 So, to recapitulate what I said the last time, we have a situation where you have a system with a Hamiltonian which is this, the function of time which we took to be H naught minus some A f of t, the prescribed function here. And then the question was what happens to thermal averages or average values observables, arbitrary observables. And we found that if you took any observable B and asked what its average was, it is B with respect to the equilibrium ensemble in the canonical ensemble, for example, plus delta B of t and we got a formula for delta B of t which read average is equal to an integral from minus infinity to t d t prime, the force integrated over the entire history of this force times a response function phi A B of t minus t prime. This response function is a certain correlation function in equilibrium. So we discovered that phi A B of tau for instance, we put t minus t prime equal to tau. This quantity, the response function was equal to an intake, was equal to the equilibrium expectation value of A of 0 with B of tau, either the commutator or the Poisson bracket, part from some numerical factor and that is what this was in equilibrium. So this average means something different. This is the average in the presence of the external force and if I write an equilibrium here it means it is in the absence of the external force like this case here. We could also write this in several ways. We could write this as A of minus tau B of 0 in equilibrium or if you like you could also write it as equal to the trace. By the cyclic property of the trace, we could write it in a number of ways. One of them is rho equilibrium with A of 0 B of tau just by the cyclic property of the trace or if you like rho equilibrium with A of minus tau B of 0 and so on. Or this B could also come here because by the cyclic invariance of the trace. So we have a number of ways of writing or computing this quantity depending on what is convenient. Now let us ask what does it imply? What does this thing imply for us? Well if you took a general, we already saw, we interpreted this to mean that the response is causal, linear, cost order in this force, causal because of this and retarded because the time argument depends on in the elapsed time since the effect, since the force acted at whatever time T prime. So given this closed formula, this formal expression here, we can now proceed to analyze this in a little more depth. The first thing you notice is that while this is true for an arbitrary f of t, an arbitrary force history because it is a linear response, the response to a sum of two forces is the sum of the responses, individual responses because of the superposition principle. Now we know that any arbitrary force, any time dependent force can be resolved into its Fourier components, into frequency components and then you find the response to a given frequency component and you can add up all those responses to get the full response. This is the whole point of Fourier analysis. So it is very convenient to do that and let us see what happens. You did that because it will give us a handle on what the system does for an arbitrary f of t in terms of what it does for various harmonics. We need a Fourier transform convention and I am going to stick to one because it is very easy to make numerical mistakes otherwise. So let me write down a Fourier transform convention here on this side. It does not matter which one you choose but we need a consistent one. So if I have a function f of t, I resolve it into Fourier components, then the Fourier components are resolved as minus infinity to infinity d omega e to the minus i omega t f tilde of omega and this implies that f tilde of omega equal to an integral minus infinity to infinity d t e to the i omega t f of t. So this is my Fourier transform convention. Once and for all, we will stick to this and if I make a mistake please point it out because we have to be careful with these signs. One of one appears with a minus, one with a plus, the 2 pi appears here rather than here and so on. Now of course there are many other conventions. Any one of them is just as good as the other. Very often they use 1 over square root of 2 pi here, 1 over square root of 2 pi here to make it symmetric. This could be a plus and that could be a minus and so on but we will stick to this convention. So when I go from a function, when I Fourier transform a function of time, it is 1 over 2 pi d omega is a measure and f tilde of omega is just d t. So we will try, I will try my best to stick to that convention. Then let us do the following. Let us ask what happens if I take a single harmonic and apply the force. So the force that I am going to apply is f of t equal to some amplitude f naught. Later I will do this with different amplitudes and that will be my f tilde of omega times e to the minus i omega t, some fixed omega for some given. So let this be so and then we ask what the response is. The corresponding response is delta B of t equal to an integral minus infinity to infinity. This is an f naught which comes out of the integral times d t prime e to the minus i omega t prime phi AB of t minus t prime and now let us set t minus t prime equal to tau. That is the obvious thing to do. So this becomes equal to f naught integral d t prime is minus d tau but this will run from infinity to zero here. So I will switch that zero to infinity d tau e to the minus i omega t prime is t minus tau phi AB