 So, today we will just discuss this quiz because I have taken a cursory look through all your answer books and I have not yet marked them, I have not graded them but I thought I would do this so that you get some idea of what this whole thing is all about and in the process we will be able to all make some side remarks, some peripheral remarks as well. So, we will go through this very slowly. The first question said the density operator of an isolated system in thermal equilibrium is delta of H minus E. So, in the micro canonical ensemble or an isolated system in thermal equilibrium the idea was rho equilibrium was delta of E minus H and the question is this true or false where E is the total energy of the system. Since it is an isolated system the energy is a constant, E is a constant and certainly you are on the energy shell which is what this thing is all about with some finite resolution. So, this is the density operator at least formally, formally it is so. Interesting question side remark that one can make is we know in a classical system which has got many constants of the motion for instance if this is a system with central forces present and no external force on a system of particles the total angular momentum of the system is constant. The total linear momentum of the system is also constant. So, strictly speaking you have access only to those micro states where not only is this true not only is the total energy fixed but also this is constant a constant of the motion and this quantity is also a constant of the motion. These are also called Galilean constants of the motion. So, the question you should ask is why do we focus on just the energy? Why should not we say that it is the intersection of the energy surface with the surface on which this is constant and that is constant. So, really it is only those micro states that are accessible and not all micro states on the energy shell. Do you see the point here? What would you say is the answer and yet if you look at every book on statistical mechanics in the micro canonical ensemble it is just the energy shell that people worry about. How come they are not worried about the other constants of motion? So, variable right I mean you are doing your why not I mean we know let me draw it schematically. So, you have this big phase space and on this phase space if you fix an E there is a sub space of it on this E is constant on this surface on this hyper surface. But you must also look at the intersection of this with the surface on which capital P is constant capital M L is constant and so on but you never do that you only talk about the energy hyper surface. So, what do you think is a reason for that? There are not too many such constants of the motion basically there are just 10 Galilean constants of the motion for an arbitrary set of particles interacting by two body central forces unless there are special symmetries in the problem. The point is the answer is that when you do statistical mechanics you are talking about a system with a given state of motion. So, you are already saying the state of motion is given to you such as capital P equal to 0 for instance you are in a state where in a system or in a frame where the total momentum is 0. So, the state of motion is already prescribed to you implicitly or tacitly which is why you do not worry about the other constants of the motion. All that you worry about is the total energy because you can increase or decrease the energy by opening up the system and letting it interact with the environment. So this is true but you should also be aware that the conditions under which this is so. The next question was each Cartesian component of the velocity of a Brownian particle obeys the famous Langevin equation in one dimension which is V dot plus gamma V equal to 1 over m 8 of t where this 8 of t is a 0 mean stationary Gaussian white noise the usual white noise we have been dealing with although I have not said so explicitly it is also a Markov process we have not talked about that in detail at the moment but you have a stationary Gaussian delta correlated Markov process that is what this noise is and the proposition is that the velocity V of t is a stationary Gaussian random process is this true or false it is true. The Gaussianity property is transmitted it has driving V so this is the driven this is the driving force and this is the driven variable the output variable and the Gaussian nature of the probability distributions of eta goes over into those of V as well. As for stationarity this one we saw is stationary it is very much stationary if you did not have the noise then you had a runaway solution for the mean square velocity and then that was not stationary at all so this stationarity is certainly true the presence of this ensures that you do have this stationarity. If you did not have noise but you just had a process for instance in an abstract way you have a process Xi dot equal to eta of t so I give you a random process Xi whose derivative is Gaussian white noise or Xi is the integral of Gaussian white noise it is called the Wiener process and the Brownian particle satisfies such an equation here. This process is not stationary because what happens now is that Xi square Xi of t minus Xi of 0 whole square this average given this thing here is proportional to t whereas we know that if you must have a process that is stationary we know that it is mean mean square mean cubed all moments and so on must be at least independent of time moreover the 2 point correlation must be a function only of the time difference the 3 point correlation must be a function of 2 time differences and so on that is violated if you did not have this friction term so this introduces regularity into the system it by the way makes the V process also Gaussian makes it also stationary process as well as