 So, yesterday we started with a simple model and I was just begun looking at this model where you have an oscillator in a fluid. So, basically it is the model of an oscillator in a fluid which satisfies the following equation. There is x double dot of t for any coordinate, any Cartesian coordinate x8 plus m gamma x dot of t plus omega naught squared x of t equal to the ensemble average on the right hand side in the presence of some external force. The random force averages to 0 and what is left is 1 over m f external of t in this fashion. And the question we were asking is, what is the average value of x of t equal to under these circumstances in general? And we are trying to do this via linear response theory. So that was the idea. Of course this is an inhomogeneous equation for x of t and therefore you can solve it by the Green function method. The Green function method says that x of t is an integral over whatever this inhomogeneous term is up to time t multiplied by the Green function and that is all you need here. So the Green function g satisfies g of t minus t prime satisfies the differential equation d2 over dt2 plus gamma d over dt plus omega naught squared acting on g of t minus t prime equal to 1 over m times a delta function of t minus t prime. It satisfies this differential equation, the same thing as an operator on the left hand side and then you have a delta function instead on the right. So we can solve this, the simplest way to do this is by Fourier transforms here, right? So let us define g tilde or let us define g of t equal to, we must be careful about a Fourier transform convention. The convention I used was this is an integral from minus infinity to infinity d tau e to the i omega tau g of tau and of course g of tau is 1 over 2 pi minus infinity to infinity d omega e to the minus i omega tau g tilde of omega. Now remember that the response function itself, the response function that we had was phi of tau. So we know that the generalized susceptibility chi of omega in this problem which I have not yet introduced for this problem. This we know in general is 0 to infinity d tau e to the i omega tau phi of tau which can also be written as equal to integral minus infinity to infinity d tau e to the i omega tau phi of tau theta of tau, the step function theta of tau. It is anyway 0 for tau less than 0 so I could write it in this fashion. That is the generalized susceptibility. This is the green function. We already have seen that this is g of tau. We have seen that already earlier that you multiply the response function by this to maintain causality and you get the retarded green function here. So it is clear from this formula that chi of omega is the same as g tilde of omega. It is just the Fourier transform. So it immediately follows that chi of omega is the same as g tilde of omega. So chi is the one sided Fourier transform of the response function but it is the actual Fourier transform of the green function. So if I compute the green function here by whatever means I have actually found chi of omega. And then I invert the transform in order to write down what the solution itself is. And after that I superpose all these harmonics to find what the solution is for a general frequency. It is a very simple problem but just to show you how this proceeds, let me do this. Let us do this explicitly. So let us try to solve this using Fourier transforms and I define the Fourier transform according to this equation. If I differentiate this once with respect to tau, I pull down a minus i omega inside. So derivative with respect to t is the same as multiplication of the Fourier transform by minus i omega. Therefore it follows from this that minus omega square minus i omega gamma that takes care of these two terms. This is minus i omega whole square is minus omega square minus i gamma omega plus omega naught square. Of course this quantity acting on g tilde of omega is equal to 1 over m because delta function has a Fourier representation 1 over 2 pi d omega e to the minus i omega tau. So the 2 pi cancels out on both sides and you just end up with this expression. Or chi of omega that is the same as g tilde of omega. This is the same as this is equal to minus 1 over m times omega square plus i gamma omega minus omega naught square. So that is our expression for the Fourier transform of the green function or the generalized susceptibility in this problem. It is clearly an analytic function of omega. You can define this for actually complex omega. It is an explicit analytic function but it has singularities when the denominator goes to 0. And the singularities because this is a quadratic in omega is just a product of two simple factors and therefore the singularities are two simple poles, okay. So this can be written as equal to minus 1 over m times omega minus omega minus times omega minus omega plus where and now the locations are of course the roots of this quadratic equation and therefore omega plus minus equal to minus i gamma over 2 that takes care of this portion. So if you have an under damped oscillator you can write this as plus square root of omega naught square. Pull out the half that cancels it to minus gamma square over 4. So there