 Let me remind you that, to summarize a little bit of where we were last time at the end of the hour, we're in the process of looking for representations of classical rotations. These are actually proper rotations, all of which are built with improper rotations that are on it. Representations of proper rotations, in terms of operators, you, which act on the overspace of some quantum system that represent the rotations. And the requirement that these operators use should satisfy the three properties listed here. First of all, they should be unitary. Secondly, the U operator corresponding to the identity rotation should be the identity operator. And thirdly, the U operator should reproduce the multiplication law of the classical rotations. These are just the listed requirements that we wrote down. Now, as you might imagine, the U operators that we eventually get are going to depend on the specific nature of the system, whether it's this fin system or relativistic quantum field theory, it's going to make a difference. However, there's an awful lot of the properties of these U operators that can be derived independently of that, but actually only depend on these representation properties or requirements that are listed here. So we're going to explore those first and then later we'll look at how specifically these operators would utilize in specific systems. Okay, so to begin with, let me relarge that the classical rotation can be written in axis angle form and that this also has an exponential form, which is even the thin kind of in-hat dotted into this vector of three by three matrices that are writing a script J on the board. And it appears as a sense of the quantum in the notes, like this. And you may notice that this axis angle parameter parameterization actually only depends on the product of the angle times the axis unit vector. If I draw a picture here, here's the unit vector in-hat. And by extending it, I get the axis of the rotation of the geometrical ideas of the angle theta is used as the right-hand rule giving of the rotation. But since the answer only appears in this product that suggests that we make a definition of that vector of an angle, something called a theta vector, it's just defined as theta angle scalar times the constant unit vector in-hat. And if we do this in this theta vector, it's along the direction of the axis like this and it has a magnitude just equal to the angle of rotation. It's just a notation, that's all. And so if we do this and let's just define an alternative notation for these rotation operators as R of the vector of angles theta and this is the same thing as theta vector dotted into this written J vector by three matrices like that. So it's simply a notation. All right. Now as long as we're talking notation, let's also talk about the corresponding unitary operators. Let's let U, U is supposed to be parameterized by R. So let's write R in axis angle form, R of an atomic vector. And let's just define this in kind of an obvious notation that U would then have to come to theta. It's in axis angle form for the unitary operators. Or if we write the R in terms of this vector of angles theta vector up here this becomes U of R of theta vector. And again it's kind of an obvious notation. Let's write that as U of theta vector. This is just a notation, that's all. All right. To go back to the problem here, the main problem is we like to find these unitary operators subject to these three constraints given the fact that we know about the classical rotations. The main strategy for doing this is to work on infinitesimal rotations first. So let me remind you that if R is an infinitesimal rotation in fact I'll write it out in this theta vector notation here because we have an exponential series for it if we expand it out it looks like this. So this becomes the identity plus theta vector dotted in with our script J, J-vector, and make these those higher order terms. So that's the infinitesimal angle version of this. Now as far as the U's are concerned let's look at U of theta vector. We don't really know anything about these things yet except that they're unitary. But we can, certainly if the angle is smaller it's frequently free to expand this in a Taylor series. So let's write this as U of zero the direction term is given is sum from K equals one to three of theta sub K to three components of theta I believe my signal looked better three components of theta times the derivative of U with respect to theta sub K evaluated at zero plus higher order terms like this just extending the Taylor series. Now you may remember we did something similar to this in the case of translation operators and in fact I'm going to follow the same general pattern that we used earlier with translation operators. First of all as far as the view of zero is concerned that's equal to one it's kind of obvious that if the operator is near identity then it's got to be close to the identity operator so the rotation is near identity and U has to be close to the identity operator and power mechanics which we just call one here. As far as the first correction term is concerned it involves these coefficients which are the derivatives of U with respect to theta evaluated at zero. One thing to notice about these derivatives they don't depend on theta they're actually just constant operators. This is quite different from U over here which is a function of three angles. So