of tau which is equal to an integral from zero to infinity d tau e to the i omega tau at term phi AB of tau. This whole thing multiplied by f naught e to the minus i. So this immediately says that if I apply the force f naught e to the minus i omega t in practice of course I apply a real force but technically we could have this as a complex number f naught I do not have to say I do not impose a reality on f naught it could be complex. So this is of the form cos plus sin or something like that. We will take real imaginary parts later but if I write a force of this kind then the response is the same force multiplied by a function of just one variable and what is that? This depends on omega because tau is integrated over it is a dummy variable. So this is a function of omega it of course depends on what say what is b but it is a function of omega. So this says this implies that the response delta B of t given this force the response this is equal to some function chi AB of omega times the force. In other words the response is just the same is a force multiplied by some function of omega depends on omega alone some attenuating factor where chi AB of omega is this integral zero to infinity d tau e to the i omega tau phi AB of tau. Is this the Fourier transform of the response function? No it is not it is an integral from zero to infinity it is the one sided you see we have not even defined this quantity for negative tau formally we can do so but right now we are not interested in it because this integral is cut off at t it is because it got cut off at t that when I change variables the lower limit became zero. So this zero here is a direct consequence of causality this quantity here is because of causality the force at a later time cannot affect the effect in earlier time. So it is like the one sided Fourier transform if this had been minus infinity then I have argued that by analogy with this formula I have said it is the Fourier transform but this fellow is just the one sided Fourier transform the integral is running from zero upwards here. This thing here just as this is called the response function this is called the generalized susceptibility corresponding to a cause A and an effect B in other words you put up the system through A and you measure some other quantity B which could be A itself in some cases. So common examples are you apply a magnetic field you measure the magnetization you apply an electric field you measure the polarization you apply a stress you measure the strain and so on. So this is a very very general framework we will look at many examples but it is a very very general framework and this quantity here carries all the information that is carried in this response function. Once you give me this chi of omega and I know what it does for a particular frequency it looks like this this is the response for a general force all I have to do is to sum all these fellows with this weight factor as a function of frequency times an f tilde of omega as the amplitude for frequency omega and I integrate over all frequencies and that is the general response in principle. So the susceptibility carries all the information but it is crucial to note that it is a one sided Fourier transform it is going to have important consequences very crucial consequences. Now the question is can we write this in terms of a Fourier transform itself can we write this chi in some fashion as a Fourier transform itself. So let me do that but before that there is one more thing I want to point out which is that the real there is no reason why this quantity is real at all. In general it is complex and we will see shortly that it cannot be either purely real or purely complex purely imaginary it has to be complex in general because we will attach physical meaning to the real part in the imaginary part. Now if you take the real part of this it is the real part of e to the i omega tau provided this fellow is real and this chi here is in terms of physical quantities. So these are all observable quantities all the operators are Hermitian then phi AB of tau is actually a real quantity. So the real part of this has a cosine and the imaginary part has a sine function here and therefore the real part of this chi is a symmetric function of omega. So real chi of omega equal to real chi minus symmetric function on the other hand the imaginary part chi of omega equal to minus imaginary part for real omega and so far our frequency is real just like time the frequency is real. Very soon we are going to make an excursion into the complex plane we are going to go to complex frequencies and then come back for physical quantities to real frequencies but for real values of omega or real which are the in fact physical frequencies are not only real but positive non-negative we will come to that but at the moment as a mathematical exercise for real values of omega you see that the real part of chi is symmetric function the imaginary part is an anti-symmetric function you can combine both by saying chi star of omega equal to chi of minus omega because if I take the complex conjugate here this is real this becomes minus that is the same as chi of minus omega that is a useful symmetric property we will make use of it. Now could I possibly write this in terms of this response could I write it in terms of Fourier transform itself the answer is yes because you see you can always write this as minus infinity to infinity dt prime f of