are two singularities at these two roots and they are simple poles in the omega plane. Now as long as omega naught is bigger than gamma over 2 plus or minus those are the two roots of this quadratic, okay. As long as omega naught is greater than gamma over 2 this is a real number here and therefore the imaginary part is negative. But we already know that the singularities of this green function with this convention for Fourier transforms have to lie in the lower half plane in omega. The upper half plane it has got to be analytic, okay. So you have two poles in the omega plane. This is real omega and let me just call it omega 1 and omega 2 and the two poles. Let us give this a name. This is just the shifted frequency of a damped oscillator, right. So it has got physical dimensions of frequency. This is equal to omega undamped. Let me put a u there to show that it is undamped because only then is this omega, this is a real number. So the, sorry, the under damped. When you have an under damped oscillator equilibrium is reached in an oscillatory fashion. If it is over damped it will just damp out exponentially. So at least in the under damped case, let us write this down, omega naught greater than equal to less than gamma over 2 corresponds to under damped, critically damped and over damped. These three signs respectively relate as you know very well from very elementary treatment of a damped harmonic oscillator or an LCR series circuit. You know that if omega naught is sufficiently large and the friction is sufficiently small then you have an under damped case. So let us first look at that case here. Where are these poles then located? Well, it is minus i gamma over 2 plus or minus omega u, which is some real number. So there is one pole here and there is another pole here and this is omega u, sorry, this is omega plus and this is omega minus. By the way, what happens to these poles if you start going from the under damped case to the over damped case? You can do that in two ways. You can either fix a gamma and change omega naught or you can fix an omega naught and change gamma in general. So the path of these poles will depend on what you are varying about parameter you are changing. For definiteness, let us say that I fix a gamma, damping is fixed and I change this omega naught. So I start with very high omega naught frequency much bigger than gamma over 2 and then it is like this, symmetric about the imaginary axis. And then I start decreasing omega naught towards gamma over 2. This pole will move down like this and this other fellow will move down like this because gamma is fixed. All that is happening is that omega u is getting smaller towards gamma over 2. At omega naught equal to gamma over 2, the 2 coincide at this point. And then beyond that, this quantity becomes imaginary. So you have minus i gamma over 2 plus a small imaginary part or minus a small imaginary part. The plus sign therefore will go upwards in this fashion along the imaginary axis and the minus sign will go downwards along this imaginary axis. And finally as omega naught reaches the value 0, this goes away and you have plus or minus i gamma over 2. So this pole goes and hits the origin here. It is still approaching it from below, from the lower half plane and the other fellow goes and sticks at gamma, minus i gamma. So even in the other 2 cases, the over-damped and the critically-damped, it is easy to see, it is fairly trivial to see that no singularity ever crosses this and goes up into the upper half plane. In fact, they approach even the real axis only asymptotically. There is really no singularity for any finite non-zero value of omega. So this justifies our assertion that will turn out when I said that the singularities of generalized susceptibility would not lie in the upper half plane. It is completely analytic there. What would happen if gamma was 0? No damping. What would that correspond to? Then you have an undamped oscillator. This goes away. We saw there were problems with the undamped Langevin particle when you did not have the friction term. Things were not in equilibrium and so on. Some such things should show up here too. Suppose you do not have this at all, then you have just this oscillator, the plane undamped oscillator, then exactly the same thing will go through except you have omega squared minus omega naught squared without this term, which would mean that you actually have a pole here and a pole here on the real axis. Now what does that physically correspond to? It says that this generalized susceptibility becomes singular at two real frequencies plus or minus omega naught. And what it actually physically means is that if you do not have damping and you apply an external force with exactly the same frequency as a natural frequency of the system, then in the absence of damping, the amplitude will become arbitrarily large. It will just diverge. As we know, if you if you push a swing at exactly its natural frequency, you go on kicking it and there is no damping. In principle, the amplitude will become unbounded. So that is what happens here in this trivial example. But in all physical cases and this is a lesson for us in general, we will see why this should