 Ac rwy'n meddwl, wedi gwneud y cyfnodi ar ein gweld y ffaith hwnna ar ei ddweud, yn ydyn ni'n meddwl ar y hyn o'r hwn iawn i'r ddweud, ac efo'r cyfnodi hynny'n gyhoedd o'u bwysig ar ei ddweud a'r hwn i'r ddweud, ac mae'r ffaith ymddydd i'r gyd-ferïl, ac mae'n diodel ar y ffaith erbyn yn ei odd. Yn çaen, mae'r ddweud ond y merogol y ddweud mor hwnnw'n credu eu ddweud. Ac y ceisio, mae'n tynn nhw'n diolch yn ddweudio, ac yn mynd yn ddiddordeb cael gwyd-dod, mae'n ddiddordeb cael gwyd-dod. If I go from here, I pick up a minus sign if I simply invert the two operators j and v. V's are just an arbitrary vector, it could be x, it could be p, it could be j, it could be whatever. Felly, dwi'n gwcwp'i ddoeddwyr, rhaid, rhaid gydweithio eu sain mewn hynny. Ond nid ddweud, rhaid i wneud eich ôl fighemau gwneud, fel swyfo ddysgu'i ddweud o'r cyfaradau ymddio'r campryd. Dwi'n gwcwp'u ddweud o'r cyfaradau. Now I can say to myself, well, so I now have this formula, from this formula, I've derived this formula, except that what here is called I, which is the index on J and the first index on Epsilon, is here called little J, and what here is called J being the middle index on Epsilon and the index on the vector operator is here called I. The mere relabelling of what appears at the bottom, so these two formulae are both correct. So we just need to pull together, we really have everything now, we just need to pull together the results that we have and just calmly understand the physical significance of them. So we've discovered that these operators like the momentum and the angular momentum are associated with displacements. They generate, they're the generators of displacements, the momentum generates U of A, which shoves your system, it doesn't literally shove your system, it makes a new system that's translated, that's the same as the old one, except its location has been incremented by A. The angular momentum operators make a new system which is the same as your old system except they've been wedlocked. They rotate it around the origin by some angle. And we have already seen that when the momentum operator commutes with a Hamiltonian, when one of these observables commutes with a Hamiltonian, we have a conserved quantity and we have good quantum numbers and things like that. So what's the connection? Okay, so let's say if P commutes with H, what does that mean? That very easily implies that U of A, the translation operator, is going to commute with H because this operator is E to the minus I A dot P on H bar is a function of P and therefore it commutes. It commutes with any function of P, so if we have, sorry, if this equals naught, then this equals naught. Now what does that tell us? To see what it tells us, we have to think about the unitary operator associated with H because H is an observable and it's associated with the unitary operator. E to the minus I H upon H bar T, which we're going to call U of T. What's this operator? Well, we already know that psi at time T is equal to the sum A n E to the minus I E n over H bar T times E n, don't we? This is our standard expression for solving, our standard means of solving the time dependent Schrodinger equation is to decompose the given state into a linear superposition of energy eigenstates at any particular time, for example, at T equals 0, and then evolve by multiplying each term in this series by this exponential here. But we can see that this could also be written as E to the minus I H T upon H bar times sum A n E n, right? Because when this linear operator looks at this sum here, it passes, it's a linear operator, so it can be distributed, pass through these A ns and look at each one of these things, then H looks at it, I can get E n and it says, aha, that's better. It's my eigencat E n, it returns E n times the number E n, and there you go, you have that. So this operator U of T is a crucial operator, it's the thing that evolves you forward in time, any state. This is nothing new, this is just a repackaging of old results. So the unitary operator associated with the observable time moves you forward in time. It carries you from today into tomorrow or whatever, right? So let's just repeat this. If P comma H equals 0, then that implies, and indeed is implied by, that the unitary operator U of A commutes with the unitary operator U of T, right? This is the thing that moves you forward in time. This is the thing that makes you a new system shoved along the bench by A. So what does that tell you? That tells you that take the state of your system and evolve it in time and then shove it along the desk and you will have exactly the same state as if you take your system, shove it along the bench and then evolve it in time. It says that whether you let it evolve here and then move it to its point where you want to have it, that will give you the same results as if you move it now and let it evolve over there. So the physical implication of this simple equation is that physics is the same here as there. Now that's not always the case. If that were a clock and we let it evolve on the floor until tomorrow and