 How do we erase the congenital tendency to this magic word adjust? We saw last time that the general response function phi AB of tau which is of course defined as the commutator of Poisson bracket as the case may be in the canonical ensemble. We saw that this quantity here could be written in terms of the matrix elements of the individual operators A and B. For instance, in the quantum mechanical case, we assume that there is a complete set of states given by the eigenstates of the unperturbed Hamiltonian H0 and in that basis this quantity got written as 1 over ih cross, the summation over n and m. If I recall right, it was Anm Bmn, the matrix elements of these operators multiplied by the Boltzmann factors this times the actual time dependence which was e to the minus i omega nm tau. Check if these all the factors are there. I am writing this down from memory from what we did last time but I believe it is okay as it stands. These were actual matrix elements, this quantity for instance Anm was phi n A in the Schrodinger picture if you like, phi m and likewise for B. So that immediately led to a representation for this fundamental quantity, the spectral function. So we are going to regard the spectral function, the Fourier transform of this response function to be a fundamental quantity and then everything will be expressed in terms of that spectral function. So spectral function phi AB tilde of omega is of course defined as an integral from minus infinity to infinity d tau e to the minus i omega tau phi AB of tau and this if you use the fact that this quantity here is where the tau dependence is sitting then this is equal to d tau e to the plus i omega tau that was my Fourier transform convention. So this is equal to 1 over ih cross summation n, n A nm Bmn times a set of delta functions in this fashion. So it just consists of a whole lot of delta function spikes that is what the spectral function is and we know already what the representation from here it follows, they follow a spectral resolution of the susceptibility itself. So that implies that chi AB of omega equal to minus i limit epsilon goes to 0 from above so epsilon goes to 0 an integral from minus infinity to infinity d omega prime phi AB tilde of omega prime divided by omega prime minus omega plus i epsilon. We had this relation between the Fourier transform of the response function and the generalized susceptibility. So this will imply that this quantity here is equal to the i n minus i will cancel so if I put this in then it is equal to well yes when I do this I have to be careful because I did this integral d tau e to the i tau times omega minus omega mn that is 2 pi times the delta function. 1 over 2 pi is times e to the i whatever is the delta function so there was a 2 pi there. So this fellow is minus 2 pi over h cross the i cancels out summation n, m a n m b m n e to the minus beta a n divided by omega prime minus omega n m minus i epsilon. So wherever omega appears I just replace it using the delta function I replace it by omega n m and that is it. So this is the representation for the susceptibility itself. So again everything is expressible either in terms of chi or in terms of this quantity. We will use this referentially because there are going to be various conditions put on this and it is the simplest way of expressing those conditions is to use the spectral function. Now the first question what kind of function is this chi as a function of omega? Yes of course chi is a function of omega. Thank you. Yes absolutely thank you. Absolutely it is got to be a function of omega so omega n m the minus sign can go away. So let us remove this and write it as omega minus omega n m plus i epsilon. Thank you. So it is got poles this susceptibility has poles at all the transition frequencies but this representation tells you explicitly what kind of analytic function this quantity is. You see it is a function of omega which is analytic in the upper half plane explicitly. The physical susceptibility for real frequencies is the boundary value from above because as epsilon goes to 0 you should put still put the limit equal to 2 pi over h cross limit epsilon goes to 0 of this guy. For real omega for real omega this is the representative but if I remove this i epsilon then it defines an analytic function in the upper half plane and what is happening is that the physical susceptibility in the omega plane is the boundary value of this function without the i epsilon as you come down from above. As you can see you give a small positive imaginary part to it and then let it go to 0 so it is really the boundary value from above. So what one can do is to define a function of omega for complex omega okay using this so we could define some function let me let me use some other symbol for it. I want to do it in the lower half plane as well at the same time. So I want to argue that let us let us call this k for want of a better symbol k A B of omega. Let us define this to be minus i integral d omega prime phi A B tilde of omega prime over we can write it as minus i is still there so omega prime minus omega omega complex let me define such a function. This quantity is defined for all real values of omega prime through this integral through this guy perfectly well defined after all the argument of a delta function has to be real otherwise it makes no sense. So all this is for real omega having got to this stage at this stage I say here is a function of omega real function integrate that function over all omega prime all real omega prime with the weight factor 1 over omega prime minus omega okay it makes sense as an integral for all complex values of omega but not real values because as soon as omega hits a real value there is going to be a singularity in the path of the integration along the path of integration. So it is defined as long as omega is not real