 I recall that we were looking at the case of one dimensional motion in a non-linear oscillator situation where the Hamiltonian function of q and p was something of the form p squared over 2 plus q to the 2r over 2r where r is a positive integer and the case r equal to 1 corresponded to the simple harmonic oscillator. We had computed what the action was for motion in this potential and we discovered that the action which was an integral over p dq for bounded motion oscillatory motion in this potential was proportional to a certain power of the energy of the oscillator itself which is e to the power r plus 1 over 2r this is the result we obtained. Of course if you translate this back to action angle variables the way we had defined it earlier this implies that k the Hamiltonian as a function of the action is proportional to the action to the power 2r over r plus 1 which immediately implies that the frequency omega of motion which is defined as ?k over ?i is proportional to i to the power r minus 1 over r plus 1 if you therefore define a degree of nonlinearity alpha say and that is defined as ?log i over ?log omega then this is equal to r minus 1 over r plus 1 and notice that r equal to 1 which is the simple harmonic oscillator alpha is 0 no nonlinearity and as r becomes larger and larger the nonlinearity tends as r tends to infinity to the value of unity. So it is a very useful indicator of the degree of nonlinearity if you like of an oscillator of this kind not a universal measure of any by any means not a universal measure or anything like that but a very useful one in many contexts. We also saw that the semi classical quantization of this system is immediate it follows at once because if you recall in semi classical quantum mechanics semi classical quantization corresponds to writing i which is integral pdq equal to n times Planck's constant where this is an integer and if you take that along with this this at once implies that n is proportional to the energy level e sub n to the power r plus 1 by 2r or en is proportional to n to the power 2r over r plus 1. So tells you something about the level spacing for quantized motion in this potential is it the other way about but ah yes of course this is log omega over log i that is this degree of nonlinearity the way the frequency changes as a function of the action thank you. So semi classically we find that the energy level the nth energy level is dependent on the quantum number n in this one dimensional problem according to this relation here valid for n much much bigger than unity which is where the semi classical rule is valid and again you notice that when r is equal to 1 so here r equal to 1 which is simple harmonic oscillations we know that en is proportional to n itself that is a level spacing which is equi spaced equally spaced energy levels and this is exactly what the harmonic oscillator does in quantum mechanics of course the exact relation for en is n plus half times h cross omega and the half arises from so called zero point motion it is the ground represents the ground state energy of the oscillator but we are not going to get into quantum mechanics here just to point out that this semi classical argument is immediately immediately leads to this result that en is proportional to this power of n for this whole family of potentials notice also that as r tends to infinity we end up with en going like n squared and that's precisely the level spacing for a potential which rises more and more steeply infinitely steeply namely a particle in a box so this is exactly the same as the level spacing for a particle in a box in a one dimensional box so that limit to is correctly obtained from this semi classical formula one can go further and actually try to find the correction to this n corrections to other this quantity here etc but we are not going to get into that right now so so much for a little digression on semi classical quantization which follows from the arguments we've been giving here now let's go back a few steps and ask what's the reason for having integrability in a Hamiltonian system so we go back to n degrees of freedom n freedom integrable Hamiltonian and ask what's the underlying physical reason for the existence of n constants of the motion f1 through fn in involution with each other yes yes yes the way we've talked about it so far is a Hamiltonian system the phase space variables come in pairs so we identify n degrees of freedom and independent degrees of freedom which means that you must specify n numbers q1 through qn for me to tell me completely the configuration of the system in real space like and in addition you need n conjugate momenta to complete the description of the system in phase space in other words to describe the dynamics of the system so I call number of degrees of freedom the same as the number of generalized coordinates that I have number of independent degrees of freedom in the case of more general dynamical systems first of all the phase space doesn't have to be even dimensional this pair wise structure the Poisson bracket structure is not necessary at all and I don't distinguish between different kinds of variables I just call the whole set dynamical variables 1 to n or as many as there are yes yes yes absolutely yes absolutely it's the same it's the same as the degrees of freedom if I give you for instance in statistical mechanics when you discuss monatomic gases diatomic gases and so on for a diatomic gas you