then read it and then moved it up here. We wouldn't have the same situation as if we let it evolve until tomorrow up there because clocks, the gravitational potential down on the floor is lower than it is up there, so a clock up there will evolve faster than a clock down there. So it is by no means obvious, it's not necessarily the case that it doesn't matter where you conduct your experiments, that evolving it in one place and then shoving it and then moving it somewhere else is going to give you the same results as shoving it somewhere else now and then allowing it to evolve at that other place. This being the same here and there is a statement about the homogeneity, so when this is the case it's a statement about the homogeneity of space. And physicists are of the view that ultimately physics has to be the same here and there and the reason that the clock evolves on the floor in a different way from on the table is not because of any homogeneity of space but the fact that there's a dirty great planet here where 8,000 miles or whatever it is from the centre of the Earth is the relative movement of the Earth and the clock which has changed the circumstances, not the inhomogeneity of space. So we're completely wedded to the concept that fundamentally space is the same everywhere and therefore fundamentally this should be the case if your system is isolated. We say that in other words we say that when this principle is not observed the reason it's not observed is your system, Johnny, isn't isolated. In the case of the clock on the floor or there it's obvious what the not isolatedness is, it's the dirty great planet. But in other circumstances it might be more subtle but we conjecture that you will be able to find something which is effecting it which is violating its isolation. Ok, so where are we? This commuting of operators is associated with something being conserved, that something is momentum. It's also associated with a statement about invariance of physics under translations. So we have a sort of set of ideas like this commuting P with H. P,H is connected to conserved momentum which is itself connected to uniformity of space. This is the same thing as symmetry under translation. And this is a set of three sort of separate things which are tightly connected by mathematics and basic principles of physics. Similarly, if it's the case that say JZ,H, so this is the generator of rotations around the Z axis, if that's equal to 0, sorry I need to have a 0 equals here, then that is associated with conservation in classical physics that's associated with the conservation of angular momentum which is why we want to call this the angular momentum operator, which is associated with the isotropy of space. So it's clearly the case that a compass behaves differently if you orient it east west or north south. It's because on the surface of the earth, on account of the earth's magnetic field, the physics of space is not isotropic from the perspective of a compass needle. And it's associated with, and as a consequence of that, it's angular momentum operator will not commute with the Hamiltonian of the compass needle. And its angular momentum will not be conserved, that's why it swings to and fro around the north pole when you let it go, the magnetic north. And on the momentum thing, just remember what Newton said about bodies moving in a straight line, etc. He fundamentally said isolated bodies have conserved momentum and so there already he was in fact connecting the conservation momentum to the isotropy of space, he had that concept of an isolated body. So in general, we're always interested in finding these operators, these observables which commute with the Hamiltonian and in general it's hard to find, so in general it's hard to find the operators that we don't have a system unfortunately for finding operators that commute with the Hamiltonian. The best system we've in fact got is to look at the uniformity of the physics to say to ourselves, can I see any reason why the system should be different, the behaviour of the system should be different if I rotate it or if I translate it or do some other thing too, to it. So observables commuting with H. But here's an example when you can spot one. If you have n particles that interact with each other and nothing else, then you have that the Hamiltonian of this system is the sum P i squared over 2 m i summed over particles. So this index here enumerates the particles plus that's the kinetic energy of each particles summed over particles makes the Hamiltonian of the whole system and then that's time, that's going to that we have to add the potential energy of interaction between the particles which will be the sum over pairs of particles which we can get by saying that J is less than I. These are the vector positions of the particles. So there's some interaction potential between these particles. They interact in pairs in this picture. So this would be the Hamiltonian of these particles which are interacting in some arbitrary way with each other as long as they interact in pairs. And what we can say is that