 So, to recall what the point we had stopped at last time, we discovered that the correlation the response function phi AB of tau which was formally equal to the equilibrium expectation value of A of 0 B of tau in equilibrium. We wrote this in a number of ways in terms of trace with rho equilibrium with A of 0 bracket with B of tau and so on. We had several representations for it but we discovered that this became equal to in classical mechanics it reduced to just the correlation between A dot and 0 and B at time tau in equilibrium this was classically quantum mechanically we had a somewhat more complicated formula and quantum mechanically it was equal to trace well it was equal to an integral from 0 to beta B lambda and then trace of rho equilibrium times e to the lambda H naught A of 0 A dot of 0 e to the minus lambda H naught and then a B of tau like this and this was in quantum this was the quantum mechanical case. So it is like some kind of Ademard transform with the e to the lambda H naught here minus lambda H naught and then integrated over lambda from 0 to beta in this fashion and then you take the trace with this fellow here. It is very convenient this whole combination appears over and over again so it is very convenient to give it a small notation introduce some notation but notice that in the classical case where things commute with each other this will cancel against that you just have a dot of 0 this gives you an integral beta and that gives you this factor and you are back to this okay. So classically this formula reduces to the classical formula but this is the more general formula here. So it is convenient to introduce what is called a canonical correlation so let us define x, y in equilibrium with arbitrary time arguments whatever be the time arguments of these dynamical variables if x and y are observables operators in the quantum case define this to be equal to 1 over beta times an integral 0 beta d lambda and then trace row equilibrium x sorry e to the lambda H naught x e to the minus lambda H naught y. So define with a semicolon here I do not want to put a comma because that stands for things like Poisson brackets and so on or a commutator but define this x, y for arbitrary time arguments of the observables x and y it is just defined in this fashion here. Then this thing here in the quantum case becomes equal to this thing here becomes equal to beta times as you can see it becomes a dot of 0 semicolon d of tau by definition. So this whole rigmarole with row equilibrium which is e to the this quantity remember is e to the minus beta H naught normalize such that the trace of this quantity is 1. So this makes sure that both the classical and quantum cases look alike in the classical case the semicolon is just deleted it is just gone it reduces to just the product okay in the quantum case you have to order these operators in this careful fashion. Now the advantage of doing this is that this response function has a very compact formula. So it immediately tells you that this is a very important formula that the generalized susceptibility corresponding to these observables chi AB of omega has the compact form integral 0 to infinity d tau e to the i omega tau phi AB of tau but that is beta times a dot of 0 B of tau. So that is a very compact formula and then the job reduces to computing this number in a sense this summarizes all of linear response theory the derivation of this formula here where the semicolon bracket has this very specific meaning okay. Now what is the advantage of doing this well this thing here has a lot of interesting properties to start with it is stationary immediately follows that it is stationary because let us see what happens to it this thing here I want to show that x of any arbitrary t naught y of t naught plus t equilibrium we want to show if it is stationary it means you can subtract the same time argument from both time arguments here same constant and you get the same answer so this will be equal to x of 0 y of t naught t alone that is stationarity so this is equal to I urge you to try and show that this is so what would you do you would go back to the definition you would go back to this thing here put in the time arguments in these places and then we know what x of t naught does for instance this quantity here is equal to e to the i h naught t over h cross x of 0 sorry t naught over h cross e to the minus i h naught t naught over h cross that is the meaning of x of t naught the Heisenberg picture driven by an evolution governed by the unperturbed Hamiltonian. So for this quantity substitute this in the bracket in the definition of this semicolon for this quantity x of t naught put in this quantity here because we finally want to show that something involves x of 0 alone then for the y part put in the same thing but with t naught plus t in the exponents so you get a long big expression here and then remember that h naught commutes with itself so e to the a h naught commutes with e to the b h naught where a and b are scalar numbers so you can move those two brackets around those two factors around and then use the cyclic invariance of the trace whether operators commute or not trace a b is trace b a always so you should take certain packages certain parts of this and blocks of these operators and you can transfer it to the left you may do that you should get this back again so all the h naughts will go away