 Right, so we have been looking at linear response theory from a specific point of view namely we decided to calculate corrections due to a small applied time dependent perturbation to the Hamiltonian. We calculated corrections to physical quantities like b some arbitrary quantity b by computing things in terms of the change in the density matrix. In other words if you recall what we did was to say that we start with some b and the expectation value of this b was equal to a b equilibrium plus a time dependent part which we call delta b and that would of course be t and we computed averages and the statement was that this to first order in the applied force was computed by computing first the change in the density operator. So in other words we use the fact that this was b with respect to rho equilibrium plus delta rho of t and we know b rho equilibrium was the equilibrium value time independent and then we computed this using the Liouville equation to first order etc. But you could equally well have done this by working in a more active picture in other words by working in terms of b of t itself in terms of rho equilibrium. So this means you would go to the Heisenberg picture with the full time dependent Hamiltonian correct to first order in it and then compute averages with respect to rho equilibrium. We did not explicitly do this we use this all this time. So what I would like to do is to make contact with what you would be more familiar with in the context of normal quantum mechanics and then we will put in the thermal fluctuations and average over things etc. We know the answer already we know the answer is given the Hamiltonian H of t equal to H not minus a f of t we know that the change delta b of t on the average this is equal to an integral from whenever you switched on the force which we took to be minus infinity in the most general case up to t dt prime f of t prime multiplied by the operator part of it and the operator part of it was this response function phi a b of t minus t prime that was the result that we got and for this we formulas among others for instance we discovered that this could be written as a of 0 comma b of t in the Heisenberg picture induced by the time independent part of the Hamiltonian trace with respect to the equilibrium density operator. So we had a formula like this and this we further simplified in the classical case in the quantum case and so on and so forth. What I would like to do is to now establish contact with the usual way of doing quantum mechanics we will do this in the Heisenberg picture just to clarify to you what we mean by going to the Heisenberg picture and so on. So it will become explicit in what I am going to do so let me call this contact with quantum mechanics with the usual perturbation theory formula in quantum mechanics. So I repeat we are just going to do something which will explicitly tell you how the commutator was up out here for this phi and we will re derive this formula from a slightly different view point. This will make it more familiar this will go through more familiar ground instead of the Louisville operator itself. So again I start with this and when we do not put any subscript on any Hamiltonian it is supposed to be in the Schrodinger picture. So this is a time independent there is no explicit time independent dependence in this operator it is a physical observable and we will take this to be the physical case a is a Hermitian operator. So the Hamiltonian is a Hermitian Hamiltonian if it is not I have the problem of writing this plus its conjugate dagger to make the Hamiltonian Hermitian but for simplicity of notation we will take this to be Hermitian operator here and the Hamiltonian to be Hermitian we will also take this to be real so that the hermiticity of the Hamiltonian is ensured right. Then we want to compute what the full system does and what would you normally do it is a whatever state you start with if you start with a state for example of the system then the Schrodinger equation would apply to the way that state evolves in time so formally one can write down a Schrodinger equation for the evolution of the state. Now it is convenient to introduce a time development operator u which is labeled by 2 time indices at t and an initial time t0. So I am going to take a general initial time t0 whatever be that etc. In general I take t0 to be the time introduced on the force and we will set it equal to minus infinity eventually but I want to make sure that we have a general formalism in other cases where you switch it on at finite times and so on so let us start with t t0. This takes you from the state of the system at t0 to the state of the system at time t and it satisfies of course the equation d over dt equal to h. In this case it is time dependent u of t. So this is the equation satisfied by the time development operator u. Because this is a Hermitian Hamiltonian we know that this time development operator must be unitary operator so that probabilities are conserved and so on. That is all we know about it to start with. You also have a boundary condition on this operator which we will write down subsequently little later. Now I repeat again there is no explicit time dependence here. So if you did not have this then the problem is utterly trivial, completely trivial. Then the solution to this is with the time independent Hamiltonian is just the exponential of this Hamiltonian multiplied by time. But the question is what is it with the time dependence present here. And we have to be careful because there is no guarantee that h at time t commutes with h at any other time. Two operators at different times of time need not commute the same operator at different instance of time need not commute at all because the Hamiltonian gets in between if you like. So what would one do? One would like to find the solution to this but first let us see what happens in the absence of this fellow here. So if h prime so this is what I called h not plus h prime of t. If h prime is absent then of course you