 Right, so let us start with reviewing a few of the key formulas that we had come up to before we took a temporary break and this had to do with the way the spectral function entered in linear response theory. So just to recall to your memory what the fundamental formulas were, we have a response function Phi AB of tau which as you know was the equilibrium expectation value and let me write the quantum case for definiteness. This is equal to the commutator of A of 0 with B of tau over i h cross and this immediately led us to a formula for the commutator itself the equilibrium expectation value of the commutator itself in terms of the spectral function. Now I think there is a factor of 2 pi missing in one of the formulas we derived earlier so to fix this properly so that you can actually get correct the numbers correct the factors correct. Let me go back and remind you what our Fourier transform convention was if you recall we said if you have a function f of t you expand it in a Fourier series as follows. So f of t is 1 over 2 pi an integral minus infinity to infinity d omega e to the power minus i omega t f tilde of omega and correspondingly f tilde of omega is an integral minus infinity to infinity d t e to the plus i omega t f of t. This was our Fourier transform convention I guess this is the one we have been using throughout so we should be consistent about it right. Then we discovered that the Fourier transform of this quantity is defined as Phi tilde so the definition of so in other words this is equal to 1 over 2 pi an integral minus infinity to infinity d omega e to the power minus i omega tau Phi tilde of omega and that was the meaning of the Fourier transform here a d of omega and we discovered there was a spectral representation for this quantity here in the sense that it had to do with the summation over matrix elements of A and B in the Schrodinger picture for instance over a basis set of states of unperturbed Hamiltonian H naught multiplied by a difference between two Boltzmann factors and then multiplied by delta functions at all the transition frequency of the system that is what this quantity had as a representation okay. So the first follow we derived was what we wrote down was essentially that the commutator of A of 0 with B of tau this quantity in equilibrium was equal to i h cross times Phi which is i h cross over 2 pi integral d omega e to the minus i omega tau Phi tilde AB of omega. So that was the representation for this quantity here the commutator itself. This immediately led us to an expression for each of these terms A of 0 B of tau and the other way about and if you recall let me write those down. So A of 0 B of tau in equilibrium were equal to the same thing but i h cross over 2 pi and integral d omega minus infinity to infinity of course e to the minus i omega tau and then we had this factor Phi AB tilde of omega we exploited the fact that this quantity here as far as the omega dependence was concerned had the sequence of delta functions. We use that to write the rest of wherever this omega mn appeared we wrote it as omega and we could pull it out of the series and that led to this representation 1 minus and similarly we found that B of tau A of 0 in equilibrium was equal to i h cross over 2 pi integral d omega minus infinity to infinity again the same thing e to the minus i omega tau time. Now we had an extra factor here e to the beta h cross omega Phi tilde AB of omega divided by 1 minus e to the beta h cross omega. So that when you took the commutator the 1 minus this cancelled against the denominator and you go back to this here perhaps I left this factor out I am not sure if you put this 2 pi and we did put this factor in okay. There was another place where I think we did leave out a 2 pi factor and we will come to that in a short while. So the anticommutator so this says A of anticommutator shortly plus meaning the anticommutator equilibrium this is equal to i h cross over 2 pi integral d omega e to the minus i omega tau minus infinity to infinity and then of course you have this pi AB tilde of omega. So you have 1 plus divided by 1 minus which was equal to so we wrote 1 plus e to the beta h cross omega over 1 minus e to the beta h cross omega was equal to minus cot hyperbolic beta h cross omega over 2 minus because this fellow has a minus here this is e to the minus whatever it is. But then we also discovered that the average energy of a harmonic oscillator of natural frequency omega at temperature T in thermal equilibrium that we discovered we call this we call that e beta of omega was defined to be the average energy of a harmonic oscillator at temperature T averaged over thermal fluctuations this was equal to h cross omega over 2 cot hyperbolic beta h cross omega. So if you multiply this you get 2 over h cross omega is this fellow and therefore minus this fellow is minus this. So we can rewrite this now this quantity here pardon me yeah I am just going to write it explicitly in terms of this fellow here. So what we have is 1 plus over 1 minus which is minus cot hyperbolic which I am going to write as 2 over h cross omega. So this is equal to i h cross over 2 pi over h cross omega and then a 2 divided by h cross e beta of omega divided by omega this is a variable so you have to have it inside the integral you cannot pull it out and is this okay or is there a sign error did I do they cannot be oh so it is a 2 pi sorry i h cross over 2 pi. So this was i h cross over 2 pi and there was a 2 over h cross right so this goes away so it is i over pi minus i over pi so this is 1 over i pi and that is even correct okay. So that is a nice compact formula useful to remember remember that this quantity remember that e beta of omega as t goes to 0 goes to h cross omega over 2 and for large t t tends to infinity this goes like k Boltzmann t which is the classical value the average energy is the average of the kinetic energy plus the average of the potential and in each case the classical equipartition theorem says that it is half