 Right, so we had just started looking at linear response theory and just to recapitulate what I said last time we had the Louisville operator L which determined for any system Hamiltonian system which determined time evolution. So the statement was that the time derivative of any observable b for instance db over dt was equal to il acting on b and one has to solve this equation of motion if you like in abstract form where l on b was either equal to i times the Poisson bracket of well in quantum mechanics it was easier it was h with b divided by i h cross this fashion or in classical physics it was h with b Poisson bracket and this was the commutator whereas this is the Poisson bracket and whereas this was true in classical mechanics this is true in quantum mechanics that is the effect of this operation on any observable b but what you need to do for time dependence explicit time dependence is formally b of t is equal to the exponential e to the il t on b of 0. So one has to exponentiate this Louisville operator which is not always trivial to do in quantum mechanics we know explicitly what this stands for this fellow here it is e to the i h t e to the minus i h t on either side of b of 0 in classical mechanics it means that you have to write down you write this down you solve the Hamilton equations of motion and then you have b of q of t and p of t inside the argument. We will find it convenient to use this notation when we need to compute something we will go back and explicitly write out what these things are but till then the formal relation this expression for the time development operator is very useful okay. We also saw this property that on an appropriate space L was equal to L dagger implies that u of t which is e to the il t is equal is such that u u dagger equal to the identity operator it is a unitary operator okay and this came about the interpretation of this fact is that probability is conserved or measure is preserved or whatever okay. Now having got this expression what we would like to do is extend this to a case where this L becomes time dependent in an explicit way so this is fine as long as the Hamiltonian you are talking about is autonomous does not have a time dependence but the moment you have an explicit time dependence in h then matters become more complicated. So in particular if you notice we are interested in a situation where the Hamiltonian h of t is equal to h not which is t independent the unperturbed Hamiltonian and then a perturbation which is h prime which is a function of t where h prime of t has the form minus some a times f of t so it is some generalized force if you like and this is the operator pertaining to the system to which this force couples so that is the form of the perturbation and we would now like to solve the same problem find physical averages of any arbitrary quantity in the presence of this time dependent perturbation. So we start by saying that at t equal to minus infinity the perturbation is not yet switched on the system is in equilibrium I could start from in any finite time but if I set the stage where the perturbation is switched on to be minus infinity then it is clear that every other case is a special case I can switch it off till some time and then start it off and so on we will see what transients will do so we will assume that at t equal to minus infinity there is no perturbation the Hamiltonian is just h not and the system of interest is in equilibrium in the canonical ensemble. So we will assume that and the system of interest is in thermal equilibrium at temperature t then the systems physical quantities are all the average values are all specified by a thermal average with respect to the density operator of the canonical ensemble so for instance the average value of B in equilibrium and I will use the subscript for equilibrium this will be equal to for any quantity B this will be equal to trace B times rho equilibrium where we set trace rho equilibrium to be equal to 1 we always we are going to normalize our density matrix so that the trace of this density matrix is 1. So rho equilibrium is equal to e to the minus beta h not divided by trace e to the minus beta this part is a scalar it is just a number this is where the operator is the trace of it is by definition 1 so this is what we start with system is in equilibrium at t equal to minus infinity you switch on the perturbation and then you ask what is the value of the average value of any quantity of interest let us take such a generic quantity to be B I want to distinguish it from this A because it does not have to be A itself there could be a special case where B is equal to A you are observing that quantity which appears in the Hamiltonian perturbation but in general it is some other quantity we want to now know what the expectation value of this B is out of equilibrium away from equilibrium in the absence of in the presence of this perturbation here we will assume that this perturbation is small in a very specific sense in the sense that second order terms in this perturbation will be neglected so it is small compared to this and we want the correction to first order so how does one go about it well it is obvious that the density matrix itself will change the new density matrix will not be the old density matrix at all be some new quantity some kind so row equilibrium will now go to row of p which is equal to row equilibrium plus a delta row of p correct to first order in the perturbation right. Similarly the Louisville operator L instead of the equilibrium Louisville operator the L will go to plus let me use another term for it let me call it delta l perturbation the