have a time independent development operator d over dt and let us call that u not the time development operator in the absence of h prime okay the unperturbed problem. And what would you do there? You would say this is t. Let us take the initial time here. There is no switching on force or anything like that. So I will say that t equal to 0. The system is given to you and then I look at it at any later time. Just one time argument here. This is equal to h not u not of t with the initial condition u not of 0 equal to 1 okay and then the solution implies u not of t equal to e to the there is an h not divided by h cross so it is e to the power minus i h not t over h cross times u not of 0 which is 1 okay. So that is a very trivial problem to do this. Once you have this then I can go back and ask what does this look like the solution to this equation look like. But first let me define the Heisenberg picture for you. So the Heisenberg picture we are talking about is a transformation from the Schrodinger picture to a time system which has time dependent operators. The time dependence comes from h not alone as I have been saying. So any operator B now becomes time dependent this is defined as e to the power as u not dagger of t B u not of t. This is the definition or if you like I write a Heisenberg equation for it. This is solution of i h cross d B over d t is equal to h not B with h not. That is the Heisenberg equation of motion. I solve this and I get this out here okay and this is equal to e to the power i h not t over h cross B e to the minus i h not t over h cross. So if I write an operator without any time argument at all by that I understood. We mean the Schrodinger picture operator which coincides with the Heisenberg picture operator at t equal to 0. Is that clear? So yeah we are still doing the case when h prime is 0 because we are going to switch on h prime and I am still going to do this first and after that take care of h prime separately okay. You will see as we go along what that entails. So we are still doing this and we are going to always continue to define a time dependence of an operator like B via this. But of course once I switch this on, once this is switched on then there could be a time dependence in the operator coming from the fact that you in the time development operator is you and not you not. I have to take that into account I would like to do it to first order in F. So that is the problem really. Now of course you can also ask what is the general solution to this for an arbitrary h of t to all orders arbitrary h of t. It is not an exponential, it is not an exponential because this follow is explicitly time dependent. So you cannot trivially do that. You need an integrating factor as you know and even classically even without operators along the line when you have a thing like d y over d x plus p of x y equal to q of x. If this were a constant this would be a trivial problem but if this were a function then of course you have an integrating factor to take into account. You have to do an definite integral of some kind put correct boundary conditions and things like that. Now it becomes much harder in the case of an operator because of this commutativity problem. You have a formal solution available it is called a Dyson time ordered exponential but we are not interested in that here. We are interested in things to first order in F. So let us see what to do. So is this clear whenever I write this I mean this always and we will use that notation. Of course when you switch this on B has time dependence because of this 2 this portion 2. I will call that B total of t the total time dependence if you like for lack of better notation but I would like to retain B of t for this guy here. And notice that in earlier derivations and linear response theory all the arguments of the function of the various operators refer to this kind of thing. They all refer to this with a nation not alone but I am now making this thing explicit here. You could ask what sort of picture is this where I take the time dependence of the operator to come from the unperturbed Hamiltonian but then I am including the interaction in some way. It is called the interaction picture but I am trying to do this without using that terminology but it will become obvious because I need only first order. Now what should I do for this? I have to now deal with this full problem. So let us use an Anzart's or trial solution and that would be let us put u of t t0 to be equal to u0 of t saying that this fellow will go on evolving due to h0 all the time multiplied by u prime of t and t0. That is a trial solution. It may work, it may not work. We will see what happens. We know this. We know this function. We know u0 of t is this guy here explicitly. We need to put an equation for u prime. So what happens? This is equal to h0 minus a f of t in this fashion u0 of t u prime t0 just substituting for h and substituting this Anzart's for u. On the left hand side I have ih cross d over dt so you have du0 t over dt times u prime of t t0. That is the first term in the derivative here plus ih cross u0 of t d over dt u prime of t t0 and that is equal to this fellow here. That is equal to h0 u 0 of t u prime of t t0 minus there is an f of t which comes out a u0 of t u prime of t just substituting. But look at this term versus that. We already know ih du0 over dt is h0 u0 of t. We already know that. So this term says this term cancels against that even though they are operators. You operate on both sides of this equation. You operate with u prime of t t0. You get the same answer. So this term cancels against this term and we have an equation for this alone. But u0 of t is a unitary operator. It is e to the minus ih0 whatever it is. So this implies that ih cross du over dt u prime of t and t0. This is equal to minus f of t and then e to the power let us move this u0 to the other side. That becomes u0 dagger because u0 is unitary. So this becomes e to the power ih0 t over h cross a e to the minus ih t over h cross on