there is some imaginary part. So in the omega plane this quantity defines an analytic function okay makes sense everywhere except on this so there is some kind of cut here in the omega plane no no no for every real omega there is going to be it is going to hit it right it is another matter what he says is right because ultimately phi itself has support only at these points. So even though you are integrating over all omega prime phi itself vanishes in between those delta functions if you like but you see you take a very large system with a very large number of energy levels so close to each other that is practically a continuum then practically everywhere in omega you are going to hit a singularity but now this analytic function here this master function if you like has a boundary value as you come down from above it also has a boundary value as you go down from below and there is no reason why these two should be the same at all what is happened is that chi retarded this is our retarded susceptibility chi AB of omega real real omega this quantity is real that is the physical susceptibility is equal to limit from epsilon goes to 0 from above of this guy. So instead of omega you say omega plus psi epsilon positive imaginary part and then you take the limit as you come down from above and the physical retarded susceptibility is the boundary value of that function from above this thing here defines an analytic function for all complex omega but because you cannot cross the real axis there is no guarantee that the function you get from above and the function you get from below are the same in fact they are not they are not the same. So similarly chi advanced we looked at the retarded green function but mathematically you can also talk about the advanced green function AB for omega and that is real too is the limit as epsilon goes from above K AB of omega minus I epsilon. So now you are approaching from below this guy is guaranteed to be an analytic function of omega holomorphic in the lower half plane. This follows analytic in the upper half plane the physical retarded response for real frequencies is the boundary value from above of this master function K and the other one you may want it for some applications in general you do actually there are cases when you do then it is another analytic function it comes from below here and whenever you have an integral like this you can ask what is the difference between this and that. So you really have to ask what is the discontinuity of this function as you across this cut and what would you say well the discontinuity let us do that just for fun although that is not what I am interested in right now but just as an exercise in analytic functions if you say the discontinuity K AB of omega and this is real by the way equal to limit epsilon goes to 0 from above K AB of omega plus I epsilon minus K AB of omega minus I epsilon and now the only place where this I epsilon appears is in the denominator right. So in one case you have so this thing is equal to let us write it out minus I integral minus infinity to infinity the omega prime by AB tilde of omega prime and then inside you have 1 over omega prime minus omega minus I epsilon minus 1 over omega prime minus omega plus I epsilon omega is real and omega prime of course is real as well what is this equal to this is a famous formula involving these I epsilon and so on you see if I leave out this I epsilon and on the region of integration in the omega prime plane here is the point omega in the omega prime plane if I leave out this portion of it I get the principal value and then the meaning of this omega minus I epsilon means the pole is at this point in the first term the pole is at omega prime is omega plus I epsilon in the second case the pole is out here. So as I do this having the pole here is equivalent to putting the pole on the real axis and indenting the contour from below and taking half the contribution from that pole right or equivalently close the contour or whatever and you are going to get 2 pi I times the value at this point right. So from above you are going to get principal value 1 over omega prime minus omega so 1 over omega prime minus omega minus I epsilon is equal to symbolically it is this principal value plus I pi delta plus I pi because you are going to go around clock anticlockwise direction from 0 to pi I pi delta of omega prime minus omega and if you put a plus here then it becomes a minus here. Now we want the difference of the 2 so you want P plus I pi delta minus P plus I pi delta so you get 2 pi I times the delta function. So the discontinuity is straight away equal to minus I minus infinity to infinity the omega prime pi AB tilde of omega prime times 2 pi I delta of omega prime minus omega but that integral can be done it just says replace omega prime by omega and the I cancels with minus I and you have a 2 pi. So we have a very interesting result which says that this is equal to 2 pi pi AB tilde so what you really been doing is to we have written a dispersion relation for this K not a Hilbert transform but a dispersion relation for this K with this Cauchy kernel here and the meaning of the whatever sitting up there the spectral function is that it is the discontinuity of this analytic function this is the general statement. So again even this master function it is directly related to this fellow here. A little later I will show that this fellow here the spectral function here is related to either the real or the imaginary part one of the two depending on the situation of the susceptibility itself. So what we are trying to do is to get several relations we know one relation between the susceptibility and the spectral function but you can invert this and ask what is the spectral function equal to in terms of the susceptibility will turn out to be either the real part or the imaginary