 So, we were in the process of examining the velocity autocorrelation function for a Brownian particle moving in a constant magnetic field, a uniform constant magnetic field. And if you recall our final answer said that this quantity which I call the correlation function apart from an normalization by multiplying it by what happens at t equal to 0, this thing here is k t over m and then we got a complicated expression for it inside. There was of course a damping factor so there is an e to the minus gamma t and then that gets multiplied for a field B which was B times some unit vector n. This gets multiplied by n i n j plus a delta i j minus n i n j cos omega c p, where omega c is a cyclotron frequency q B over m, q being the charge of the particle. So, there was the second term minus a third term which is epsilon i j k n k sin omega c p. And this was true for t greater than 0 equal to 0. This is the expression we derived. So, if you check I rearrange some of the terms and then after simplification this is the expression you get. Now the first thing you notice about it is that gamma does not appear anywhere here at all. It appears only here. So, the effect of dissipation is taken care of entirely by this factor exactly as in the field free case. At the moment you have a field switched on you have this expression. Of course, if you switch it off you go back to just this expression with a delta i j because this is 0, that is 0, this is 0 identically and you end up with just the original expression as before. Now we would like to find out what happens for negative values of t. We derived this for positive values. We would also like to examine what is it do for negative values of t. Now recall that in the field free case we had v i of 0 v j of t in equilibrium. This was equal to k Boltzmann t over m e to the minus gamma modulus of t. The way we derived this thing here is by pointing out that this must be an even function and the way to get the evenness of this or the oddness property whatever the symmetry property of this is to put in if you like time reversal and ask what the whole thing does under time reversal exploiting stationarity. So if you recall in the field free one dimensional case we started by saying v of 0 v of t in equilibrium since this is a stationary process. So equal to implied by stationarity of the random process v. This was equal to v of minus t v of 0. This should be a delta i j. Thank you. So in the one dimensional case I said that the origin of time does not matter. So I subtract t from each of these arguments and I end up with this expression here. This immediately of course says since these are commuting variables this immediately says that this is a symmetric function of t. That is the reason we got hold of e to the minus gamma modulus c. Yeah n is 0 in that case. I mean there is no n. This is gone. Omega c is 0 of course. And this goes away. This vanishes. So I wanted to appreciate that we are looking at what happens for t negative not by doing another calculation but simply arguing that this is a stationary random process. So I can subtract any number I like from the argument of this whole function from all the time arguments and when I do that it becomes an even function of t. So the argument was then it is k t over m e to the minus gamma modulus of t using stationarity. We can do pretty much the same sort of thing and this term will become modulus t here but let us do this systematically. Let us ask what happens to this in the general case. So in general I already mentioned that even in the presence of the magnetic field the velocity continues to be a stationary random process because there is no energy being given to it. There is no dissipation involved with the magnetic field or anything like that. So if you grant that then it is immediately clear that in general v i of 0 v j of t if I call this quantity let me call this equal to phi i j of t. The reason I am calling it phi is because when I divided this by its value at t equal to 0 I call that c i j of t. I do not want to use the same symbol phi i j is. So this is equal to v i of minus t v j of 0 by stationarity I subtract t from both sides. But this is equal to v j of 0 v i of minus t which is equal to phi j i of minus t because these indices get interchanged. This did not happen in the one dimensional case. So this tells us that this correlation function c of t this matrix c that we computed yesterday is such that it is equal to c transpose minus t because of this property. It becomes a transpose and then the argument becomes minus t. So it immediately follows whether I use phi or c it does not matter they just differ by a multiplicative constant k t over m. So this immediately follows that c i j of t plus c j i of t these are the this is the even part of the tensor c i j of t divided by 2 of course. The minus part is the odd part of this tensor. So given any second rank tensor t i j I can write t i j plus t plus t j i over 2 that is the even part and the odd part the anti symmetric part is t i j minus t j i over 2. So this is the symmetric part of the tensor correlation tensor this is the anti symmetric part. And by this property this is equal to c i j of t plus or minus c i j of minus t using this property. It follows therefore that this correlation tensor c i j of t is such that its symmetric part is an even function of t because that is an even function of t and its anti symmetric part is an odd function of t. That is a general property as soon as you have a correlation matrix for some random variable and it is stationary it follows that