this side. That is u0 and then a u prime of t. This is h0 of t u prime of t which is of course this quantity here is what we have called a u of t. For any operator that is the definition of the Heisenberg picture operator as induced by h0. So this is equal to minus f of t a of t u prime of t t0. Now what is the boundary condition on this fellow here? Boundary condition u prime of t0 t0 equal to 1 because this is due to the switching on the perturbation. And it is switched on at time t0 equal to at any t0. So at that instant of time it is the identity operator. It starts with that and then it goes off to whatever it is. No, no. I am actually going to put t0 equal to minus infinity. Right now it is arbitrary, completely arbitrary. I have defined my Heisenberg picture operator to coincide with the Schrodinger picture operator at t equal to 0 as is conventional. No, no, no, no. Because there is always this evolution due to h0 going on. So I want to keep track of that. And I simply want these 2 pictures coincide at some arbitrary t equal to 0. I call that the origin of time. So this is the boundary condition I want to impose. And that is a differential equation except this is an operator. This fellow is an operator and we have to be careful. This is a number. It is some function. It is a C number, classical number. So we need to solve this equation using this condition and there is this ih cross sitting here on this side. Now what is the way to solve this differential equation? It looks like I have exchanged one problem for another. This is again time dependent and it is an operator. And we do not know how it commutes with that. We have no idea at all at the moment. So we have not done anything significant except that now I am going to say let us convert this differential equation to an integral equation which will incorporate the boundary condition that u prime of t0 t0 equal to 1. So this will imply, this thing implies that u prime of t t0 equal to 1 and let us move the ih cross down here plus i over h cross integral from t0 up to t because that integral whatever it is at t equal to t0 will vanish and then you will get this equal to 1 on this side. So this is the correct inhomogeneous term to take into account that boundary condition times d t prime f of t prime a of t prime u prime of t prime t0. Why is that? Because if I first of all if t0 t equal to t0 this goes to 0 and u prime of t0 t0 equal to 1 which is our boundary condition. Second if I differentiate both sides, if I differentiate this fellow and multiply by ih cross I end up with precisely this because the only t dependence is sitting here and I use the formula for differentiation under the integral sign. There is no t dependence here so there is no partial derivative. There is t dependence here and you are supposed to set the integration variable equal to this limit and differentiate the limit with respect to t which is 1 and if you do that you get f of t a of t u prime of t t. So the integral equation is exact no approximation made here etc and it will tell you how this Dyson series is generated as you can see because now I solve this by iteration I say this is small so I will do first order, second order, third order etc and what is this to first order to 0th order in f it is equal to 1. To first order in f it is an integral again with this quantity here with the second integration variable etc right. So by iteration I can get the solution the exact solution but I am not interested in that I am interested in that to first order. So therefore u prime of t t not is equal to and this means to first order the perturbation this is equal to 1 plus I over h cross integral t not to t dt prime f of t prime a of t prime times 1 because that is the solution to 0th order and there is already an f here. So if I retain this to 0th order in f that is good enough plus higher orders which I am not interested in and what is u of t prime it is u not of t prime a u not dagger of t prime a u not of t we do not care we do not care at all we do not care what it is and we do not care what t not is either. So now let us look at the case that we are interested in so let us set t not equal to minus infinity that is the most general case I can of course put a theta function so that this ensures suppose this f is switched on at some finite time t not I can always put a t prime minus t not here a unit step function here that take care of it. So this is the most general case so what does u become even before I do that what does u of t not t t not after all that is what we are interested in what does this become this quantity is u not of t just sitting here multiplying this whole thing times 1 plus i over h cross integral t not to t d t prime f of t prime a of t therefore u of t minus infinity is this so we have our exact expression to first order the correct expression to first order and notice how the Heisenberg picture as a has already got the Heisenberg picture evolution in it and the full time development operator has a correction this is the uncorrected one unperturbed part but then there is a perturbed part which is the product of this u of t with this integral all the way here. Now let us look at what the time dependence of some physical quantity like B is so let us look at what is B and I want the total of t so let us put a total because I have already used the symbol B of t for this quantity so I have already said B of t has been defined by mean to be u not of t dagger B and this is the B Schrodinger or B of 0 but now I want with this full thing and what will this be well if the state vector or the states of the system are governed by the time development operator u then the operators are governed by u dagger operator u so this of course is by definition u of t minus infinity B u of t minus infinity with the dagger here where I found this quantity to first order but had I found it to full glory to all orders this would have been the exact time development of this quantity of any quantity here pardon me yeah we are going to check