part and we will see how. But for the moment this is how you show that A the physical retarded susceptibility is the boundary value from above in the omega plane of a certain analytic function of omega master function which has this spectral representation. The spectral function itself is the discontinuity of this across the real axis the retarded the advanced green function is the boundary value from below of the same master function here okay. So it actually for the price of one you solve two different kinds of problems two different boundary conditions but we are not going to get into that at the moment. Let us come back here to this and ask the following question where do these formulas what do they reduce to if you had for instance we have seen there is a temperature dependence let us see what we can do say about that. So I need to again write down expectation value of let us write this commutator down explicitly V of tau minus V of 0 A of tau sorry V of tau A of 0 in equilibrium this was equal to left hand side was the commutator was the response function and we had a representation for it and you have tell me what were the factors this is equal to i h cross I divided by h cross. So let us put that back n, m A and m B m n A to the minus beta E n E m E to the minus i omega n m tau we are writing all these representations for tau greater than 0 okay. For tau less than 0 you would have a certain symmetry property which will come to in a short while but for the moment let us keep tau positive pardon me 1 over i h cross times the commutator was the response function and that is equal to this fellow here when it is the commutator divided by i h cross that is the response function that was equal to this series here. So I just brought this across on this side what happens at absolute 0 of temperature as you go to 0 temperature what should what do you think should happen this would correspond to beta going to infinity in other words you switch off thermal fluctuations and then you should be back to quantum mechanics at 0 temperature. There is no i h bar on the right hand side yes oh both should be yeah right right right oh yeah yeah it is okay yeah because I wrote the response function as 1 over i h bar times this I multiply by that i h cross this goes here you are right absolutely dimensionally as she says it should be just a perfect okay. Now what happens to this as t goes to 0 beta goes to infinity what happens to rho equilibrium remember the density operator in the canonical ensemble was e to the minus beta h naught but normalize such that trace e to the minus beta h naught was 1 normalize to that always. So what would you say is the density operator what is the spectral representation of the density operator itself we are now assuming that the system is describable by a complete set of states in some Hilbert space so it is really not the formalism as written down here is really not the most general one because I have made a specific representation I have said that this system has a Hilbert space there are systems which where you cannot talk about them and describe them in terms of state vectors in Hilbert space there is some density matrix and that is the end of it but we have made this assumption that here we actually have a Hamiltonian system it has got a nice Hilbert space complete set of states and it is slightly perturbed from equilibrium. So what is rho equilibrium equal to as an operator as an operator it is e to the minus beta h but now I am saying let us represent that operator in terms of the ket vectors phi n in terms of basis formed by the Eigen states of h naught. So it is clear that what you must do is write e to the minus beta h divided by trace e to the minus beta h naught that is what that is what the density operator is in abstract form now trace rho equilibrium is guaranteed to be equal to 1 because I have divided by this quantity. Now let us write this out in the basis formed by the phi n so this is equal to in the denominator it is clearly equal to summation over states n states labeled by n or the collection of quantum numbers times trace so this is phi n e to the minus beta h naught phi n that is the denominator obviously. What is the numerator what is the numerator numerator has got to be an operator so if these fellow yeah if these fellows form a basis for all states in the Hilbert space there is also a basis for all operators right for instance the unit operator is just sum over n phi n ket phi n bra so every operator should be writable in that form if the operator commutes with the Hamiltonian h naught then you will have only the diagonal terms right otherwise in general if you have an abstract operator a you should be able to write this as phi n phi n and then some matrix element here what matrix element do you have here that is what you call a and m and there is a summation over n right that is what you mean by this operator okay just as when I write a 2 by 2 matrix a b c d I mean a times 1 0 0 0 plus b times etc and that is the outer product that is right so it is just the same thing over again and therefore what is this equal to it will have only these projections it will not have it will not have phi n phi m here it has got to be diagonal and then a summation over n of course and this state this projection is weighted by the corresponding energy that is of course the representation of the density matrix in the basis formed by the eigenstates of h naught. So notice these are operators this is just a number so that is what my row equilibrium is what happens to this as t goes to 0 or beta goes to infinity so let us suppose that all your n yeah so you can see yeah what does it do why should it be only the ground state but I am not assuming that the ground state energy is 0 why should I assume that it is 0 it is bounded from below so we have we have assumed here tacitly