 Ok, mae gennym ni yw y gallwn yma yn gyflau Paulyw. Mae'r ystafell yn dweud y cyfrinol, ac mae'r rhaglau. Mae'n gweithio'n gwybod, mae'n gweithio'n gwybod. Felly mae'n gweithio'n gweithio'n gweithio'n gwybod yma yn gyflau Paulyw yn gyflau'n gweithio'n gweithio'n gweithio'n gweithio'n gweithio'n gwybod. ac mae'n gofyn yw'r ddweud i'r ddiweddol cyflawn yma. Mae yna'r ysgrifennu ei ddweud, oherwydd rwy'n cael eu cystechau, ac mae'n rhaid i'w ddweud ar y ddweud o'r ddweud o'r ddweud o'r ddweud, am yr hyn o'n gwneud o'r gwaith clasical yma, ac yn ystod y boddyn ni'n meddwl i'ch gwneud o'r magnonau. Mae'r bwrdd h-bars yn ddweud o'r llaw o amgrifnol, yn oed o'r ddweud o'r maen i gael o'r ddafï. Nid yw'r cydodd, yr ystyried yr ysgrifennu am gweithio gyda gyda'r cydodd? Mae'n ddysgu'r ysgrifennu, mae'n ddysgu'r ysgrifennu, y gallai gennych gyflyniadau sy'n ddysgu'r ysgrifennu, sy'n ddysgu'r ysgrifennu sy'n ddysgu'r cydoddau sy'n ddysgu'r ysgrifennu. Diolch yw y maen nhw'n leisio mlynedd o e s M teimliadau Ys Z S M. Ewch chi'n adeilad o'ch syniadau ar hyn o'ch ei ddysgu nifernau, mae'n trefiad i'ch gwahanol i ddod yr E S Z, ac felly mae'r ddau a'r ddau o'ch teiml ac yn awg o'ch dyfodol wrth uwch ar herfionmau yma, ac mae'n rhaid i'ch gilydd rhoi i'ch peisio i ddau o'ch ddau i ddau, roedd mae'n rhanion ar E s i'ch ddau. Ac yw'r oedd o'r amlwg ym Mhwysgrôl i ddwylliant. A dych chi oedd oes eisio'r Sx, fy mod i ddwylliant eisio Sx byddwch eisio a phot-dwylliant Sxょ'r Mhwysgrôl, pan mae gen i'r ragwyr nesaf y cerdd, ond mae gen i'r ragwyr cerdd yma ym Mhwysgrôl a os mwyn o'r ragwyr yn oed, a oed yn cerdd arall. So yna fydd loser o'r matrix i gyngenol. y ddweud y ddweud eich part yn y ffagorau am y parlymaetriad, ond y ffagorau yn ei ffagorau'r part yna, ac mae hynny'n bod y ffansion Alfa... ysbyty'r S-1 yw'r rolyf M, felly Alfa, y M, yw'r Rhyw Yn M, ac yw'r Rhyw Yn M, mae'r ddweud... yw'r ddweud ymwysliwch, ac mae'r Rhyw Yn M, yw'r rhaid. hawn here. Just some square root. So, for example, it is very straight forward to pick a large value of S. In this diagram here that should be up there, is for the case of S equals 40. Then for this large value of S to have your computer find the Igen values, sorry, not the Igen values, we know the Igen values, out of these three matrices I can construct if I take n is equal to for example nothing sin theta, cos theta. So this is a unit vector which lies in the, this is theta, this is the Ez direction, this is the Ey direction. felly dyna'r unid ffactor n. Felly, mae'r unid ffactor wedi'i gwneud hynny, mae'r unid ffactor wedi'i gwneud hynny, ddyn nhw'n gwneud y matrix o'r ysbyn ar y dyfodol, mae'n nx, sx plus ny, sy plus nz, sz. Felly, rydw i'r ysbyn ysbyn ysbyn ysbyn ysbyn ysbyn ysbyn ysbyn ysbyn, mae'r rau llawer y dyna sy'n dyfodol, ma'n ei gref i er mwynwyr i fynd i'r dyn nhw, sgwm ffodol I, llunio i sylwmiad, ac rydw i ddych yn ceisio eu dŵr i'ch creu, ein pinion sydd ei gaeli, oherwydd mae'n un cyffredin Seinre, mae'n ddyn nhw i dydyn sydd yn ddwyf yn un, ddyn nhw i'r un, rydyn nhw i'r wneud, dyna ddyn nhw i'r un, dyna ddyn nhw i ddyn nhw i ddyn nhw i'r un, A1 thru AS. Mydai ychydig wedi fy ng kaloiri BoDeryll прин........ The State in which you are guaranteed to get the value S. Now I'm supposed to have the maximum possible value for the angular mentum… In the direction of N. You're expressing that state as a linear combination of states with different amounts of angular mentum版w y Z axis. Yn rhan o rhan o'r ddiadau o'r fforddiwn gwahanol mewn gwahanol, dyna yw'r fforddiwn eich rhae, i meddwl bod ein zos嘲i un i ffodol ei ddull i ddull i ddweud wedi ffodol ei ddull i ddweud mwy ffodol ei ddull i ddweud, O'r ffordd mae'r pethau ffaldio'r Mercedes-Benz-Benz C-2S'u rhai, erbyn fytrwch chi ddatgol maith o'r ffordd, a y tro yna bod, mae'r FFees 40 yw 81, unrhyw o'r gemynydd y gwyloedd yn 80. Felly, os gallwch chi'n hyffordd, a'r rhif wythbaeth yw ffaldio'r foto'r ymd CEO, mae'n rywun ar gyfer 3 rwyfydfa ffordd, three different values of cost of cost the of theta so this is for cost theta is point five this is for cost theta sorry minus point five this is some other value of cost theta can't quite read it and so on right so so for this if you if you take this value of cost theta which is minus 60 degrees then the then these a's which of course are complex numbers have moduli that look like this so there's they're non-zero in some interval around here which is to say so what does that mean physically what are these numbers this is the amplitude that if you would measure along the z axis so first get your system into this state your system being in that state we would understand it to say that its spin is in the direction of theta what's then is the this is this becomes the probability to measure that it has s units of angular momentum along the z axis this becomes the amplitude to find that you have s minus one units along the z axis and so on so if the angular momentum so if the angular momentum vector really were a direction of theta how much would we expect to find along the z axis so classically in this state theta n we expect sz to return s cos theta s cos theta