let's look at these first derivatives and making the box to separate things off here let's make a definition I'll put the definition up here we'll define a new set of oscillators which I'll call it capital J sub K this is a Roman K and not a script K because there's a different font of the nodes capital Roman J script J this is defined as minus i h bar times the derivative of U with respect to theta sub K evaluated at zero it's just proportional to this first Taylor series of coefficient that's all. In fact what we've done in defining J sub K is to first of all split off the factor of minus i the reason we do that is that it makes J sub K commission. This is something I'll come back to in just a moment but that's what the minus sign does is it makes the commission and secondly we split off the factor of h bar which is really just conventional which you'll recall we did something like that also in the translation operators as well however splitting off the h bar means you can see that the view is dimensionless and V would be theta is dimensionless and theta is dimensionless it means that J has the same dimensions as h bar which is action or otherwise it has dimensions of angular momentum but in fact we will call this the angular momentum of the system. Now in doing this we're following rather closely once again of what we get in the case of translation operators you'll recall we define the linear momentum of the quantum system as being the generator of translations and likewise here we're defining the angular momentum as the generator of rotations in the same sense the commission operator that's involved in near-identity symmetry operation. I'll come back to why this is called angular momentum in the moment but I'll tell you a few more arguments for right now this is just a substitution but if we apply this to this u of theta here this expression here what we find is that for small angles we find that u of theta vector is equal to identity minus i over h bar and theta vector dotted into now this moment J which is our vector of operators or if I go back to the axis angle notation using where it is theta vector is theta times n hat this just becomes 1 minus i over h bar times the angle theta times the axis n hat dotted into J plus i over terms. Alright Now in the case of the rotations the classical rotations this small angle representation of the classical rotations is really just the first term of the Taylor series and if you sum it up you get the exponential which you see up here but the question is can we do some similar summation here right now all we have for our u operators the first term of the Taylor series can we carry out a summation of the entire series and the answer is yes it follows the same pattern we used earlier for the right of this exponential notation actually we used it at least a couple of times before once in connection with the translation operators then we go through it now to derive an exponential form for u of n hat comma theta when theta is not necessarily small for the higher order terms different words like this let's take u of n hat comma theta and let's look for a differential equation for it which will write as d v theta starting out as we asked for the derivative of this vector theta by the definition of the derivative this is the limit as epsilon goes to 0 of u of n hat comma theta plus epsilon minus u of n hat comma theta divided by epsilon a limit form for derivative now the first factor of u of n hat theta plus epsilon because of the fact that the u operators reproduced the multiplication law in classical rotations this means that this is equal to u of n hat comma epsilon times u of n hat comma theta because that's the same rule in classical rotations of A so you can factor out a u of n hat comma theta out to the right side and this becomes the limit as epsilon goes to 0 of u of n hat comma epsilon minus identity divided by epsilon that whole limit multiplied on the right by u of n hat comma theta like this now as far as that derivative excuse me, as far as this limit is concerned we can easily get that from our small angle formula this formula here is valid at this point the formula's value in theta is small so writing epsilon instead of theta I've got a valid expression here for this I put this in as a factor 1 it kills the 1 here divided by epsilon and it killed the theta which is bigger placed by epsilon and what's left over is nothing but minus i over h bar times n hat divided into 2j the angular momentum and so to make a little space here to get a differential equation that says u of n hat comma theta with respect to theta is equal to minus i over h bar n hat divided into j times u of n hat comma theta of course the initial conditions is that it has to equal 1 and theta is equal to 0 so we integrate this and what we get is I'll put it right over here because the solution which is the u of n hat comma theta is equal to e to the minus i over h bar times n hat divided into j is the exponential form for the unitary unitary operators which represent represent rotations of nth assemble form now this isn't a solution yet to our problem our problem is applying the u operators that correspond to the given classical rotations and we haven't solved the problem we just merely re-expressed it there should be a theta here theta times n hat divided into j we haven't solved the problem exactly because we don't actually know if the j's