation dissipation theorem relating the dissipation in this equation with the strength of the fluctuations driven by the noise. The question is, what is that for this process? Not very hard to do. All you have to do is to find out what is the correlation here and how does it get related to this quantity here. The corresponding relation here turns out to be m m k Boltzmann t gamma bar of omega equal to an integral from 0 to infinity dt e to the i omega t eta of t0 eta of t0 plus t. That is the second fluctuation dissipation theorem in this problem. Now you can see how this is going to reduce to this, okay. Because in the case where this quantity is integral e to the i omega t gamma of t, one sided Fourier transform. To go to the ordinary Panjama equation, you have to assume this to be a delta correlated quantity. So with capital gamma times a delta function, if you put delta of t, then t is equal to 0. This goes away and you get capital gamma which is on this side. And on this side you are going to get m k t gamma bar. Now the way to go to a constant is to put this memory kernel gamma of t, little gamma of t to be equal to constant gamma times a delta function. So when gamma of t goes to gamma times delta of t, no memory, then you get the ordinary Panjama equation from the generalized Panjama equation. If you put that in, this term here becomes gamma but with the factor missing because this quantity here when I say delta, it got to run from 0 to minus infinity to infinity to pick up the full contribution of the delta function. Otherwise you pick up half the contribution and when you put that back here, you get this relation. So that is how from the general second fluctuation dissipation theorem, you end up with the special case for the ordinary Panjama equation. Now the question asked is, it is inconsistent in the general Panjama equation, generalized equation to have a delta function as a correlation of the noise, to have this delta correlated? Yes indeed, because if you put in a delta function of t here times a constant, this integral becomes just that constant but on this side it has got a function of omega and there is an immediate inconsistency. You cannot have a function of omega equal to a constant because certainly we know that if this is non-trivial, this is some function of omega and that is gamma bar. So on this side you have got a function of omega and on this side you have got a constant which is inconsistent. So this is how we see that delta correlated noise is inconsistent with the generalized Panjama equation. This has got to have a finite correlation time. In fact that is the reason for introducing the generalized equation in the first place because we said look, in real life no noise is purely delta correlated. There is always some finite correlation time. So that is the reason why we have to introduce a memory kernel for consistency. So the statement as it stands is true. The next question says, the dynamic mobility of the particle vanishes for infinite frequency as omega tends to infinity. So this too is a very straightforward question because from this equation, we derive the fact, this equation implies that the dynamic mobility in this problem is 1 over M times gamma bar of omega minus R. Whatever be this, it is clear that this quantity here becomes infinite as omega tends to infinity and this just vanishes and physically that is expected because the dynamic mobility gives you the average velocity amplitude for a given sinusoidal amplitude of some sinusoidal frequency omega. As omega tends to infinity, this becomes so large that the system cannot respond. It does not, its inertia prevents it from responding. Therefore it should go to 0 which is what is happening. Without or without this term, it still goes to 0. So this statement is certainly true. The next one is a little tricky. It says in this model, the generalized Langevin equation model, the equal time correlation between the velocity and eta, the noise eta is not equal to 0 but the equal time correlation V of t with the acceleration V dot of t is equal to 0 and the question asked is whether this is true or false. So again within the context of this equation, the question asked is does this imply that V of t V dot of t in equilibrium equal to 0 but V of t with eta of t in equilibrium is not equal to 0 and you are asked is this true or not. It is true, it is true because the whole point about fixing the causality and stationarity conditions in this problem, well the first problem was in the ordinary Langevin equation, while we discovered that the correlation of the velocity is stationary, the velocity is a stationary random process that V of 0 V of t in equilibrium is e to the minus gamma modulus t, we ran into the difficulty that stationarity was apparently violated because if you took V square of t average, that is equal to k t over m, Maxwellian and you differentiate both sides. The right hand side has no dependence on little t, the left hand side you differentiate and then set t equal to 0 or something like that. Then you immediately discover that this must be equal to 0. For a stationary process, any stationary process xi of t, xi of t xi dot of t must be