is the projection of a vector of length s pointing in the direction of theta it's that it's the projection of that down the x axis down the z axis sorry so and what what you're finding here is that these these amplitudes peak around the place where classical physics would say this is the answer and the quantum physics is saying well you you have a chance to get all these answers with probabilities which are given by the square of these numbers so quite strongly and as you change theta so as you change the vector you change the state the input state you change the direction of your spin in you you change the place where these amplitudes peak so that's only for s equals 40 and classical objects have s of 10 to the 30 or whatever and as you get more and more as s becomes bigger there are more and more of these dots along this line here there are here 81 dots I suppose 81 numbers have been calculated right because there are 81 components in the vector by the time you've got to 10 to the 31 dots you'll find that they you know that they're really completely peaked around here so that's how we out of this quantum mechanical stuff we recover at high spin the classical idea that things point in some definite direction and you can go on to show that the that the expectation value of sy which classically should be should be s sin theta is indeed s sin theta and what's more the uncertainty you can work out the rms you can work out what the expectation value of sy squared is and you find that that's essentially the same as the expectation value of sy itself squared in other words there's no one there's very little uncertainty at high s in what you will get for sy so so these these amplitudes what's happening here is in quantum mechanics we have to calculate a whole series of numbers which are the components to describe so to describe the the the spin state of something we have to construct in the case of spin a half two numbers in the case of spin one three numbers and so on two s plus one numbers we have to calculate being the amplitude to find the various possible answers on sz if you would make the measurement s set what we're doing essentially is recovering the probability distribution for the sz measurements which in classical physics is a delta function glitch at s cos theta but in quantum mechanics we don't our probability distributions are not delta functions there's some kind of spread out things and you're seeing what they are there but as you go to higher and higher spin amounts the probability distributions narrow down around the direction of spin which classically so in classical physics we say the direction of this spin is given by the Euler angles or by the by the polar angles theta and phi we just have some completely definite come on you stupid thing we have some completely definite direction and what whereas in quantum mechanics we need a whole load of numbers because we're defining a probability distribution in classical physics it is strictly speaking a probabilities probability distribution but it is a delta function and all we have to do is specify the the centre point of the of the of the delta function probability distribution and we do that with just two angles and in quantum mechanics we need a load of different numbers to spell out the whole probability distribution properly now the other thing i wanted to say on this topic of you know relating quantum mechanical levels of anglimentum to classical levels of anglimentum is uh is the importance of this so we know that s squared has uh the total anglimentum operator has e values s s plus one which is clearly greater than s squared and remember s this thing came into the world as the maximum value of the anglimentum around the given axis so and how much greater this is than this depends on the value of s so when s is a half we have s s plus one is clearly equal to three quarters which is three times the quarter squared sorry a half squared right this is the maximum value and that's telling us that you can have you always have a third of your spin down each of the three axes uh if you have a spin half particle and the most you can ever know is whether the one component is pointing this way or that way but we never remotely get the spin properly aligned with one axis because there will always be two units of anglimentum somewhere other in the in the plane orthogonal to that chosen axis uh so when we have s is one we have s s plus one is equal to two um which obviously is two times one squared so now the the amount of anglimentum we can have down one axis is twice is is a is a whole half here it was only a third now it's become a half of the total anglimentum and each each orthogonal each direction uh in the perpendicular plane has has less than in the direction that you've chosen to align the anglimentum with as you go down to large values of s you have that s s plus one is practically equal to s squared because s because obviously s squared is going to be by definition bigger than s um and that means that we can get essentially all of our anglimentum pointing down a given axis so the important message is from this that we're