ask how many degrees of freedom does a diatomic molecule have yes yes ah the question is whether in chaotic motion the concept of degrees of freedom appears or not of course it does it has absolutely nothing to do with a kind of motion identification of the number of degrees of freedom yes I believe so I believe so unless there are other reasons to believe that external forces are present or there is a time dependent perturbation acting on the system or anything like that but if I give you a collection of molecules and you assume Newtonian mechanics to hold good and the system is isolated they act with forces acting upon each other caused by themselves then I don't see why it's not a Hamiltonian system why do you say I can't formulate the system I assume let's assume put a model on it let's assume that Newton's equations are valid let's assume that there is a certain potential energy between two molecules a certain distance apart once I have a model of that kind I have a Hamiltonian system with a very large number of degrees of freedom no doubt the motion is in general chaotic it's very irregular it's not integrable yes it's assumed to be Hamilton yes indeed of course in real gases you have many other complications for instance you might have to bring in quantum mechanics you might have to solve the entire problem quantum mechanically that's a separate subject in itself you're not going to get into that here but otherwise yes it's a Hamiltonian system the kind of motion that a system has whether it's regular or integrable or ergodic or chaotic this has nothing to do with the identification of the number of degrees of freedom absolutely nothing to do with it so let's look at an end freedom Hamiltonian system for which the Louisville-Arnold criterion tells us that the system is integrable if you have n constants of the motion f1 to fn in involution and then we saw that a transformation to action angle variables is possible and once you make this transformation the Hamiltonian becomes independent of the angle variables it's a function of the action variables alone and then you can integrate the entire 2n set of equations that you have all 2n equations for prescribed initial conditions at least in principle you can do this now you could ask what's the physical reason why this system is integrable why these systems are integrable what's the physical significance of these f1 through fn of course the action variables which we talked about which lead to the natural frequencies of the system the omega i are certain combinations of these f's so in that sense there's already some physical interpretation for these f's but can we think of this in a slightly more physical fashion is there something much more immediate and the answer is yes the existence of these constants of the motion is related to some hidden symmetry in the problem a certain dynamical symmetry in the problem for instance if I looked at two harmonic oscillators and the spring constants are the same in the two perpendicular directions then you would immediately tell me that the Hamiltonian is rotationally invariant you make any rotation in the xy plane and the Hamiltonian doesn't change at all so there's a certain symmetry in the problem and the existence of symmetry is linked to integrability whenever a system is integrable whenever you have these constants of the motion there exists a certain dynamical symmetry in the problem now let's try to understand what this dynamical symmetry is please go a little bit into this it's a vast subject by itself but let me at least give the rudiments of the subject let's go back and ask what is a canonical transformation actually do to a system so we start with a system with variables Q P and you make a canonical transformation to a new set of variables capital Q and P what does this do what kind of transformation is it or if you like if these Q's and P's are combined into a phase space variable x 2n dimensional vector then these could similarly be combined into a 2n dimensional vector xi these quantities are functions of these quantities which preserve the Poisson bracket structure in other words we know that Q i Q j is equal to P i P j is equal to 0 and we know that Q i P j is equal to delta i j now what's meant by a statement like this what's meant for example by this statement here what I mean by this is this implies if I write out this Poisson bracket explicitly it implies that the summation from k equal to 1 to n over all degrees of freedom delta Q i over delta Q k delta P i P j over delta P k minus delta Q i over delta P k delta P j over delta Q k this quantity is equal to delta Q i over is equal to the delta function that's the definition of the Poisson bracket in the original coordinates and if the Poisson bracket structure is preserved it means you have a relation of this kind for every value of i and j similarly we could write these down as well now is it possible to use this matrix j which we had introduced and write this in simpler form well yes indeed because you already know that I can write any Poisson bracket I could write any Poisson bracket in more compact form if I use the j matrix what would that be it would say take this quantity Q i take its derivatives take the transpose of this constructed into a row vector put the j in between and then the column vector from the derivatives of this and that would be equal to delta i j so can we combine all these three relations into a single relation