be so. In all physical cases, the generalized susceptibility will have both a real and an imaginary part. We did not have gamma at all. It is pure real. We are in trouble as we will see. So system can never be either the susceptibility can never be either purely real or purely imaginary for reasons which will become clear as we go along. All the physical examples of susceptibilities that we know of will have this property. For instance, you put light through a medium. What would the corresponding susceptibility be? It is either the dielectric function of the medium or the refractive index if it is a transparent medium. Now in general, the refractive index is a complex number. The real part of the refractive index will measure the amount of dispersion. But the imaginary part will measure the amount of absorption. And you cannot have pure dispersion without any absorption. Just as if you give me a piece of resistance, just a resistor, it has got a self-inductance. So there have or you give me an inductor, it has got a resistance. So it has got to have a reactive part as well as a resistive part always. And that is exactly what we are seeing here in this simple example. Alright now back to how to compute this quantity. So let us try to calculate what the response does although it is fairly straightforward here. We got till here and now let us compute it. So what we have is chi of omega or g tilde of omega and therefore I use this formula. Now I have to find g of tau equal to 1 over 2 pi and there is an m there in chi. So this is equal to minus 1 over 2 pi m and integral from minus infinity to infinity d omega e to the minus i omega tau divided by omega minus omega plus in this fashion. That is the integral you have to do. Now this sort of integral is fairly, this is the straightforward integral to do but here is where the power of contour integration comes in because this kind of integral where you have a trigonometric function here and then a rational function down here. It is geared, contour integration is tailored to do these integrals very easily. Now what would one do? Notice that if you write omega as well the contour of integration to start with in the omega plane is minus infinity to infinity there. There is a pole here at omega plus, there is another pole at omega minus. To do this contour as a contour integral you have to close the contour. Remember this is minus r to r in the limit in which r goes to infinity. That is what is meant by this integration. So you keep r finite and then you close the contour and then let r go to infinity. Now if you write, if you go on to take an excursion into the complex omega plane and you write omega as omega 1 plus i omega 2 then e to the minus i omega tau becomes e to the minus i omega tau. Omega 1 tau and then minus i times i omega 2 so it becomes e to the omega 2 tau. This is an oscillatory term but this fellow blows up for positive tau if omega 2 is positive. So for positive tau you cannot close it in the upper half plane because this contribution would be non-zero. But you can close it in the lower half plane because this contribution from the semicircle is 0 in the limit in which the radius goes to infinity. So you can add a well chosen 0 to do the integral provided you keep track of the fact that when tau is positive you have to do it in the lower half plane and when tau is negative you do it in the upper half plane. So you need to do this in this fashion for tau grade less than 0 and you need to do this for tau greater than 0. But you see if tau is less than 0 this contour does not enclose any singularity at all. So you can shrink it down to a point and it disappears. Therefore this integral has got to be equal to minus 1 over 2 pi m times the theta of tau because we know it is 0 when tau is less than 0. So that factor has emerged automatically as you can see the green function must have the theta function and whatever multiplies this is the response function. So theta of tau and then what? Now in the lower half plane when you close the contour you pick up contributions from the poles with a minus 2 pi i it is in the clockwise direction. So there is a minus 2 pi i times e to the power minus i omega plus tau from the omega plus part. So you multiply by omega minus omega plus and take the limit as omega goes to omega plus. So you got omega plus minus omega minus which is twice omega u. So there is a twice omega u plus e to the minus i omega minus tau and now you multiply by omega minus omega minus take the limit. So you got omega minus minus omega plus which is minus 2 omega u. So you get a minus sign and then 2 omega u. So let us pull that out and you have this. So this 2 pi goes against this 2 pi. So this is equal to i over m times 2 omega u eta of tau times this guy here. So this is e to the power minus i times omega plus or minus is another minus i gamma over 2. So the minus sign skill and the i square gives you a minus 1. So this is minus gamma tau over 2 times e to the minus i omega u minus e to the minus omega u that is equal to minus 2 i sin omega u tau. So now we are in good shape. The i cancels against this, this cancels, the 2 cancels and you left to adjust that. So we have our final answer