familiar with this regime where we can get something to point in a well-defined direction uh but the atomic world works in this regime where there's always loads of anglimentum in in the directions that you haven't been working on okay so i now want to turn to a new topic which is the addition of anglimentum the last thing we have to do with anglimentum so this is a very important topic for um atomic physics because atoms contain um i mean the simplest atom hydrogen already contains uh a proton that carries a half each bar of spin an electron has the same amount of spin and then the electron may have orbital anglimentum it may have anglimentum by virtue of its orbit around the proton so a generically hydrogen atom contains three units of anglimentum and uh we want to know so what are the states of the atom in which the atom has well-defined anglimentum so we're going to study uh and this is an application of the machinery that we introduced i guess early this term to discuss composite systems this is a classic uh this is an application of the of our theory of composite systems so if you're feel unsure about the theory of composite systems please go back and have a look have another look at it because this is what we're going to be applying it's all that stuff about Einstein bodalski rosin et cetera yeah what underpinned that because what we're going to do is we can we're going to consider two gyros we're going to have uh we're going to have gyro one has j one so it has has has total j j one j one plus one so it has m m lies between minus j one and j one so that's the rate at which this gyro spins is fixed by some server motor or something right it's and it's spinning at this rate and we're going to have gyro two um which obviously is going to have total anglimentum squared is this so so this will be m one and m two is going to lie between minus j two and j two so we've got these two gyros uh there might be objects belonging to a navigation system and we're going to stick them inside a box and they're not going to talk to each other they're going to have there's going to be no Hamiltonian there's going to be no coupling physical coupling between these two gyros at all but we are going to put them in a box close the lid and then say so what are the states in which this box has well-defined anglimentum and they will turn out to be what we will find is that when the box has well-defined anglimentum if you open the lid and and ask what happens if i look at the anglimentum of gyro one i will get a variety of different answers it will be uncertain what i'll find for gyro one and gyro two will have an anglimentum that will be correlated with gyro one so when the anglimentum of the box is well-defined it has a definite amount of anglimentum and it's pointing definitely in some well you know the amount parallel to z axis is is definite when you look inside the box you'll find it is uncertain what the anglimentum of the bits are and we'll explain physically that it's a physical necessity that's the case that's not mysterious uh but we'll i hope make it make it evident that that's so right but the moment we're going to address this kind of mathematical problem we know that the states of the box oh no no no sorry so the states we have we have two sets of completes of complete sets of states complete sets of states are going to be j one m there's a family like that and there's a family j two m two so right this should have an m one shouldn't it since we know what what the total anglimentum of the first gyro is the only thing to discuss is what its orientation is and if i consider the set of states like this j one m one uh with m one ranging between minus j one and j one that's a complete set of states for the first gyro this is a complete set of states of a second gyro in other words um we will be able to write any state of gyro gyro one as some some uh a a m one j one m one et cetera right so one of the states what what's a complete set of states of the box it's the set of states j one m one j two m two right we we discuss that that if we have a system a and a system b a complete set of states uh is obtained by taking a member of the complete set of a and motoring it by a member of a of any other member of the complete set of b uh if you take linear combinations of those you get everything in other words the box can quite generally the state of the box can be written as some sum of b uh m one m two j one m one whoops it's meant to be a pointy bracket j two m two and we want to find the states of the box any state of the box can be written like this where these numbers uh for suitable choices of these numbers these amplitudes what we want to do is find the states of the box which are eigen functions of the box these angular momentum operators and do you remember but when we discuss these things these composite systems we had that um that you added the operators of of systems of subsystems and it uh and you multiplied there and you multiplied their kets that's how it worked so we want to consider now what the relevant operators are well we're going to have for gyro one we have j one squared we have