H is invariant. This expression is manifestly the same if X i goes to X i plus A. If you simply add a vector A to all the locations of the particles and as you shove the whole system along by a vector A, then the arguments of all of these interactions stay the same and you don't affect the momentum so H is invariant. And what does that tell you? That tells you that the generator of this transformation is going to be a conserved quantity. So this transformation, well, it implies conservation of the generator which is going to be the total momentum being the sum of the momenta of the individual particles. Which of course we recognise as the total momentum of this system is going to be conserved because action and reaction are equal and opposite back to what Isaac Newton. These points you haven't seen probably made in this way before, but I would like to make the point that they are actually very fundamental points of physics. They are not peculiar, they are not special to quantum mechanics. In classical physics all of these statements remain true. It's just that when you do elementary mechanics you don't have the machinery at hand to see the connection between symmetry and conserved quantities. These are really very basic points which are true in quantum mechanics but they are also true in classical physics. But we have now the apparatus so we can see these things rather more clearly than we can in classical physics. So we now have time to cycle back to what I skipped which is motion in a magnetic field. This is a particularly important topic because an awful lot of quantum mechanics was developed historically. It was developed by sticking atoms into magnetic fields. It's obviously also an important topic in the sense that we use magnetic fields in an awful lot of devices and people also now stick their crystals in magnetic fields to see what happens. So it's still an important way of probing systems when you're trying to understand systems whose physics is based on quantum mechanics and it's very important to understand how this happens. And there's a fundamental difficulty we have to address up front which is what we need to know is how to modify the Hamiltonian. Because in quantum mechanics you put the physics into the Hamiltonian. The Hamiltonian tells you what forces are acting, what the system consists of, it encodes what the physical laws are for your system. So if you switch on a magnetic field it must be that it's changing the Hamiltonian somehow. So the question is so how is it changing the Hamiltonian? And if you take the view that h is equal to p squared over 2m plus v, because you've got some particle, then you're in trouble because there's no magnetic contribution in the potential energy because the Lorentz force never does any work. The Lorentz force v cross b is perpendicular to v so v dot v cross b identically vanishes and the Lorentz force never does any work. So it can never contribute to the potential energy of your system and therefore you can't look for magnetic contributions in here. So it turns out that because magnetism is a relativistic correction to electrostatics, right, fundamentally that's what it is, and I think it's one of them. I'm always amazed and I don't think I really understand why it is that our electoral devices overwhelmingly use this, you know, your vacuum cleaners, your disk drive. I mean we make the electricity in fact using a relativistic correction to electrostatics. We do almost nothing with electrostatics. There are a few electrostatic, a few scientific instruments use electrostatic drives but it's almost an unused, you know, Coulomb's law is almost unused except to get the electrons to go down the wires in order to generate this relativistic correction because they're moving at a slightly different speed from the ions in the wires. Anyway, but it is a relativistic correction to electrostatics and so in order to find out how to change your Hamiltonian you really need to do relativity, that's the proper place to look and I'm not going to be able to derive this for you. I'm going to be able to tell you what it is which is what we need to do is replace that P by P minus the charge on the particle. So when we have a magnetic field we put it in by replacing P in our original Hamiltonian by P minus QA where this is the charge on your particle and of course B is the curl of A. So A is the magnetic vector potential that generates the magnetic field. So I'm not going to be able to justify this because to explain why this should be so we need to do some relativity which is way out of scope for the quantum mechanics. But what I should do with this is use Ehrenfest theorem to convince you this gives you that this gives us the classical equations of motion in a magnetic field and ultimately only experiment can tell you whether this is right or wrong. So let's use Ehrenfest to recover classical physics out of this. So what we have is ih bar dbdt of the expectation value of xi. What's that? That's equal to the expectation value of xi,h. We're going to, we'll drop this because we're not really interested. For