which says that g of tau equal to e to the power minus gamma tau over 2 sin omega u tau divided by m omega u multiplied by theta of tau from this factor. So we have a closed expression in this case. Therefore this part of it is equal to phi, the response function. And you can see in the underdamped case the response reaches the value 0, the response function reaches the value 0. So if you plot as a function of tau, you plot phi of u, this is started because this is linear in tau, it is going to start at 0, no instantaneous response as you would expect. Then it is going to climb up to a peak and then go down in this fashion. And the envelope goes like e to the minus gamma tau over 2. So it is exponentially damped but there is an oscillatory tendency here because it is underdamped, okay. What would you do if it is overdamped? Well if it is overdamped omega u becomes imaginary and then sin of i theta is related to sin hyperbolic theta apart from an i, etc. So you can actually write down, it will be exponentially damped, it will just die down exponentially without the oscillatory behavior. I leave you to work out the critically damped case, etc. They are all trivial extensions of this. So we have closed answer in this case, completely closed answer. In fact there is one more step and that is to write down what the actual average value is which is easy to write down now. So you can write what is x of t equal to, this is an integral from minus infinity to infinity into t dt prime times f of t prime phi of t minus t. And we have an expression for phi of tau, this is phi of tau. So you can plug it in and that is it. For arbitrary force history, you can write down what the exact displacement is, average displacement is at any time. So this was a toy problem, baby problem because it was just the harmonic oscillator damped but it gives us a lot of valuable lessons. It corroborates the fact that the susceptibility is analytic in the upper half plane. It also shows certain symmetry properties of this thing, of this chi because I leave you to find out, to check that if you took this, we know that chi of omega and this problem is equal to minus 1 over m omega squared plus i gamma omega minus omega not squared. And it is a trivial thing to verify that real chi of minus omega is, that is immediately obvious because this goes up on top and the real part is going to involve this on top. The imaginary part is going to involve this with the mod squared of these fellows. So it is immediately obvious, the real part is symmetric and the imaginary part is anti-symmetric in omega. Yeah, exactly. So together these two correspond to saying chi of omega, chi star of omega equal to chi of minus omega, which is obvious here. So from this relation these two follow, for real omega, for real omega. Actually there is a symmetry property for complex omega as well. What is that property? You do not have to put an omega star here, out here. Does it mean you know chi in the lower half plane? You have to be careful here because she is saying that chi of minus omega is related to chi star of omega star. We have an explicit expression for chi here so there is no problem but in general in the more complicated problems you would know in the omega plane you would know that it is an analytic function here and we may not even be able to discover what is the singularity structure here because it would not be anything as simple as this in general. So all manner of B's could lurk there and we do not know. So we do know that is analytic up here. Now what would you say if omega were complex and you have a symmetry like chi star of omega star equal to chi of minus omega. Yeah, we should know. Now what is the symmetry property? What is it saying? If you start with some omega you could put a star here as well. It does not matter. If you start with some omega here where is omega star? It is down here. But where is minus omega star? Both the real and imaginary parts have to be flipped. So it is here and you are safe. You are really safe because it is not going into the lower half plane at all. We do not know anything about it in general. So it is relating a chi here with a chi here and everything is okay in general. So one has to be careful to check this all the time that it is not giving you an information. This representation alone is not giving you more information. We will see what the implication of this is. The implication of the fact that chi is well behaved everywhere in the upper half plane and in general on the real axis as well has serious implications, important implications and we will see what this means. In a nutshell what it will mean and we will derive this is that the value of chi at any point is determined in a self-consistent way as an integral over chi over the rest of the real axis. So it is trying to say that an analytic function does not sit in isolation. At every point because you say this function as derivatives of all orders these functions have a highly constraint. You cannot really arbitrarily specify this function as you please. Once you specified it in a small region you specified it everywhere okay. We will get back to that issue. But now at the moment let us go right back to