j one z and we have j one plus and j one minus the raising and lowering operators where this is equal to j one x plus or minus i j one y and of course we will have the same kit of operators for the for the second gyro that's for gyro two and then for the box we will have we will have j squared which will be j one vector plus j two vector squared and we will have j z which is equal to j one z plus j two z and we will have j plus minus which is equal to j one plus minus plus j two plus or minus so we add the operators belonging to distinct systems uh of here because it's a square you know right this thing should be the vector operator belonging to the box squared so we add the individual the the vector operators belonging to the individual boxes so we have to do a bit of these these are fairly straightforward we have to a bit of footwork on this so let's find out let's let's expand this we have that j squared for the box is equal to j one plus j two dotted into j one plus j two and this is not i mean this there's nothing funny going on here because the operators belonging to distinct systems another another thing we covered in the whole in the composite system discussion operators belonging to distinct systems always commute so we can multiply this out just as if they were ordinary weren't operators but just ordinary boring vectors and find that this comes to j one squared plus j two squared plus j one dot j two twice over this is because j one i comma j two j commutator vanishes operators belonging to distinct systems always commute well this is fine this is in our list of operators but this is not in our list of operators right j one dot j two is not up there so we need to we need to we need to write this in terms of things that are up there so we say j one well okay no i want to get an expression for that in terms of the things already written up here and what i do is i say let's consider j one plus times j two minus that is j one x plus i j one y j two x minus i j two y which is going to be j one x dot times j one x times j two x plus this on this will give me a j one y j two y which these are two of the components of the elements that are buried inside the j one dot j two but i get other stuff unfortunately which is i get plus i j one y j two x minus j one x j two y so this i want this i don't want but we can get rid of this by arguing that if i write down j one minus j two plus so reverse the plus and the minus this will everything will carry across the first two terms will will emerge but what will happen here is that this will become this minus sign will migrate from here to here because i've changed where the minus sign happens here so i'll get minus i j one y j two x plus sorry minus j one x j two y so when i add these the left sides these pesky terms that i don't want will go away and i will have that j one plus j two minus plus j one minus j two plus is equal to twice j one dot j two minus j one z j two z right because these two taken together make j one dot j two minus the z bits right which are inside here so now we have what we want which is an expression i now go back to this j squared here and replace that with stuff to do with j plus and j minus so i now write that j squared is equal to j one squared plus j two squared and then i want i want this so i take this i take plus j one z j two z plus j one plus j two minus plus j one minus j two plus so this disgusting mess on the right expresses j squared the total and momentum operator of the whole box in terms of operators whose action upon the states of the box i know that's the key thing what i've been doing here is getting an expression where i know what every one of these operators does on those states those states of the box j one m one j two m two right i do not know what j x or j y does to those things it makes a disgusting mess but i know what every one of these operators does to those things that's what i've that's the purpose of this algebra okay so now now a little physical argument suppose you've got your first gyro pointing in the z axis sort of aligned with a z axis and you've got your second gyro appointed aligned with a z axis then you'd think that your total angular momentum would be will be the sum of the angular mentor of the two gyros because they were both parallel to the z axis you would argue they were parallel to each other and you'd have the total angular momentum in the z direction so what we do now is we investigate j one j one j two j two this physical argument suggests that this is the object uh j one plus j two comma j one plus j two so this is a state of the box in which it has this much angular momentum and all of it pointing down the z axis on the grounds that if you take two gyros both pointing in the z direction surely you've got a box with surely their angular mentor just adds so we want to show that this is the case physically it seems reasonable physically is it true we we check that it is true by applying the relevant operators to both sides right so if i if i do j z on this i'll just say j z on the l on the left hand side what do i get i get j one z uh j two z plus j two z right because