a Markov process. Continuation of that is that V of 0 V of t correlation is a decaying exponential function of t we know that is true because we know that V of 0 V of t this quantity we discovered was k Boltzmann t over n that is the t equal to 0 value and it is multiplied by an exponential gamma modulus t so the correlation drops off on both sides so indeed it is a decaying function exponential function of mod t continued the displacement of the particle X of t minus X of 0 is a stationary random process so the proposition is that this quantity this random variable integral from 0 to t dt prime V of t prime this is said supposed to be a stationary random process or more generally if you wrote t naught here this is t naught this is a stationary random process but surely this thing is a function of both t naught and t there is no guarantee that is a function of t minus t naught at all definitely not and it is in general not a stationary process in fact it is the integral of the velocities the displacement and we know it linearly increases with time so it is definitely not a stationary random process. Next again for all positive t the mean square displacement is given by X square is 2 dt so the proposition is that in this case X of t minus X of 0 whole square the question is is this equal to 2 dt 2 or false it is false it is only 2 in the diffusion limit right large how large greater than gamma inverse the time scale in the problem we know that the exact result here is this so if I call that capital X X of t X square of t we know this is of the form k Boltzmann t over m gamma twice and inside you have m gamma squared gamma t minus 1 plus e to the minus this is for t greater than equal to t. So only for gamma t much much bigger than 1 does it go to the diffusion limit that is the diffusion regime so that was straight forward enough. Next we have the velocity of a Brownian particle of mass m and charge q in an applied uniform magnetic field the Langevain equation is given and the statement is the different Cartesian components of v are also uncorrelated to each other for t greater than equal to 0 is this true or false it is false because they get tied up with each other due to the magnetic field which mixes up components in fact we computed this you notice that the v should have been a bold face in the dot so the equation is v dot plus gamma v this quantity here was equal to the cyclotron frequency v cross the direction of the magnetic field plus 1 over m a vector valued noise a vector valued noise white noise in this case we know what the velocity correlation does we know that v i of 0 v j of t by the way the fact that it is in the magnetic field does not change the stationarity property the velocity is still a stationary random process in fact it is still diffusive just with the modified diffusion constant so this quantity is k B t over m that will appear and then e to the minus gamma t for t greater than 0 and then if you recall there is this portion which comes from this part comes from solving the problem in the magnetic plus or minus I do not remember I think minus epsilon i j k n k sin whatever it is it is not proportional to delta i j so it is clear that there are cross correlations here entirely because of the magnetic field then for the same problem in the magnetic field the mean square displacement is a formula given for it and the statement is whether the displacement r of t minus r of 0 whole square this is just the dot product with itself whether this thing here is 2 k Boltzmann t over m gamma times phi gamma square plus omega c square over gamma square plus the question is whether this is true or false recall that what is happening here this is x square plus y square plus a square if you like if you feel this along the z direction then the x and y portions get inhibited they get attenuated by the z displacement remains as before so this still remains 2 k Boltzmann t over m gamma gamma but inside you have times t and then you have the same thing and then 1 plus gamma square over gamma square plus omega c square from the displacement in the y direction or x direction plus 1 in the y direction so that gives you this in the diffusion limit once again in the diffusion limit so all that is happened is that the diffusion has been inhibited somewhat if you did not have magnetic field omega c is 0 then this cancels out and gives you 3 3 2s are 6 and you indeed have r square goes like 6 d t because 2 d t is a portion between x square y square and z square on the other hand when gamma is small or negligible compared to omega c and this becomes very very large then you still have diffusion but you would not have any motion in the x y direction at all that means where this displacement will actually tend to 0 and it will diverge only along the z direction and therefore it goes to 2 k t over m gamma t so you should check this you should check these limits always just to make sure that you get the right answer the correct physical answer again continuing the same problem given that the initial velocity is v 0 the statement is the conditional average velocity v of t conditioned upon the initial condition v 0 is a vector processing around the direction of the magnetic field with angular frequency omega c cyclotron frequency while its magnitude decreases with time like e to the minus gamma t true or false this is true the statement is true you solve the problem exactly this is precisely what you would expect and what you get so the mean value v of t we want the mean value of this thing but suppose you did not have any noise at all you did not have any thermal noise at all just had a magnetic field switched on and then you have a magnetic field in this direction n and then initial velocity of this particle v 0 like this now what is this going to do the magnetic field does not give any