on the j's are defined in terms of the u as you see up here are the first derivatives but this is a great improvement over the original problem because the original u operators have three continuous parameters here and now you see this continuous parameters have been the dependence on them is indicated explicitly right through the exponent and what's left over are the j's which are just three operators three only, three operators are only that are independent of any parameters so this changes our strategy and we're going to find the u is what we're going to do is first try to find the j's there's only three of them and they don't depend on the theta's and we find them and we form this exponential and that will give us the u's so in order to do that we need to know what the properties of the j's are something to replace this list of properties of the u's here we need to know what the properties of the j's are so that we know what we're looking for in finding the j's well one of the properties of the j's is the prohibition I mentioned that up here but I didn't prove it let me just indicate in a little bit of space here how the proof goes it's exactly the same as what we did in the case of a momentum operator when we were working treating momenta as generator translations what we do is we just go back to the infinitesimal version small angle version of the operator you require a few times the diagonal of the identity and the direct consequence of that is that jk is even the jk dagger I won't actually go through the details of this because it's exactly the same as what we did earlier with the translation operators but in any case one of the properties of the j operator is the angular momentum is that these are these are prohibition operators now so let's summarize here what the properties are of the angular momentum the angular momentum is a vector of operators so the first one is that the prohibition there's three of them they don't depend on angles or axes or anything, they're just fixed operators maybe another one dimensional let's go back to number three number three maybe the most important one is that the unitary rotation operator is expressed in terms of the angular momentum by this relation in the i over h part theta and in that order for number two it turns out what we're going to want is one of the commutation relations of the angular momentum operator what to fill this in perhaps you recall we did translations we used the properties of the translation operators to deduce the commutation relations of the momentum and we found momentum operators had to agree with one another we're going to do something similar to that now we're going to use the properties of rotations to find the commutators of the angular momentum to find them this way the story of this goes back to what we did earlier on my summary board what we did earlier, classical rotation to remind you in studying the commutativity of classical rotations we look at this product of a product of A, B, A inverse, B inverse where the A and B are rotation operators R1 and R2 each of which has an axis and an angle like this, so they're exponentials of anti-symmetric matrices going into an obvious notation and what we did was we expanded out the series for this four-way product and we got the result to look like this so in particular the first rotation is about the x-axis and the second rotation is about the y-axis and if the angles are small then the first order correction of this product is a rotation of the z-axis because it's got a cross product and the z-axis is there Is there a sine arrow there? Should that be jj equals positive value at the bottom? Yes, the part right excuse me I'm going to go over the minus side the minus side appears here but not there yes, thank you very much Yes, absolutely alright and by the way that follows the same signs that we did in the case of linear momentum now yes, so this matrix C especially for small angles is an interesting way of examining how rotations don't commute now, in order to find the quantities of the linear momentum operators which is now on the board decided number two that we're interested in here number two is to concentrate on the rotation operators that correspond to these rotations here and the rotations C so let's do it this way maybe before I do that I'm going to do one more thing which is to take this result here and let's write this in the form it's obviously a small this is obviously a new identity rotation let's write this in the form of identity plus let's call it phi times m hat rather than j because if we do that this is the beginning of the series of the rotation operator about the axis m hat by angle phi and you can see just by this expression with that one we see that phi times m hat is just equal to thing one theta two times m hat one cross m hat two and as I mentioned so for example if m hat one is x hat n hat two is y hat then m hat is z hat and phi is the product of the two small angles that's an example of this alright now I guess since I've got this stuff here I'll do the calculation on this board and then I'll go back to our summary for the results on so the strategy here then is to look at the unitary operators that correspond to this product we're going to look at u of c and we're going to compute this in two different ways one way is to just write it because the unitary operators have to reproduce the multiplication law