j total z is the sum of the z operators of the of the gyros operating on the right hand side which is j one j two j one j one j two j two so the way these composite operate system operators work is that this looks at this and we get j one because this is an eigen function of this operator with this eigen value times j one j one this stands idly by j two j two uh so that's that and then i have plus this looks at that and produces a j two j one j one standing idly by j two j two produced as the eigen ket so indeed we get j one plus j two times what we started with j one j one j two j two so that confirms that this object is an eigen function of this operator for the box with the expected eigen value uh yeah there probably isn't that because this came across onto this side and we wanted a two yes thank you very much there is a factor and this was about to be important isn't it there is a factor of two there because we wanted a two j one j two from up there and we had twice this which came onto this side of the equation so so that's that's that now we check j squared what does j squared do when it's applied uh uh it's going to be j squared on this so j squared i want to do j squared on on the right side and j squared we've discovered is j one squared plus j two squared plus two j one z j two z plus j one plus j two minus plus j one minus j two plus all that disgusting mess has to operate on that operates on j one j one j two j two well this operating this is an eigen ket of this operator with eigen value j one j one plus one and it will then return this and we'll find that this gets returned so i'll just stick it in the back as a common factor similarly this one looks at that and produces j two j two plus one times itself then j one z looks at this and produces a j one times this um and j two looks at this and produces a j two times this so now we have a plus two j one j two and that's the action of this operator on this product then j one plus looks at this tries to raise this trailing j one to j one plus one but you can't because because we're already at the top so it kills it so the j plus operating on this kills it and it doesn't much matter it does not matter what j two minus does to this because it's multiplied by nothing similarly when this j two plus operates on this it kills it trying to raise that j two to one more so so the action of these two operators on this is to produce nothing and i can close the bracket just there so so j squared actually this really should be on the right hand side j squared on the right hand side produces this bracket times this ket which shouldn't have been written so far to the right and we can now rearrange this because we got two j one j twos i can take one of those j one j twos and and and deal with it by putting it inside there so i can write this as j one j one plus j two plus one so i've to this bracket i've added a j one j two that's one of those and the other one i put inside this bracket by writing it as j two times j one plus j two plus one so this one this this j one produces a j one j two which is the other one of those so this is how much i've got of j one j one j one j two j two and now i can immediately see that this is j j plus one of j one j one j one j two j two where where j is j one plus j two so that proves that the thing it proves the conjecture we started with that this object is an eigen state of the box with the eigen value with a total angular momentum eigen value j one j two so that this this establishes which establishes so we've proved by hard work that j j sorry j j this being a state of the box is equal to j one j one j two j two where j is j one plus j two that was rather hard work the next bit's easier um because we can now apply the minus operator the j minus operator to both sides of this equation and on the left side we'll get some multiple of of of j j minus one on the right side we'll get something more interesting so now we apply j minus which is equal to j one minus plus j two minus to both sides j minus applied to j comma j produces there's a square root here which turns out to be j j plus one minus minus it should be m m plus m m minus one but m is j so minus j j minus one times j comma j minus one so i've applied my my the box is tipping operator a lowering operator whatever you want to call it that tips its angular momentum away from the z axis and we get this multiple i'm appealing to the stuff we showed when we started on angular momentum for this for this square root times the state tip where it's where we've got we're tipped at one unit away from the z axis so that's what we get on the left hand side um so this is the lhs on the rhs we have that j one z uh plus j two no no not z minus minus take your one minus j two minus this sum is the same as this operating on j one j one j two j two which we proved is the same as this what does that give me it gives me the square root so this j one minus interrogates that it reduces a square root let's evaluate this square root sorry this square root simplifies because we have a j squared and a j square with a minus sign and a j and a j with a plus sign so this