and the answer is yes not surprisingly it turns out that all these relations can be combined let me do that here this whole set of relations is equivalent to saying that the acrobian matrix of the transformation which let me denote symbolically as delta psi over this delta x by the way this stands for the matrix delta Q p over delta Q p this matrix transpose j times delta psi over delta x this quantity is equal to j itself it is easy to check that all these relations are summarized in the single line by this equation this matrix equation it is not very difficult to verify that this is so I have used this symbolically just to tell us immediately that it is just the Jacobian matrix corresponding to the canonical transformation now what does this tell us this quantity is a 2n by 2n matrix it is transpose with a j in between times the matrix must be equal to j itself the same matrix j and recall that this j was equal to 0 the unit n by n matrix minus that and 0 it was a 2n by 2n matrix of this kind what would you call a matrix such that it is transpose times the matrix itself is equal to the unit matrix what do you call a matrix which obeys this condition matrix transpose times the matrix square matrix equal to i itself well if it is real then it is an orthogonal matrix this is the definition of an orthogonal matrix the inverse of the matrix is equal to its transpose that is an orthogonal matrix if the matrix is restricted to real entries then of course the unitary matrix is an orthogonal matrix because there is nothing to complex conjugate but in general the matrix could have complex elements in our cases here all these are real variables therefore we do not have any complex elements so a matrix of this kind is an orthogonal matrix can you give me an example of a transformation of coordinate say which is represented by an orthogonal matrix every rotation absolutely right every rotation in physical space is an or represented by an orthogonal matrix in three dimensional Euclidean space any 3 by 3 orthogonal matrix of unit determinant represents a physical rotation of the coordinate system the simplest of these of course as you are well aware is if you took the x y axis and you went off to x prime and y prime at an angle alpha then this rotation in the x y plane about the origin is represented by a 2 by 2 orthogonal matrix whose structure is is what what is the matrix representing this rotation absolutely it is just cos alpha sin alpha minus sin alpha cos alpha and so on in three dimensions you can generalize this to higher dimensions so orthogonal matrices with unit determinant with determinant plus one represent physical rotations that matrix there which represents a canonical transformation in the 2 n dimensional phase space is not orthogonal because there is a j sitting here and a j sitting there but by now we have got used to this j appearing everywhere in Hamiltonian dynamics this is pseudo orthogonal in a certain sense such a matrix is called a symplectic matrix let me write that down a matrix m transpose j m equal to j so this is some where m is a 2 n by 2 n matrix is called a symplectic matrix so what is the lesson we learned from that equation there canonical transformations are represented by symplectic matrices just as rotations are represented by orthogonal matrices of unit determinant canonical transformations are represented by symplectic matrices therefore in exactly the same way that the study of orthogonal matrices tells you everything you need to know about rotations in exactly the same way the study of symplectic matrices tells you everything you need to know about canonical transformations. So this is an algebraic approach to the study of canonical transformations. Now this kind of equation has remarkable properties we will see in a second the first of which is the following if I took the determinant on both sides what is the determinant of j, j was defined by that matrix the determinant of j is plus 1 we check this out so determinant j plus 1 it turns out from this equation it follows if I took the determinant of the left hand side the determinant of a product of matrices is just the product of determinants and the determinant of m transpose is the same as the determinant of m. So this tells you at once that this implies determinant m is equal to plus or minus 1 because the square is equal to 1 it turns out that you can show without too much difficulty although it is a non trivial exercise that the determinant of a symplectic matrix is in fact plus 1 this is why I mentioned earlier that canonical transformations also preserve orientation they do not just preserve the preserve a number of things among other things preserve orientation but this is reflected in the fact that the determinant of a symplectic matrix is plus 1 just as orthogonal matrices form a group the product of two orthogonal matrices is also an orthogonal matrix every orthogonal matrix has an inverse and they form a group in exactly the same way the symplectic matrices of a given order form a group of matrices this is a subgroup of the set of all 2n by 2n matrices which are non singular so let me write that down the group of all 2n by 2n real matrices so we are restricting ourselves to matrices which are real with real entries 2n by 2n non singular what is a non singular matrix the determinant is not equal to 0 in other words there is an inverse for the matrix so the group of all 2n by 2n non singular real matrices is