the even part of the correlation tensor the symmetric part of the correlation tensor is an even function of the argument and the anti symmetric part is an odd function of the argument. We will use this property later on. This does not require you to know what the function is at all. I have only used stationarity at that end. We will use this property. It is a crucial one. We will use this symmetric property but you can see how this is how it actually is tallying with what goes on here. We need to write down what this thing is for t less than 0 and let us write it down in the following way. What does this do? So this for t less than 0 less than equal to 0 v i of 0 v j of t is equal to none of this gets changed k t over m e to the minus gamma. Modulus t, I have to put a mod here exactly as in the field free case multiplied by whatever is inside here is going to be the time reversed value of whatever is in the bracket for t greater than 0. So you have to reverse time from t to minus t. What happens if you do that? You get n i n j plus delta i j minus n i n j cos omega c t and then what do I get? The next term it is not just setting t to minus t that is time reversal but when I reverse time and let the system run backwards I should also change the sign of the magnetic field. It is an externally applied magnetic field so under time reversal I have to change its sign as well. You can see this happening even in Newton's equation for a charged particle. If you look at what the Lorentz force does the equation is f equal to d p over d t equal to q times v cross b. This is an external applied field. Now I make the time reversal transformation t goes to minus t then p goes to minus p because this is m d r over d t and r does not change position but the momentum changes sign. Whatever was going in this direction now goes in the backward direction. So this changes sign, this changes sign. Therefore this does not change sign d p over d t and if the equation has to be the same under time reversal invariance in other words we impose the fact that Newton's equation for a charged particle is invariant under time reversal. This is an imposition from outside. If you impose that condition then since v changes sign, b must change sign. So does the electric field change sign? No because if you put in an electric field also it is q times e and that does not change sign because this side does not change sign. So this cannot change sign under time reversal but this will change sign, b will change sign. You can kind of physically understand it by saying in heuristic terms. Look finally every magnetic field is produced by some current loop of some kind. If there is time reversal then that current flows in the opposite direction and it produces a field in the opposite direction okay. So b must go to minus b which means that n k must go to minus n k. So under time reversal goes to minus n sub k. So this term remains minus epsilon ijk nk sin omega c. And this expression is true for all t. We have taken care of this here for negative t. So that is the exact expression for the correlation function for all t. And now you can verify explicitly that this property is satisfied. So check out that the symmetric part of the tensor is an even function of t and the anti-symmetric part is an odd function of t. That appears by inspection because this tensor, the even part, the symmetric part comes from here, here and here. This part is anti-symmetric in i and j. That is an odd function of t. This part is the symmetric part of the tensor and that is an even function of t. If that did not happen, I would really be in deep trouble with worry about what is going on. So this gives you a check on what the reversal properties are. It helps. That symmetry is valid. It is applicable here too. It is a general statement. So it better be true here too. Now given this, the next step is, well we can proceed in many directions but the next step we could do is the following. We saw that for a single free particle in one dimension, we saw that the diffusion coefficient got related, in position got related to the velocity autocorrelation. So we saw in fact that in that case, we saw that x of t minus x of 0 whole squared went asymptotically for very large t to 2 dt and we got an expression for this d. It is k t over m gamma in this large gamma model but we saw that from first principles, if I just took this displacement to be the integral of the velocity and then impose the fact that the velocity is a stationary process, I ended up with a statement that d was equal to an integral from 0 to infinity dt v of 0 v of t, the equilibrium autocorrelation function of the velocity. So we got this explicit formula. Now the question is what happens now in this problem? What would happen in this case? Well you can kind of see intuitively that the portion that is not going to be, that is going to be unaffected would be the one that is along the direction of the field, along the vector m because there is no magnetic force in that direction at all. But in the perpendicular or transverse directions, there is a force which tends to make this particle go in loops, go around the direction of the field in cyclotron orbits. So it is diffusing, it is being kicked around but every time it is kicked around, it is still trying to curve back on itself. So I would expect the diffusion coefficient to be less in the longitudinal, more in the longitudinal direction unaffected from the free particle case and in the transverse directions I had expected to be a little smaller. The fact that it is diffusive, there