except the one that involves this time argument y and you should get this expression here you could also further write this as x of minus t y of 0 you subtract anything same argument same constant can be subtracted from the two time arguments in this two two point correlation so the first important property of this quantity is stationarity that helps a great deal the second property is symmetry and this follows in the following in the following way so let me show this so the first property is stationarity and the second property is x independent of what time argument you have regardless of what time argument you have x of anything y of anything else so let us let us call it x of t 1 y of t 2 independent of what t 1 and t 2 are by the way this is a function of t 1 minus t 2 alone this quantity is equal to y of t 2 x of t 1 so there is the symmetry property which is a huge help because it says you can actually commute these fellows around when you have this susceptibility when you have this response function here for this this formula for Phi AB you can it does not matter which order you put this now how do you go about showing that well we really have to go back here and show that this is equal to y x independent of time arguments so let us start at this point and this is just a mathematical trick so it is not very let us let us write it out here x semicolon y equilibrium equal to 1 over beta that is its elements 0 to beta d lambda trace e to the minus beta h naught e to the lambda h naught x e to the minus lambda h naught y that is this quantity x semicolon y I want to show its y semicolon x so clearly I am going to exploit the cyclic property of the trace bringing something here and putting it on this side so how do you go about it I need these lambdas to act on this y but remember that these two factors do not these two do not commute because they do not commute with x necessarily of course these two commute these two do not commute and so on so h naught x and y in general are operators which do not commute with each other how would you then produce this how would you get this across this fellow here electric pardon me change of integration variable exactly so set let us set beta minus lambda to be the variable of integration then this is a minus d lambda prime if you put lambda prime is equal to beta minus lambda the integration runs from 0 to beta here so the minus sign cancels and so this thing is the same as 1 over beta integral 0 to beta d lambda trace e to the minus beta h naught e to the instead of lambda I should write beta minus lambda just changing variables of integration I remove the prime after n change the variables so these factors go away you have e to the minus lambda h naught here and now move this across to this side so you have e to the minus beta h naught y e to the lambda h naught sorry so I move this whole block as it is e to the lambda h naught y and that is gone this is why semicolon x so there is a change of variable here change integration variable and that gives you the symmetry property so you can see the great advantage of defining this semicolon bracket because it is almost behaving classically you can put these two fellows in either order you do not care it is stationary it is symmetric and once you have that notation in place then actually you can play around with it as if these are classical objects provided you have a semicolon in between and it is got another very important property the third one which is the following I expect I expect that this if a and b are observables they are represented by Hermitian operators if they are physical observe real physical observables I then expect that if I apply a real force the response should be real that quantity so I want certain reality properties of this whole thing I want I would like to show that the response function must satisfy certain symmetry properties in order that the response to a real force be real okay so let us do that in slow steps let us first take what happens if you take the complex conjugate of this semicolon bracket so let us find x semicolon y equilibrium star this is equal to 1 over beta 0 to beta d lambda e to the my trace trace of this fellow pardon me oh yeah this is plus and this is minus thank you and I need the Hermitian conjugate of this so that is equal to 1 over beta integral 0 to beta d lambda trace then of course when I take the Hermitian conjugate of a product of operators they appear in the reverse order so now we are looking at what happens if a and b are x and b are wire Hermitian so x equal to x dagger y equal to y dagger these are physical observables so once I have this all I have to do is to reverse the order so it is equal to y y dagger is the same as y h0 is Hermitian the Hamiltonian so e to the minus beta h0 is Hermitian that is this fellow and then x and then this fellow and then this guy now the same trick as before we change variables of integration so this is 1 over beta integral 0 to beta d lambda trace y e to the minus so I am going to instead of lambda I am going to say if lambda prime is beta minus lambda so lambda is beta minus lambda prime so minus lambda is minus beta plus lambda prime and then this h is e to the beta h0 e to the minus lambda h0 is there an extra beta h0 somewhere yes that is okay it is okay now this factor cancels against last because it commutes through it is not here so where are we now yes yes now