t prime phi ab of t minus t prime multiplied by step function theta of t minus t prime but this is the unit step function I presume I is this the notation you use for it or do you use any other notation sometimes people say I write u or h or whatever but this is equal to plus 1 t greater than 0 so that is my definition of the theta function the unit step function I will retain that. So if I put that in then this integral runs all the way to infinity because it gets cut off at t less than t prime and I want to give a name to this so let me call it g ab of t minus g ab is a green function because this is how you write the thing in general you would say I apply a perturbation and I am looking for a response whatever equation it satisfies we do not care some dynamical equation after I take averages the average value of this correction to expectation of b is an integral over the force history over all histories times the green function here we will write a specific model down where you will see what the what g of t looks like this fellow looks like so if you do that then you could also write this formally you could also write this as an integral from minus infinity to infinity d tau e to the i omega tau phi d of tau theta of tau just by putting in a theta function there but this is what I have called the green function so this is equal to integral minus infinity to infinity d tau e to the i omega tau g ab that is a Fourier transform so it immediately says that this quantity is a Fourier transform of this g ab g ab tilde if you like so where does that get us well it actually gives us some information because this is chi Fourier transform of g ab which is the Fourier transform of phi ab of tau times theta of tau by definition but what is the formula for the what theorem tells you how to find the Fourier transform of a product of two functions of t the convolution theorem so it is clear that this must be equal to an integral from minus infinity to infinity d omega prime the Fourier transforms of each of these so phi ab tilde of omega theta tilde of omega sorry of omega prime theta tilde of omega minus this is the convolution theorem where I have to define these quantities where phi tilde ab of omega is the Fourier transform of phi ab now chi is the one-sided Fourier transform but we now want the full-blown Fourier transform so we go to this expression here the Fourier transform is this integral minus infinity to infinity dt e to the i omega t phi ab of t notice this requires now a knowledge of phi ab of t for t less than 0 so we have to find that somehow what we will do is to compute phi ab of t and ask what it does as you reverse time negative t so we will be able to define what phi ab of t is for all t we will do that by and by so this is equal to that and theta tilde of omega equal to an integral from minus infinity to infinity at least formally dt e to the i omega t theta of t now you can write this fellow as 0 to infinity dt e to the i omega t which does not exist which does not converge because e to the i omega t oscillates as t goes to infinity and as a Riemann integral this does not exist but this does not bother us because we know how to define Fourier transforms of distributions theta of t delta of t and so on are not ordinary functions they are distributions right. So what I need is a Fourier transform a Fourier representation for this quantity here what would you do I would like to find a representation as a contour integral of some kind how would you do that there are many representations but we want something which looks like this representation here what would you do well let us start with the statement that delta of t and I want to keep track of that kind of convention e to the minus i omega t so it is 1 over 2 pi integral minus infinity to infinity dt e to the minus i omega t delta of d omega that is the well known representation Fourier representation for a delta function right it also says delta tilde of omega is 1 that is what this is trying to tell us okay. Now given delta of t how is the theta function related to the delta function it is the integral of it right and it is clear that if you got a delta function in t at t equal to 0 if you integrate from minus infinity up to any point less than 0 the answer is 0 you integrate beyond t equal to 0 and the answer is 1 so it is precisely the theta function the integral of the delta function if I differentiate if I integrate this quantity I must integrate this with respect to t so the conjecture is that this quantity theta of t is equal to an integral 1 over 2 pi minus infinity to infinity d omega e to the minus i omega t divided by minus i omega formally it is the integral right so let me write it as i and put a minus sign here over omega but that is not a great idea because this integral does not exist it is singular and it passes through omega equal to 0 and this fellow is singular so it is not yet there we have to fix this problem what would you do you would shift this pole at omega equal to 0 away from the real axis and then you would find out what the boundary value is in the limit right. So we want an answer we want to fix it in such a way that theta of t must be equal to 1 for t greater than 0 and 0 for t less than 0 right so let us look at this contour integral in the omega plane the integration runs from minus infinity to infinity there is a pole at the origin if that pole is displaced downwards by i epsilon instead