the moment we're not interested in V which would contain for example the electrostatic interactions, the interactions with the electric field if there were any. Let's just, let's not worry about it. Let's just take it that what we have is a particle moving in a magnetic field. So I want to take this to be the Hamiltonian to keep life simple. So what is this? This is 1 over 2 m expectation value of xi,p minus qA squared commutator, psi. Now we know how to take our commutators. This is at the, why does xi not commute with this? It doesn't commute with this because it contains the momentum operators. Well in particular it contains the ith momentum operator. And this is going to be 1 over 2 m if we're pedantic. We could do it more quickly than this. This will commute with, this is, there are two of these coupled together. It'll commute with the first one. We should do the commutator with the first one. X is going to commute with A, we're going to have that xi comma A equals naught because this vector potential is a function of X. The vector potential depends on X, it varies with X, therefore it is a function of the operator X. So X is going to commute with it. So the reason X is going to commute with this bracket is because it contains Pi. So we're going to have xi comma P, the vector P dotted into P minus qA, close brackets. That's one of the two terms and then unfortunately there's another term which will in fact be identical. But just to be pedantic let's keep it right. This is going to be P minus qA dot xi comma P. So what we've done is regarded this as P minus qA times P minus, dotted into P minus qA, which is a product. We've used the rule of doing the commutator on a product. Do the commutator on the first term, leave the second alone, that's that. Then leave the first term alone and do the commutator on the second one. Then the commutator is on these brackets reduced to merely xip because of that. So this dot product could be written as a sum over K of pK dot pK minus qA K. And this commutator is going to be nothing except when K equals I, when this will be an ih bar, and we'll discover that dbdT of xi is equal to, this ih bar is a mere number. It'll commute with this so we don't need to worry, this term is going to generate the same as that. So this gives me a one over M of P minus Pi minus qAi. So this, what this is telling us is that the classical velocity because the rate of change of the expectation value of the position is what we would call the classical velocity, the expectation value of the velocity if you like, is not equal to the momentum over M, it's equal to the momentum minus qA over M, or alternatively it's telling you that Pi is equal to MVI, oh sorry, these, whoops, there are expectation values here. This is, there are expectation values here, right? This was always expectation value. So what we're discovering is that the expectation value of the momentum is equal to mass times expectation of velocity plus q times expectation value of the magnetic vector potential at the location of the particle. And there's a problem on the problem sets that tries to convince you, this is obviously, this is the momentum of the emag field. Point is that if you move a charged particle, you are moving it's electromagnetic, you're moving it's electrostatic field, the electrostatic field causes the magnetic, the combination of an electric field and a magnetic field endows the, makes for a momentum flux in the now electromagnetic field and there's a calculation which makes it look as if, I think it probably is broadly true that the moving charge, to get a charge moving you not only have to give it momentum but you have to give the fields some momentum so this really is the total momentum. But the thing is it's not, the field, the particle is not on its own, it's not the only repository of momentum, the electromagnetic field is also a repository of momentum. So we have this non-trivial relationship now between momentum and velocity and again this is not something special to quantum mechanics, we've derived this in quantum mechanics but it's a known result in Hamiltonian mechanics, those people who've done S7 may have encountered this formula, I'm not sure whether it goes quite that far. Let's have a look at the other equation of motion, we should have a look which is harder, dbt of the expectation value of Pi, so is going to be psi Pi, H, but what's H? It's Pk, I'm going to write it out now, Pk minus Q Ak squared summed over K commutator sticking another psi. This is the commutator of the momentum with the Hamiltonian where I've now written out the Hamiltonian in reasonably gory detail. Over 2m, over 2m, I'm missing a 1 over 2m, am I not? 1 over 2m, in order that that's, 1 over 2m times this is the Hamiltonian. Why does Pi not commute with this? Well obviously Pi commutes with Pk, that's not a problem. Pi doesn't commute with this because this is a function of X. So when we work this out, we get a 1 over 2m, psi big bracket, we're going to have this thing commuting with that, so we'll have a minus Q Pi, Ak. This is going to be summed over