the general representation for chi and see whether we can simplify the expression. So let us go now to the canonical what is called the canonical. Well actually it should be called the Kubo because the Japanese places Kubo first pointed out that you could write a representation for this correlation function valid for both the classical and quantum cases and it involves a certain specific operation that you need to do to simplify the correlation function. So recall where we had started. We started by saying phi AB of tau was equal to in general the equilibrium expectation value of A of 0 with B of tau where this bracket stood for either the Poisson bracket or commutated divided by ih cross depending on whether it was classical or quantum in equilibrium. But you could also write this as the trace over the equilibrium density matrix of rho equilibrium with A of 0 times B of tau. We could also write it like this by the cyclic invariance of the trace. Now notice that I am using the cyclic invariance of the trace and it requires certain assumptions on the operators. We will assume that till we trip up that these conditions are met actually. But remember that trace AB is equal to trace BA but under these circumstances but it does not mean that AB is equal to BA. The operators themselves will not commute and we must be very careful about that in general. So we have this equation here. Let us look at it classically and see what this actually means. Whether we can make this a little simpler or whatever. Now if you go here this actually means A of 0 classical is equal to Poisson bracket of rho equilibrium with A of 0. What I mean by A of 0 is that the time argument in the dynamical variables the Q's and P's is at T equal to 0. And the values of Q's and P's at any time T is found by nominally solving the Hamiltonian equations of motion and since we are talking about it in equilibrium there is no external force here to start with. The whole thing is done in the canonical ensemble. Rho equilibrium is E to the minus beta H naught the unperturbed Hamiltonian and this refers to time evolution under the unperturbed Hamiltonian. That is what appears in this equation. Now this is equal to if I wrote this out in terms of P's and Q's this is equal to delta over delta Q equal to minus beta H naught delta A over delta P at 0. Let me just drop that argument for a minute minus delta equal to minus beta H naught over delta P delta A over delta Q. It stands for this summed over all degrees of freedom in the system. So this I should put Q sub i, P sub i and sum over i. This is understood. So if you like to write it as summation over i. Now I should use this bracket but if I differentiate this this is the same as differentiating E to the minus H in same as differentiating H naught pulling out a minus beta times E to the minus beta H naught. So this can be written as E to beta times E to the minus beta times that. And then I have a delta H naught over delta Q i delta A over P i and similarly H naught over P i. But that is equal to the Poisson bracket of H naught with A which is equal to beta times E to the minus beta H naught the Poisson bracket of A with H naught. This is A at 0 always. Let us even put this in. But in classical mechanics I know that A Poisson bracket of A at 0 H naught is just A naught. It is the time derivative of this dynamical variable at t equal to 0. So if that A is the position this stands for the velocity. If it is the velocity it stands for the acceleration and so on. So this is equal to beta e to the minus beta H naught A dot of 0. It is supposed to differentiate and then put t equal to 0. And we put that back here. So this fellow has turned out to be equal to beta times trace E to the minus beta H naught A dot of 0 B of tau. I restore this and I written this down. But what is this equal to? This is the equilibrium autocorrelation function correlation of A dot of 0 with B of tau. So we have an extremely compact formula that tells you what to do in this case. This has become equal to, so I want to retain this equal to in the classical case beta times A dot of 0 no comma, no comma. So what has happened is that the average value, the equilibrium statistical average value of a commutator or a Poisson bracket has become just beta times this 1 over k t times the average value of A dot with B of tau. If this is a mechanical force A is x for one dimensional motion for example. And we have been computing the susceptibility, the position at some time t later on. So B is also x. So it says if you want to find the response function to find x at a later time you need to know how the velocity at an earlier time is correlated with the position at a later time. Of course because the velocity at an earlier time will tell you what the position is at a later time. So it is not surprising that you ended up with this expression, okay. So this is one crucial result, classical but quantum mechanically is another story. So now let us see what happens. So this won't work, it is not equal to this. So I again start with this fellow here. So in the quantum case q m, A and B need not commute and they need not commute with H naught itself in general. When they commute things become