k t per degree of freedom actually it is half k t this is also for kinetic term also. So what is the correct statement of that equipartition theorem just as a side remark for every quadratic term in the Hamiltonian you get half k t because it is a Gaussian integral then and then you get a half k t right. So this quantity here will tell you the exact answer but in between for intermediate temperature but those are the limiting cases against which we are going to check various things as we go along. Now this immediately proved by the way there was also a relation between the susceptibility and the spectral function itself and that is probably where we left out factor of 2 pi so let us fix that remember that let us let us forget about the subscripts for a while so remember that we had phi remember that we had chi of omega was an integral from 0 to infinity d omega e to the i omega tau phi of tau sorry d tau this was the definition of the generalized susceptibility right and the question was how is it related to the full the full Fourier transform of this fellow so the argument was that we could write this as theta of tau from minus infinity to infinity and you would not be making a mistake there so this says this is equal to the Fourier transform of the quantity phi of tau theta of tau just to recapitulate what we have already done so we need to find the Fourier transform of this product here but that is equal to if I call this phi tilde the Fourier transform that is our definition this theta tilde this is equal to an integral of what we now use the convolution theorem and I think that is why the 2 pi got lost what is the convolution theorem say so this is equal to it is a Fourier transform right so this is equal to an integral d omega prime minus infinity to infinity phi tilde of omega prime theta tilde of omega minus omega prime right divided by 2 pi there is an extra 2 pi here because the way we did this was notice where the 2 pies are going by our convention when you go from T when you write T in terms of omega there is a 1 over 2 pi so you are going to have to do that here and here when you computing the Fourier transform so there is a 1 over 2 pi whole square and then you get an exponential which will kill one of the frequency terms so when you integrate over time you get 2 pi delta function and the 2 pi will go against one of the 2 pies leaving a delta function you do the other omega integral and then you end up with this so there is a 1 over 2 pi and this is the factor I think we left out because I was playing around with this and I am not getting I will tell you where the inconsistency came up and where I discovered that this probably was there was a factor missing but it is trace to this point okay so now all that remains is to find out what is this quantity and what is that quantity all right so let us do it again so we do not make mistakes this time we need a representation of theta of T this is 1 if T is greater than 0 0 less than 0 and our argument was it is equal to minus infinity to infinity d omega with omega we would like to have e to the minus i omega t so e to the minus i omega t times theta tilde of omega 1 over 2 pi that is the definition of this fellow here but what is this equal to you could also write this as an integral from minus infinity to infinity d omega e to the minus i omega t divided by and you need to have the contour closed in the lower half plane because there is a minus sign here so if I close it in the lower half plane I am going to get a damped exponential down there so the pole should be in the lower half plane so this has got to be omega plus i epsilon and when you compute it you have a pole at this point and that is your contour integral and you are closing this by adding this well chosen 0 you are going to get minus 2 pi i times the residue the residue is 1 sitting out here right when you put omega equal to 0 you just get 1 there so what is this going to have what is the factor that should multiply minus 1 over 2 pi i so when I multiplied by minus 2 pi i this is going to give you 1 that is going to give you e to the epsilon t and epsilon goes to 0 you get 1 so that is the representation therefore if you compare it with this you can read off what this fellow is so it immediately tells us that theta tilde of omega equal to let us write this as plus i over 2 pi so by comparison theta tilde is equal to i times divided by omega plus i epsilon this is 2 pi that goes in the definition of theta tilde so now we are set let us put that in here and therefore it says we will remember this now erase all this so we have chi of omega equal to 1 over 2 pi and integral d omega prime phi tilde of omega prime minus infinity to infinity and then you have to multiply by i so it is i divided by omega minus wherever omega appears you put omega minus omega prime plus i epsilon or if you like this is equal to 1 over 2 pi i and this is the way it is convenient to use it d omega prime phi tilde of now let us put all the a b's and so on omega prime divided by omega prime minus omega minus i epsilon I took a minus sign out and brought this i downstairs this was the 2 pi we missed you wrote minus i times the rest of it but this 2 pi is important and you will see why in a minute there is so many of these factors around but this is the place where there was a mistake in here okay now once we have this representation then we are in good shape go back to the other statement we made now and look at this we derived dispersion relations for this etc now look at phi tilde a b of omega this is equal to by definition this is equal to an integral from minus infinity to infinity e to the power i omega t phi a b of t by definition since we use tau let me use that for the time difference d tau e to the i omega tau phi a b of tau I write this as cos plus i sin immediately so you can write this as cos omega tau plus i sin omega tau yes recall that this quantity phi a b of tau both classical and quantum cases this was equal to beta times a dot of 0 b of tau where