energy to this particle all it does is to change the direction of v 0 so it starts processing around and what is the formula for it it is a rotation formula rotates by the amount omega c t in time t so this is equal to clearly it must be equal to e to the m omega c t on v 0 if you did not have any friction before you took any average this is all that would happen this is the rotation matrix remember that m i j in three dimensions this is actually once you write the generator of rotations in three dimensions so what has been written is j dot n i times j dot n is equal to this precisely what it is in three dimensions I have used here for the three generators of rotations you have used those three by three generators the representation the original representation of rotations by 3 by 3 orthogonal matrices when you do that you end up with this answer here this is the matrix and then the angle through which you rotate is omega c t you exponentiate it you end up with the finite rotation so it is this fellow here you have to exponentiate this on and this is a function of course of t as well as omega naught and since we know that in this case we know that if you square if you cube this matrix you get minus the same matrix we have already seen that so this is equal to i plus m sin omega c t plus m squared 1 minus cos omega c t this guy acting on v naught and you can write this down explicitly so what is it going to be proportional to there will be a portion along n there will be a portion along v naught and a portion proportional to v naught cross n the portion along the original vector itself along v naught v v naught itself the portion along n would involve v naught dot n times unit vector n and then the perpendicular portion will be v naught cross n with a sin omega c t multiplying so this is some finite some vector which you can compute if you took this average out here after putting in the noise what happens what happens to this guy all that happens is as it is going around it is being hit buffeted and it is making the average go to 0 so all that happens is this is equal to e to the minus gamma t times that that is the only effect of the magnetic field so this is all that happens you still have a diffusion in the velocity space you still have a mean which is going to 0 a variance which is increasing from the initial delta function it increases till it hits the equilibrium value which is given by the Maxwell distribution finally so we can even write the distribution down we can even write down what is p of v comma t given v naught this follow apart from some normalization factor which we will write in a minute has got to be so the following it is got to be the exponential of minus it is still a Gaussian retains that property so it is v minus e to the m omega c t v naught e to the minus gamma t that is the mean value v bar of t squared divided by twice the variance so it is twice k Boltzmann t one minus e to the minus 2 gamma t and it is k t over m so this is m out here the normalization does not care about what this average is because you are going to shift this out anyway when you do the integral so it will just be m over 2 pi k Boltzmann t 1 minus e to the minus 2 gamma t this to the power 3 halves for each dimension you get this to the power half that is it you have not derived this I have just said that it is a Gaussian and once I assert it is a Gaussian then if you accept that it is completely determined by its mean and its variance that is a normalized distribution this is the three dimensional analog of the Onstein-Ohlenbeck distribution but with a magnetic field present so physically it is very appealing it is very simple I will at some stage try to derive it is called the Fokker-Plank equation we will write it down we will write this what we will do is what I will do is by a shortcut I will show that every time you have a stochastic differential equation for the random variable of the Langevin type you have an equation a differential equation for the probability distribution directly so one can solve that in principle and get this result moving on next one was just a definition says consider a classical Hamiltonian system with n degrees of freedom and a given Hamiltonian and L is the Liouville operator statement if F is a function of the dynamical variable such that the Poisson bracket of F with H is 0 then the Liouville operator acting on F must give you 0 identically so this is obvious because you are already given that this is true but we know that L acting on F is equal to I times H with F so that is of course equal to 0 identity it is just the definition of the Liouville operator. Now what is the reason for using the Liouville operator our reason was that the formalism looks a little simpler both classical and quantum it looks exactly the same but of course you need to exponentiate this Liouville operator and that is not a trivial matter in general the classical case much harder as we saw all the proofs I gave were all for the quantum case because I was using the cyclic invariance of the trace but if you try to do this with phase space functions then we run into we have to integrate by parts it is a little messier to do so oh yes it says here the next proposition for a dynamical system with Hamiltonian H of q, p that time development operator e to the iLt is unitary whether the system is classical or quantum it is true is a unitary evolution remember that in classical mechanics in Hamiltonian mechanics if you solve the Hamilton's equations of motion so you solve Hamilton's equations of motion the given Hamiltonian and given initial conditions so what you are doing is to go from q naught p naught at any time t you solve the Hamilton equations of motion then along the phase trajectory you are really going to qt p right I put subscripts but really this should be inside the