denoted by GL the general linear group of order 2n over the reals so this stands for the general linear among such matrices the symplectic matrices form a subgroup of such matrices in other words the product of two symplectic matrices is again a symplectic matrix all 2n by 2n for a given order so it is a subgroup of this and this group is called the symplectic group and it is denoted by SP and it is a subgroup of GL therefore for a given dynamical system with a given number of degrees of freedom n the study of its canonical transformations amounts to the study of the symplectic group of the same order of order 2n now a great deal is known about the properties of such matrices so great deal is known about the symplectic group and what it implies what its generators are and so on and so forth therefore we have a fairly good idea of what the symmetry possessed by a system should be a Hamiltonian system should be under a canonical transformation a Hamiltonian flow goes to a Hamiltonian flow measure is preserved what do you need for a symmetry of the system though what do I mean by the dynamical symmetry of a system what would you say is given a dynamical system what would you say is a dynamical symmetry of the system one possibility is to look at its Hamiltonian if it is a Hamiltonian system and ask what kind of transformations leaves the Hamiltonian unchanged that is one possibility but of course in the case of a Hamiltonian system you need more than that you actually must make sure that the transformations of coordinates of the variables is also canonical so that the structure of Hamilton's equations is unchanged the most general way of defining what the dynamical symmetry group of a dynamical system is is to say a set of a transformation which leaves the solutions unchanged maybe takes one solution to another and it mixes up different solutions but the solution space the set of solutions of the dynamical equations should remain unchanged that is what I would call a dynamical symmetry of a system the full set of solutions must remain of the dynamical equations must remain unchanged if I accept that then it is easy to see what I mean by the dynamical symmetry group of a Hamiltonian system let us write that down by this I mean the dynamical symmetry group of a system is a set of transformations of its phase space variables set of transformations that leave the solution space pardon me that is sorry I did not get the question what is F no not necessarily not necessarily there is nothing to do with the Hamiltonian system right now we are talking about a very general statement I am simply saying when I have a dynamical system described by a set of differential equations and I ask what is meant by a symmetry group of the system the most obvious statement is that it is a set of change of variables of the system such that the solutions do not get changed the set of solutions does not get changed that is what I would mean by dynamical symmetry group now let us come to a Hamiltonian system and ask what is the dynamical symmetry group of a Hamiltonian system what can it possibly be well what do you need for a Hamiltonian systems solutions to remain unchanged first of all I certainly need the following I certainly need the Hamiltonian system the Hamiltonian flow to go to a Hamiltonian flow certainly need that this is done by the set of canonical transformations but I need more than that I would like to leave the solution space unchanged in other words the Hamiltonian itself should not change so that the equations of motion are exactly the same in the new coordinates as they were in the old coordinates right so not only do I need the set of canonical transformations but I need a subset of this group of canonical transformations which leave the Hamiltonian itself unchanged in other words I should not change from H to K H must remain unchanged that does not always happen with all canonical transformations therefore the dynamical symmetry group of a Hamiltonian system for a Hamiltonian system the dynamical symmetry group is the subgroup of canonical transformations that leave H itself unchanged so that not only do you go from a Hamiltonian flow to a Hamiltonian flow but moreover it is the same Hamiltonian and then you are guaranteed that the dynamical equations do not change and therefore the space of solutions does not change either in general therefore this dynamical symmetry group of a Hamiltonian system is a certain subgroup of sp2nr whatever be the number of degrees of freedom that also could be a group it may not exist you may not have much symmetry in the problem at all and that is in fact what happens in general but then in those cases where the Hamiltonian is integrable it turns out that you have a dynamical symmetry group and in fact that group is generated the infinitesimal generators of these transformations that belong to the dynamical symmetry group are related to the constants of motion f1 through fn so yes yes of course yes in a group every element has an inverse so this is certainly true canonical transformations have inverses that was our first premise that going from the small q's and p's to the capital q's and p's was actually invertible this was our premise that the canonical transformations we have talked about are global canonical transformations namely they apply in all of the phase space concern of course you could have a local canonical transformation which is not applicable in all of phase space