equal to 0 on the average, that is the meaning of stationarity but the Langevin equation violated it. The question is whether it is true in this equation and the answer is yes, it is true, this thing can be established rigorously, this is certainly true but is this true or not? Now in the ordinary Langevin equation we took this quantity, the force on the right hand side to be uncorrelated to the velocity at the same time arguing that the force determines the acceleration at any given time and the velocity comes a little later after integration, so this quantity is certainly 0 by causality but that led to a problem, that led to an inconsistency with stationarity. To fix that we took this problem and said look, stationarity has to be rigorously obtained here because equilibrium tells you that this velocity is stationary but causality has to be written more carefully, that the effective random force is not this but a portion of it which comes from the history here and the assumption of causality was that the velocity is uncorrelated with that effective force at a later instant of time. When you did that carefully then it turns out this quantity is not 0 at all but rather it has got a finite value which also we computed I think, so somewhere along the line we had this here, so it turns out that in this model if you compute this thing here, this turns out to be equal to m times an integral from 0 to infinity dt prime, gamma of t prime, v of 0, v of t prime. So at a given instant of time, this correlator is not 0 but rather depends on the memory kernel in this fashion. So it is an integral over the velocity correlation at all previous instance from 0 to infinity weighted with this memory function and it is not identically 0. So the statement as it is made is actually true but this is a little trickier to evaluate but it is straightforward to evaluate it after you make the proper causality assumption here. Then finally the last one on this set of questions is in the same generalized Langevin equation as tau tends to infinity suppose the memory kernel decays to 0 like some power of tau, inverse power of tau. So you are given, you are given that gamma of tau goes to 0 as tau tends to infinity like 1 over tau to the power alpha where 0 is less than alpha less than equal to 1, I believe it is less than equal to 1. The statement is does the static, the static mobility of the particle vanishes in this case. So the question asked is is mu at omega equal to 0, equal to 0 as a consequence of this? Yes or no? Yes, it is true. The reason is very straightforward. We know that mu at omega equal to 1 over m gamma bar of omega minus i omega, so mu of 0 is just 1 over m gamma bar of 0 and this integral is integral 0 to infinity d tau gamma of tau. If this dies down like 1 over tau or 1 over square root of tau or any power of tau less than 1 in the denominator then of course this integral blows up, it diverges. So this tends to infinity and therefore mu of 0 vanishes, this quantity therefore tends to 0 because this memory is so strong, this dissipation, this is like the effective drag is so strong that the system does not move. That is what is meant by very slowly decaying memory, okay. So much for the generalized Langevin equation. The next set of questions has to do with linear response theory. Given that A represents a physical observable of a quantum mechanical system and you have asked, the statement made is phi A A of tau equal to minus phi A A of minus where recalled that this quantity stands for the expectation value in quantum mechanics of A of 0, A of tau over i h cross inequality over the canonical ensemble here. So true or false? The statement is true and why do you say that? Because under time reversal you are asking what happens under time reversal and we know in general you have phi A B of tau in general is equal to epsilon A dot epsilon B phi A B of minus tau where epsilon A dot is a time parity plus or minus 1 but you know yeah there is a dot here which is equal to minus epsilon A epsilon B phi A B of minus tau because A dot and A have opposite time parities like position and velocity or velocity and acceleration. So in this case A is equal to B phi A A of minus tau equal to minus phi A A because epsilon A whole square is 1 whether it is plus 1 or minus 1 does not matter epsilon A itself the square is always 1. So this response function is an odd function of tau and that helps us answer the next part of it. It says that the corresponding generalized susceptibility and spectral function are related according to real chi A A of omega is equal to half phi A A tilde of omega. So you are asked is real chi A A of omega equal to 1 half phi A A tilde of omega. Is this true or false? Well because of this this thing implies that it is phi A A tilde of omega is actually twice 2 I times the imaginary part of chi A A of omega. Had this been a symmetric function this would have been twice the real part but because it is anti-symmetric it is twice the imaginary part. But now we know that the imaginary part of the susceptibility is an odd function of omega while the real part is an even function of omega and in this case the spectral function is proportional to