extra portion which comes because of this portion of the Hamiltonian base prime part again l perturbation will be a first order so now how would we compute a thing like this well you take this now the quantum mechanical language makes these things explicit you can see what will happen these operators to start with at t equal to minus infinity they are in the Schrodinger picture before you have actually switched on a time dependent perturbation okay and once you switch it on they change with time and the change in time of any physical quantity can be done one of two ways either you have a time dependent density operator and you trace and you take expectation value with respect to that time dependent density operator of a time independent operator the Schrodinger operator or you make the operator have time dependence to first order in the perturbation and compute averages with respect to the equilibrium density matrix this would not matter it does not matter which one you do depends on what is convenient you but at a formal level we could write the following we could write delta over delta t rho of t equal to minus i l so let me call this equal to l total l total row that is the equation satisfied by the density operator this here is called the von Neumann equation in quantum mechanics or the Louisville equation in classical mechanics so this is equal to minus i l equilibrium plus l perturbation acting on row equilibrium plus delta rho which is a function of t formally but this is equal to minus i l equilibrium acting on row equilibrium minus i l perturbation acting on row equilibrium minus i l equilibrium acting on delta rho and then a last term minus i l perturbation acting on delta rho there are four terms at a formal in a formal sense but this term is of higher order in the perturbation because this is perturbation and that is also perturbation this is clearly of at least second order in the perturbation so if I drop that term then this is an approximation correct to first order what can I say about the first term this involves here a Poisson bracket or whatever of function of H H not acting on H not row equilibrium is a function of H not right and H not commutes with itself so we know delta rho equilibrium over delta rho is 0 because we know that that must satisfy this a trivial statement but let me write it out I mean formally delta rho equilibrium over delta t equal to i l equilibrium on row equilibrium and row equilibrium does not have any time dependence at all this is 0 by definition this involves H not the exponentiation etc this involves H not and they commute to each other so this thing is 0 classical or quantum it does not matter incidentally going back to equilibrium statistical mechanics it is because this is 0 that we can assert and this involves H not that we can assert that this must be a function of H not right I mean the equilibrium density matrix in the canonical micro canonical whatever ensemble is a function of the unperturbed Hamiltonian or this what function it is cannot be given got from this equation this serves as the equilibrium statistical mechanics density operator serves as a kind of boundary condition on this equation at t equal to minus infinity it is some function of H not and what function it is depends on the physics depends on the ensemble and it is therefore put in not from this Louisville equation but from equilibrium statistical mechanics you need to derive that so that is the standard derivation in equilibrium statistical mechanics for us we assume this is given this quantity is given by the canonical ensemble so this term is 0 it is identically 0 so that gives us a first order differential equation for this quantity in which this thing here acting on delta rho is something we know we know what this L equilibrium does right so I can write this down that equation down as delta over delta rho sorry delta over delta t delta rho of t plus I L let me write this carefully yes L equilibrium on delta rho is equal to minus I L perturbation on rho equilibrium is it right yes okay right so this function is known to me this quantity is known to me and this term here this delta rho is unknown to me and this is the first order differential equation notice that there is time dependence here this is a function of t there is time dependence here yes this acts like some kind of source this term here is time independent delta rho has time dependence here and we can write the solution to this equation. So formally delta rho of t equal to this is an inhomogeneous equation for delta rho so the answer is a function is a particular integral plus a complementary function but now I am going to choose the complementary function such that delta rho is 0 at t equal to minus infinity that is the boundary condition delta rho because at t equal to minus infinity the density operator was the equilibrium density operator therefore delta rho was identically 0 together this now implies a solution delta rho of t equal to only the only the particular integral the complementary function has gone to 0 by the boundary condition so it is an integral from minus infinity up to t dt prime times integrating factor e to the pdx or whatever it is the integrating factor here is e to the minus il equilibrium of t minus t prime times t minus t prime sorry so I better write it not to avoid confusion t minus t prime l equilibrium and that must act on this fellow so it must act on minus il perturbation and what time argument should I put in t prime of t prime rho equilibrium so notice exactly as we expected this portion is involves the exponential of this density of this level operator but the portion that