it is a simple matter take y to the end so take this fellow and take it all the way to the end so this is equal to so the third property is reality the response function therefore written in this form is stationary symmetric and real real for Hermitian operators that immediately implies a certain symmetry property here so you will see what the consequence of it the consequence is that so this quantity is real if a dot and b are Hermitian instead of x I call it a dot instead of y I call it b we have just seen that is real if a dot and b are Hermitian okay now what appears in the Hamiltonian if you recall is this remember our perturbation or our total Hamiltonian was equal to h was h0 minus a f of t this was the operator that appeared not a dot a dot came because of manipulations in between so a is certainly a Hermitian operator I apply real force to the system h0 is a Hermitian operator a is a physical observable and the Hamiltonian has to be Hermitian so this is Hermitian here but what is the guarantee that a dot is Hermitian how do I know that if a is Hermitian the operator a dot is also Hermitian I mean this I mean this sounds like common sense right because if after all if a is the position a dot is a velocity and that is as real or as physical as the position itself but what is the guarantee in general that for some arbitrary observable a which is Hermitian represented by Hermitian operator what is the guarantee that the operator a dot is also a Hermitian operator yeah exactly so you would write a dot the operator a dot at any time is equal to this is Hermitian that is Hermitian so is the commutator Hermitian pardon me anti Hermitian commutator of two up Hermitian operators is anti Hermitian of course because a b minus b a becomes b a minus a b when you take the Hermitian conjugate right that is saved by this guy that changes sign to therefore if a is Hermitian a dot is Hermitian okay the commutator of two physical observables represented by Hermitian operators has to be anti Hermitian and you put another I there it becomes Hermitian okay so we are in good shape this says this is real if we are Hermitian operators. Now if I apply a real force to a real system to a system I expect a real response on the other hand I know that I know the following I know that if I apply a force f not e to the minus i omega t to the system this was my general force expression for a particular frequency omega right so the force was this implies a response chi a b of omega f not e to the minus i omega t that is the response that is how we defined the generalized susceptibility it was the quantity that attenuated the force which was applied with one particular value of frequency any frequency component if the force had an amplitude f not the response was the same thing multiplied by this generalized susceptibility right. So if the force and it is a linear response so if the force is f not star e to the plus i omega t if that is the force then what should the response be you can read it off from here the force is got a term sinusoidal oscillation e to the plus i omega t right so the response has to be chi a b of minus omega f not star e to the i it has to be so because it is linear response I can put whatever amplitude I like whatever frequency I like and compare with this if I add these two fellows if I add these two forces then what do I get immediately implies therefore if the force is f not e to the i omega t plus f not star e to the minus i omega t e to the i omega t over 2 I take the half the amplitude in each case must imply response chi a b of omega f not e to the minus i omega t plus chi a b of minus omega f not star e to the i omega t divided over by 2 but this object here is real it is a real part of so this whole thing it says the real part of f not e to the minus i omega t so that is a real force the response has to be real so this must be the real part of some complex number we almost there here is f not here is f not star here is e to the i minus i omega t plus i omega t so this would be real so this is equal to this part equal to the real part of chi of omega chi a b of omega f not e to the minus i omega t provided if and only if this part is the complex conjugate of this chi is no other way if and only if chi a b of minus omega equal to chi a b star of omega right you see the argument all I have used is a superposition principle and the definition of the generalized susceptibility that leads me to the conclusion that if I apply real force the physical force then the response is real if a and b are Hermitian if and only if this condition is satisfied for real omega but let us look at what the complex conjugate is this implies that chi a b of star of omega this quantity here at a complex conjugate this must be equal to beta times integral 0 to infinity d tau e to the minus i omega tau right times this quantity phi a b this was equal to apart from this beta this fellow here was equal to phi a b of tau right so this is equal to out here a dot of 0 b of tau equilibrium complex conjugate it because I am taking the complex conjugate so this however is equal to chi a b of minus omega provided this is equal to its own complex conjugate provided that is real but we just saw that for Hermitian operators that is real so the physical reason why you want that response function to be real is that it ensures that when you are dealing with susceptibility involving Hermitian