is no question because this integral exists because of this factor and these are just oscillatory terms here. So the question is what is the generalization of this formula? This is the famous Kubo Green formula, a special case of it and when we do linear response theory in some detail, we are going to work this formula out, we are going to explicitly prove a much more general version of this formula for all kinds of susceptibilities or response functions. But let me state the result here. What happens now is that the diffusion becomes a diffusion tensor, the coefficient is a set of coefficients, diffusion tensor denoted by D i j because I also have to tell you where does this fellow come from, what does this D i j look like and it comes from the fact that the probability density in position obeys in the analog of the diffusion equation, a little more complicated than free diffusion and that involves a set of coefficients which are summarized in this diffusion tensor here. We will do this. When we write the general formalism down of linear response theory, we will come back to revisit this problem. But the answer can be written down here and you can see some physics in it so we can see what they end up with. This tensor is defined as the symmetric part so it is in half D t phi i j of t plus phi j i where phi is the autocorrelation. So this fellow here stands for D i of 0 E j. We can therefore compute it. We can go back here and compute it. We need only the symmetric part of this tensor therefore this does not contribute at all and so only this portion contributes and this is an easy integral to do. First of all, this term is immediately obvious it is e to the minus gamma t that is just 1 over gamma. This term is e to the minus gamma t cos omega t which is gamma divided by omega square plus gamma c so omega square plus gamma square that is it. So we can write the answer down explicitly and turns out D i j this thing here is equal to A B T over M gamma as before multiplied by that portion remains as it is n i n j plus delta i j minus n i n j gamma square over gamma square so that is the answer. As I said this will not contribute at all because it is the anti symmetric part of that tensor and does not contribute. Now can you interpret this? For instance you can see directly what is going to happen. For instance suppose n was equal to 0 0 1 so the field is in the z direction and we have the x y plane there is no force in the x y plane sorry there is no force in the z direction because the field is in the z direction. Then n i the only term that contributes are the diagonal terms as you can see and now what will happen if it is 3 D 3 3 that is going to have a contribution 1 from here and this portion cancels out and it is the original diffusion points. On the other hand if you look at 1 1 for example in this case then this term is 0 does not contribute that portion is 0 this gives you 1 and it gives you gamma square over gamma square plus omega c square. So the diffusion coefficient the longitudinal part equal to k Boltzmann t over m gamma as before but the D in the transverse direction is equal to D longitudinal multiplied by this term so it is therefore attenuated it is decreased. If omega c becomes much larger than gamma square then you can see the decreases by a factor gamma square over omega c square that is exactly what this says you do not have to go to a special case like this. This thing here is a projector in the direction of n and that thing there is 1 minus the projector so this is from you know from ordinary vector pardon me it is the projector on the transverse direction because well all you have to do to verify it is multiply both sides by contract it with n j or something like that and it immediately tells you d i j n j just as an n i here because n j n j is 1 when you sum and this is obvious what you have there. So you know when you do elementary vector algebra in three dimensions you will learn about the dot product and the cross product between two vectors and so on but there is also a tensor product of two vectors where you do not write anything in between that is called a dyadic in ancient literature just another word for a tensor of rank 2 but when they did vector analysis people use this term dot in the beginning. So this quantity here is such that it is a two headed object it is such that no matter what vector you operate this on whether from the left or right you produce the component of the vector in the direction of n because if you take this vector and dot with a then there is a component of a along the unit vector n multiplied by the unit vector n so it gives you how much of that vector points along the direction of n. So you have an arbitrary vector a and this is your unit vector n and this fellow is actually measuring this this quantity is measured here right. So this is the projector it is the projection operator if you took three orthogonal directions and projected them out then you get the identity operator because when you apply that to any vector you have taken all the three orthogonal components of this vector and added them up so you get the original vector itself. If for instance you took this E x E x plus E y E y plus E z E z that is a projection operator which when applied to any vector will give you the vector itself. So what should this be? This should be equal to the identity operator in any linear space you sum up all the projection operators and you get the identity operator. This is like saying that if I have an abstract space with eigenvectors with an orthogonal basis phi and phi