we have not looked at that possibility at all we are all only talking about global canonical transformations and they are certainly invertible in fact since you raise the point a canonical transformation is a cement is a symplectic transformation so it satisfies this relation here and of course what does this imply at once let's take this m across to the other side by applying the inverse operator so it immediately says m t j is equal to j m inverse and let's bring the j to this side by applying the inverse of j therefore I get j inverse m transpose j is equal to m inverse but I know that j inverse is minus j j inverse is equal to j transpose is equal to minus j so in fact I know that m inverse is equal to minus j m transpose j for a symplectic matrix therefore the inverse exists no reason why it shouldn't this determinant we already saw was plus 1 so these transformations indeed form a group the point I am making is that the dynamical symmetry group could be much much smaller than this group than the group of canonical transformations could be much smaller in general because I also pointed out that in cases where the system is not integrable you don't have any symmetry at all but when you do have some special symmetry the system could become integrable this is the idea let me for example go back to the two dimensional harmonic oscillator and ask what the symmetry group could be this is not easy to identify it's not a trivial task to identify the dynamical symmetry group of a system which we know to be integrable in general one has to do a little bit of work to do this so let's look at the two dimensional oscillator which is h of q 1 p 1 q 2 and p 2 and this was p 1 squared plus q 1 squared over 2 plus p 2 squared plus q 2 squared over 2 if I took an oscillator which has exactly the same frequency in the 1 and 2 directions this is the isotropic oscillator here this Hamiltonian has a great deal of symmetry that symmetry group the group of canonical transformations is the symplectic group in two degrees of freedom so this is sp4 over the reals what's the group of transformations that leaves this Hamiltonian unchanged what would you say is a group of transformations that leaves this Hamiltonian itself unchanged notice that I can just take out the factor half and write this in this fashion so I should write here 2d isotropic isotropic to mean that it's got exactly the same properties in all directions as you can see the potential energy is q 1 squared plus q 2 squared which is invariant under rotations in the q 1 q 2 plane so it has circular symmetry what would you say is the symmetry or the group of transformations of these four variables each of which runs from minus infinity to infinity that leaves h unchanged we can switch variables but there is a big huge continuous group of transformations what would leave this unchanged this combination unchanged if I have x squared plus y squared plus z squared in three real variables x y z what group of transformations leaves is unchanged all rotations about the origin leave it unchanged in the three dimensions x y and z what leaves this unchanged all rotations in phase space in this four dimensional phase space all possible rotations about the origin would leave this unchanged what would that group be the symmetric group of h itself what would this group be it's the group of rotations in four dimensions four Euclidean dimensions what would that group be it's a group of four by four matrices but what sort of matrices would these be they have to be orthogonal what should the determinant of the matrix be plus one so volumes are left unchanged and that group is a group of orthogonal transformations in four variables called oh four but because the determinant is one you write an s here to write special or unimodular determinant plus one on the other hand this stands for orthogonal so on the one hand the group of canonical transformations in two degrees of freedom is sp4r the set of all four by four matrices with real elements which are symplectic which satisfy that condition on the other hand this particular Hamiltonian is left unchanged by the group of all four by four matrices with real entries which are orthogonal and which have unit determinant it's clear that for the dynamical symmetry group of this problem the set of transformations in phase space that leaves this Hamiltonian unchanged and takes the Hamiltonian flow to a Hamiltonian flow doesn't change the Poisson bracket structure is that set of transformations which belongs to both this group as well as this group in other words the intersection of these two groups so all matrices which are both symplectic as well as belong to this so forth that set of transformations that special subset of canonical transformations is in fact the symmetry group of the system so as you can see it's not easy to identify the symmetry group of a dynamical system because not only must transformation be canonical but it should also preserve the form of the Hamiltonian in this case since the Hamiltonian was so simple I could do this directly without too much trouble I could just identify it by inspection this looks like the surface of a sphere in four dimensions and therefore I could immediately write down a so forth now that's group which is the intersection of these two groups is a different group altogether it's smaller than either of these groups and it's the group of turns out to be the group of all two by two unitary matrices