that we are going to check that yeah it will be as you can see because I have to take the dagger of this it will become this fellow go on the right hand side and then this would become a minus and this is Hermitian so to first order it will be second order you got to check order by order consistently but it is the full thing is the full thing is as long as the Hamiltonian is Hermitian so the model of the story is that if the Hamiltonian is time independent and Hermitian then e to the power i Hamiltonian times t over h cross is a unitary operator but the statement is even if the Hamiltonian is time dependent as long as it is Hermitian the time development operator is unitary it is a complicated time ordered exponential but it is still unitary and you can show that order by order so now let us see what this fellow does so this therefore implies that B total of t equal to u dagger so it is this fellow goes to the right hand side and we have to write 1 minus i over h cross integral minus infinity to t dt prime f of t prime a of t prime because a equal to a dagger I have taken that to be the case if it were not the case then you have to add a portion to the Hamiltonian which will make it Hermitian so you have to add minus a f of t minus a dagger f star of t so that the sum becomes Hermitian quantity sometimes you would see in books plus h c to show that it is Hermitian conjugate has been understood to be added but the fact is that I am now going to look at the simple case where a is Hermitian just for notational simplicity so the minus here and you have this multiplied by u naught dagger so this is u naught dagger of t and then the quantity B and then a u which is u naught of t on this side and then 1 plus i over h cross integral minus infinity to t dt prime f of t prime a of t prime okay but we have to keep this only to first order for consistency there is a second order term here due to this f times that f but we have to throw that out because we do not have the answer correct to second order we do not have the time development operator correct to second order so keeping that second order term would be inconsistent which is equal to this times this will just give you a B of t the Heisenberg picture B of t governed by u naught evolution and then you need this times that and then this times that so let us put this times that first plus i over h cross integral minus infinity to t dt prime f of t prime times what well this is always a B of t so this portion is B of t I am multiplying 1 with B of t so B of t and then an A of t prime and in the opposite order it is A of t prime with B of t okay. So let us do the following let us write this as minus 1 over i h cross is that correct I have to be careful here I want to be careful so let us not mess around with these signs so let us take this term first so minus i over h cross integral minus infinity t dt prime f of t prime A of t prime B of t that is this times that with the minus sign and take it the other way there is an x i over plus i over h cross so I will put a minus sign inside the bracket so let us this write this as plus over i h cross which is equal to B of t plus integral minus infinity t dt prime f of t prime and then a commutator of A of t prime with B of t over i h cross but this is precisely what we call the response function well it is actually t minus t prime that is not hard to show that is really hard to show you put in those e to the i h s and etc are not quite not quite so let us not do that yet let us not be in a hurry so we have this expression here and now the last step is to say B total of t average value now we are going to take averages and with respect to what should we take the averages with respect to row equilibrium because we have already put in all the time dependence in the operators we solve the equation of motion for the operators by writing a time development operator acting on the operators and finding out what that does explicitly and computing to first order so this average is now with respect to row equilibrium that is equal to average value of B of t in equilibrium row equilibrium which I call equilibrium plus integral minus infinity to t dt prime f of t prime times the expectation value of A of t prime commutator B of t in equilibrium over i h cross by the way this is time independent B equilibrium this part of it is identically equal to B equilibrium it is time independent why is that why does that happen because yeah because what you need here is e to the minus beta h naught you have to take a trace of e to the minus beta h naught B of t but B of t e to the i h naught t over h cross B e to the minus i h naught t over h cross by the cyclic invariance of the trace move this portion across to this side and of course it commutes with itself it cancels against this and you have trace e to the minus beta h naught B which is equal to B so this portion of it is harmless it is just you can compute it at any time it does not matter in the Schrodinger or Eisenberg picture does not matter it is just the equilibrium average so this part is what we defined as delta B of t by definition this was the change and we are back to the formula we had for delta B of t so delta B of t this quantity is exactly precisely so this we can write now as integral minus infinity to t dt prime f of t prime phi AB of t minus t prime because the expectation value of A of t prime B of t commutator equilibrium is equal to and that we did we explicitly did this we wrote out this quantity and put in all the e to the minus beta h naught and took it across it so that is how we did that is one way of now explicitly deriving so what we what have we done we have done first order perturbation theory in the Eisenberg picture it also tells you how to generate the higher orders so in principle if you are interested you can go on to the next order and find out what is the correction etc. So again go back to the integral equation for u prime and work it out work out the solution to the second order consistent a lot of algebra it would not