this must be u of r one times u of r two times u of r one inverse which is the same as u of r one dagger times u of r two and u of r two dagger like this however r one and r two are written in axis angle form like this so this becomes the same thing as u of n hat one times theta one times u of n hat two times theta two times u of n hat one times theta one dagger times u of n hat two times theta two dagger and we now have an exponential series for all four of these so we can multiply this out the first one becomes one h bar theta one times n hat one dotted into angle momentum operator j the second order term is minus one over two h bar squared times theta one squared times n hat one dotted into angle momentum of j quantity squared plus higher order terms having this on the second order that's for the first factor here and then this gets multiplied times the other three factors and I will bother to write out all the details which should be easily filled in the series for them and you do the multiplication and you carry out the algebra out to the second order and in the first order you get one and in the first order you get zero because all the first order terms cancel and in the second order what you will find is the minus sign is one over one over h bar squared times theta one times theta two times the commutator of n hat one dotted into j with n hat two dotted into j plus higher order third order terms first non-management terms in the second order now on the other hand see itself here as a near identity of the rotation which we have written here in terms of this n and phi stuff so the answer must be one minus i over h bar m hat times for this one phi times m hat dotted into the commutator because it's a near identity rotation operator the phi times m hat is theta one theta two times theta one cross n hat two so this becomes one minus i over h bar theta one theta two times n hat one cross over n hat two dotted into the angle of momentum plus higher order terms like this so as I say we evaluated this u over c in two different ways what we're using is just the representation laws up there that's the only thing we're using these two terms that the order of theta one and theta two have to be equal to one another you can think of this as a double Taylor series in these two variables of theta one and theta two so in setting these two terms equally we're going to cancel the theta one and theta two and in the rest let me just multiply both terms by minus h bar squared and the result will be this this is the commutator of n hat one dotted into j with n hat two dotted into j is equal to i h bar times n hat one cross n hat two dotted into j so for example if we let n hat one be x hat and n hat two be y hat then n hat one cross n hat two is z hat and we see the commutator of j x with j y is equal to i h bar j c or more generally to put it into a form that we'll use we'll go back in the summary here in the answer guess but this becomes i h bar times x l i j k j c k and thus we can write commutations for the angular momentum in this manner so to recapitulate a little bit where we are with this so to go back to the step to summarize the step we just went through we use the we use the commutation properties in the classical rotations plus these representation rules in order to derive the commutation relations of the generators of rotations in one mechanism and this is what we get the standard way of deriving the angular momentum commutation relations in an introductory course is to start with the orbital angular momentum which is r cross p it's usually r cross p for connective potential problems then you work out the commutators of that in order to get the commutators of x and p to get these standard relations then later on when you come to spin spin should have the same commutation relations because it's another angular momentum that should be the same as orbital and it's used here we're following a different logic we're deriving the commutation relations in the angular momentum essentially from the properties of rotations this gets to the essence of the matter really in a better way because these commutation relations really follow from the geometry of Euclidean space it really comes from nothing else it goes all the way back to the commutation relations of these j matrices and in the classical rotation which in turn are related to their orbital rotation matrices so that's the real meaning of lying behind is it's a geometry of three-dimensional space another thing to say is that the angular momentum even in classical mechanics is not always r cross p as you saw in this monopole problem so it isn't even general to do this this is a general definition of the angular momentum as a generator of rotations and it applies not only to problems with orbital mechanics but also to potential problems but also problems with spin problems with magnetic fields for example you're going to apply to relativistic problems you can get the angular momentum electric magnetic field all the same way it's the same about procedures all the way through alright now in terms of how this fits into our strategy of trying to find unitary operators which represent rotations what we've done is we've simplified the problem considerably but now instead of trying to find an arbitrary u depending on three parameters we only need to find just a three-vector of omission operators and satisfy these commutation