with determinant plus one so let me not write that down in explicit terms and get into group theory but let me just write the result down and say the intersection is SU 2 which stands for the group of all two by two unitary matrices with determinant plus one so it's a set of transformations which can be put into one to one correspondence with this set of matrices and this requires some mathematics to prove which we won't prove but I'm going to simply assert that this is so the reason I do so is because I'd like to identify the generators of this group and show you what the physical meaning of these generators is you see what are the constants of the motion here first of all we know it's integrable we had two constants of the motion which were in involution with each other and what were those two constants one of them was F1 which we could take to be just this the energy of the first oscillator and F2 was one half 2 2 squared plus P2 squared what else do you think is a constant of the motion here it's clear the system is integrable we know that the frequency ratio is unity in this case the same frequency for both these oscillators so on the surface of this torus we talked about the last time when you go around the torus this way once you also go around this way once exactly once so the motion is periodic completely period there's just one period unit frequency these two are also in involution with each other now to describe the trajectory you need one more constant of the motion which is an isolating integral so that the trajectory doesn't wind itself around on this torus completely but rather is a discrete curve is a curve is isolated curve for each set of initial conditions what else do you think is constant in this problem let's think a little physically Q1 let's say is the x axis and Q2 is the y axis and you've got a problem where you have an isotropic oscillator the same spring constant in both directions and the potential is circularly symmetric if I write this down in circular in plane polar coordinates this is just one half r squared so what do you think is constant in such a motion well here's this particle in the Q1 Q2 plane well the actual shape of the trajectory will depend on initial conditions it will depend on what the phases are what the initial values of Q1 and Q2 are that's certainly true what do you think is a symmetry I mean what what else is constant what sort of force is the particles here and it's attached by spring to the center what kind of force is exerted by the spring in what direction is this force it's always radial therefore this is a central force problem it's certainly a central force problem what do you expect is constant in a central force problem the angular momentum that's right there's no torque on this particle the angular momentum is therefore a constant of the motion we guaranteed that now what's the angular momentum in this planar problem you just have two variables here's x and here's why and this is p sub x and that's p sub y what's the angular momentum well if you use this formula r cross p since you have reduced to a plane everything is in a plane this cross product is essentially a single number what's r cross p there's only one component to it and what's that equal to not p1 p2 q1 q2 what's think in three dimensions if you have r cross p in three dimensions what's the z component of r cross p what's the third component of the angular momentum which I'll call L or since I know that it's the third component if I'm looking at a three dimensional problem I mean just call it j3 what's this guy equal to yeah it's equal to q1 p2 minus let me put a half here for a reason it should become clear in a second I'm guaranteed that this quantity is a constant of the motion in other words the Poisson bracket of j3 with H is zero that's not hard to verify so indeed this is a constant of the motion and now it turns out that the following quantities j1 j2 and j3 which we've already written as q1 p2 minus q2 p1 j1 is a quarter of q1 squared plus p1 squared minus q2 squared minus p2 squared and this is a half q1 q2 plus p1 p2 it turns out that each of these quantities is a constant of the motion so indeed it turns out that ji jk sorry ji with the Hamiltonian is zero i equal to 1 2 3 so we have other constants of the motion these two together with this third one the isolating integral actually specifies the trajectory completely the moment you put these three quantities equal to constant then in this four dimensional space you found the trajectory completely because the intersection the mutual intersection of surfaces on which this this and this are constant specifies a curve and that's indeed the phase trajectory for any given initial set pardon me j1 is yes yes I'm not saying they're independent I'm going to come to the significance of the j's a little later but I'm saying f1 f1 plus f2 is the Hamiltonian itself of course for integrability you just need two constants of the motion which are independent of each other and which are an involution they are represented for instance by f1 and f2 but I'm actually finding a whole lot of other constants of the motion for describing the motion you actually need three isolating integrals in this problem and they are represented for example by f1 f2 and f3 and j3 but in addition these combinations are also constants of the motion sure this is just f1 minus f2 apart from some constant factor this is something else all together and this