n over n this is equal to the identity operator. In Dirac notation this is the way it is written but it is the same statement out here. So it is clear that written in tensor form this fellow is being written as n i n j in tensor form and the complementary part of it if you for example take n to be E x the rest of it is this fellow here which is 1 minus this guy here. So 1 minus n n is the projection in the plane perpendicular to n of any vector the transverse components which you may further resolve into two orthogonal components but it does not matter this is the complement. This multiplied that must be 0 when you project in some direction and project in the orthogonal direction when you operate the two it must be 0. So this fellow written in tensor form is of course delta i j minus n i n j. So that is what we have been doing all along this part is a longitudinal part this is the transverse part now it is crystal clear immediately. So this is the longitudinal part and this guy here is the transverse part of this tensor that is just a general decomposition of rank 2 tensor in 3 dimensions okay. Having said this we will come back to this problem we will come back to this problem much later after we do linear response theory look at some other correlation functions like position velocity etc etc. But let me write one more answer down purely on few physical grounds we argue and let us write this answer down and that is the actual distribution of the velocity remember that in the one dimensional case I wrote this distribution down by saying it is the Onstein-Ohlenbeck distribution I argued that it is still a Gaussian I asserted that it was still a Gaussian and therefore it was determined by its time dependent mean and variance I wrote those down and that was just an Onstein-Ohlenbeck distribution can we do the same thing here well I again assert that the distribution asymptotically as t tends to infinity goes to the Maxwellian distribution nothing happens it remains at that temperature and goes to the Maxwellian distribution the question is how does it do so and that question can be answered fairly straight forwardly. So in 3D free particle free Brownian particle namely no for this fellow the conditional density V of at time t given an initial velocity V 0 this is just the generalization of the Onstein-Ohlenbeck distribution written component wise and multiplied together 3 Gaussians and you end up with m over 2 pi k Boltzmann t 1 minus e to the minus 2 gamma t that is how the variance goes this whole thing to the power 3 halves exponential of minus m into V minus V 0 vector e to the minus gamma t whole squared that is how the mean drifts to the origin there is a vector here divided by 2 k Boltzmann t times this quantity for the variance that is the 3 dimensional Onstein-Ohlenbeck distribution okay. Now the question is what is this in the presence of a field what would this be I am already telling you that this remains a Gaussian that the asymptotic distribution is still the Maxwellian that the dissipation is taken care of entirely in this factor gamma t that this will not change this does not change none of this changes how will the field show up what will it do. So think about it in this fashion whatever be V 0 this distribution is finally going to go with this kind of square Gaussian in this fashion here. So when you start with some vector this velocity and as t tends to infinity the average velocity is going to shrink to 0 this is the mean instantaneous mean for a given V 0. So clearly this vector is reducing in length due to dissipation that feature is still going to be true right but as it reduces in length due to collisions this mean value what else will it do it will rotate around the direction of the field okay and how does it do so. So it will rotate with a cyclotron frequency about the direction of the field right and how would you implement that I already wrote down an expression we when we derived this correlation function I already told you that we exponentiated the generator of this rotation matrix which was m omega c t and that is how it relaxed and that gave you terms proportional to cos omega c t sin omega c t etc. So in the presence of the field exactly the same thing goes through and this is now intuitively clear what I am going to write down will be intuitively clear p is present then p of V p V 0 is equal to the same thing but you do not get this here you get a certain u here which will of course depend on t and V 0 and then there is an e to the minus gamma t that is the effect of the dissipation and this u of t V 0 at any instant of time it is rotating around the direction of the field it starts at V 0 at t equal to 0 and rotates around precesses around the direction of the field. So that comes about by e to the power m omega c t acting on V 0 take this to be a column vector and take this rotation matrix and since it is time dependent there is a t dependence here and that is it that will tell you how if you started with V 0 this vector and this is the direction of the field how this vector precesses around it that is what this portion does simultaneously its length is coming down according to this factor. So it is this thing here square and that is the answer but we can write this down because remember m cube was minus m so the exponential can be computed and not surprisingly this is equal to the identity operator plus m sin omega c t plus m square cos acting on V 0 so that is what and you can work this out I give you to figure out simplify this write down what it is I am not 100 percent sure about the