relations and then when we find them we just form this exponential and this gives us our operatives which reproduce the classical multiplication laws of rotations now that doesn't mean that they're rotational operators in a physical sense so what we then have to do is to check and see that it makes sense physically anyway that's the strategy that we use for finding these rotation operators now at this point I'm going to I'm going to carry out that actual strategy in the simplest possible system which is that of cases of the spin one-half system of course you know something about spin one-half systems already from your undergraduate course but I think for most of what I'm going to say you need to think back to the story that you went through earlier we use the experimental data and the possible quantum mechanics and we affect derived and hollow matrices so let me remind you of how the matrices satisfy the commutation relations of sigma i sigma j is equal to twice i f4 by jk sigma k these are almost the same as the commutation relations that we require for the angular momentum and so what we do is we guess that for a spin one-half system the momentum should be identified with h bar to two times sigma and the reason we do that is because if we make this substitution then these commutation relations are satisfied and they're also permissioned so they satisfy properties one and two and if that's true then what we get for the rotation operators is putting an axis angle four and for a spin one-half system would be equal to e just plugging this h bar to sigma here into j of the h bar to cancel it becomes e to the minus i theta over two times the impact value of the sigma well this is an exponential which was on homework number one and the answer if you expand out the series becomes cosine theta over two minus i times the impact value of the sigma times sine theta over two and that when we box the whole thing one-half system this turns out to be rotation operators now I have to say I'm confusing a little bit the operator with the matrix which represents the operator and that's in a frequency basis but that's simple simple thing you'll not be confused about here are some really matrices on the right-hand side but on the left I've got the operator so I've lost some of that difference so this is the matrices now probably the first question to ask is whether it makes sense physically in terms of a rotation so to discuss this let's go back to the magnetic moment vector operator which appeared in the Stern-Berlin apparatus if we have a specific quantity of state and we take the expectation value the state side here belongs to the previous basis of the sigma-hat system then what results is just a number the meaning of this is a vector of operators so what results is a vector of numbers just ordinary numbers now let's apply a a rotation operator to the state side let's call this the state side prime and if we form the expectation value of the magnetic moment operator with respect to the state side prime like this this gives us another vector of ordinary 3 numbers and so what should be the relationship of the expectation value of the magnetic moment in the unrotated state versus the magnetic moment in the rotated state well it's obvious that if this is reasonable physically this should be the classical rotation multiplying the unrotated vector in other words expectation value should transform by the classical rotations this is R this is really not room here this is really R of impact on the theta with the same axis and angle as it appears in this u should be that well if this is so all I need to do is to substitute inside prime is equal to u psi in here and then also let's simplify things a little bit let's notice the view is proportional to some constants proportional to the poly matrices sigma as we found in earlier work and so the result again is this psi times u dagger sigma times u psi should be equal to the rotation matrix multiplying psi sigma psi I want to understand the notation here when I say the rotation matrix this expectation value is the ordinary 3 vector of numbers so I'm just multiplying the matrix times the vector of the usual sense of matrix multiplication if that's not clear well let's see before I go to that let me just say that this equation has to be true for all sides so we can strip the sides from both sides so we get a requirement is that u dagger sigma u must be equal to R times sigma if you explain more exactly what this means it means that u dagger times one of the components of the sigma u i u must be equal to sum of j of R by j sigma j so as I said the significance of this equation is I haven't shown this as true yet but this is what we expect in a reasonable definition of the unitary rotation algorithm should satisfy this result so what we need to do is to check to see whether our tentative our provisional definition of rotation on here actually does satisfy this result well this is straightforward to do so to write this out if I take u dagger else and if it was changing the minus side to the plus side so what we get here is we get cosine theta over 2 plus i times impact dot sigma times sine theta over 2 times sigma i like this times cosine theta over 2 minus impact dot sigma times sine theta over 2 this is equal to the question mark we want to check in sum of j of R by j times sigma j so this is the formula we need to check now the left hand