involves the perturbation is linear in this quantity as we expected because we dropped this term we dropped this term otherwise we would have had to exponentiate this term and since it is time dependent you cannot even write the solution that easily you have to write a time-ordered exponential or something so the formal expression is very complicated but to first order in the perturbation it is very simple now let us simplify that so this gives you in the next stage let us write down what this fellow is remember that l let me write this properly yes remember that the level operator acting on any function or operator was actually given by i times you have to now remind me what it was here h with f or h with f divided by h cross that is what the level operator did the reason I am messing around is because the this thing here is Hermitian so that is the reason I have to take out this i factor to make sure the time development operate is unitary and it is a nuisance keeping track of this i so what does this become minus ilf becomes just this you agree it becomes just this okay so let me define a bracket so that I stop carrying this symbol around all the time let us define define the bracket of 2 quantities let us call it a and b to be by definition equal to either the Poisson bracket or the commutator of a b over h cross let me just use this round bracket as a symbol to denote either the Poisson bracket or the quantum version of it which is the commutator divided by h cross okay so this l with minus i l with f is h with f so what happens here so let us let us write this out so it says delta rho of t just a bit of notation I want to keep all the i factors right otherwise we run into trouble later on so it is equal to this guy integral minus infinity to t dt prime e to the minus i t minus t prime l equilibrium times the bracket of minus i l perturbation times the equilibrium and that is equal to h prime with whatever this is so this is the round bracket of h prime of t prime with rho because that is my definition of this guy for l perturbation the Hamiltonian is h prime so that is the form but we already know that this is equal to minus integral minus infinity to t dt prime let us take out this guy f of t prime because that is sitting here the time dependence is in a scalar function times e to the minus i l i t minus t prime l equilibrium on a in the Schrodinger picture a rho equilibrium now let us start let me start inserting time arguments in this this a that appeared in the Hamiltonian was in the Schrodinger picture a at t equal to 0 that is when I switched on the perturbation coupled to this dynamical variable here. So this is the formal expression okay but you could also write this as equal to minus integral minus infinity to t dt prime f of t prime notice that all the operators are sitting here in quantum in the quantum case in the classical case the dynamical variables that evolve are sitting here this quantity is a prescribed function you have to tell me what is the time dependence of the force that you are applying times e to the i t prime minus t l equilibrium acting on a of 0 comma rho equilibrium what should I do next what should I do next pardon me well do I have to expand it explain what is this acting on what is this equilibrium going to act on it is finally going to give some commutator this fellow here is going to give some time dependence to quantities will it change this at all no this is the argument that is going to change this is the time argument. So what is e to the i l t on whatever is inside on any f of 0 equal to f of t by definition but now a subtlety little subtlety that arises pardon me no it is correct it is correct because no no no e to the i l let me neutral here this is true for any operator right okay now what I have is l equilibrium. So this is the time dependence as per what Hamiltonian h prime or h naught or the sum of the two h naught so really it is the Schrodinger picture operator which is going to the Heisenberg picture operator as governed by h naught. So the quantum way of writing it is easier this thing here is equal to e to the power i h naught t over h cross f of 0 tau e to the minus i h naught tau over h cross we know that is the solution to the Heisenberg equation of motion for any operator in quantum mechanics governed by a time independent Hamiltonian h naught. So I have assumed that you already know that solution okay is it is everyone does everyone feel comfortable with that or exactly without saying so I am going to an interaction picture I will explain in physical terms what this means you see you switched on a perturbation the Hamiltonian that you started with h naught you added to it an h prime of t if you did not have this quantities will still evolve with time as governed by this Hamiltonian by the unperturbed Hamiltonian and we assume that problem we can solve. Now we ask what happens if I switch on a time dependent perturbation to the Hamiltonian then it is convenient do the following when you are trying to find out the time dependence of physical quantities or their averages it is convenient to remove the known time dependence of those operators due to this part of the Hamiltonian and then you ask what is the effect of the perturbation it is like this suppose you have a system which is static and then you set it rotating with some constant angle of frequency then a convenient way of understanding the dynamics is to go to a co-rotating frame at which it in which it appears to be at rest and then if you perturb the system further there may be a little bit of jiggling etc etc and you can analyse that separately in the co-rotating frame of reference