of 0 then we ask what happens to this integral well so let us ask what is limit epsilon goes to 0 from above integral minus infinity to infinity d omega over omega plus i epsilon so the pole is at minus i epsilon e to the minus i omega t what is this fellow equal to as a contour integral it is perfectly well defined now because there is no singularity on the line region of integration on the contour of integration but now the only way we know how to do contour integrals is using Cauchy's theorem for a closed contour so we need to close the contour we must formally define it as running from minus r to r and take the limit r going to infinity but we could close the contour with a semi circle and if you do that in this fashion here then the contribution if this is r some radius r on this circle this variable omega is r e to the i theta you can see that if I have e to the minus i omega 1 plus i omega 2 t this is equal to e to the minus i omega 1 t if omega is complex like anywhere in this plane and then plus omega 2 t times e to the omega 2 t now if t is bigger than 0 this is a positive number then this contribution vanishes as r goes to infinity provided this term is negative this means omega 2 the imaginary part of omega must be negative which means omega must be in the lower half plane in other words this is the contour to choose right if t is negative you are forced to close it in the upper half plane but there is no singularity enclosing it in the upper half enclosed by the contour in the upper half plane so this is going to give you 2 pi i whatever it is if it is in the lower half plane if t is positive and if t is negative it is going to give you 0 so this is equal to so let us put in the minus 1 over 2 pi i as well minus 1 over 2 pi i this quantity is equal to there is a minus 1 over 2 pi i here and when you close it here you are going in the clockwise direction so it is minus 2 pi i times the residue 2 minuses cancel each other and you get e to the minus i t times omega is minus i epsilon and now take epsilon goes to 0 e to the 0 is 1 and you end up with the theta function so this is precisely equal to a theta function if I want a theta of minus t and I have to put a plus i epsilon move it into the upper half plane so what we have done is to find the Fourier representation for the theta function so theta of t is equal to this in the limit as epsilon goes to 0 from positive values so what is theta tilde of omega we have a formula now therefore theta tilde of omega equal to limit well I do not write this limit it is understood in the sense if you look at the Fourier transform definition its integral theta tilde of omega theta of t if you look at its definition 1 over 2 pi e to the minus i omega t times the Fourier transform and what is the coefficient of 1 over 2 pi e to the minus i omega t so theta tilde of omega is equal to minus 1 over i which is i divided by omega plus that is the Fourier representation of a theta function. Now let us go back and we said that the susceptibility chi AB of omega was equal to a convolution minus infinity to infinity t omega prime phi AB tilde of omega prime theta tilde of omega prime minus omega sorry omega minus omega that was the convolution therefore this is equal to minus infinity to infinity d omega prime phi AB tilde of omega prime divided by there is an i outside omega minus omega prime plus i epsilon wherever omega appears I have to write omega minus omega prime it is convenient to write this in order to do the integral as omega prime minus omega minus i epsilon. So let us write it as equal to limit 1 goes to 0 minus i time so the susceptibility at any frequency omega has this Cauchy kernel out here 1 over omega prime minus omega and the weight factor with which a given frequency omega prime appears is given by the Fourier transform of the response function. So this is like a spectral resolution with different frequencies of this function here so not surprisingly this quantity is defined as the spectral function. So we have introduced three different quantities all related to each other the first is the response function phi AB of t the second is it is one sided Fourier transform which is the generalized susceptibility chi and the third is the Fourier transform of the response function which is the spectral function and a weighted integral over the spectral function with this Cauchy kernel gives you this quantity here. So you see already all we have used is that the theta function has that spectral resolution. So the relation in abstract terms the relation between the Fourier transform of a function and it is one sided Fourier transform is quite complicated it is not an all together trivial relationship at all we will use this fact we will use this in here we will when we write dispersion relations for this generalized susceptibility we will use this relation here okay. So to summarize what is going on so far we have a causal linear retarded response characterized by a certain equilibrium average of a correlation function some commutator expectation value in equilibrium it is Fourier one sided Fourier transform gives you the generalized susceptibility it is two sided ordinary Fourier transform gives you the spectral function and the latter two are related by this thing here. Now let us go further and ask what will this susceptibility actually do in practice well the first thing to do and I will come back to this point is that chi a b of