be linear response theory anymore definitely but it will be second order perturbation theory and then of course we also showed that phi AB of tau this quantity was equal to the bracket that we wrote here A of 0 B of tau this quantity in equilibrium this was also equal to in the classical case just beta A dot of 0 B of tau equilibrium and in the quantum case it was equal to integral 0 to beta d lambda trace of e to the minus beta h naught times e to the lambda h naught A dot of 0 e to the minus lambda h naught B of tau which we wrote as the same thing with the semicolon here by definition so all the non-commutativity of quantum mechanics the complications arise from here from this complex expression relatively complicated expression because you have this sandwiching this is sandwich between these two operate and then all the rest of the properties of this semicolon this common this response function namely the fact that it is stationary it is real for A and B Hermitian it is symmetric function if you interchange the A and B even though they do not commute with each other etc those things follow because of the cyclic property of the trace so I thought I would do this just to show you explicitly that the same answer is obtained from perturbation theory first order perturbation theory by choosing that particular answers for the time development operator okay and what is it that you are doing in choosing that answer you are simply saying that this operator is basically the unperturbed operator but for the perturbation you got an extra piece which satisfies a separate time dependent equation and that equation can be solved by converting it to an integral equation and iterating term by term so that is one way of solving such operator equations with time dependent on inside the operator. Now there is one more thing that we need to do and that is to connect this with dissipation after all you are in a force to the system so there is some kind of energy that is increasing or decreasing some energy is being dissipated and we would like to see how the system absorbs energy if it is a quantum mechanical system it is obviously going to be able to absorb energy only if it takes you from one level to another or brings you down from one level to another one of the two so we should be able to see that explicitly. I am just going to outline the steps here and leave the rest of it as an exercise for you to work out so let us do that case and for simplicity let us look at the case where B is equal to A itself then remember that Phi AA of tau equal to the equilibrium expectation value of the commutator of A of 0, A of tau and A is Hermitian when what is it we want to compute? We have a Hamiltonian which is now time dependent so there is an H of t equal to H0 minus A f of t. We would like to find the rate of change of this the expectation value of this Hamiltonian of the internal energy of the system so we would like to find what is this d over dt H of t and that comes from the time dependence here so this quantity is minus A of t expectation f dot of t I should say A total and the explicit time dependence is sitting here and that is the quantity I want to compute but this quantity so this thing here is equal to so let us compute minus of this fellow this is equal to A equilibrium f dot of t is a term sitting here that is a harmless term you can see plus and this is the interesting part plus an integral from minus infinity to t dt prime f dot of t out here times Phi AA of what? f dot of t minus t prime f of t prime in this fashion. We have an expression for this fellow here in fact we can go all the way back and write this expression in terms of the spectral function they have an explicit expression for the spectral function in terms of e to the i omega t and things like that and you can do this integral because the time integral finally this fellow here is a function of tau and that will become 0 to infinity in tau do the integral and it will essentially give you once you put in time dependences here if you put a Fourier mode it will give you the susceptibility at some temperature at some value of the frequency. So now let us say we now put in a physical case we say f of t equal to the real part of f not e to the minus i omega t so I am trying to find out what happens if you hit the system with a force at some frequency omega which is equal to f not e to the minus i omega t plus f not star e to the i omega t over 2. So f dot of t is trivial this is equal to minus i omega over 2 times f not e to the minus i omega t minus f not star e to the i. f not is some amplitude which is necessarily got to be complex because I want the force to be real so f not plus f not star put that in put this expression in for f of t prime put this expression in for f dot of t t collect the exponentials together when you do that you are going to get e to the i omega t minus t prime everywhere sometimes t plus t prime 2 you get that term as well right you are going to get both kinds of terms in there integral is minus infinity to t change variables to t minus t prime equal to tau and wherever you recognize e to the i omega tau times phi a a of tau replace it with chi of omega and in terms with the wrong sign e to the 2 t plus t prime what should you do subtract twice t so make that t minus t prime inside and then there will be e to the 2 i omega t outside right. So that big mess the huge in my I do not I do not want to write it down here because it will be quite a large number of terms due to this fellow here and what we want to find is the average rate of dissipation so average then over time t from 0 to pi over omega which is the time period over a full cycle of this force that will give you the average rate of dissipation so are the steps clear it will give you may be a constant we do not know you will have e to the 2 i omega t you have to integrate over full period may give you 0 may not etc so in general it will disappear as you can see but at the end of the day at the end of the day you would expect this