physical operators you apply real perturbation the response also is real that is guaranteed by the fact that this correlation function even in the quantum case we do not care even there it is real guaranteed to be real otherwise we would be in trouble to lead to an inconsistency so that is good that we have seen that the reality of this quantity this response function ensures that the susceptibility has this symmetry property which is needed to ensure that the response to a real force is real now the consequence of this in turn is immediate so again now we are specializing to a and b or a dot in the Hermitian so and hence a dot chi a b star of omega equal to chi a b of minus omega I want to emphasize that this this of course we have assumed omega to be real when I took the complex conjugate of this quantity I just put e to the minus i omega tau which would only be true if omega is real time of course is real so there is no problem tau is real but we want to make sure omega is also kept real because very soon in a minute we are going to talk about complex omega by the way this implies immediately that the real part of chi a b of omega equal to the real part of chi a b of minus omega take real parts on both sides and you see that it is a symmetric function the imaginary part chi a b of omega equal to minus the imaginary part minus it is an odd function so this is an even function and that is an odd function it is going to be useful so there are these symmetry properties it is going to be useful because physically you talk about real values of omega which are also positive non-negative but we are going to talk about analytic properties in omega in the omega plane which will involve integrals over negative values of omega but they can be got rid of by using these properties so you can fold things back onto the positive real axis now yeah I very often wanted to do the opposite I want to I want to draw contour integrals and so on so we will see we will go back and forth the point is that if you know this quantity for some positive omega you also know it for negative omega by this symmetry property but now there is a very interesting property of analytic behaviour and let me do this by motivating it starting with this formula which is this one so and we have seen that this is real for Hermitian a and b and therefore a dot and b we have seen that it is symmetric and a and b exchange and we have seen that it is real symmetric and stationary now if this integral exists for real values of omega this is an oscillatory factor then in general they should go to 0 as tau goes to infinity so that the integral converges if that is so if I add a converging factor to it then the integral gets more and more convergent so if this exists if this exists and that is by no means guaranteed we do not know for sure we got to look at it if this exists for real omega then by exists I mean if it is convergent for real omega it will certainly do so for imaginary omega greater than 0 so I make omega complex purely as a mathematical exercise I know physical frequencies are real but here is a formula which is a function of omega and now I say alright very nice let me talk think about this in terms of the complex variable omega purely as a mathematical excursion there will be a reason for doing so then the question is does it exist does it make sense well in general if you are familiar with the theory of analytic continuation then you know that if something is analytic in the complex variable sense on some dense set then you actually have some continuous set of some suitable kind of set in the complex plane then you can define in general analytic continuations of this function to the rest of the complex plane or to some region of the complex plane called the domain of whatever holomorph or something but now we will look at this very heuristically if I put omega equal to omega 1 plus i omega 2 in this factor e to the i omega tau becomes e to the i omega 1 tau times e to the minus omega 2 tau so there is an extra damping factor provided omega 2 is positive omega 2 is the imaginary part of omega and if it is positive then since tau runs only over positive values you are guaranteed that this provides you with an extra convergence factor out here it is crucially dependent on the fact that tau runs only over positive values if tau had gone to minus infinity then it is finished you cannot do this so this means that the formula as it stands makes sense even for complex omega as long as the real imaginary part of omega is positive in other words as long as you move into the upper half plane. So what we have is a situation where in the complex omega plane this is omega 1 and that is omega 2 the imaginary part you assume this formula all along the real axis and you said by assumption that this formula defines for your function of omega through this convergent integral for all real omega it follows as a consequence of that assumption that if you move up into the complex omega plane throughout up here at any value of omega to find the value of this function at that value of omega all you got to do is to substitute that value of omega in this formula and it makes sense. So the formula provides an analytic continuation into the upper half plane so this thing here can be analytically continued and moreover you are guaranteed it defines an analytic function there you can therefore differentiate it