that we have a respectable system whose ground state energy is bounded from below if it is not bounded from below and goes to minus infinity then everything will sit there and take infinite energy to get it up out of there so what happens now you see notice that this thing here can be written as e to the minus beta e naught phi naught phi naught plus e to the minus beta e 1 phi 1 phi 1 plus h at time yeah divided by this fellow here and this guy they are all orthonormal so it is e to the minus beta e naught plus e to the minus beta e 1 now if e naught is less than e 1 less than e 2 less than or not which it is because it is the ground state then you pull out the factor e to the minus beta e naught all these factors are going to go to 0 as beta tends to infinity as long as e naught is greater than 0 but whatever it is you can see that you can pull out this factor it is the biggest of the lot and these fellows will have e 1 minus e naught etc which are positive quantities and therefore as beta goes to infinity they will all go away this e to the minus beta e naught will cancel against this so it is obvious that this will go to just the projection of the ground state as it should and beta goes to infinity this thing goes to just the projector of the ground state that is what is meant by saying there are 0 temperature things are in the ground state it says that all the Boltzmann factors go away and the only thing that the density matrix has left in it is the projector of the ground state there is no need to assume that e naught is 0 or anything like that e naught is just smaller than all the other ease and of course we have assumed that the system has got a spectrum bounded from below so what we have to do here in this case is precisely that go back to the calculation and this equilibrium now is just replaced by a phi naught phi naught so at t equal to 0 in the limit of t goes to 0 it goes to this they can be no reference to temperature anymore gone to 0 that is it and this is it and this quantity phi naught phi naught is normalized to 1 so that is automatically down there is gone now what you do is to insert complete set somewhere here so insert here summation n phi n phi n and ditto here insert those follows here and you are going to get the matrix element a 0 n and then a b n 0 times e to the i omega n 0 or whatever it is and you are going to get the opposite here so the first term will have a 0 n b n 0 times e to the whatever it is and then be a term which is of the form b 0 n a n 0 e to the whatever it is and the frequencies here would be just the frequencies would be omega n 0 plus and minus signs so I leave you to complete this figure out what this and just check that it goes to what you expect from ordinary quantum mechanics each time and you can write down the response function at 0 temperature alright now let us look at something more interesting which is got to do with the properties of this spectral function what I would like to do is to exploit what we have for the commutator to write out expressions for quantities which do not involve the commutator but any product of two operators arbitrary operators. So recall that we found that a of 0 b of tau minus b of tau a of 0 in equilibrium this quantity was equal to summation n, m a n m b n n times what e to the minus beta e n m times e to the minus i omega n n tau so this immediately led to the fact that phi a b tilde of omega was equal to a summation over n m a n m b n n this form there was a 1 over i h cross in the phi and then we wanted to take a Fourier transform of the response function so there was a 2 pi and then a delta function of omega minus omega n so let us take this quantity to be a known quantity and I want to write various things in terms of this various spectral representations of various time dependent quantities given this. So what should I do the 1st thing to do is to write this as equal to 2 pi over i h cross summation n, m a n m b n n e to the minus beta e n times 1 minus e to the minus beta e to the beta h cross omega n m because that puts a plus e to the beta e n minus e to the minus beta e n so that is okay times the delta of omega minus omega n now since I am always going to integrate over omega in this I have a delta function here which phi is only when omega is equal to omega n m so I can replace this omega n m by omega itself. So I end up with phi a b tilde of omega divided by and then I pull it out of the bracket to the left hand side this guy here is equal to 2 pi over h cross summation n, m a n m b m n e to the minus beta e n so what I have got is the 1st part of this fellow that is this so now I can claim pardon me the n i factor 2 pi over i h cross yeah now I want to be a little careful with the algebra here so I removed this portion if I now take its Fourier transform so I do integral from minus infinity to infinity d omega 1 over 2 pi this guy e to the power minus i omega tau phi a b tilde of omega divided by 1 minus e to the minus beta h cross omega this is equal to 1 over i h cross summation n, m a n m b m n e to the minus beta e n an integral over omega times e to the minus i omega tau times that so e to the minus but what is that equal to apart from that i h cross factor it is equal to the 1st term here came from a of 0 b of tau so a of 0 b of tau divided by i h cross is 1 over i h cross this garbage without that it is just this fellow right so it is equal to i h cross over 2 pi times this guy so it tells you that a of 0 b of tau alone equilibrium no commutator alone is equal to 1 over 2 pi i h cross over 2 pi d omega e to the minus i omega tau phi a b tilde of omega divided by 1 minus e to the beta h cross omega by sleight of hand what we have discovered I can do this much more laboriously but what we have discovered is that while we have a nice respectful representation for the commutator we can do