K. Maybe it would be better if we put a sum over K here. We've got to have one somewhere. Times Pk minus Q Ak, so that's P commuting with the first of these two brackets, and then we will have Pk minus Q Ak of Pi Ak commutator, and a factor of minus Q from here. Close big brackets sticking another of psi. So this is a disgusting mess that we have, and now we have to address the question of, so what is this commutator? What is Pi, Ak? We need it in two slots, we need it here and we need it here. We now use our rule for doing the commutator of a function of X. We used previously almost the rule for a function of P. The rule was that this is equal to D Ak by dx i times the commutator Pi, xi. The reason this is a function of xi, that's why this commutator fails to vanish, and we derived this rule quite early on, that you can, by Taylor series expanding your function, you can convince yourself that this is true, that we just have a derivative times the commutator with respect to whatever it is we're taking the derivative with respect to. This is minus i h bar, so by dx i. So we're going to get some i h bars, which we can cancel over there, so we're going to have the db dt of Pi, the rate of change in momentum, which should be equal to 4, so we're being well. This is turning out to be, yeah, sorry, yeah. So we're going to have a 1 over 2 m. No, we can take out a factor of q over 2 m. This minus sign is going to cancel that minus sign, that q I've taken outside. The sum over k has not collapsed. No, it's still there. Yep. So we're going to have a psi, a sum over k, what of, we're going to have a dA k by dx i for this one here, but we've got all the factors pk minus q ak, and we're going to have essentially the same thing, but in the reverse order, pk minus q ak of dA k by dx i. Close a break bracket and stick a matching ket of psi on the outside to take the expectation value. So, unfortunately, I cannot combine these two terms as they stand into one term because this is a function of x which refuses to commute with that p. Similarly, this one, so this thing is trapped on the left side of p and that one is trapped on the right side of p and I can't combine them. And in quantum mechanics, this is as far as I can go. I now have to, so this is a respectable, totally above-board quantum mechanical calculation. To go further, I have to say, well, look, what am I trying to do? I'm trying to recover the Lorentz force for you. I'm trying to show that the classical, this is predicting the correct classical physics. If I'm predicting the correct classical physics, I can, if I'm talking about the classical physics, each of these operators can get replaced by its own expectation value. So the issue here is that here I have to take the expectation value of a product of one operator on another operator. And such an expectation value is not automatically the same as the product of the expectation value of this on the expectation value of that. Because fluctuations in this operator may be correlated with fluctuations in that, in that sort of quantum fluctuations. But if we're in the classical limit, we don't worry about these fluctuations. We assume that they all average away to zero, like the interference pattern associated with the bullets. The fluctuations average away. We're just left with the mean value of the mean. So we now replace this product with a product of the expectation of the product with the product of the expectation values. And then these become mere numbers. So this becomes an expectation value of this operator, the expectation of the value of this number. And then of course the numbers can be arranged in either order and I can stop fussing about this. So we can, we now say in the classical limit we're specialising now to the classical limit when we can neglect fluctuations, we can write this as q over m because I'm going to combine these two terms, the sum over k of the expectation value of decay ak by the xi times the expectation value of pk minus q ak. Now we can simplify again because this expectation value of pk minus q ak, remember we showed above, we honestly showed above without any fudging, was equal to the mass times the expectation of value of the velocity where that's understood. So this thing here can be replaced by m vk. And the m's cancel. While we're about it, why don't we replace this pi with m, well with what we get from up there, I've lost it, here we go, it's m vi minus, sorry, plus qa. So this now comes down to db t of m vi plus q expectation of ai. That's using that respectable formula up there for the relationship between velocity and momentum. Yes, that's correct. And that is going to be q because the m's are going to cancel the sum over k of dak by dxi times, what did we say, it was vk. We ought to put an expectation value around everything because we are dealing now with expectation values, we've explicitly gone to the classical regime. Okay, so we're nearly there, it doesn't probably look as, what are we trying to get? I'm trying to get that mass times acceleration is equal to v cross b. And it may look as if I'm still