factor here plus or minus but you can check this out I believe it is an m but you can check this out explicitly is it a 1 minus omega c t I think so yeah it is 1 minus cos omega c t that is right for the same reason as before no m cube is minus m so that brings in the minus whether this is plus or minus this was a minus check this out explicitly but it is physically clear what is happening this is the interpretation and that is the way the once you know the distribution gets modified as t tends to infinity you can see this goes away and you are back to the max value. So this is a very very very appealing physical picture that in the presence of the field the velocity keeps doing this but because of the particles colliding against state it damps out to an average value of 0 while it is so it is some kind of helical motion which goes to 0 we have not yet written down the differential equation of which this is a solution we do that we will have a general formalism to write it down and we will be able to write this down in the presence of the magnetic field it is not very hard quite straight forward so we will do that eventually so we will reserve we will stop here with this in fact we will stop here with the Langevin model itself and now start with the formalism of linear response theory so that we can handle these questions much more generally even without a specific equation of motion like the Langevin model. So we will start now a study of linear response and again I have to say what our target is we start by assuming that so our problem is the following the general class of problems we are going to address is the following we start by saying that we have some system which is described by Hamiltonian which is time independent and the system is in thermal equilibrium at set temperature beta in contact with a heat path so the system consists of this is described by some Hamiltonian H naught unperturbed Hamiltonian and it is a function of some dynamical variables q's and p's we will also look at the quantum case simultaneously whenever there is a quantum case coming up I will say so explicitly but we will look at both these simultaneously because there is a formalism which applies to both of them the only difficulty is in quantum mechanics you know that different observables are represented by operators on some Hilbert space and these operators do not necessarily commute with each other and the whole essence of quantum mechanics is in the non-commutativity of this whole business capturing that by any classical manipulation is not possible in any simple way at all so this is intrinsic quantumness if you like the non-commutativity and we have to use operators to describe it we will take adequate precautions but to the extent possible we will develop the formalisms in a matched way and the way to do this is to say in classical mechanics we would write down Hamilton's equations of motion and then we deal with Poisson brackets and so on in quantum mechanics we would write down the Heisenberg equation of motion for observables directly and you deal with commutators instead of Poisson brackets as opposed to the usual Schrodinger equation where you write down an equation of motion for the wave function itself for the state vector itself that is the so-called passive picture but when you do classical dynamics you hardly ever do that you hardly ever say in classical mechanics when you are looking at a particle or a body moving you do not talk about a distribution of probability density in phase space or anything like that you write directly the dynamical variables and write their equation of motion down they act of picture in quantum mechanics the conventional way of doing it using the Schrodinger equation is to go and look at a passive picture. So you make a statement about the way the state vector evolves and the state vectors give supposed to give you probability densities and so on with some further manipulations so you are making a statement about how the distribution evolves and you say the dynamical variables themselves the the averages are computed with respect to that changing distribution that is the passive picture which is very natural in quantum mechanics but it is very artificial in classical mechanics but these are completely equivalent to each other as we will see when we go along. So we will mostly use the Heisenberg picture in quantum mechanics or the corresponding dynamical evolution via Hamilton's equations in classical mechanics. So the general problem I have is the following you have an unperturbed Hamiltonian but this system is now in contact with a heat path at a fixed temperature and is in thermal equilibrium. So physical quantities are evaluated average values of physical quantities like some let us call it some observable b which is in general a function of the q's and p's of the system it would have any number of degrees of freedom this collectively denotes all the generalized coordinates all the generalized momenta conjugate momenta the average value in equilibrium to avoid confusion let me when necessary write this outside. This is equal to in the canonical ensemble it is equal to formally trace of a density matrix rho equilibrium b divided by trace of rho equilibrium. This is the average value in equilibrium I presume you are familiar with this formula this is equilibrium statistical mechanics is everybody found familiar with this comfortable with this you could ask what is meant by trace of course if I write the usual quantum mechanical sense if I write finite matrices for these traces for these operators