very trivial but in general these 3 operators need not commute with each other at all and we have to be very careful about preserving orders. So let us write this out, this quantity, this thing out. This is still true, this is a general formula. So now we have phi A B of tau to be equal to trace the commutator of this fellow with B of tau on the right hand side, right. So first we got to find out what is this guy equal to. Let us do that. So this is equal to 1 over h cross the commutator trace e to the minus beta H naught A of 0, A of tau of course minus B of tau, well, alright, let us leave it like that, B of tau, sorry, this is still on the right hand side, minus A of 0, e to the minus beta H naught is okay. I have just written this out explicitly in 2 terms. My target is to pull out this e to the minus beta H naught because then I will argue that whatever is left is the equilibrium average value and I take the trace with respect to it. I am stopped from doing so because this factor is sitting in the middle here, it does not commute to this. So one solution is to put plus or minus the same factor here. So let us write this to 1 over i h cross trace e to the minus beta H naught times A of 0 minus e to the beta H naught A of 0 e to the minus beta H naught, the whole thing is inside the trace. This may not commute with that, this may not commute with that, these 2 fellows may not commute at all. In general, this is all, this is stuck at this point, but now Kubo had a very clever way of resolving this paradox. What do you do with this term here? Recall that we are working in the Heisenberg picture and I put a time dependence as governed by the Hamiltonian H naught. So when I say that any operator, let us call any operator M, M of t is equal to e to the i H naught t over H cross, M of 0 into the minus i H naught t over H cross. That is my definition of the time argument inside operators in the Heisenberg picture. I can do this formally, formally, purely formally without affecting all orders of operators, etc. By identifying beta with a time, so essentially if I say beta is equal to i t over H cross, t equal to minus i beta H cross. So purely formally, I can write this whole thing as equal to 1 over i H cross trace e to the minus beta H naught, A of 0 minus A of minus i beta H cross, little bracket here, B of tau. But this is now looking more and more like the difference between two time arguments looks at the integral of the derivative between these limits. So I could actually write this as equal to 1 over i H cross trace e to the minus beta H naught and integral d t prime from minus i beta H cross to 0, A dot of t prime where that is the dynamical variable which when integrated will give you A of t times B of tau and it is precisely this difference. But now you got to specify a contour of integration that is along the imaginary axis from 0 to minus i beta H cross that is uncomfortable, this fellow is uncomfortable. So let us change variables. Let us put minus i H cross, let us put lambda equal to, I want to convert this to a real integral and I would like lambda to be, when t prime is equal to this, I would like lambda to be beta. So how do I fix it? i t prime over H cross, will that do the trick or t prime equal to minus i lambda H cross. Then this quantity here is integral to 0 and when t prime, I have put, when t prime is minus i beta H cross lambda is equal to beta, beta and then minus i H cross d lambda A dot of, now I have to put in what is going on, A dot of what? t prime is minus i lambda H cross. So that is what this integral is, which if I am little careful, I write as 0 to beta, we get rid of the minus i cross sign and that will cancel against this. So you will permit me to write this as integral 0 to beta d lambda A dot of minus i lambda H cross B of tau and this i H cross goes, these were intermediate steps. But what is this fellow equal to? This guy here, now we are back to the real thing, it is equal to e to the power i by H cross H naught times minus i lambda H cross. So the i and minus i cancel, the H cross cancels, it is e to the lambda H naught. So this stands for, therefore this is equal to trace e to the minus beta H naught integral 0 to beta d lambda e to the lambda H naught A dot of 0 e to the minus lambda H naught B of tau and it is the trace of this whole guy. So it is the equilibrium expectation value, not of A dot of 0 with B but an integral from 0 to beta of this guy with these factors put in. Yes, yes, yes, you can pull the integral outside because the trace is over only the operators. This is going to be something called the Kubo canonical correlation, it is more complicated than the classical case and the classical case everything commutes. So this fellow cancels against this and that is it and things will be exactly the same thing that you got before. So let me stop here, even the beta factor will come out because you got to do 0 to beta d lambda and that will give you a beta factor which is sitting here. So I will start at this point and show you what this does, what this quantum correction will do a calculation of a specific case and you will see why this thing is an extremely interesting quantity. Let me stop here now.