it for each term in the commutator except that the factor the extra factor that comes is this so it is the Fourier transform not of this guy but of that divided by this factor here I am sure you can I am sure you can but I think it is a lot more laborious yeah because you know discovering this factor is a little more tricky right so what I did was I exploited the fact that inside here this thing is a function of n and m but I pull this fellow out the Boltsman factor and then I replace this by the one with omega and pulled it out of the summation altogether I think it is a shortcut now similarly for the other fellow you put the e to the minus beta in here you need this term again you pull this out but you have to kill this term so to kill this and produce em you need to multiply by e to the beta h cross omega right so it is clear that b of tau a of 0 equilibrium is ih cross over 2 pi and integral minus infinity d omega e to the minus i omega tau e to the beta h cross omega divided by 1 minus so that when you subtract the 2 this factor cancels out from top and bottom and you will be back to this guy you will be back to the representation for pi itself so we have spectral representations for both these quantities therefore we have one for the anti-commutator okay so let us see what that does so that says oh it should have erased this so it says that the anti-commutator a of 0 b of tau let me call the anti-commutator plus with a plus here I do not want to use a curly bracket as sometimes as is done sometimes because it is the Poisson bracket we use that for the Poisson bracket so this fellow in equilibrium is that plus this so is equal to i h cross by 2 pi integral minus infinity to infinity d omega e to the minus i omega tau and then 1 plus e to the beta h cross omega over 1 minus e to the beta h cross omega times the spectral function but I can pull this one out e to the half beta h cross omega then you get 2 cross and I pull the same thing out to get minus 2 cm right so this is equal to minus 2 scancel pi a b tilde of cot hyperbolic beta h cross omega over 2 so the anti-symmetric part the a of 0 b of tau the anti the commutator had representation without this guy but the commutator anti-commutator the symmetric part of the product has this representation now there is a very neat way of we will interpret these things this was just a little piece of algebra but we will interpret these things carefully does this remind you of anything does that remind you of any particular famous quantity well even simpler than that if you have a harmonic oscillator so the energy levels are n plus half h cross omega right let us ask what is the average energy of a harmonic oscillator quantum oscillator with natural frequency omega at temperature beta inverse right so let me call this average value of h oscillator is equal to h cross omega times n plus half we have to sum it is not degenerate this thing is not degenerate so you have a summation n plus half e to the minus beta h cross omega into n plus half divided by the same thing n equal to 0 to infinity now look at the e to the minus half is going to go away numerator and denominator we do not have to worry about that so this goes away and then you have n e to this fellow divided by just this fellow this is a geometric series yeah or d over d beta of minus d over d beta of this guy right so whatever way it is you can see that with an n up here you are going to get the sine hyperbolic and without it you are going to get cos hyperbolic so you are going to get this cot hyperbolic once again so the average energy of a harmonic oscillator let us call it e beta h oscillator at temperature t let me call that equal to e beta and let me put it as a function of omega for a given natural frequency of me this fellow is equal to well what is going to be the average energy at t equal to 0 half h cross omega it is a ground state energy so that is got to come out it has got to be that right what happens as beta goes to infinity to cot hyperbolic goes to 1 as beta tends to plus infinity so at absolute 0 it is h cross omega that is it okay what happens at t equal to z infinity what happens to this guy what happens to cot hyperbolic beta h cross as beta goes to 0 it diverges it diverges like what so cot hyperbolic x what does it diverge like as x goes to 0 what should it diverge like it is sine hyperbolic over cos cos has only even powers 1 plus x squared etcetera over 2 so only the sine hyperbolic is relevant what does sine hyperbolic x 2 as x goes to 0 what a sine x 2 as x goes to 0 goes to 0 like what like x goes to 0 like x so cot hyperbolic also goes to 0 like x exactly like x so cot hyperbolic therefore x goes like 1 over x so this leading term is going to be h cross omega over 2 1 over x is 2 over beta h cross omega equal to k Boltzmann t right now classically what is the average energy of an oscillator at temperature t half k t because of the kinetic energy half k t because of the potential energy so it is k t this is the du Long Petit limit or whatever classical limit so in good shape it matches so this is in fact the formula and therefore you can write this thing now this fellow down in terms of this e beta you can write this cot hyperbolic as 2 over h cross omega so this h cross goes away here and you are left with this is 1 over 2 pi i 2 goes away sometimes this is called the fluctuation dissipation here we will get back to this so this thing here is just shorthand for this thing here there is no oscillator we are talking about but it is convenient it is very convenient to use that special the crucial point is we now have spectral representations for the product of 2 operators at different time arguments all of it in equilibrium by everything is with respect to the equilibrium ensemble and from this we are now going to start extracting some physics so I will stop here now. .