becomes the square root in fact of two j j j minus one so what we're going to get now is the same situation we're going to have j one minus working on this is going to produce the square root of two j one operating on j sorry and the output will be j one j one minus one this will stand idly by whilst that happens j two j two and then we have to add that's this plus sign here the result of j two minus banging away at that whilst this stands idly by we'll get a root two j two j one j one standing idly by j two j two minus one being produced so we have that just to we can now equate the left and the right sides again we can say that the state j comma j minus one is the square root of j one over j of this johnny one no no j one j one minus one j two j two plus another square root which is j two over j of j one j one j two j two minus one so what have we shown we've shown that when the box has angular momentum that's tipped a touch away from the z axis if you open the box you there are two and look and look at the individual gyros inside the box there are two outcomes you might find you might find that the first gyro is tipped away from the axis and the second is all parallel to the axis or you might find that the first gyro is on the axis and the second is tipped away from the axis and it's inevitable that that has to be the result one of these gyros has to be off the axis but they can't both be off the axis because then we would have only we'd be two units of angular momentum down the axis short so what we've got here is correlated states of gyros the gyros have become entangled in this state in which the box has well defined angular momentum the the results of measurements of the two individual of the gyros in the box are in are have become entangled let's make a picture now to help organize these calculations because we've just begun unfortunately on what in principle is a is an extensive exercise of calculating but from now on it's almost mechanical the way to go is to uh is to is to put your original state j j up here it's why i'm putting up there because the origin somewhere like here and this is j units up from the origin and this is this is j units down right so here we have a minus j is j units down here's the origin we started with this state and established what it was that it was j1 j1 times j2 j2 then we used the lowering operator j minus to move around this semicircle to a state here that we've just constructed which is j j minus one which turns out to be a linear combination of this and this and we found out what the factors are that make the linear combination and we can now apply we can now take this state and we can apply our lowering operator on this to generate state here which will be j j minus two let's imagine doing that we won't do that but just imagine doing that if we would apply the j minus to this we would get some multiple some wretched square root times that target if then when we apply j1 minus plus j2 minus to this side we get four terms because each of j1 minus and j2 minus works on this and this so each of these two things generates two terms when j1 minus works on this we get j1 j1 minus two times j2 j2 standing idly by so let me just do this we can say that j j minus two is an amount of and we could work out what this amount I mean it's straightforward to work out what this number a is going to be but we won't do it of j1 j1 minus two times j2 j2 then when j2 minus works on this it produces some amount of j1 j1 minus one j2 j2 minus one right because because it lowers this to j2 minus one whilst this stands idly by then when we use j1 minus on this we get some more of what we've already got we get this gets lower to j1 minus one and we get some more of this which we can absorb in this b and then when j2 minus works on this that goes down to a j2 minus so we get plus another amount of j1 j1 j2 j2 minus two physically what does this say it says that if your box has its angular momentum tip two units away from the z axis if you open the box three that you may find three things you will find one of three things either that the second gyro is still bang on the axis and the first is tip two away or that each of them is tipped a bit away from the axis or that the first one is bang on axis and the second one is tip two away so it's perfectly reasonable what you see when you open the box is what in some sense if you thought about it beforehand you would have expected to see this apparatus will deliver you the numerical values of a b and c and therefore tell you the probabilities of those three outcomes so we have it we're getting complete information of what we will see if we do open the box and we can plod on like this until we're completely worn out the expressions will become you can see it looks at these expressions as we go around here are getting more and more horrific because if we apply j if we do an if we do another lowering on this apply j minus to this and j1 minus plus j2 minus to this this will have a they'll be this will give us a term j1 j1 minus three whilst and so on so we'll have more terms mercifully in the real world when you're dealing with small