quantity here is the angular momentum about the origin but now you can't have more than two constants of the motion in involution with each other that's for sure so in fact that's the maximum number that could be independent and in involution you can choose them as you please with various linear combinations I'd like to choose this because this completely separates degree of freedom one from two altogether the others mix it up in some fashion or the other of the significance of this j1 j2 j3 is that they obey very interesting Poisson bracket relations and they obey the following relation among themselves you have j i j j is equal to epsilon i j k j k this is the totally anti-symmetric symbol in three dimensions it's equal to plus one if i j k are a natural permutation or even permutation of one two three minus one if it's an odd permutation and zero and all other cases this is the Levy-Chivita symbol in three dimensions so this stands for a set of three relations j1, j2 is j3 j2 j3 is j1 and so on in cyclic permutation what is this remind you of this is exactly relations obeyed by angular momentum components if you write r cross p in three dimensions and ask what is a Poisson bracket of Lx with Ly or Ly with Lz or Lz with Lx you get exactly this set of relations so in some funny fashion angular momentum algebra in three dimensions is related to the dynamical symmetry group of the two-dimensional isotropic oscillator and the reason is group theoretical it's purely algebraic it turns out that this group is generated by three combinations of q's and p's which are precisely these three quantities there if you like the generators of this group SU2 so you see what's happening is to summarize things here's a system which is completely integrable it has two degrees of freedom so we have two constants of the motion in involution with each other which are functionally independent of each other and those are f1 and f2 since the motion is periodic rather than quasi periodic we need a third isolating integral another algebraic function of the q's and p's that's provided by this quantity j3 for instance which has the significance now of being the angular momentum about the origin so we have the energy of the first oscillator the energy of the second oscillator and then the angular momentum about the origin in addition there are these combinations j1 and j2 such that j1 j2 and j3 are all constants of the motion but they're not in involution with each other they can't be too many of them instead they obey a certain algebra the Poisson bracket of any two is a linear combination of the same quantities in this case just the third one with the appropriate sign that algebra represents something much deeper this algebra the existence of this this set of relations implies that there's a certain dynamical symmetry group in the problem which happens to be the intersection of the symplectic group sp4r of canonical transformations in this problem with the symmetry group of the Hamiltonian itself which happens to be SO4 and that group can be put into one to one correspondence the set of transformations with the group of two by two unitary matrices which is this group here and this group has three generators it happened to be the same as that of angular momentum in three dimensions and that's the reason why you have these combinations j1 j2 j3 so as you can see dynamics and the algebraic structure underlying integrable equations they're very closely linked with each other and this two dimensional oscillator gives us a simple model in which to understand the origin of these symmetries in this case of course if you go to three dimensions write the oscillator down in three dimensions that's a much bigger group the symmetry group is much bigger the canonical transformations would be sp6 r and then you'd have to look at the intersection of that or the subgroup of that which also leaves the Hamiltonian unchanged and that's a much more complicated group turns out that happens to be SU3 in that case and so on in fact the n dimensional isotropic oscillator has a symmetry group which is the same as a set of transformations the set of transformations it's a symmetry group is the same as the group of n by n unitary matrices with determinant plus one a very useful relationship in many applications but I don't want to get into that right now I'd like to go back and give you another problem which also you're familiar with where there's an extra symmetry we'll see where this comes from yes if the generators is three it's exactly the angular momentum algebra this thing here no it will not be for example the 3d isotropic oscillator the dynamical symmetry group is SU n SU 3 in fact the general statement is the n dimensional isotropic oscillator n dimensional isotropic I shouldn't call it n because this number n has been used for degrees of freedom okay it's the same number actually n dimensional is SU n no it doesn't obey this algebra it obeys a more complicated algebra but the question you could ask is how many of these are there that's something we can directly answer how many do you make how many generators do you think there are in in SU n well let's start from you could start from one by one matrices then we could go on to two by two three by three etc the least trivial case is the the simplest non-trivial case is SU 2 so let's look at that we want to look at all unitary matrices two by two matrices which have unit determinant I want to look at