this is defined as in the in the classical case is just the product in the case this is an integral over from 0 to beta and then there was this Adama transform of this operator on either side the sandwich between e to the lambda h not e to the minus lambda h not and then you take trace with e to the minus beta h not okay now by the way as an aside we did all those manipulations showing the reality of this correlation the response in the case when a and b are Hermitian we showed it was symmetric this quantity was symmetric regardless of the time arguments of these quantities whether they were Hermitian or not we showed also that the reality and symmetry and the stationarity of this quantity regardless of what these time arguments were throughout we manipulated by taking the cyclic property of the trace with e to the minus beta h not you could now ask I never normalized it I never put anything in the denominator I assume trace e to the minus beta h not was one but we know even in the oscillator case it is not you have to explicitly divide by it so does it make any of the conclusions wrong does it invalidate no because that is a c number it is a pure number on both sides all the manipulations we did were with the operators you can divide by a c number and then get back to the correct answer right away so all by carelessly wrote or rather wrote in a slip short way e to the minus beta h not as row equilibrium instead of writing that divided by trace of the same operator it does not matter you can put that number and it will still distribute so the final formula this thing here involves taking a trace with respect to a normalized density matrix you use that fact now the statement was Phi AB of minus tau was equal to epsilon a epsilon b Phi AB of tau with a minus sign where epsilon a if you like a parity of a and epsilon b is the time parity of b in the sense that you ask what happens to these operators under time reversal how do they transform position for example do not be changed momentum will change linear momentum will change what about angular momentum that will change too it does not matter what sort of angular momentum it is any angular momentum must transform in exactly the same independent of the origin what about more complicated combinations what about what about the Hamiltonian itself does not change so the point is you are looking at those Hamiltonians which do not change under time reversal because p squared over 2 m does not and therefore every other part of the Hamiltonian also should not then and that case you have a well defined epsilon a well defined epsilon b and this whole thing is either plus 1 or minus 1 so this quantity is either symmetric or anti symmetric okay you can artificially construct cases where it is not but then you have to be careful those those response functions would not have well behaved symmetry property well defined symmetry properties but in the cases that they do if this is symmetric so we can now write this down immediately if Phi of tau equal to Phi of minus tau and now we will assume that a and b are Hermitian operators so these things are real Phi a b is real quantity the chi here has then appropriate transformation properties okay so we have shown the reality of this quantity then you can see that this portion will be 0 identically because it is an odd function and this is an even function so this says Phi a b tilde omega is equal to twice integral 0 to infinity d t cos omega d tau cos omega tau Phi a b with a factor of 2 and what is that equal to it is the real part of our susceptibility because the susceptibility is defined as 0 to infinity of that integral the real part of e to the i omega tau is cos so this is equal to twice the real part of chi of omega so there is a direct relation between the spectral function and at least part of the susceptibility now let us write that other relation we derived down so that because I need to write this down this was equal to okay here we are we need to keep this in mind we are going to put this in here and see what happens on the other hand if Phi of tau Phi a b of tau equal to minus Phi a b of minus tau these are a b then Phi tilde is equal to only this quantity here and it is 2 i sin whatever it is this equal to 2 i integral minus infinity to infinity sorry 2 i integral from 0 to infinity d tau e to the i omega tau Phi a b of tau which is equal to 2 i times the imaginary part chi a b of omega sin sorry tau sin omega tau that is 2 i times the imaginary part so in either case you have a relation between the spectral function and either the real or the imaginary part now let us see how that tallies with this fellow here we have a formula which says chi equal to this fellow out here directly now let us look at the case where Phi happens to be a symmetric function then we know that this quantity here this thing here can be replaced by twice the real part of chi so what is it actually saying it is saying that and we now must check consistency so in the symmetric case Phi a b of tau equal to Phi a b of my tau in this case if Phi a b of my tau in this case we must have chi a b of omega must be equal to what we are doing is right 1 over 2 pi i an integral minus infinity to infinity d omega prime and then twice the real part of the same object the twice goes away so real part of chi of omega prime divided by omega prime minus omega minus i epsilon a b the question is does it make sense or not well we should use the representation for the terms of the principal value plus i pi delta if there is any justice in the world all the factors will come out right equal to 1 over i pi principal value minus integral d omega prime real part chi of a b of omega prime over omega prime minus omega plus i pi delta because there is a minus sign here with respect to the variable of integration so this pole is in the upper half plane and therefore you get plus i pi delta so plus 1 over i pi i pi times an integral minus infinity to infinity d omega prime real part of chi a b of omega prime a delta