some way from that, but it's not so bad actually. So what we have on the left here is the rate of change of obviously the velocity and the vector potential evaluated at the location of the particle, not just anywhere else but at the location of the particle. So suppose we have a static field, static b field, so that means that the partial derivative of a with respect to time can be taken to vanish. If this thing were non-zero, it would generate an electric field, and that you know, right, a time varying magnetic field creates a, by Faraday's law, creates a curly e-field and that leads to a more complicated equation of motion. So we're interested in static b fields so that the rate of change of the magnetic vector potential at any given point is zero, but this time derivative is not zero because the particle is moving and sensing the vector potential at different locations. So what we have is that d by dt of ai is equal to, what's it equal to? It's equal to the, by the chain rule, it's equal to dxk by dt, sorry that should be a total derivative, dxk by dt times dai by dxk. So the reason that this quantity is changing is because the place where we're making the measurements is changing at this rate and this is the rate at which a changes with location. So what we now have is that m dvi by dt, mass times acceleration is equal to q, there's going to be a factor q on the right, I write down the terms I've already got, sum over k dak by dxi vk and then I'm transferring from the left side this times q, right? So I have, and it's going to become minus, this is vk and this is dai by dxk. Well I suppose strictly speaking this bracket should be here because that summation sign is over, is over both these summits here. Now this is actually equal to q v cross b i'th component, well because it's v cross, ask yourself what this would, I mean just, I claim that this is true. Let us see whether it is true. In order to expand this vector triple product we would say it was this thing dotted with this thing in the direction of that thing. So if I expand this I get q, this thing dotted with this thing that means a sum over k of vk, vk ak, direction of this thing gives me a nabla i because I'm trying to calculate the i'th component. And then minus this thing dotted with this thing direction of that thing. So that's a vk nabla k ai. It's a little bit of a complicated vector triple product because this is a differential operator and it is operating only on this, never on this. So that's why I've written them in that form. It's this thing dotted with this thing direction of that thing but this is only working on that and then it's this thing dotted with this thing direction of that thing that's nice and easy. And I think you can see that this term is this term and this term is this term if you move that around in back. These two terms are the same. So we have indeed recovered mass times acceleration is equal to the Lorentz force in the classical limit. I think that's really all that I want to do. That's all I want to do. That justifies provisionally the use of this Hamiltonian, where was it? P minus qa all squared over 2m being that change in the Hamiltonian introduces a magnetic field into the physics. And we will use that when discussing atoms down the track and if you look at the back end of chapter 3 you can see there are some quite entertaining things you can do with the motion of a particle in a uniform magnetic field when it turns out that you can recycle you can recycle the physics no, you can recycle the formalism and the mathematics that we did for the harmonic oscillator you can recycle it for this uniform magnetic field case the basic principle is that if you have a uniform B field and a non relativistic particle moving in a uniform charged particle moving in a uniform B field you can have the orbits of circles the particle circles around in this uniform B field with some radius that depends on its speed if you have a fast particle it goes round you know we have we have mv squared over r is equal to q v vb so we have that that v over r is equal to qb of m is equal to the LeMau frequency so the the angular frequency which the particle goes on its orbit depends on the strength of the magnetic field and the charge and the mass but not on the energy it doesn't depend on how fast you're going so fast particles go on big circles and take the same time to go around to slow particles so you have a characteristic all the motion is at some characteristic frequency and that is reminiscent of a harmonic oscillator and allows us to there's a fundamental underlying physical reason why we can solve the problem of motion the quantum mechanical problem of motion in a uniform magnetic field using the apparatus of the harmonic oscillator so I think you should have I hope some people will have some fun looking at that in the vacation it is very good quantum mechanics it's very important physics but unfortunately we are not going to have time to cover it in the lectures but magnetic field will be important in the context of atomic physics okay so that's it until next term