you have done that you can always go back to the original frame by undoing this rotation. So this is what is the meaning of the interaction picture when you have an interaction or when you have an interaction with an external force of this kind then it is convenient to go to a frame of reference if you like or a picture or a representation where the time dependence due to this routine evolution is already there is known it is like going to a rotating frame of reference okay and then you look at the perturbation alone. So what I have been doing is to take these operators which are really in the so-called Schrodinger picture and then saying let us introduce a time dependence here define in terms of L equilibrium here okay. So it stands for this whenever there is a time argument in any operator it stands for this with H naughts out here. So with that understanding we can now write what this is this will not be affected by this operator at all because you will have H naught and here you have e to the minus beta H naught and then e to the minus i H naught tau and since H naught commutes with itself you can bring this across here and give it gives you just e to the minus beta H naught. So the Hamiltonian itself will not change your time that is what it says when you go to the Heismberg picture right. So it does not act on this but it acts on this therefore you can write this as equal to minus an integral from minus infinity to t dt prime f of t prime bracket a of t prime minus t rho equal to in this form where by a of t prime minus t I mean precisely this e to the i L tau on with tau equal to t prime minus t t prime is always less than t that is correct that is correct yeah that is correct we will see this explicitly yeah so this is t prime minus t definitely but it does not matter this argument this formula is true whether tau is positive or negative it does not matter and now let us go right back that is delta rho of t. So now we want to know what is the change in any physical observable so if you give me an observable b any observable b then b in equilibrium equal to trace rho equilibrium times b by definition because trace rho equilibrium is 1 now what is the change in b in the presence of a perturbation so either I can compute b in terms of delta rho sorry in terms of rho of t this must be the same as b of t computed in rho equilibrium whether I put the burden of the time dependence on the density operator or on the observable does not matter but I must do 1 of the 2 I have just finished doing this because I found what is delta rho because this quantity equal to b equilibrium plus delta b of t average value and it is this quantity that I have written this thing here is what I have written by evaluating this in terms of this is rho plus delta rho of t b of t I have written by definition equal to b plus delta b but now the argument that I am making is that this quantity is the expectation value of b with respect to the change in the density matrix the extra portion by definition because I do not know how to compute this as so directly so I chose to do this by changing finding the change in the density operator instead so therefore home all we have to do now is to say okay b average therefore delta b of t equal to trace of b with respect to delta rho of t this is all I have to do but all these fellows come right through the trace so this is equal to minus integral minus infinity to t dt prime f of t prime and then come the operator parts so I have to take the trace of what of this quantity with respect to delta rho of t which is sitting here so this is trace b of 0 it is what that quantity is with respect to this guy here a of t prime minus t rho if I write this by definition to be equal to the following physical quantity minus infinity to t dt prime f of t prime times a portion which does not involve the perturbation explicitly this is found from the unperturbed Hamiltonian this is found from the unperturbed Hamiltonian this fellow is the original operator that is weighted with this force so what we have is a situation where the change in the physical quantity at any time t the first order change is an integral over all force histories from minus infinity to t gets cut off at t because of causality the force at a time t prime greater than t cannot affect the response at a time earlier than t right at a time t times a quantity which is completely independent of the perturbation itself depends on a depends on b and depends on the unperturbed Hamiltonian so let us call this phi a b of t minus t prime I have deliberately defined it as t minus t prime although what appears there is t prime minus t it does not matter we will see in a minute what would you call this function well it is some kind of a response function of the way the variable b responds to a perturbation which involves the variable a if a and b change to some other variables you get a different response function but the crucial thing the important thing is that this fellow does not involve the perturbation at all that is gone that is here first order in the perturbation okay and it is very physical now it is a retarded response because if we can prove that this fellow is a function of t minus t prime then the response at time t depends on a earlier force applied at any time t prime not dependent on t and t prime separately but how much time has elapsed between the cause and the effect so it is retarded response to start it is linear response because it is first order in f and it is causal because the effect gets cut off at time t so we really have causal linear retarded where this quantity is defined to be minus this guy here but now we