 chickens. So let's get underway. We were talking about Spinner Half, the most important type of spin, yesterday, and we got this far. So any state as regards its spin, its orientation should be expandable as a linear combination of the state plus which means you are certain to get a plus a half if you measure the spin along the Z axis 100 oes. The will be these coefficients here, and the complex number here. The amplitude to measure plus half on S-Sed, or the amplitude to measure minus half on S-Sed and we're calling these, it's obviously a handy notation to call that thing A and this thing B. And then what we want to be able to do is write the result of using some ysgolwydd o'r ysgolwydd wedi'i gael sydd. Roeddwn i weithio adeil y cyfnodol i'r newid gan maen nhw er mwynhau a mwy i'r gwerthio hynny bod ysgolwydd ystafell gyda'r sefydledig sydd o ganddi-ganddi-ganddi sydd. Fe'n gwneud amdano dwi yn gwneud hynny iddyn nhw'n gyflaen amser C a D o'r ddyn nhw'n meddwl i'r g beautol yw'r gweithlo. if on the right we put in the two numbers that characterize psi on the right, we get out on the left the two numbers that characterize phi after we've multiplied by this matrix of four complex numbers being the expectation value of the relevant whatever operator we're trying to use between the plus states, the minus states and then these non-classical off diagonal bits on each side. And we said, I think we finished by saying that if i is z, in other words if we're interested in the result of using sz on a psi, then this matrix is very simple because sz on plus is simply a half of plus. So we get a half appearing here, we get minus a half appearing here because sz on minus is minus a half times minus and we get nothing appearing here and here because plus and minus are orthogonal. So we have this diagonal matrix which is no accident, it is simply the matrix that contains the eigenvalues of sz down its diagonal because we use as basis vectors the eigencats of sz. We made that choice and the result is that the matrix representing sz is diagonal with its eigenvalues down the diagonal and this matrix is conventionally written as a half times this matrix which is called sigma z and it's called a Pauli matrix because Wolfgang Pauli introduced it into physics although it was known to mathematicians. The matrix is like this. Okay, so more interesting is if we ask ourselves what's the matrix for sx. So the matrix from sx is going to involve things, well we're going to have for example plus sx plus, this is a complex number, we want to know which complex number and the secret is to write s, of calculating this, is to write sx is a half of s plus plus s minus where s plus minus, so the matrices, sorry, are the operators that we already introduced in the context of j and l to reorient the angular momentum either towards the z axis or away from the z axis. So these are sx plus and minus i times sy. So this operator was introduced in the form of j plus minus but remember spin and total angular momentum have the same commutation relations, the same behavior in every way. So these ladder operators are this and obviously if you add s plus to s minus you get to sx because the sy terms cancel, so this is definitely the case. So this thing here can be written as a half of plus s plus plus, sorry, sorry, well this is what I'm trying to calculate, s plus s minus plus. S plus tries to raise this to an even larger value, this is plus a half, it will try and raise it to plus three halves and the message value is allowed because the total spin is only a half, so it kills it in the process, therefore this one is zero. S minus successfully lowers this to minus but minus is orthogonal to plus so this is zero, so this element here is zero and that's the top left corner of the matrix for sx is zero. Similarly, exactly the same reasoning would need you to conclude that the bottom right hand corner is zero and the non-zero elements occur off diagonal so if we look at plus sx minus we're looking at a half of plus s plus minus plus s minus minus. S plus raises minus to plus, successfully s plus on minus is exactly one times plus so this number here is equal to one and minus tries to lower this and kills it in the process and therefore this is equal to zero so this off diagonal element is in fact equal to a half. We know that the bottom right hand element is the complex conjugate of the top right hand element because s is a Hermitian operator so we know now that the matrix is sx is represented by the matrix half of nothing one one nothing also known as a half of sigma x, the Pauly matrix, this is the Pauly matrix sigma x and when we do the same thing to find out what sy is we write this as a half of plus sorry one over two i of s plus minus s minus because if you take the difference of sx plus i y and sx minus i y you will end up with two i sy so we have this and what do we get this s plus raises this to minus to plus so so plus s plus minus again equals one so therefore this is equal to one over two i also known as a half minus a half minus i over two so the matrix representing sy is going to be is going to be a half of one minus i i sorry one nothing nothing