side looks kind of complicated but it's got a sigma here a sigma there and a sigma there so the worst that happens is the product of three sigmas a lot of terms only have two sigmas some terms that only have one this cosine times that cosine times that sigma okay so you can use the rules of the sigmas like sigma i sigma j is delta i j plus i s 1 i j base sigma k the multiplication rules of the sigma that allows you to take a product and reduce it to a linear term likewise you can take a cubic term and reduce it to a linear term so there's algebra to do all of that which I'll leave as an exercise for you so when you're done what you find is the flow of the following you find you find the a work of the vector notation here as you see here rather than the index form which you see in the line below the work of the vector notation is what you find you find the left hand side becomes cosine theta times sigma plus 1 minus cosine theta times in hat times in hat dot of big sigma plus sine theta times in hat cross sigma that's the result of the algebra and the question is is this equal to r after one sigma rotating sigma the u of the character is the axis angle form so I should really write this as r again on the theta times sigma and the question is this true well it is true because I'll remind you that way back when we started looking at classical rotations we worked out a formula what the rotation does in vector in these three terms and they reproduced exactly by this provisional definition of rotations just in the half systems and so it checks it works and that means that it satisfies the physical requirements that a rotation operator should have in spin on half systems namely that expectation values of vector operators should transform these vectors where the expectation values transform by the classical rotations alright so the main result here is the one that I got boxed at the top this is now Melvin Krump's a satisfactory definition of rotation operators for spin on half systems I guess the annual momentum based on the connotation relations and then just working it out the two most important results on this upper board here well aside from the summary of copies of J when is the explicit representation of the rotation operators to still have systems here but also there is this expression over here for what happens when you conjugate sigma by a a unitary rotation operator let me repeat that because this is an important result I'm making this list of unit times of in hat of in hat and theta so u in hat theta dagger times sigma times u of in hat on theta is equal to r of in hat on the theta multiplied by sigma this is an important formula that gets used over and over again this is the adjoin formula for spin rotations just to give it a name I'll remind you the adjoin formula we had in the case of classical rotations was this if we took an a hat dotted into our script J vector this is an anti-symmetric matrix and conjugated by a rotation this is the same thing as rotating the vector this has the geometrical meaning that the axis of the rotation transforms the vector this is the first term in the expression for an infinitesimal first correction term in the expression for an infinitesimal rotation geometrically it means as I said the axis of the rotation transforms the vector I want to show you how this adjoin formula looks like this one to do that I want to do two things first of all let's take both sides of this equation some of the ordinary 3 vector a so the dot appears here over there on the vector on the right hand side let me also take the u and replace it by u inverse so that r gets replaced by r inverse so if I do this on the left hand side I get u of n hat on the theta without a dagger times a dotted in the sigma times u of n hat theta dagger people with a dotted into not r sigma but rather r inverse I replace r by r inverse so this is a formula that's just equivalent to the line of Bosch now when you have a dot product like this and you have a rotation you won't find one side of the dot product you can shift it over the other side if you replace the rotation by its inverse just because the inverse is the transpose it won't be one side of the dot product the other will be one side of the dot product so this is the same thing as r and n hat come together without the inverse so I move y a dotted into the sigma when I write it like that you can see that it has the same structure as we see it appear except the j the generator of the classical rotation is replaced by the sigma and in conjugating the sigma we need to use the u operators because that's a quantum generator but the sigma is a vector transformed by the classical or the axis transformed by the classical rotation I threw that to show you why I'm calling both these formulas adjoint formulas they really convey very simple integration now there's one problem which is left about speaking still spin one half rotations there's one problem which is left and this arises if we think about an axis let's say here n hat another angle a negative rotation like this and we want to apply a rotation about the axis n hat by an angle 2 pi so there's a complete rotation of an angle around 2 pi and let's apply this to some quantum state which means a two component spinner well look at the middle of the border there data is equal to 2 pi cosine of theta over 2 is cosine of pi which is minus 1 and sine of theta over 2 is sine