values of j there comes a point at which this the lowering operator j1 minus will simply kill this because for example if j1 were the number one this would be j1 minus one and when the might when the lowering operator worked on that it would try and lower this to a number more negative than that and it would kill it so the expressions get more and more complicated as we go down here and it turns out that when you go along here they start to simplify because you get more and more of the lowering operators killing their targets and the you get you get sorted out and you'll find that you arrive down here at j comma minus j you will find that this is simply what it has to be physically but you will discover that it is j1 sorry j1 minus j1 times j2 minus j2 that is to say you will find automatically that that when the box says its angular momentum in the minus direction open the box there's any one thing you can find which is that both gyarrers are pointing in the minus z direction and it's worth doing that not in general j but it's worth going all the way around for example for j for each of the j's are equal to one and one of the j's equal to one and one equal to a half say it's good to see that that happens so in order now to complete the setup there's one more thing we have to do which is uh well strictly speaking we should do some state counting I suppose uh why don't we do some state counting um so the number of basis states the number of basis states of of the contents of the box is 2j1 plus 1 times 2j2 plus 1 this is the number of ways we're allowed to to orient the angular momentum of the first gyro and for each such orientation of the first gyro this is the number of ways you can orient the second gyro so that's the number of of possible states of what's in the box um the number of states of the box um should be the same because because whether it's our choice to either think about the whole box or to think about what's in the box so there should be as many states of the box as there are what's in the box and how many have we got so far we've got 2j1 plus j2 plus 1 right because this is j1 plus j2 and going around this circle we get 2j1 plus j2 states and that's much less than this if if j1 and j2 are big so we haven't got enough states and uh it's intuitively evident that if you have two gyros in a box their angular momentum don't have to be parallel to each other they can be inclined they might for example be anti-parallel in which case you would have only you think j1 minus j2 of angular momentum so what's the problem here is we've got all the states in which the two gyros are parallel to each other despite despite what you might think by looking at these expressions here right remember this is just remember these these these gyros have angular momentum other than what's appearing in the z direction they've got angular momentum in the x y directions as well so this may look as if the two gyros are not parallel to each other but they are and there's a problem in the problem in problem set five about hydrogen which makes which illustrates that point okay so these are all parallel to each other so what we need is the states which are not parallel to each other uh and uh the way to go is to say is to is to find what the what the expression is for the state for this state which is going to be the state j1 well sorry j minus one j minus one because the two gyros are not parallel there's a bit of cancellation of their angular momentum uh but all the angular momentum of the box is parallel to z axis that's this state here this state is a linear combination we calculated in fact it's a linear combination of where of this state and this state and i argue on physical grounds that this state should be another linear combination and it must be orthogonal to this because this is an eigen function an eigen state of the j total j squared operator with an eigen value different from from this so i now argue that the j minus one this is a state of the box j minus one is a linear combination of j1 j1 minus one j2 j2 and j1 j1 j2 j2 minus one it must be a linear combination of these two and and we have to choose a and b so that it's orthogonal to the state we've already got so comparing above you can see by inspection the condition that j uh that j j minus one j minus one j minus one equals naught that equation implies that a is equal to minus the square root of j2 over j and b is equal to the square root of j1 over j if you put in these choices for a and b you you found a state which is orthogonal to that and then we can you can if you're if you're a skeptic you you've you've got a well-defined state you can apply j squared to it and show that it produces you the expected eigen value and it's trivial to see that this thing has an eigen value j1 minus plus j2 minus one for the jz having got this we can apply the j minus operator mechanically to find this state and this state and so on all the way around down to here and this is how we construct the states of the box so we better talk a bit more about this it's time is up we better talk a bit more about this on Wednesday but we've got the main ideas