the imaginary part. So it is clearly an anti-symmetric function and therefore this statement that this is not true instead you have this statement here. So that is the correct statement okay. So the spectral function is therefore an odd function because it is proportional to the imaginary part of chi. So this of course immediately implies phi A A. So that takes care of this part and then there is set of questions on the Boltzmann equation. The first of which says let f of r, v, t denote the phase space density in the one particle phase space mu space of a dilute classical gas. We will assume that the molecules of the gas only undergo binary collisions with each other. Then the statement is at every instant of time the gas can only be in a state in which f satisfies the Boltzmann equation. Is this true or false? We are not saying the gas is in equilibrium. We are saying that it satisfies, we are not saying there is f of r, v, t and in the absence of an external force this thing goes in equilibrium to f equilibrium of v which is independent of time. And in an absence of an external force this quantity is equal to n times the Maxwellian w of v. We have seen that. Even if there is an external force present which is dependent only on the position of the particle, a conservative force then this quantity here goes to the Maxwell Boltzmann distribution. So this will go in equilibrium as t tends to infinity to e to the power minus phi of r over k Boltzmann t times w of v, right? Apart from some normalization factor such that the whole thing is normalized. The integral of this over d3r, d3v should be equal to the number density. So this constant will fix that normalization, okay. Now the question is at every instant of time is the Boltzmann equation satisfied or not? We are not saying that at every instant of time the phase space density is the equilibrium density. We are not saying that at all. The equilibrium density could happen in between. It could occur. It is one of the possibilities. The gas could be in a state where it is exactly where this density is exactly the equilibrium density. And as t tends to infinity we know that it attains that, okay. And now the question is at every instant of time is the Boltzmann equation itself valid or not? Well, no. In general no because you see the Boltzmann equation is valid if and only if the assumption of molecular chaos is valid. If and only if in mu space we have said that at some point are inside the cell, the centre of a cell in mu space we have assumed that the probability that you have a particle in this r with velocity v, v1 and t, probability density is this. And the probability density that you have 2 of these guys with velocities r1 and velocities v1 and v2 is the product of these densities. What does that mean? That means there is no correlation in velocity space at all, okay. This would certainly not be true if there are recollisions. Something hits against this particle, it goes and hits another particle and comes back and hits this. Then this kind of assumption is not valid anymore. You can easily see that it builds up a correlation between these 2 particles, okay. That is immediately clear. Take the simplest possible example, put everything on a line, let us say the particles are equal mass and put a wall at one end. Particle A is sitting here, particle B comes and hits it, bounces back, hits the wall, comes back and hits A. Surely the velocity of the projectile the second time is dependent on what happened the first time. So this is no longer true. It is only true at some instance of time. And at those instance the Boltzmann equation is valid, okay. So the answer is false. The Boltzmann equation holds good, only at those instance where the molecular chaos assumption is valid. In between there are all kinds of states possible, which do not satisfy the assumption. No, it says let f of r, v, t denote the phase space density in the one particle phase space or new space of a dilute classical gas. We assume that the molecules of the gas only undergo binary elastic collisions with each other. That is it. That means no energy is lost. And then the statement is that every instant of time the gas can only be in a state in which f of r, v, t satisfies the Boltzmann equation and that is not true. That is not true. The next statement says since the Boltzmann equation is the first order partial differential equation in t, delta f over delta t equal to something or the other, it follows that r and v comprise a 6 dimensional Markov process. Because as you know for a Markov process the master equation says that the density, whatever be the conditional density, satisfies a first order differential equation in time. What that equation is depends on the actual process, the transition rates and so on. But it satisfies a first order process in time. And not only that it satisfies the chain condition. So it is even if you like a non-linear integral equation just like the Boltzmann equation. But is the Boltzmann equation a Markovian master equation? What do you think? So if you recall you had delta f over delta t plus there were terms like v dot gradient to respect to r, f that does not bother us. Because this