the diagonal elements will be nothing for the same reason that they were with x also known as a half of Pauly's matrix sigma y. So that's where the Pauly matrices come from. There's simply the matrix representations of the spin operators in a basis in the base when you choose as your basis the eigen vectors the eigen ket of sigma z. So let's use these use this apparatus to do something slightly interesting it's an excellent exercise both in practicing getting experimental predictions out of this abstract apparatus and also we learn something interesting about how how the orientation of atomic scale things behave somewhat counterintuitive arrangements. I don't think this computer is going this system projection system is going to work today for some reason so okay so the point is that so the point is that a spinning charged body is a magnetic is a magnetic dipole. I think that's kind of plausible so that so electrons neutrons protons except well sorry not neutrons electrons protons being spinning charged bodies have little magnetic moments they are little magnets. So if you put a magnet in a B field you have this is the energy of a magnetic dipole in a mag field. So there's a minus sign here which says that the energy is lowest when the magnetic when the dipole is aligned with the magnetic field. So when this dot product is positive the energy is lowest so that's why magnets compass needles and whatever align with the magnetic field. That also means that if a magnetic dipole is aligned with the field it the energy will drop as it moves into a region of bigger field because it'll this will become a more negative number. Whereas if it's anti aligned with a magnetic field then its energy will increase if it moves into the magnetic field because this will become a this will be negative and the two minuses will cancel. We have a more positive energy so since things tend to move in the direction that minimizes their potential energy we have that magnets aligned with B will be sucked into a region of stronger B. So a magnet dipole aligned B so that means that mu dot B greater than nought is sucked into a field. If the field strength varies spatially which it often does the what particles which have their fields dipoles aligned will be sucked into B and similarly the other ones will be repelled. So the anti aligned whoops aligned a poles will be repelled from a region of high B. So that was the physics that Stern and Gerlach exploited in 1922 in experiments which astounded the world. They found themselves they made themselves a magnet or should we call this north and we'll call this south. They made themselves a magnet which had pole pieces one of which was pointy and the other of which was flat or even well I think it was flat. But you could it could also be concave like this and then you can imagine how the field lines run the field lines run like this somehow. I'm not doing a very good job of it my diagrams are usually rather rubbish. The point is that here we have a crowding of field lines which means we have high B near knife edge. So I have a nice picture of this but the computer isn't willing to show it because this is the end view of a long of a long thing. So this is like the point of a knife right we're looking end on a point of a knife and this is just this is a table somehow. So if you if you have some particles with some spin coming in here and name it right so that they're heading for this. Well they're heading a bit below this region of high magnetic field like this. Then the ones that have their spin aligned this way into B are going to be sucked into the region drawn attracted by the region of high B near the point of the knife and move on up here. So this is the particles which have mu dot B greater than north and particles with anti aligned with mu dot B less than north will come down here. Of course these all grotesqually exaggerated in fact you'll have a very subtle curvature and then you'll have a straight line. So we get the particles deflected either way. So what they did was they took silver atoms because silver atoms turn out to be spin a half particles coming in here. Then what and they found which surprised them and everybody else that half of their particles half of their silver atoms went off this way and half of the silver atoms went off that way. So when they they detected the atoms on a screen over here they got two blobs distinctly separated. The quantum mechanical interpretation of this is that as these atoms when the atoms are in here they sorry I haven't said that mu the magnetic moment is equal to some number the gyromagnetic ratio times the spin operator. So when they're in here the magnetic field is as it were measuring their component of spin in the direction of the magnetic field. That's what you say to yourself. So there are only two answers possible either you'll get plus a half or you'll get minus a half for the value of this. And therefore mu will be either a half G in the direction of B or it'll be minus a half G in the direction of B. If it's so and the half for which it's plus a half G will be deflected that way and the other lot will be deflected down here and there you go. So