of pi to zero so if you have impact on a 2 pi this is plus 1 on the other hand if I take u of n I come to zero this is of course equal to plus 1 and if I square the first line u of n hat of 2 pi u of n hat of 4 pi that becomes plus 1 so there's a problem here because we were imagining it's been a race now we were imagining we started out by trying to find a mapping from the classical rotations into the association this is what we call the representation of the classical rotations in terms of unitary rotation operatives well r of n hat comma 2 pi is equal to r of n hat comma zero which is equal to the identity and we were requiring that u of identity should be equal to plus 1 so u of n hat comma 2 pi should be equal to plus 1 in an association like this but what we're fighting is is a good n hat comma 2 pi is equal to minus 1 so although we follow the rules that were listed by this set of demands that were imposed the answer doesn't actually work it actually doesn't work because of this most of it works but this doesn't so we have to address this issue there's two aspects to it let me address the mathematical aspect first what's going on here is that instead of finding an association between a classical rotation and a corresponding unitary operator which acts in our quantum system in the case of a spin line hat system we're finding that a single classical rotation which is named in the identity actually corresponds to two spin rotations plus 1 and minus 1 there's a way of looking at this if I take this phase this is a classical rotation this is the group SO3 then I have the identity here it's one of them then let's over here take the space of spin rotations space of matrices that look like this and we're going to mark some of the border and what we're finding is instead of this rotation corresponding to a specific spin rotation which would be the intensity plus 1 we're finding it actually corresponds to two of them we're over at two operators and so instead of having an association that works this way this isn't really right it's a double value function in effect it's more appropriate to look at it as going the other way to say that there's not an association between classical rotations and spin rotations but the other way given a spin rotation would be finding a classical rotation to set the spin rotations by the way is a set of two by two matrices that look like this and these matrices all have a property that first of all unitary but that was by construction also the property that they're determined as plus 1 you can just check that by taking the determinant of that equation writing out the column matrices in any case what that means is that all of these spin rotations belong to the group SU2 this is mathematically language for the space of unitary matrices which have a determinant plus 1 that's just what it means the unitary SU2 matrix is in fact the matrix of this format is a spin rotation so there's a one-to-one correspondence between spin rotations and SU2 matrices and that's in fact what this set of spin rotations is is the group SU2 and it does not stand to the one-to-one correspondence of the classical rotations it stands to the two-to-one correspondence and thus although we started out by saying that U is a function of R it's actually more appropriate than U so if you have any U I can find a specific R by the way, if I take any other R besides the identity this will map over here into two rotations and they're U and minus should be always different by sign plus a burn sign so as R is at the R of U this is the same as R of I it's possible to actually find an explicit format for R as a function of U this is done in the notes we won't use it so I won't derive it but I'll put it for you as R IJ is one half a trace of U dagger times sigma I times U times sigma J and you can see from this formula that if I take both U and minus and this is quadratic U if I take U and minus you could give it the same R that's like two years by signing back to the sub R you can also show in this formula that R of U1 times R of U2 is equal to R of U1 U2 in other words although we started out by looking for unitary operators that reproduced the multiplication loss of classical rotations what we end up getting in the case of the spin on half system is spin on half rotation operators such that the classical rotations reproduce their multiplication loss the other way around in an important sense the group SU2 is the more fundamental group and then SO3 is the derivative group this is a really invested in quantum mechanics and it also has as a virtues even in purely classical problems actually interesting this SU2 group was known to Cayley and Klein in the 19th century even before quantum mechanics came about with this double association people would do computer programs to pretend to integrate the equations of motion of gyroscope like samples with spacecraft or something that's effectively free gyroscope not even how its attitude is a functional time they don't use SO3 they use SU2 for that even though it's a pure classical problem it's much easier in a computer program okay so this is double value representation now the question is what is this one the question is some of you see what it says it says it's anomaly normally what it says is that if in an app system like an electron and I rotate it by an angle of 2 pi the electron state doesn't return to itself but it rather