have to write out what this is what is this response function so let us write it Phi a b of tau let us put t minus t prime equal to tau this here is trace b of 0 a of minus tau rho equilibrium you can rewrite it in many ways so let us see first how we should rewrite this in one particular way this equal to trace b of 0 let us do the quantum case because it is easier to write but it is exactly true in the classical case as well so this is a of minus tau rho equilibrium minus b of 0 rho equilibrium a of minus tau I have written out the commutator explicitly making sure that I do not take this b across because none of these fellows may commute with each other so especially in quantum physics I have to be very careful about the order even in classical physics you know that the Poisson bracket of a with b is minus that of b with a so I am going to be very careful with the order and now we use the cyclic property of the trace so in this term I regard this as one unit and this is one unit and move this across here so this is equal to trace b of 0 a of minus tau rho equilibrium minus a of minus tau b of 0 rho equilibrium trace a plus b is trace a plus trace b of course but I can take out the rho equilibrium common here and it is the commutator of b of 0 with a of minus tau so we have a very very compact formula which says phi a b of tau is equal to trace the product of the commutator a of minus tau there was a minus sign somewhere there was a minus there is a minus here there is a minus here right so this traces this is a minus minus minus everywhere so this is equal to trace the inner the Poisson bracket of a of minus tau with b of 0 rho equilibrium with a plus sign because I inverted this it became minus here and plus here so it is the Poisson bracket or commutator of a of minus tau with b of 0 that appears but what is this trace of something times rho equilibrium it is therefore the equilibrium average of this quantity right so this stands for this is by definition equal to a of minus tau b of 0 the Poisson bracket or commutator bracket in equilibrium so it is a remarkable statement it says the response to an applied force is dependent the response function which characterizes the response to an applied force is dependent only on the equilibrium average the thermal equilibrium average of this commutator bracket the elapsed time is appearing here this and now in some sense you can see that I am computing the average of the quantity b what is appearing here on this side is the average in equilibrium of the correlation of b with a in some sense anti-symmetry is suitable to get you this Poisson bracket or commutator so this is the remarkable thing about causal linear retarder response that we will see that fluctuations govern this quantity because you are taking an average average over what over thermal fluctuations in equilibrium so the says the fluctuations are governing the behavior of the average fluctuations in equilibrium govern the behavior of the average away from equilibrium in the presence of a perturbation so it is a profound statement it looks like a very simple derivation but it is a very it is it is first order perturbation theory plus statistical physics that is all we have done but it is already a very very deep statement here but you can write this in other ways you can also write this as equal to if you take the sequence equal to trace a of minus tau b of 0 rho equilibrium is e to the minus whatever it is so this is rho equilibrium minus b of 0 a of minus tau rho equilibrium trace of this fellow now write this fellow here yes it is true classical and quantum yeah this is a Poisson bracket I am doing only the quantum case here so that it is easier to understand otherwise I have to do classically you have to do integration by parts etc right much messier notation so all the while I am saying that we do the quantum case and then I let Planck's quantum go to 0 things will go to Poisson brackets etc at the pain just as we showed that L was a Hermitian operator in quantum mechanics it was trivial in classical physics it was a little harder to show the same thing is true here I just want to show what this will become now what you have to write is this fellow you write as e to the you have to yes you have to divide by h yes definitely yeah so this is e to the power i h0 tau over h cross with a minus sign a of 0 e to the i h0 tau over h cross with a plus sign on the right hand side this fellow here you put that in here and use the fact that this is a function only of h0 then you can move this factor out to the right hand side because it commutes and then by cycle covariance you bring it to the left hand side here where it will hit this B so you will have e to the i h0 tau by h cross B of 0 e to the minus i h0 t tau over h cross times a of 0 here so it will transfer the time argument to this quantity and d tau here so I do not know that yet it is not is time independent but I do not know what these quantities will do yeah you are right finally it arises from the fact that h0 is time independent and therefore these correlations will all become stationary they will become independent of the time argument but this is an explicit demonstration of that so this quantity here and I will stop with this this is also equal to expectation value of a of 0 B of tau in equilibrium okay which is a little more physical to understand because this says this is whatever tau appears here is due to the unperturbed Hamiltonian and its correlation with the original Schrodinger operator a is what is giving you in thermal equilibrium is what is giving you the response function okay so we will start at this point next time.