omega which is 0 to infinity d tau a to b i omega tau phi a b of tau or let me just write t this quantity here if it exists for real omega then certainly it is going to exist if you multiply it by some function which is a decaying exponential of t because if this integral exists then you multiply it by a number which is less than 1 and you integrate especially if that thing goes to 0 as t tends to infinity it is going to converge even better. Now if omega becomes a complex number in other words you move off the real axis in the omega plane so far we have defined it on the real axis for real omega but if you give an imaginary part to omega if that imaginary part is positive then this is going to have an i omega 2 multiplied by this which is going to be e to the minus omega 2 t and if omega 2 is positive as it is in the upper half plane the integral exists and converges. So this function as an purely mathematical device defines a susceptibility not only for real values of omega but also for all complex values in the upper half plane. So this provides an analytic continuation of this function to the upper half plane definitely and the same representation is valid. On the other hand if omega acquires a negative imaginary part there is no guarantee this integral converges. In fact you will see that in that case you have an increasing exponential and this integral will not converge unless this dies down fast enough. So in general it will not converge here this representation is useless as far as the lower half plane is concerned but it is analytic in the upper half plane that is obvious by inspection. There are some further technical niceties but this heuristic argument is good enough for our purposes. The upper half plane this function is well defined we are going to make an excursion into the upper half plane and derive dispersion relations for this after we do some examples. What can you expect in the lower half plane well you cannot say much from here because this will be a representation that does not even converge anymore. It is like saying I give you the geometric series 1 plus z plus z squared all the way to infinity this is equal to 1 over 1 minus z provided mod z is less than 1 but if mod z is greater than 1 substituting that value of z inside the series will just give you infinity and if mod z is exactly equal to 1 then depending on what value of z you are actually choosing on the unit circle you can get all kinds of values from the infinite series. For instance if you choose z equal to plus 1 you get 1 plus 1 plus 1 which goes to infinity if you choose z is minus 1 you get 1 minus 1 plus 1 minus 1 which oscillates between 0 and 1 so you can get whatever you like depending on what the value of z you substitute it is but for all mod z less than 1 in the interior of the unit circle you can substitute a numerical value of z and you are guaranteed that the series converges to the correct value of the function okay. You may or may not be able to continue it outside the unit circle in this trivial case you can because the function 1 over 1 minus z makes perfect sense for all z except z equal to 1 where it is singular. So we are in that position we have an integral which makes sense everywhere here and therefore defines an analytic function of this omega we do not know what is going to go on downstairs but the point is that you know that if you have an analytic function of a complex variable that is analytic everywhere has no singularities whatsoever including the point at infinity that function has to be a constant there is no function which is analytic everywhere including the point at infinity right. So in general I expect that in physical problems and this is the lesson that you must bear in mind I expect this quantity to have singularities in the lower half plane now you could ask why not the upper half plane and what would you say it is entirely a matter of the Fourier transform convention had I chosen the opposite convention had I put a plus here and a minus here then the generalized susceptibility would have been converging would have been analytic in the lower half plane in omega and the upper half plane would be may have singularities and so on. So this is the reason why we have to be careful we choose a Fourier transform convention and I stick to it okay. So the one I have chosen is the conventional one and in this convention with this plus sign here this is an analytic function in the upper half plane yeah this is now it is entirely possible that for real values of omega also find a die down so slowly at infinity because this follows just an oscillatory function that this will not exist may not exist at all. Now that is a careful case we have to handle carefully I have made the implicit assumption that this exists this integral exists as a Riemann integral for real values of omega but this is an assumption may not be valid and we will give an example to show you that under physical conditions this will always happen. The best way to do this of course there are pathologies there are cases where it is singular at the origin and things like that and then I will use a trick to define this. The trick used is in general that you include a small e to the minus st here and then analytically continue in s to 0. So you find the Laplace transform of this fellow which will converge even if I increases and does not increase does not converge to 0 at infinity and then you make an analytic