sort of term appears even when you write down the Langevin equation in phase space. When we wrote down the Kramer's equation for instance, we got terms like this. We got v streaming terms plus f over m dot gradient to respect to v of f. This is equal to on the right hand side delta f over delta t collision. And if you recall this was a complicated integral which involved quadratic squares, I mean products of 2f's at 2f's minus final minus initial and so on. We had a very complicated collision term here. So this is by no means a Markov process of any kind. It is a much, much more complicated thing than that. Finding what the correlation is between the autocorrelations and so on in this case is very tricky because you have to essentially solve this Boltzmann equation. And notice there is no explicit randomness which is brought in anywhere here. In principle you are supposed to know this on the right hand side and solve itself consistently. So it is certainly not a Markovian master equation by any means. However if you recall when we looked at the so-called single relaxation time approximation, in the case when you linearize the Boltzmann equation, now we did not clearly explain what are all the assumptions which you need to do the linearization. But if you did, it is a small departure from equilibrium and on top of it you make the single relaxation time approximation. Then under suitable cases the velocity relaxed like the for like a Kubo-Anderson process. It relaxed like a Markov process. So that is a special case of a special case. In that case you can mimic it but that is a crude approximation to the actual Boltzmann equation. This is not a Markovian master equation. And it does not say how strong the collisions are. It just says v1 and v2 go to v1 prime v2 prime. It does not say put any bounds on how big this quantity should be. It does not put any bounds on how big the relative velocity v1 minus v2 should be at all. Could be drastically different. Yes, yes. No, detailed balance only involve time reversal in variance for the dynamics, right? So in general what you do is for Markov processes if detailed balance obtains then you can write down the equilibrium distribution by using the detailed balance condition just as we could do so here too by inspection of that collision integral. But there is nothing which says detailed balance should be there because in the Markov process I can specify the transition rates, probabilities between different states of the system quite arbitrarily subject to some ordinary conditions. But the fact is they are completely arbitrary. There is no reason why detailed balance should apply at all for a general Markov process, okay. Whereas here the physics is saying that it does apply because ultimately the collisions, the dynamical process of the collisions is time reversal invariant. Whatever is happening there is assumed to be time reversal invariant, okay. So that is when some physics has been put in about the molecular level process of scattering and then you get this detailed balance coming out, popping out, okay. And finally consider the gas in the absence of an external force. The statement is for any given distribution f of v, t, there is no dependence on r. The Boltzmann H function is defined as f log f integrated over v. Statement, H of t is a monotonically decreasing function of t, true or false. So let us be careful. You define the Boltzmann function as integral d3 v f of v, t log f of v, t. And is it true that this quantity satisfies d H over d t? Alright, less than equal to. Is this true? Exactly. You see the definition, this quantity, you give me any distribution and I will define this quantity, the H function. What we have proved is that if this satisfies the assumption of molecular chaos and therefore satisfies the Boltzmann equation, then you can show this. So what it is saying is that if at instant of time t, if at this instant of time, some particular instant of time, say t0, the Boltzmann equation is satisfied, then the slope, local slope at that point is negative. So this H function will go like this. But it is not saying that at all these instance of time the Boltzmann equation is valid. The local instantaneous slope to one side is negative. And then you can also show by clever argument that if at this instant of time the Boltzmann equation is going to be satisfied, then at an immediately preceding instant the slope is positive. So this function actually fluctuates in this crazy fashion. It is not monotonic at all. What we do know and what we can establish is that at some of these peaks, at local peaks, at points where the Boltzmann equation is satisfied, the H function is at a local peak. The converse is not true. We are not saying that at all such peaks the Boltzmann equation is satisfied. And certainly it is not satisfied here or here or here. So it is only satisfied in a statistical sense. Then there are other questions about how frequent are these points, how dense are they and so on. Those are subtler questions. Okay, there is a lot of very intricate stuff involved there. So the theorem is very clear. And the statement as it stands is false. H of t is not monotonically decreasing