at the end of the day you have a stern girlar filter you put in the particles with they've just come out of some oven. You've heated up some silver in an oven, made some silver vapour, allowed it to diffuse out of some holes, collimating slits and that kind of stuff. So it's coming along here with some thermal velocities and out of your filter you have a load of atoms which have their spins in this case up on Z and the ones that come out here are in this state. So it's a machine for it's a practical device for creating silver atoms which are in this state. Now you can play some entertaining games by installing another stern girlar filter. So let's just block these off, stop them being a nuisance, stick in another stern girlar filter here and now let's measure the, let's measure S sub N. So let's measure the spin along some unit vector N and let's take, so we're going to have this to be the X direction. We're going to have this to be the Z direction and what the Y direction will have to be out of the board. And what we're going to do is we're going to take N is equal to nothing, sin theta, cos theta. So N is going to be a vector which if theta is nothing is just in the Z direction and if theta is pi by 2 it's in the Y direction and it can be allowed to scan between these directions as we vary theta. And what we want to do is calculate which fraction of the atoms will survive will get through the second filter. So this is the filter F1, this is the filter F2 and you want to calculate the probability that an atom gets through both filters. Or let's focus for the moment on the probability that an atom that has got through the first filter gets through the second filter. So the probability that you pass F2 given that you passed F1 in quantum mechanical language is plus a half on N, given that you're, well we'll just say plus on N, given that you were plus on Z. So this is the state that you're in, up there it's just called plus, but now I put in a Z to distinguish it from this which is in the direction of N, that this is an eigencat of Sz with eigenvalue a half, this is an eigencat of S sub N with eigenvalue a half. And this pair of things makes me the amplitude by the basic dogma of the subject for the probability of this outcome. So I need to mod square this and I've got the probability that I want. So we can work this out, we can get this complex number as soon as we know how to write plus on N as an amount of plus on Z, plus an amount of minus on Z, right? Because, so if we get this number and this number then we have the probability that we want is going to be mod a squared because a star is going to be exactly that number. So out of this catch you could get the bra you want up there by complex conjugating it, you'd have an a star, bang in with plus on Z and you'd pick out a star. So the probability we want is just mod a squared. So that's our exercise to find a and b and we'll be all done. How to find a and b? Well what's the point about, what's the point, what's the defining characteristic of that cat? It is that it is an eigencat of this operator with eigenvalue a half. This defines N. It's totally characteristic of these sorts of calculations, a wide range of quantum mechanical calculations that this sequence of arguments. I want a certain complex number, it will involve some cat. Ask yourself what is the defining characteristic of the cat, it will usually be because it is an eigencat of some operator. Now we have a well defined mathematical problem. Find it because what is Sn? Sn is equal to a half of nx sigma x plus ny sigma y plus nz sigma z. Sort of a dot product between the unit vector n and the vector made up of the three palli matrices. Well nx is zero so basically we've got an ny, we agreed was going to be sin theta and this we agreed was going to be cos theta. So at the end of the day it is a half of, now sigma z we've got up there it's got one in the top left hand corner and minus one in the bottom. So I get a cos theta and a minus cos theta appearing on the diagonal because of sigma z. And this has got a minus i in the top right hand corner so we get a minus i sin theta appearing there and its complex conjugate has to appear down here. So this is the matrix that represents Sn where theta is defining the direction of n. Now all we have to do is say that this matrix cos theta minus i sin theta i sin theta cos theta on AB is equal to AB. This vector has to be an eigen ket of this matrix with eigen value one in order that it's an eigen ket of Sn with eigen value a half. Because the original expression was Sn on this equals a half of that but here is a half I can cancel on the two sides. So I'm looking for the eigen ket of this operator with eigen value one. Notice I don't waste my time finding out what the eigen values of this operator are. I know that because this is a matrix that represents a spin operator Sn I know before I start that the eigen values are plus and minus, well this one plus and minus a half of this one plus and minus one. So we don't waste time finding out what the eigen values are we just get on and solve these equations. There are two equations here but because we're looking at an