function of omega prime minus omega which is equal to the real part to write that term first real part of chi a b of omega that is this portion the i pi cancels out minus i by pi times a principal part of integral minus infinity to infinity d omega prime real chi a b of omega prime over omega prime minus but together with the minus sign and the p over pi this is precisely the imaginary part with the minus sign so recall the dispersion relation which we already derived which was real part of chi y equal to p over pi integral d omega prime imaginary part of chi of omega prime over omega prime minus omega and imaginary part of chi equal to minus p over pi integral d omega prime real part of chi of omega prime over omega prime so there is an extra minus sign so this says the rather obvious fact that this is real part of chi a b of omega plus i times imaginary part and we have checked that it is indeed correct so all the factors are in place now there is no mistake anywhere says chi equal to real part of chi plus i times imaginary part similarly check out what happens if it odd function again you will discover that there is an identity for the imaginary part and the real part will be represented in terms of an integral over the imaginary part make sure the factors are right in that case too all right so I went through this lengthy exercise because this factor of 2 pi bothered me and I really could not figure out where we lost this factor and it was not in the relation between the phi and the chi the real or imaginary parts but rather convolution appear okay now let us get back to where we were and let us go back to this representation that we had and see where it gets us these are called moment relations so let me write this down quickly let us begin with this expression a of 0 b of tau rather let us write this as yes okay a of 0 b of tau in equilibrium was equal to we wrote an expression for it explicitly which was equal to an integral from minus infinity to infinity d omega prime e to the minus i omega prime tau d omega minus i omega tau phi a b tilde of omega divided by 1 minus e to the beta h cross was there a factor here was there a factor outside there was an i h cross because we brought this i h cross across I think there was also a 2 pi right divided by 2 pi so if I leave out factors of 2 pi keep me honest because we do not want to go through another of these things okay so now suppose you set tau equal to 0 then you get an expression for this quantity first of all if you looked at the commutator itself a of 0 b of 0 equal to time commutator equilibrium equal to i h cross over 2 pi in that case this factor was not there you simply had this and nothing else so it is equal to just the integral minus infinity to infinity d omega e to the minus i oh this is gone pi a b tilde of omega. So the equilibrium value of the commutator is something you may be able to compute if you given these operators at equal times you can compute what the commutators are using canonical commutation relations very often. So there is some function of x or p and so is this you can compute this commutator and you are guaranteed that is equilibrium value is equal to the zeroth moment of the spectral function okay. Now you can write this in terms of all the matrix elements so you are essentially getting some kind of some rules all the time. Now what would this be that was for the anti commutator this is the commutator it is just the commutator just the commutator yeah I am just doing the commutator part we will see what the use of the anti commutator is in a minute you can actually guess what we are going to do for the anti commutator what the use is going to be but look at what happens with the commutator so what is a dot of 0 b of 0 what would this be again there is an ih cross over 2 pi and an integral minus infinity to infinity d omega what you have to do to produce a dot you have to differentiate with respect to t prime right. So do is to start with this by the way is the same as a of t prime let us write it here this is the same as a of t prime b of t and this fellow stood for t minus t prime so if I differentiate so the factor I have is e to the minus i omega t minus t prime if I differentiate with respect to t I get d over dt brings down a factor minus i omega and after that I will put t equal to 0 if I differentiate with respect to t prime this brings down a factor i omega and we need to keep track of time so I differentiate with respect to either t or t prime and then set t equal to t prime and by stationarity this is the same as 0 so what is this equal to this is going to bring down a factor i omega outside so it is i omega times pi so apart from this i factor it is the first moment of the spectral function and if you recall what the representation of the spectral function was in terms of all those omega mn delta functions you can carry out this integral and you get omega times those matrix elements must be equal to whatever this quantity is and you can easily see that this is also equal to minus a 0 with b dot of 0 now this game can be played many times over so what is the expectation value of a double dot of 0 b of 0 remember that these are all operators what is this equal to I differentiate twice with respect to this so I bring down and i omega twice so this quantity is equal to a dot of 0 I could have b of 0 equilibrium with a minus sign and that is also equal to without a minus sign a of 0 b double dot of 0 a dot b dot so I do once this once that the sign changes if I do both there minus i omega whole squared and the minus goes away that is equal to i h cross over 2 pi and integral over d omega minus infinity to infinity i omega whole squared times pi a b so apart from the i factors it is the second moment of the spectral function so you can see we can knock off the whole lot of moment relations now and each time you do a dot you got a different dynamical variable but now you can write that dot as a commutator with the Hamiltonian H naught so you really creating a whole