function of t. It is a rapidly fluctuating function of t. And finally consider the gas in absence of an external force. Let F equilibrium be the equilibrium phase density and V1 plus V2 goes to V1 prime, V2 prime as usual. It is an elastic scattering process between two molecules. Then statement, F equilibrium is unique and satisfies the detail balance condition. Such that F of V1, F of V2 is equal to the product F of V1 prime, F of V2 prime. True or false? True. This satisfies by the system is in equilibrium. Okay. So at this stage detail balance is valid and it is a necessary and sufficient condition for equilibrium density. So that condition actually as we saw determines the equilibrium distribution completely because what you do is to take logs and then it says log F of V1 plus log F of V2 is log F of V1 prime plus V2 prime. So log F of V must be something which is a function of all the quantities that are conserved in the collision process such as a constant, the velocity itself for equivalence particles, the sum of initial velocities and the sum of the kinetic energies. So that is what necessary and sufficient condition. For proving the necessary part of it, we have to go through the H theorem. The sufficient was inspection. Just by looking at the fact that this condition satisfies the collision integral vanishes and in the absence of an external force, F is time independent explicitly. So that part was straight forward. So the statement is true. Okay. Then we had a few fill in the blanks. The first of which said that we have a system satisfying the large y equation and you are asked to find the velocity correlation time in terms of the memory current. So what is integral 0 to infinity dt V of t0, V of t0 plus t, equilibrium divided by V square equilibrium. This quantity has dimensions of time. This is by definition tau, the velocity correlation time. And the question is in the ordinary Langevin equation this was gamma inverse. The question is what is it in the generalized Langevin equation? Pardon me? Yeah, this quantity is equal to 1 over gamma bar of 0. And the way you establish that is to go back to the first fluctuation dissipation theorem which essentially says that if you multiply this by e to the i omega t, then you get this quantity is equal to V square equilibrium which is kt over m as usual divided by, divided by gamma bar of omega minus i omega. All you have to do is to set omega equal to 0 and divide by this quantity. So that immediately tells you this implies that tau velocity is 1 over gamma as indeed you can see that in the case of constant friction this will go to 1 over gamma. But again I repeat in the generalized Langevin equation this quantity is not an exponential decaying exponential something much more complicated. You can easily see what it will be. Instead of this e to the i omega t imagine this to be the Laplace transform. So certainly this is true e to the minus st of this is gamma bar of s where this is the Laplace transform plus s. Therefore this quantity is the inverse Laplace transform of this expression. Had this been a constant that is just e to the minus that constant times t but this is some complicated function of s we do not know what it is in general. So the correlation function can be very messy, very very messy indeed. You have to invert this Laplace transform that is not a rational function of s in general. Then the next one was to find the Fourier transforms of A of 0 B of t commutator equilibrium the Fourier transform of this fellow divided by the Fourier transform of the same thing but with the anticommutator. Now we know that this thing here was equal to the spectral function on the right hand side phi tilde of A B e to the minus i omega whatever it is. So the Fourier transform was essentially i h cross times the spectral function. The Fourier transform of this had the same thing except there was a cot hyperbolic become minus sign. So this ratio will become minus tan hyperbolic beta h cross that is all you have to do is to read off the Fourier transforms in each case and it is equal to this. Then the equilibrium mean square value of A squared is related to an integral of the spectral function according to according to what? Now you have a formula for A of t B of t prime or A of t prime B of t whatever it is expectation in terms of the spectral function a spectral representation. All you have to do in that is to put A equal to B and t equal to t prime right. So you end up with A squared equilibrium therefore cannot be independent on time. This is equal to i h cross I do not know there is a 2 pi there must be 2 pi somewhere over 2 pi integral minus infinity to infinity d omega phi tilde of A of omega 1 minus e to the beta h cross that is it yeah yeah you get a apparently different expression. So what happens if you use A of t so you have one expression for A of t prime B of t then you have another expression for B of t A of t prime then you have a third expression for the sum of the 2 which is really the full return. In every one of these except in the commutator where if you put A equal to B and t equal to t prime you get 0 equal to 0 except for that all of them will give you expressions