eigen value problem only one of them. These two equations are linearly dependent upon one another only one of them contains useful information. The other one repeats that information. So we merely need to look at the top equation and it says that a minus one sorry sorry sorry a times brackets one minus cos theta. So if I'm going to get a cos theta equals a on the right hand side so if I go on the right hand side we'll have a cos theta a into one minus cos theta is equal to minus i b sin theta. In other words we're going to have that b over a which is all that I can, it's only the ratio of a to b that I can determine out of this. The absolute values have to be determined from a normalisation condition are equal to b over a is equal to one minus cos theta over minus i sin theta. And we can clean this up a bit if we use some half angle formulae because this on the top is twice the sin squared of theta over two sin theta is twice sin theta upon two cos theta upon two. So we can cancel a number of things, the two's cancel, one of the sin theta's cancel and we end up with sin theta over two over minus i cos theta over two. So I can write now that a b is equal to cos theta over two i sin theta over two. So if you work out the ratio b over a of these two I think you will get that because this minus i can be put as an i on the top. Moreover this thing is correctly normalised, it just happens. So in principle I would now need to deal with the normalisation, I've only been calculating the ratio of the components. I want mod a squared plus mod b squared to come to one but it jolly well does by good fortune. This is the complete bottom line, this gives you the probability that we pass f2 given that we passed f1 is actually equal to, we said it was going to be mod a squared is therefore cos squared theta upon two. Does that make sense? If theta is equal to nothing then the second filter is also measuring the z component of angular momentum and the output from the first filter is guaranteed to return plus a half for the z component of angular momentum so this probability must be one and indeed cos squared of nothing is one. If the theta is pi then the second one is plus a half, then n is pointing in the minus z direction so getting plus a half in the direction n is equivalent to getting minus a half in the direction z but we know for certain that we're going to get plus a half in the direction z so the probability of this happening is zero and indeed cos squared of, if I put theta equal to pi I'm looking at cos squared pi upon two which is nothing so that makes sense. If I put theta equal to pi upon two then we're measuring then the n direction becomes the y direction and we're measuring in a direction which is orthogonal to the z direction and then you would think that knowing what components of the angular momentum in the z direction was couldn't possibly affect the angular momentum in the y direction so you would expect that there was equal probability at passing the second filter that is to say of getting plus a half for the spin along y that plus a half on y and minus a half on y be equally likely by the symmetry of the situation and indeed cos squared of pi upon four or cos of pi upon four is one upon root two to cos squared of pi upon four is a half and that makes perfect sense as well so this formula predicts the kind of thing that you would expect. Okay, suppose we now have, we won't do this in all detail but let's just sketch it out suppose we have now another filter so we have f1 as before we have f2 as we've just calculated now suppose on the output of f2 we include f3 right, so this one is going to measure in the theta direction as said this one let's say this one has its axis in the phi direction also in the xy plane right so you measure first of all the spin on z then you measure on the unit vector cos theta nothing sin theta cos theta sorry then you measure and then those that return plus a half in that direction you measure in the direction nothing sin theta sin phi cos phi suppose we do that so the probability of passing f3 given that you passed f2 is going to be we'll call this vector n and we'll call this vector m say we'll use this notation this will be a half on phi a half on theta so the output from this filter definitely has particles with plus a half component of angular momentum in the direction defined by theta and I want to know the amplitude that those particles have will definitely give me a plus a half if I measure in the direction defined by phi the answer to that is according to the dogma of the theory it's that and I can expand that into here I can slide the identity operator taking the form of plus on z plus on z plus minus on z minus on z we've slid identity operators in many times before in more complicated contexts so this thing that we're doing here is going to be a half phi sorry that's a blunt end a half phi plus z plus z a half theta plus a half phi minus z minus z a half theta now these complex numbers we already know we just calculated them right this was a which we used this was b which we didn't use but we got it written down up there it's i sin theta upon 2 so this one here is cos