algebra of all the operators in the system and you are finding their equilibrium expectation values of various commutators here and if you know the left hand side then you have constraints on the spectral function on the right so the famous f some rule the Thomas Kuhn Rice some rule and so on these are all special cases at finite temperature of this so in the exercises I will give some of these examples what I mean to say is this quantity for instance could also be written as follows instead of the dot so a dot remember is equal to commutator of a H naught divided by i h cross so this quantity here is also equal to the expectation value 1 over i h cross whole squared of the commutator of commutator of a with H naught with H naught all at the same time so all the multiple nested commutators can all be discovered once you have this therefore you have a lot of constraints the various kinds of some rules now does that is it of any use well yes because look at what the representation of this quantity was look at what we said about phi chi a b tilde of omega this was equal to 1 over 2 pi i have I got this i right now an integral minus infinity to infinity the omega prime phi a b tilde of omega prime divided by omega prime minus omega no I mean so let us write this as i over 2 pi and absorb the minus sign here so let us write this as omega minus omega prime plus that is the advantage of late I give you did that it show you an error immediately that you miss something of course if you say go to error it goes to some random place the previous error or maybe it introduces an error because late I can be self correct auto correct late I they are all very important so nowadays I always say any errors which remain are entirely due to the importance of auto correct not due to me so if you now ask what can we say about this you see this quantity here has data functions at all the transition frequencies now you could say alright let us look at very high frequencies significantly above all the transition frequencies of the system then the contribution here omega prime will be this will fire at all the transition frequencies so if omega is much bigger than all of them you get an approximation to the susceptibility at very high frequencies. So formally formally you can write this as i over 2 pi you pull out an omega so this is equal to a summation n equal to 0 to infinity let us even absorb this some constant over omega to the n plus 1 sorry c n to the n plus 1 where c n equal to this i over 2 pi integral minus infinity to infinity d omega prime phi a b tilde of omega prime times omega prime why the omega prime omega prime to the power n I did a binomial expansion of 1 over omega minus omega prime pulled out an extra omega and then these are the coefficients it does not have to be but if you look at in cases where the matrix elements corresponding to those frequencies if they are small then you can throw those out because remember this delta functions are all weighted remember that what happens in the spectral function is a series like this omega n m but this multiplied by a n m b m n right you need this to be non zero for that frequency to contribute but if you say alright these things die down as very large values or whatever these quantum numbers are then you can get the satisfactory high high frequency expansion of the susceptibility where these quantities these moments are known in terms of equal time commutators explicitly for instance the most famous of these is this first moment some rule which you come across in quantum mechanics but what we have now is the values of these quantities at finite temperature that is important for instance we know that if h is equal to p square over 2 m plus some potential v of x and you ask what is x with h this of course is going to give you a commutator with this which is p over m of course but now if you ask what is x with x over h this is equal to 1 over m apart from some i h cross factors times some constant because x commutes with any function of itself the only thing is this and you have this quantity now this x with h is like x dot is like a velocity so you got a position commutator of position with momentum and that is a constant you can therefore find out what this is going to be a number here but on the other hand this is going to be give you a non trivial quantity on the left hand side so there is a big you can play this game various systems you can find out what the high frequency expansion is now what is the use of the anti commutator I will put some of these in the exercises but what is the use of the anti commutator well because you have a of 0 b of tau with a plus sign if you put a equal to b at equal times then you get mean square values of physical quantities so you would for example get what is this quantity for an oscillator or what is p squared etc in equilibrium you can therefore compute at finite temperature now you need to use this formula with that cot hyperbolic etc of course you know what this is for a harmonic oscillator we know that at the high enough temperatures half m omega squared expectation of this goes to half k t and at 0 temperature it is going to go to half the ground state energy if you like but now in between what does it do a finite temperature this is going to give an exact answer so in a nutshell the spectral function has this formal representation in terms of the transition frequencies of the system but it also provides a great deal of information about various due to some rules it is highly constrained and you get x results for mean square values expectation values of commutators nested commutators and so on so in a given practical example one will have to of course make approximations but the formalism is quite clear what remains for us to do in this business is to relate this quantity relate the formulas we have derived to the rate of dissipation we have got to make connection with the Fermi golden rule in first order perturbation theory I will do that and then we will take it from there.