there is only one expression right. So what is happening yeah they will all give you exactly the same if you simplify this. If you simplify this thing here this I pull out a half h cross omega here and then this becomes minus sign hyperbolic or whatever it is this side right. So they will all give you exactly the same expressions. One of them has an e to the beta h cross omega on top this one does not have it at all but you can convert one to the other but be little careful you also have that e to the i omega whatever it is right. So in doing this conversion you may have to change omega to minus omega so it is just a little piece of algebra but you should satisfy yourself that all of them will give exactly the same expression. I can write that in several ways then they came these questions on diffusion which I think could might have been a little tricky because we did not spend too much time on this when I discussed this topic. So the first question says to find for diffusion in 3 dimensions we have the Gaussian's fundamental solution of the diffusion equation and then you want to find the probability distribution of R squared which is also a random variable. Now that is straight forward because all you have to do is first you find the probability distribution of little r which is 4 pi r square times the fundamental distribution and then so let us suppose that you have p of r, t equal to 1 over 4 pi dt to the 3 halves e to the minus r square over 4 dt. From here you go to let me call it phi of r, t the distribution of r itself this is just equal to 4 pi r square it is this density you integrate over angles and you get the distribution pdf of the radial distance okay. Then you want to go to rho of r but we know that rho of r dr equal to phi rho of r and t if when little r is between r and r plus dr capital r is between r and capital dr capital r then this equation obtains in this case both are positive so we have no problem it is being mapped only once so if you plot it is a one to one map okay then therefore there is no complication about folding and things like that. So this immediately implies that this quantity so it says take this quantity write eliminate little r and write it in terms of capital r and put this Jacobian factor in which is equal to 1 over 2 root r should not forget that and then you get something like square root of capital r e to the minus capital r by 4 dt so that is the density that part is trivial. The next question is what is the Fokker Planck equation satisfied right and the last part of it says what is the Langevin equation satisfied by it. Now there you have to be careful to find the Fokker Planck equation is not very trivial what would you do? We will discuss the answer after I have gone through the answer books just to find out what people have done. What would you do for what? We know the Fokker Planck equation for this that is just the diffusion equation. What should you do to get the Fokker Planck equation for this? So this quantity is delta p over delta t equal to d del square p that is the Fokker Planck equation just the diffusion equation. What should I do to get the Fokker Planck equation for this? In this equation put p equal to 5 over 4 pi r square and then solve write it as an equation for 5. So that is the Fokker Planck equation for 5. What should you do for the Fokker Planck equation for this? Change variables once again do the same thing. Now use this use this relation here and then get you can do it in one step of course you can do both these in one shot right. Crucial point is each time you have to keep track of the Jacobians. This one is the phase phase factor and this one is the Jacobian factor you got to keep track of that. So you got the Fokker Planck equation for rho and then you got to use this correspondence backwards in order to get the Langevin equation for capital R. That will have a drift term and it will have multiplicative noise. So it is not the ordinary diffusion equation. Would I have just done this by starting with the Langevin equation for little r and then going to capital R? After all if I get some equation do not ask me how equal to something on the other. Can I simply say r equal to r square implies, can I put this in and get the Langevin equation for this? No. No. This is where the tricky part of white noise or the Wiener process comes in okay when you have multiplicative processes. But whereas going through this route of this correspondence between the Fokker Planck and the Langevin equation systematically is a shortcut to get the Langevin equation for capital R or little r for that matter which are non-trivial statements because this so-called white noise that we talked about is not a well-defined object mathematically. You have to use what is called a Wiener process and then you have to use Ito calculus which is somewhat different from normal calculus. So we will not do that but I will mention after I go through your answer books we will come back and I will mention but the right answer here is okay. So that this last portion would have been a little time consuming the rest I think was very straightforward etc okay. So we will start next time with critical phenomena.