continuation. So in principle you can do analytic continuation to define the response function but we look at the specific cases to see what happens. Now quickly you can see that if you consider the following problem so let us consider a simple harmonic oscillator in a heat path. So we have a single particle a massive particle an oscillator we already did Langevin dynamics for this for a free particle but we will now assume that it is a simple harmonic oscillator in one dimension and this is immersed in a bath of particles and it drives fluctuations on this particle. So what does one do in this case well either I could write the Langevin equation down as a random equation stochastic equation make some assumptions about the noise or I do not do that I simply say look let us compute what this problem does in terms of thermal averages which is at attitude we have taken here right. So the equation that I write down would be the following m times x double prime of t average value plus m omega naught squared x of t equal to in the absence of an external force this fellow would be equal to some internal force which is provided by the friction in the problem right. But I could add an external force to this whole thing then I need a model for this guy and let us take this approach that the average force on this particle depends on the average velocity of this particle the usual friction term then this term this becomes exactly what the Langevin equation gives it becomes plus m gamma x dot of t plus m omega naught squared x of t is equal to whatever force you apply on the system f external on t. So I am applying an external force and applied force on the system and the system is in this heat bath it is being buffeted around by the particles of the heat bath and because this force is present the system has been disturbed from equilibrium I want to know now what its response function is what is the generalized susceptibility and so on. Now given an equation like this the way you solve it is of course by using green functions so you immediately say x of t equal to an integral from minus infinity to infinity d t prime f external of t prime times a green function you could also write it in terms of a generalized susceptibility and so on we will compute that point. Now the fact that you have t minus t prime here is a reflection of the fact that the system is time translation variant it is the same parameters right from the beginning so there is time translation variance this is g is a function of t minus t prime out here. Now what is the equation obeyed by this g of t minus t prime clearly you must apply the derivative operators twice derivative first derivative etc etc and that must give you on the right hand side f external of t which will happen only if this is a delta function on the right hand side. So it immediately follows that d 2 over d t 2 plus gamma d over d t plus omega naught square on g of t minus t prime equal to 1 over m delta of t minus t prime this immediately follows that is the usual green function. I have used a boundary condition here and that has tacitly assumed that at t equal to minus infinity there was no external force system was in thermal equilibrium that is the condition we have been imposing all the way all the time that is why I have only a particular integral here which depends on the source and no complementary function here because this integral runs from minus infinity I must pull out causal functions out of this this must get cut off at t by causality it must come out of this directly what would one do to solve this equation you do Fourier transform because Fourier transform converts differentiation to multiplication right. So if I define g tilde of omega is equal to by that equation here integral minus infinity to infinity d tau e to the power i omega tau g of tau and use a similar representation for the delta function then you get an equation for g tilde of omega you put that back and do the inverse Fourier transform and you are going to get g. I leave that to you as an exercise we meet again give me the answer and we will start from that point since time is over but the theta function must appear naturally this thing is physical so it is got to cut off at t here. So remember this quantity here is phi of t minus t prime times theta of t minus t prime that must appear when you solve the problem automatically by the way what phi is this what is a and what is b in this problem we are finding the average value of x so b is x that is for sure. Now the equation of motion said it is f external of t so it is clear it would come from a potential which is minus x times f external of t if you differentiate that with respect to x and put a minus sign you are going to get this. So in this problem this is really x x this is g x x I have not written it explicitly but that is what it is. I could have said what is the average velocity then I would get the x v and so on. So now you see how to identify a and b in this simple example it is the displacement we are asking for so this green function will be x x but you can also see intuitively that there is going to be a relation between g a b or phi a b and phi a dot b or phi a dot b dot or phi a b dot depending on what you differentiate etc. So there are going to be relations between these functions as we will see. So compute this quantity and verify that it has singularities only in the lower half plane in omega let me stop here.