then it is very clear you take diagonal elements and sum them up but what do we mean by it classically what I mean by this is this rho equilibrium is e to the minus beta Hamiltonian of q and p that is the equilibrium density matrix density operator and this stands for integral dq integral dp over all the q's and p's over all of a space times b of q p e to the minus beta the Hamiltonian H naught of q p divided by this fellow here is dq into dp e to the minus beta H naught of q and p this is the normalization factor this average statistical average and the idea is that in the canonical ensemble the weight factor in phase space is e to the minus that Hamiltonian over kt beta will always stand for 1 over k Boltzmann t the inverse temperature always I will try not to use beta for any dynamical or any other variable or parameter. So this is what is meant by the trace in this in the classical case all integral over all phase space this fellow here now this term this fellow here is in fact just a normalization factor for the probability so you could submit in the definition of the density operator by saying this divided by this number here is in fact my probability distribution then it is a normalized probability density function there is a name for this what do you call this in equilibrium statistical mechanics it is the canonical partition function this fellow here this integral is z canonical it is not a function of the q's and p's because they are integrated over it is a function of the temperature this parameter. So putting a system in contact with the heat bath drives fluctuations into it and we do not need to know what this heat bath actually is made up of we do not need to know its details what are the degrees of freedom and so on the entire ignorance of the heat bath is put subsumed in a single parameter called the temperature so this is the canonical of course if you do this in a finite volume and things like that with a fixed number of particles then it is a function of those variables as well microscopic variables but we are going to look at a case where I put this equal to 1 so I put this trace row I am going to put trace from now on I will take trace row equal to 1 I will redefine my density operator in such a way it is this divided by that number so that the whole thing is traced out to 1. Now remember that when you do quantum mechanics I am not assuming that the system is in any one pure state so when it is in a finite temperature you cannot associate with it in general of pure state namely a state vector in a Hilbert space you can only associate a statistical weight to it and that is given by the partition function by this either by the density operators so this is what we are going to do so this is what it is in equilibrium and now for the problem the problem is that we are going to say that this fellow gets perturbed the perturbed Hamiltonian is some H of q p and possibly t which will in general be H naught of q and p plus the perturbation which I am going to regard as small in some sense and it involves an operator A of q p times a time dependent c number force so f external of t let me explicitly say it is an applied force and a is some dynamical variable pertaining to this system to which this force coupled through which the force is coupled to this system couples with the external force now you could ask why this minus sign well suppose it were a mechanical force right when the force on the object is f external of t then I associate quote unquote a potential with it which is minus x times this f external if it is one dimensional motion then minus d over dx of this fellow is supposed to give me the force and of course that will give me f external and this term has the dimensions of an energy quote unquote a potential energy even though there is a time dependent force here so to keep track of that I have a minus sign here no other reason and to take care of the fact that this could be a very general kind of force not necessarily a mechanical force coupling to the displacement I have a form like this in general. So a refers to the of course you could say there could be several forces maybe I have a i f sub i that is a generalization which is trivially taken care of so this is the general form that we have but in some physically measurable sense this is supposed to be small compared to that and now I want to know what happens to averages so I take any other quantity b and I ask b equal to what this will of course be a function of t so I would like to know what function of t it is so in general this fellow is going to be of the form b equal to b in equilibrium which is what it would have been using this rule here plus a delta b which is going to be a function of t some average value and the whole point is to find this so this is our target we will choose all kinds of a is here all kinds of b is here b could be a itself you could ask what happens to a itself so that is the general problem and we are going to work to first order in this perturbation so everything will be to first order in this external force and then the assumption is if this is very small first order term alone then this is infinitesimal or first order alone which is why I put a delta here that is the rational for this okay and the target is to find this under very general conditions this could be classical it could be quantum mechanical this could be horribly complicated we do not care we are going to try to find a formalism which will tell me what this quantity is so here onwards that is our target to do this let us see where it takes us okay.