any number of times as you know an analytic function of a complex variable is a very special kind of function it means it satisfies the Cauchy-Riemann conditions between the real and imaginary parts it means that every derivative of this function exists and is also an analytic function satisfying the Cauchy-Riemann condition there is a Cauchy integral formula for this and then you also know that the line integral of this function around any closed contour is provided the contour stays entirely in the region of analyticity and does not enclose any singularities the answer is 0 so is everybody familiar with these theorems okay. So once we have that in place then it provides a very powerful handle it says if you start anywhere and you draw close contour like this C the integral of d omega let me drop this subscript a b all the time because it is true for any a and b as long as whatever properties you have established so far are true so let us for convenience simplicity notation drop this and just call it chi of omega this over this contour C is equal to 0 provided C does not get anywhere into the lower half plane provided it remains in the region of analyticity of this function which is the upper half plane and the real axis by assumption that is a consequence of Cauchy's theorem integral formula or whatever you call it it is a consequence of the fact that this thing is a nice analytic function. So now suppose you start with some value of omega on the real axis which we are interested in let us suppose I start with some fixed value of omega here and I want to choose another symbol for the integration symbol so let me call this omega and let me call this real omega prime and this is imaginary omega prime I want to make sure and this is certainly true now I can distort this contour as I please provided I never come down below here and what I would like to do is to use this property to derive a formula and that is the target for chi at this point this physical point in terms of an integral over chi over the rest of the real axis okay and the way to do that is to single out this point here how do I single it out but I should ensure that chi of omega becomes a residue of some function of omega prime at this point so I should therefore divide by omega prime minus omega. So let us consider the analytic function chi of omega prime divided by omega prime minus omega in the omega prime plane as a function of omega prime this fellow is analytic on and above the real axis this provides a simple pole at omega prime equal to omega so this function itself is analytic everywhere in the upper half plane except for a pole at this point and it has a simple pole out here as long as I do not cross that pole or hit that pole I can distort this contour as I please provided I do not go into the lower half plane so let me start distorting it without changing the value of the integral so I start by saying an integral over this contour C d omega prime equal to 0 because it is an analytic function if C looks like what I drew earlier but I can make that C look bigger by writing it like this I can make this go on the real axis because it is still analytic there but for clarity I have shown it a little above I should avoid this pole so I go above it and stay in the region of holomorphy and I go to this side and the answer is still 0 I would like to do this till I extend this to infinity and take an infinitely large semicircle am I allowed to do that how do you know how do you know because I am going to have to argue that this contribution I would like it to go to 0 is that going to happen is that going to happen I have assumed that it exists but I have not said anything about what it does as omega prime goes to infinity along any direction pardon me chi is independent of this no phi anymore right I am just saying this is an analytic function I am only focusing on this function as a whole phi is gone I mean it is represented by this so I am just saying this is an analytic function of omega prime in the upper half plane with a pole at omega prime equal to omega on the real axis everywhere else in the upper half plane it is analytic so am I allowed to say that if I write a contour integral from minus r to plus r this is plus r and then this big semicircle it is still valid this is still true can I take r equal to infinity am I allowed to do that I have the exponential where is the exponential factor that is been used up in showing that this is analytic that is gone so when is the contribution from that semicircle let us write it out let us call this contribution from the semicircle let us call this contour gamma big gamma so this quantity is integral minus r to omega minus epsilon plus an integral from omega plus epsilon to r plus an integral over gamma of the same thing chi of omega prime the omega prime over omega prime minus I am going to let r go to infinity so this integral runs minus infinity to infinity except for that little interval around omega and then I would like this to be 0 is that guaranteed think about it till tomorrow since we have run out of time think about otherwise it would not be useful because it would still involve integrals over some complex values of omega I want to restrict it to real values physical values of omega right so I want this contribution to vanish and you have to tell me what further assumptions needed to do this so we will take it from here tomorrow.