theta upon 2 this one here is i sin theta over 2 but we also know what this is because this is going to be the same excuse me we have a complex let's just ask ourselves carefully exactly what is b b is actually the complex conjugate of this sorry these need complex conjugate signs can we remind ourselves actually where we are on this I'm now worried about whether I'm dealing with a complex some of these need complex conjugate signs what exactly are a and b they were defined just to get this right what we said was that a half on theta was equal to a plus z plus b minus z that's what we said that was the definition of a and b so what is this this thing here is plus on z a half on theta yeah so so what I said originally was correct there are no stars here so that's just for note right now back to this this is the complex conjugate of this is essentially the same as that with theta replaced by phi so we know that this will be the complex conjugate of this with theta replaced by phi this is in fact real so this is going to be cos phi over 2 similarly this the complex conjugate of this is the same as that with theta replaced by phi so I now have to write down the complex conjugate of that which is minus i sine phi over 2 so that's what that comes to so the probability that we get through f3 given that we got through f2 is going to be cos squared phi over 2 plus because that minus sign and that i and the pair of i's make a plus sign sine squared theta of phi which is also known as cos phi over 2 minus theta over 2 I think so does this make sense it tells me that I will if phi over 2 if phi is the same as theta I'm certain to get through that's good if if the difference in the angles is pi upon 2 then I have a chance sorry we have failed to mod2 the whole thing maybe there's muttering about that so this got expanded to this and this whole thing needed a mod2 and this needed a mod2 and we were doing various calculations and then this needed a mod2 and this needed a mod2 so it just became cos2 so when the angle is so all this tells us is that it had to tell us we would have been worried if we hadn't discovered this that the probability of getting through the third filter given that we got through the second filter should depend only on the difference in the two angles and indeed should go like the difference divided by 2 yep gosh yeah sorry you're completely right right so let's go back to this line here this was cos pi over 2 cos theta over 2 plus sin pi over 2 sin theta over 2 that's what it is and then we have to do a mod2 of it excuse me we have a formulae intrig which says that this combination of cosines is the cosine what's in here is actually the cos of pi upon 2 minus theta upon 2 and then we had to square it sorry now we can learn something we can make a little get a little physical result here by considering the case that theta is equal to pi on 2 pi is equal to pi what does that mean is that N is equal to E sub Y the axis of the second filter is equal to E sub Y you're measuring the spin in the Y direction the axis of this one we'll call it M is then equal to minus EZ so what's the probability of passing F3 given that you passed F1 and that's the same as the probability what that is physically is the probability of eventually having your spin being measured to be in the minus Z direction given that as you emerge from F1 you had your spin in the plus Z direction so we already had this ok right what is that well it's the probability of passing the second filter given that you passed the first times the probability of passing the third filter given that you passed the second therefore it's equal this probability was a half in this we already discussed that in the case that that theta was pi upon 2 so we were measuring in a perpendicular direction to the direction associated with the first filter this probability came out to be a half we felt that was natural because we've seen that it depends on the difference of the two angles and the difference in the two angles here is pi upon 2 so it's times a half so it's a quarter so a quarter of the particles which emerge with their spin quote unquote in the Z direction are found eventually to have their spin in the minus Z direction and this is concrete evidence that the second filter hasn't just measured the spin of the particle it's redirected it so this is a manifestation this result is a manifest so the probability of just doing F3 given F1 and no second filter is zero so putting in the second filter the intermediate filter affects the result and that's the realignment so we should talk briefly spin a half is far and away the most important case but let's just briefly talk about spin 1 and make the point that everything that we've been doing here generalises to arbitrary spin there's nothing we've been doing here which is really peculiar to spin a half so in the case of spin 1 we have that psi can be written as an amount of 1 so 1 1 if you like plus an amount of 1 nothing plus an amount of 1 minus 1 so there are three complex numbers needed to define the orientation of a spin 1 particle and for example a W boson or a Z boson are particles with spin 1 photons also have spin 