the set of all these matrices and I permit complex entries okay we'll come back to what a generator is of a group since I have to tell you how these groups are generated what I mean by it is let's come back let's come back and tell you what a generator is I don't want to have a digression within a digression so what's what's the what's the number of parameters that you need to specify a 2 by 2 matrix start with a 2 by 2 matrix with possibly complex entries we count the number of real parameters always 8 parameters because if the matrix is a b c d and a and b are all a b c d are complex numbers there's a real pardon and imaginary part of these so in general 8 real parameters are needed now I start putting conditions on these matrices what's a unitary matrix a unitary matrix is one where you dagger you is equal to the identity matrix so this is unitary the dagger stands for the complex conjugate transpose the Hermitian conjugate of this matrix so I take the transpose and then do a complex conjugate of this matrix and I insist that this be true so what I'm insisting upon is that a b c d multiplied by the complex conjugate transpose so this becomes c star b star a star and d star b equal to 1 0 0 1 I insist upon this how many parameters are left now how many conditions does this give me gives you four conditions so how many conditions are left how many parameters are left four parameters are left I now insist that the determinant of the matrix B equal to plus 1 that's what makes it s how many parameters are left 3 and therefore the number of generators is 3 in other words any element of this group is found by taking certain special matrices multiplying them by parameters and exponentiating these matrices and that gives me a finite element that's called a generator the group so there are three generators and that's exactly the number we found here I'll illustrate what's meant by generator in a minute but now what do you think it is for SU n what do you think it's going to be for n squared it's n squared minus 1 therefore in three dimensions it's 9 minus 1 which is 8 generators and therefore there are 8 constants of the motion which obey a certain algebra among themselves more complicated than this considerably more complicated than this but that provides you with a symmetric group now let's come to the question of what I mean by generator and let's go back and do a little bit of elementary algebra here so I start with the simplest example of a rotation in two dimensions in a plane and let's write the equations down directly so I start with the XY plane and I make a rotation about an angle alpha to write new coordinates X prime and Y prime and of course I have X prime Y prime is equal to a certain matrix acting on XY this matrix here depends on the parameter alpha and it tells me X prime is a certain linear combination of X and Y and Y prime is a certain linear combination of X and Y and it's well known of course that we have it's let's call this matrix which represents rotation by an angle alpha let's call it R of alpha now what are the properties of R of alpha what does a rotation do it keeps the origin unchanged it's linear it's homogeneous in other words zero remains zero the origin remains unchanged and distances don't change now does this coordinate system become a left-handed coordinate system what's right-handed remains right-handed so this means that this matrix R satisfies R transpose R equal to I it's orthogonal which keeps distances unchanged and the determinant of R equal to plus one therefore this R of alpha is an element of a group of matrices which are orthogonal two by two matrices with real entries in this case and with determinant plus one how many parameters are needed to specify a rotation one so this is a one parameter group and there's one generator in three dimensions you have three Euler angles in general therefore you have three generators for SO3 in n dimensions Euclidean space how many generators do you need for specifying rotations why n are there are n axis or n angles and this is precisely the point where I want to stop and we'll take this up that answers not right because that tells you something about the nature of rotations itself let me redefine a rotation a rotation is a linear homogeneous transformation which keeps the origin unchanged for instance which is orthogonal distances are not changed and the determinant is plus one so right-handed system remains right-handed system the reason you say n is because you assume that every rotation is a transformation about some axis and there are n axis but this is not true because if I think of four dimensions then of course I could have a rotation which changes in the XY plane but leaves both the other two coordinates unchanged or if I look at two dimensions there's no third axis it's about a point so an axis need not be identified with the rotation that's an accident of three dimensions it's an accident of odd dimensionality we'll come to that so the number of generators is the number of independent orthogonal planes you can find how many planes can you find in n dimensional space orthogonal planes like the XY YZ etc. NC2 which is n times n-1 over 2 that's the number of generators so I'll explain with this is a good example to explain what's meant by generators for some elementary group theory and then we'll take it from this point so let me write that down SON I'll explain what's meant by generator next time using this example