bracket for this guy and for each degree of freedom this is true now we know this is true q naught p naught equal to 1 to start with these are canonically conjugate variables and we know that at any time qt pt is also true equal to 1 so time evolution preserves the Poisson bracket it is also orientation preserving because this transformation it turns out is what is called a symplectic transformation a canonical transformation and it preserves orientation that is it does not change right handed to left handed coordinates and so on in phase space so it is in fact a transformation whose which is measure preserving which is canonical etc etc I leave you to find out what is the generating function for this transformation what do you think is a generating function every canonical transformation has a generating function not necessarily unique generating function for a given canonical transformation you may have more than one generating function what do you think is the Hamiltonian itself so you have to establish this you have to establish this and then you discover of course that the equations of motion themselves are the equations which tell you how to get qt and pt from the generating function they are exactly the same in structure we cannot be standard gen canonical transformation the simplest one which is non trivial is as you know the one that takes you q to p let us write it the other way is the one that says q equal to p and p equal to minus p so the Poisson bracket of this with that is minus the Poisson bracket of this is one the determinant of this transformation is one so it is measure preserving and it is canonical the generating function for this depends on whether you want a type 1 type 2 type 3 type 4 generating function are you familiar with the types of generating functions what type would this transformation be if I say it is a type 1 then I must have a generating function f 1 which is a function of little q and capital Q I leave you to check that if you choose this generating function you are done it generates this transformation but you could also choose it as type 4 and it is a simple matter to verify that that is also generate that also generates precisely this transformation so the type of a canonical transformation is not uniquely specified for the transformation you may have more than one generating function for the same transformation now the reason I said this is a standard the basic canonical transformation first of all it is linear so the Jacobian is all constants and things like that the matrix of this the Jacobian matrix is precisely this which is the matrix that you need when you write the Hamilton equations down in the form of simple gradient or something like that the symplectic the symplectic matrix itself so if m is the Jacobian of any canonical transformation then m times a matrix omega times m transpose is equal to omega where omega is the this matrix is 2 n by 2 n matrix with 0 identity minus identity 0 and the statement is Suresh's statement is that omega itself m equal to omega itself will generate we will satisfy this condition now the reason why this is regarded as a fundamental canonical transformation is that there is a very deep theorem due to Darbu which says that anywhere in phase space in an n diamond degree of freedom systems phase space anywhere locally you can find a canonical transformation which looks exactly locally in any phase space any Hamiltonian system so that is the significance the fact that there exist this is the same statement as saying Suresh would say it means that there is a symplectic structure to the phase space this fellow acts like a metric and such a transformation is possible so anyway this is big machinery of canonical transformations. In quantum mechanics however you talk about unitary transformations you know that when you have a time independent Hamiltonian then the time development operator u of t for a given t naught this guy here this time development operator is such that it when it acts on any observable this gives you e to the i h naught t over h cross e to the i minus i h t so this time development operator is equal to this guy here and h is Hermitian so this is unity automatically so the whole idea of writing the Louisville way of writing things is that you write you identify this u of t with e to the i l t whose action on any observable is to produce a conjugation with respect to this operator here. So if you like in the quantum case this l is a super operator it is acting on operators on Hilbert space and it is producing a thing like this again you can show that in the space of these operators on which l acts as a super operator this is a Hermitian operator that is what we just we demonstrated by defining an inner product in the space of the operators and so on. Classically there is no such extra structure here this is in just as a function it acts on functions of a space but this is a partial differential operator and it involves exponentiation so formally writing both classical and quantum mechanics down as unitary time evolution is a very simple and compact way of dealing with matters at least in the formal sense so that was the purpose of the exercise there. Then the next one was the response function a of 0 b of tau bracket either Poisson or a Hermitian bracket where a and b are physical observables can never be negative for any tau greater than equal to 0. So the question asked is whether this thing can be negative so we are asked to find out if a of 0 b of tau equilibrium and recall that this thing here x y this stands for either x comma y or x y over I h cross in quantum mechanics this is classical this is quantum mechanics and the statement asked was whether with respect to the equilibrium density operator e to the minus beta h naught whether this quantity can be negative or not for tau