quantity this quantity to be proportional to mod f naught square that is not too surprising right time something which involves the susceptibility and I want you to show that it is proportional to chi imaginary times constant there is some omegas and stuff like that now you can come you can write this down in terms of the spectral function and a is Hermitian so the spectral function is going to involve weight factors a n m b m n those are going to now become a n m b m n and then there will be summations and things like that so those are called oscillator strings and then there would be all the transition frequencies of the system so that the term which corresponds to absorption and then the term which corresponds to emission they are both there in quantum mechanics so there is always stimulated emission along with absorption and finally you will discover that you have got time dependent first order perturbation theory the Fermi golden rule but with a finite temperature so there are always things weighted by the Boltzmann factor so compute this I am going to give this write the answer down in the problem sheets and then you will see check against what you have done so this kind of brings me to the end of formal linear response theory the rest of it will be in the form of exercises which we will put in the notes but I would like you to go through this derivation step by step all over again and complete this complete this piece of algebra I would do this except for the fact that they did not let it carry over to next time and it is a lengthy slightly lengthy piece of algebra but you see the strategy that is involved and once you do this for one frequency you know the answer for all frequencies because you superpose for an arbitrary force history okay that kind of completes what I wanted to say about linear response theory we still have to make contact with another way of doing this whole business which is to introduce a random force into the system and then compute various quantities once you have a stochastic model for the system including dissipation now that is not easy to do in quantum mechanics because including friction and quantum mechanics is a non trivial task we have sidestep that task but the difficulty with that are the more powerful very powerful way of doing things is that you have to make some assumptions about the nature of the stochastic force that you have and when you have quantum mechanics you also have to worry about commutativity and so on makes it more complicated the formalism of linear response theory on the other hand gives you a very general framework for solving such problems plus points and minus points this is more general than the other case but the other case is very specific and it will go further in term once you make a detailed model of the stochastic force or whatever is driving that fluctuations then you can go little further here there is a fluctuation dissipation theorem here too and that is the relation between we already know in this case kind of intuitively you can see we have a formula for this quantity for A of 0 B of tau in equilibrium this was our response function and we have a formula for its Fourier transform which is essentially phi tilde of omega right if you like that is like taking care of this quantity here is taking care of the response of the system that is what the response is when you apply the force on the other hand the fluctuations themselves would be given by things like what happens under pure thermal fluctuations not the commutator but the anticommutator would give you a measure of the fluctuations especially if you set A equal to B then you get a mean square values that will give you fluctuations the variance of that quantity etc and you know that there is a relation which says that if you took the anticommutator here we could see that this went like an integral d omega phi A B tilde of omega times this quantity this e beta omega over omega whereas this fellow was just the spectral function the Fourier transform of the response function right which measured the dissipation in the system if you like so there is a connection which says that the power spectrum of this quantity which is Fourier transform is related to the power spectrum which measures the dissipation namely phi tilde by this formula one of them is phi tilde and the other is phi tilde times this quantity so that is the generalized fluctuation dissipation theorem you have not used any model for the random force or anything like that but it is a relation between fluctuations and dissipation because the dissipation is governed finally by this quantity which comes from the response function and the fluctuations are governed by the power spectrum of this one the Fourier transform of this quantity so if you like that is the fluctuation dissipation theorem in this problem without any reference to the stochastic force or anything like that which is the way it appears in the Langevin model here so this is intrinsic to it we have not assumed any specific model of stochasticity and yet the thermal fluctuations are getting connected to the dissipation in the system so I will write that expression out explicitly in the handout so you can see and check what happens and you can also compute now things like the mean square displacement of a harmonic oscillator quantum mechanical oscillator at finite temperature at very high temperatures this would be essentially k T and very low temperatures it would be related to h cross omega but in between temperatures it comes from using these some rules here for the spectral function so some of these things I will give us exercises for you to check out so with that we stop here and the next thing we are going to do is to go back backtrack a little bit and fill in a gap we had not filled in earlier which is to connect up the Langevin model with the corresponding differential equation for the and for the probability distributions themselves so things like the Fokker-Planck equation and so on and we take it from there next time.