1 but they have certain pathologies because they are a rest mass so it's as well not to include them in this discussion so we have that and the consequence of that is that if I have a spin operator si working on a psi that maps to a matrix problem where we write this as a b c d times 1 1 plus e times 1 nothing plus f times 1 minus 1 so this is represented by three complex numbers d, e and f this is represented by a, b and c and there will be a matrix relation between these we will have that d, e and f are equal to a matrix which we will make with plus sorry we will have 1 1 the total total angular momentum quantum numbers so let's just call it 1 si 1 and then we'll have 1 si nothing and then we'll have 1 si minus 1 and so on and so forth and here we will have nothing si 1 nothing si nothing so we have a 3 by 3 matrix operating on a b c this is how we will concretely do our computations and we need to know so we'll have 3 matrices 1 for s x, 1 for s y and 1 for s z s z will be the easy one to do s z will be the matrix of its eigenvalues down the diagonal so it will be 1 nothing minus 1 and nothing everywhere else which follows immediately from the fact that s z on this produces 1 times this etc etc etc and when we want to work out what we want to do for s x so when we want to work out 1 s x 1 we'll replace that s x by a half s plus plus s minus s plus will kill this s minus will lower this to nothing which is orthogonal to this so we'll have a 0 in this slot when we put s x in here we have a half of a plus plus minus minus will lower this to minus 1 which is orthogonal but plus will raise it to this and it will produce in fact root 2 we'll have that s plus operating on nothing will turn out to be root 2 times 1 so s x will be a half of nothing root 2 nothing we'll get nothing in the right thing because s plus can raise minus 1 to nothing but it can't drag it all the way up to 1 and s minus of course kills minus 1 so we get a matrix that looks like this and we will get for s y a matrix that's most handily written as 1 over root 2 this is more easily written as 1 over root 2 of nothing 1 nothing 1 nothing 1 nothing 1 nothing just taking out the factor of root 2 and this one is most easily written I mean is handily written just the same way we derive it nothing minus i nothing minus i it's a emission matrix so here goes i here goes i nothing nothing so these are sort of these are the generalisations of the these for a spin 1 problem and it's worth doing some Stern-Gerlach type experiments thought experiments with the spin 1 systems to just see what the differences are I did want to talk about the okay let's just briefly talk about this let's go all the way to spin s which is much greater than 1 right so in the classical regime we want to understand how out of this can we recover the classical situation that if I hold up a piece of chalk it has a well defined orientation none of this probabilistic this thing and that thing and the other thing you can see where the damn thing points we have to recover this out of this probabilistic apparatus and the way to do that is to imagine what these spin matrices look like for spin n to construct them everything we've done carries over absolutely straightforwardly we have that s n well for s z in this case is going to be s s minus 1 s minus 2 down to minus s along the diagonal it's going to be the matrix of the eigen values of s z and diagonal s x is going to be here we will have the state s s x s then here we will have s s x s minus 1 and so on s x s minus 2 and if you want to apply this in classical physics it will be 30 something by 10 to the 30 something it will be enormous but nearly all the numbers will vanish because well this number we already know is equal to 0 this number vanishes because this you replace by a half of s plus plus s minus s plus kills this s minus lowers it's something orthogonal to this this will be non-zero because s plus will raise that to s which will couple to that and in fact this will turn out to be alpha of s minus 1 so when s x works on this we get a horrible square root which I'm calling alpha of s minus 1 times s so that's what this will come to this will come to nothing because we'll have s plus that will raise this to s minus 1 which is not good enough and s minus will lower it which is useless so this is equal to 0 and everything else is going to be equal to 0 so this matrix is going to consist of non-zero numbers just above the diagonal nothing's down the diagonal so let me write this out this is going to be on the diagonal precisely nothing above the diagonal we will have alpha of s minus 1 which is easily worked out it's the square root here we will have alpha of s minus 2 here we will have alpha of s minus 3 s minus 3 and so on down the diagonal and just below the diagonal we have the complex conjugates of those these are in fact real numbers and therefore we have the same numbers and nothing's everywhere else so this is a very simple matrix it just has two non-zero diagonals and we can work with it and we can now do things like suppose we have no time so it's time to stop sorry but I think it probably is worth I'll finish it off tomorrow