greater than 0 c can it be negative we have seen that it is actually real if a and b are Hermitian which is the case we are dealing with it is actually real but does it have to be negative or positive definite there is no such constraint that it should be positive definite it very very a common instance would be if you plot it as a function of tau if you plot it phi of tau for some arbitrary a and b it could go like this it is sort of physically expected it will die down as tau becomes large but it could oscillate to change in sign but yeah you can change the sign of a instead of a you look at minus a so that immediately says so it is no no such requirement on it at all in fact in the magnetic field case you know that there are these signs and cosines and they will of course cross the axis and go down next the Kubo canonical correlation a dot of 0 b of tau semicolon reduces to the classical product in the in the classical limit and the answer is yes of course it does so we call what we had we had a of a dot of 0 b of tau equilibrium if you recall this was an integral from 0 to beta d lambda and then a trace row equilibrium which was e to the minus beta H0 this fellow here times e to the lambda H0 a dot of 0 e to the minus lambda H0 b of tau the trace of this whole product of course in the classical case things commute this goes across here cancels the integral just gives you a factor beta and you have the expectation value with respect to row equilibrium of a dot of 0 b of tau so classically that semicolon is just deleted finally the generalized susceptibility is the Fourier transform of the response function with respect to the time may be true or false it is false what is it the Fourier transform of green function so it is the Fourier transform of theta of t times phi a b of t that is certainly true but just the introduction of this theta makes a huge difference at once in fact there are immediately relations between as we will see the spectral function and this and let me write all those down so chi let us forget about the a b chi of omega is integral 0 to infinity dt e to the i omega t phi on the other hand this fellow here is also equal to integral minus infinity to infinity dt e to the i omega t theta of t this is the green function and that is the Fourier transform the Fourier transform of phi itself is phi tilde of omega this fellow is integral from minus infinity to infinity d t e to the minus d to the i omega of t phi of t in this fashion compare these two there is an integral going all the way from minus infinity to 0 plus 0 to infinity here and that is missing here so this was our spectral function for which we wrote down representations and so on. Now there is a relation between these two we already know it we already know that phi we know that phi of omega equal to an integral from minus infinity to infinity d omega prime phi tilde of omega prime over omega prime minus omega minus i epsilon in the limit as epsilon goes to 0 from above and there is some factor here i over 2 pi or something like that i minus i over 2 pi there is no now look at this thing here I can write it as cos plus i sign definitely so this could definitely be written as minus infinity to infinity dt e to the cos omega t if phi of t has a definite parity if it is an even or odd function of t then one of the other of these two fellows vanishes on the other hand this fellow here if I rewrote this this is equal to integral 0 to infinity dt cos omega t plus i sign omega t phi of t this is the real part of chi and that is the imaginary part of chi after you do the integral you get the real and imaginary part if phi of t is an even function this portion vanishes and this remaining integral becomes twice the real twice the integral from 0 to infinity right so if phi of t equal to phi of minus t will imply immediately that chi of omega is equal to phi tilde of omega equal to twice chi of real part of similarly if this one odd function of t then this integral vanishes here and what is left is i times this fellow here which is twice 2 i times 0 to infinity but that is equal to i times the imaginary part right so phi of t equal to phi of minus t with a minus sign will imply that phi tilde of omega equal to 2 i imaginary part right so there is another intimate relation between these two guys now I leave you to put take each of these cases put it in here and look at what is going to happen so it is starting to look as if chi is given by take the real case for first take this even case first it looks like chi is given by an integral over itself over the real part alone what is going on that looks really weird right you are telling me the full susceptibility is the integral of the real part alone this fellow but it is got to be completely consistent we have not done anything wrong anywhere remember that this fellow 1 over this guy is the principal value plus i pi delta so if you put that in an equate real and imaginary parts you will end up with a statement that the real part of chi is equal to the real part of chi and the imaginary part will be an integral over the real portion here which will give you the dispersion relation this side is in general complex so the way to resolve the right hand side into real and complex parts assuming this is real which by the way is true here immediately is simply to resolve this Cauchy kernel with this guy as principal value plus i pi delta so it will give an identity for the real part will say real part of chi is equal to the real part of chi identity but it will also say the real part of chi is the principal value integral of the imaginary part which is the dispersion relation and the same will happen in this case but the question is how do I decide this that is decided by time reversal invariance because you see finally what you have is pi a b of t equal to a dot of 0 b of