suffers a phase change which is minus 1 the question is and you might say well that's just a phase it is just a phase but it's a physical and the answer is that it is physical so here's the here's the situation of size of the electron state that are rotation by any axis by an angle of 2 pi acting on it takes you to the negative state it doesn't just have physical consequences it does it's been verified experimentally that this is true it's discussed and summarized although it's slightly more detailed than it does the best experiment is for studying this time the best experimental situation for studying this type of phenomenon it's the involves effectively drag scattering of thermal neutrons by means of the silicon crystal this was done back in the 1970s with the because of the semiconductor technology it's now possible to create silicon crystals in which the lattice of atoms in the silicon crystal is uniform without any defects or breaks and over considerable distance centimeters even though there's 10 to 9, 10 to the 8 something lattice spacing in every centimeter so from this standpoint of diffraction ratings this creates a very good diffraction rating diffraction ratings are spoiled if this spacing isn't exactly perfect in any case what you do is you have a beam of thermal neutrons coming in like this into this crystal neutrons are produced by nuclear reactors that come from the fission of uranium and in this state typically we have energies of something like the order of a million electron volts from the fission but we create thermal neutrons which have energies of 300 kelvins which is 140 of an electron volt we create them by taking the high energy neutrons and then letting them diffuse through a substance that has a lot of hydrogen and typically paraffin is used for this and what comes out of so they bang in the hydrogen that was slowed down and then you get a gas of a room temperature of neutrons and by running it through holes it can turn this into a beam now of course they have varying energies distributed by the voltage distribution but you can select out and give an energy in various ways in fact one of the ways to do it is to use the crystal as a diffraction rating and then the different energies come off at different angles that you can't control by diffraction ratings so by taking an angle of energy it turns out the wavelength of thermal neutrons is comparable to the spacing of the atoms inside the silicon crystal it's not one expert so the result is when the neutron beam passes into the crystal neutrons have a definite energy go off in definite angles like this and people did very interesting experiments in the first place just looking for interference basically testing the boys my idea is about on the waves if you have a second crystal like this then you'll get a bright scatter going back and the other angle coming back together you can recombine the beams so this is like what you do in optics with a beam splitter using hats over mirrors to recombine the beams but this is not about neutrons so what you do is besides testing things like the sled experiments there's something else you can do here and you can insert in one of these beams the region in which magnetic field can be applied magnetic field causes the neutron spin precess the amount of precession depends on how long it spends in the magnetic field and how strong the magnetic field is but the spin precession is actually a rotation out there as we'll see later on it's actually a rotation out there exactly the form that's been buried here it's a rotation let's see I guess I erased it it's the spin rotation with some axis of the direction of the magnetic field in the angle of proportional time so by adjusting the magnetic field strength you can make the angle equal to 2 pi or any angle you want for the electron passive excuse me, the electron passing through this magnetic field region if you turn the magnetic field off there's no rotation at all so by doing this you introduce by rotating by 2 pi for example, the theory is direct in this phase shift of minus 1 in one of these two beams compared to the phase shift that you'd have if there was no magnetic field and if that phase shift is really there you should see a shift in the interference fringes where these two beams come together I wanted to say that the experiments fully confirm the predictions of this formula here rotating an electron by 2 pi really does change the space I write for so this is real it's it has to be taken seriously that's the basic story in double value representation of the rotations by means of spin rotations it's one of the most important facts about spin rotations I don't have too much more to say about spin rotations, maybe a few things which I'll do next time and then the next step to summarize just a little bit where we've been where we're going to go, where we've been was we developed a general strategy for finding rotation operators it is to find the angular momentum first it has to satisfy the right commutation relations then we guess what that was for spin line half systems and worked out some of the properties the next step after we're done the spin line half systems is to solve that representation problem in generality to find all possible quantum operators J which satisfy permission operators satisfy the angular momentum commutation relations and then find that the rotation operators are the corresponding