t in equilibrium so now we are ready to talk about the time reversal property we have computed this for various cases for t positive but now how do you find what happens for t negative it depends on the time reversal properties of these kinds so this is equal to epsilon a dot epsilon b pi of t this is this will imply that pi a b of minus t is equal to this where epsilon a dot is either plus 1 or minus 1 depending on whether a dot is invariant under time reversal or changes sign detour for this but the time reversal suppose a is a displacement x it has nothing to do with t so it is time reversal properties it remains x goes to x under t goes to minus t but a dot is then velocity that is dx over dt and that changes sign so in all cases epsilon a dot can be simplified a bit and we could write this as pi a b of t equal to minus without the dot because a and a dot are guaranteed to have opposite time reversal property so this is always this product with this thing here is either plus 1 or minus 1 and you can tell which one is which for example I apply a displacement to the Hamiltonian proportional to x the displacement the coordinate that is a is equal to x in that case a dot is therefore v and if I am measuring v as I do in the Laun Jouin problem then the correlation a dot b is essentially the velocity-velocity correlation and you are guaranteed from this from this fellow here that in that case you are going to get a symmetric correlation which is what happened we found v of 0 v of t is e to the minus gamma mod t so in all cases the time reversal property of this function of this of these operators or dynamical variables will tell you whether it is an even or odd function and therefore whether this is true or whether that is true so this is something which I had meant omitted to mention but we have it right here. Next came all the fill in the blanks and the first question was of course the internal energy in terms of the partition function sorry I forgot to write trace there so we know the average value is minus delta over delta beta log z that is trivial then the second portion was the second question was this probability distribution for the conditional velocity this is in fact the Fokker-Planck equation for the velocity conditional density and now there is a little subtlety here so let me point this out so you have delta p over delta t equal to delta gamma times delta over delta v v p plus gamma k Boltzmann t over m d 2 p over delta v 2 by the way not surprisingly you would have guessed that the fluctuation dissipation theorem has been put in here so I get the Maxwellian distribution asymptotically now in equilibrium p goes to p equilibrium this portion is 0 by definition p equilibrium has no time dependence then you have an ordinary differential equation but its second order but notice there is a d over dv that can be pulled out so this will immediately imply that d over dv of k t over m dp equilibrium over dv plus v p equilibrium equal to 0 that is the second order differential equation ordinary differential equation what can you conclude from this that this must be equal to a constant it must be true at every value of v including plus minus infinity but at plus minus infinity each of these terms is 0 p equilibrium vanishes therefore the constant is 0 therefore it is 0 for all because it is a constant so this goes to 0 and that is a first order equation and as you can see the Maxwellian distribution is a solution move this constant there and then you can integrate by separating variables all right the next one was just the calculation of d using the integral you need to use the fact that d is integral v of 0 v of t for any Cartesian component it is a trivial integral and the one with the pole at 0 this I had already mentioned earlier in class so you just have to add this lambda over omega your dispersion relation then the last part and let me say this is something which presumed that you know already what happens in an electromagnetic field is that the Hamiltonian which is p squared over 2m for a free particle goes when you have a magnetic field to p minus ea over m ea squared over 2m plus of course if there is a scalar potential you have this but in a constant magnetic field you can choose that to be 0 always so that is your Hamiltonian and therefore to first order in the vector potential this is equal to p squared over 2m minus e over m p dot e classical but if you have a constant magnetic field you can write a simple representation in the Coulomb gauge for the magnetic field it is half b cross r the fact that b changes with time does not affect this it is got to be uniform magnetic field so this thing becomes and of course by the cyclic property of this triple product you can write this as b dot r cross p or you can write it as r cross p but that is the angular moment does this remind you of something this is the gyromagnetic ratio that is the angular momentum so what is this equal to it is the magnetic moment formed by the current loop right that is all that is how the particle will respond it has no spin so there is no intrinsic magnetic moment the only thing it can do is it moves in a circle creates the current and this couples to the external field that is exactly what you expect mu dot p minus mu dot p where mu is the magnetic moment is the potential energy in the presence of a magnetic field so that is precisely what would this be correct quantum mechanically where would I have gone wrong pardon me yeah so I cannot write p minus e a the whole squared the cross term would be p dot a plus a dot p I have to be careful about that then you would get pretty much the same thing the coulomb gauge it does not matter yeah exactly so that is a further thing that it really does not matter in the coulomb gauge so I leave you to verify to check that out but you get expected result.