So yesterday we had an awkwardness about this formula here because what overage arrived disagreed in the ordering here from what's in the book. And it just should reassure you that what I derived was correct and what's in the book is correct. And it's quite instructive, too, because these are important formulae, so it's good to have them in your mind. So let's just let's just understand how it comes about.
The both of these are right. If you if so, what's happened between these two is that these things have been swapped in their order. But crucially, also, these things have been swapped. So if I go from here, I pick up a minus sign. If I simply invert the two operators, j and vs is just an arbitrary vector. Okay. It could be x, it could be P, it could be j, it could be whatever.
Right? So if I swap these two operators, of course I pick up a, I pick up a minus sign, but then I can pick up a matching minus sign on the right side by swapping these two indices of the Epsilon symbol.
So that's minus i Sigma K, Epsilon, j i k v k. And now I can say to myself, well, so, so, so I now have this formula from this formula of to to derive this formula except the what here is called I, which is the index on J and the index on 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 II. So it's a mere relabelling of what appears at the bottom. So these two familiar both?
Correct. Okay. 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 with displacements. They generate they're the generators of displacements. The momentum generates. You have 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 is being incremented by a the incremental operators make a new system which is the same as your old system, except they've been rotated around the origin by some angle.
And we have already seen that when the that when the momentum operator commits with the Hamiltonian, when one of these observables commits 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 if so, let's say if P commits with H, what does that mean?
That very easily implies that U of A the translation operator is going to compute with H because this operator right is e to the minus i a dot p pon h bar is a function of p and therefore it commutes. If h commits with p commits then the function of p. So if we have sorry. If if this equals nought, then this equals nought. Now, what does that, uh. What does that tell us? To see what it tells. What it tells us.
We we have to think about the unitary operator associated with H because H is an observable and it's associated with the unitary operator. Each of the minus i h. Upon H. Bar t. Which we're going to call U of T. What's this, operator? Well, we know. We already know that IXI at time t is equal to the sum and each of the minus i e and over h for t times n that way.
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 ion states as any particular time, for example, at t equals zero and then evolves by multiplying each term in the 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 and.
Ian Wright 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 ends and look at each one of these things than H looks at it. I can get in and it says, Aha, that's my, I can get e n it returns, it returns e and times the number n and then you go and there you go, you have that. So this operator u of T is is very 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. If it carries you from today into tomorrow or whatever. Right. So. If so, let's just repeat this. If P comma h equals nought, then that implies and indeed is implied by that the unitary operates a 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 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 is this that will give you the same results as if you move it now and let it evolve over there. So this this what is the physical implication of this simple equation is that. Physics is the same. Here is their. Now, that's not always the case. If. If that were a clock. That were a clock. And we let it evolve on the floor. Until until tomorrow. And then read it well and then moved it up here.
Well, somewhere higher up here. We wouldn't have the same situation as if we let it evolve until tomorrow up there. Because. 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 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 with that other place. This being the same here and there is is is a statement about the the homogeneity. So when this is the case, it's a statement about the homogeneity of space.
It's it's a right. And we would say and physicists are of the view that that. Ultimately physics has to be the same here and there. And the reason and the reason that the clock evolves on the floor in a different way from on the table is because not because of any homogeneity of space, but the fact that is a dirty, great planet here.
We're 8000 miles or whatever it is from the centre of the earth, and it's the relative movement of the earth and the clock which has changed the circumstances, not the homogeneity 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 should be the case. If, if, if if your system is isolated.
We say that. In other words, we say that when this principle is not observed, it's the reason it's not observed is your system. Johnny isn't isolated. In the case of the of the clock on the floor there, it's obvious what the court what they're not isolated in this is it's the dirty group planet. But in other circumstances it might be more subtle. But we would we conjecture that you will be able to find something which is violating the isolate, you know, which is which is affecting it.
Which is which is violating its isolation. Okay. So where all 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 on translations. So so we have a sort of set of ideas like this commuting. Well. P with h well, so p comber h is connected to conserved. Momentum. Which is itself connected to uniformity. Of space. Which 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. We similarly if we if it's the case that say Jay Z comma H. So this is the generator of rotations around the Z axis, if that's equal to nought. So I need to have a nought 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 result of space. So it's clearly the case that. A compass. A compass behaves differently. If you oriented east, west or north south. Right. It's. Of. Because because on the surface of the earth, on account of the Earth's magnetic field, the physics of space is not isotropic from 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 compute with the Hamiltonian of the compass model 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. Magnetic north. And he is 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 he was in fact, connecting the conservation momentum to the to the eyes of space.
He had that concept of an isolated body. So 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 can meet with the Hamiltonian, the best system we've in fact got is to look at the at the uniformity of the physics, to say to ourselves, can I see any reason why the system should be different? The behaviour of systems should be different. If I rotated where if I translate it or do some other thing to, to it. So observables. Commuting with age.
But here's an example. When you when you can spot one, if you have any particles that interact with each other, nothing else. Oops into egg. Then you have the Hamiltonian of this system. Is the sum API squared over to him. I summed over particles. Hi. So this index here enumerates the particles plus. So that's the kinetic energy of each particle. Some of the particles makes the Hamiltonian to the whole system. And then that's time. That's, that's going to that.
We have to add the potential energy of interaction between the particles, which will be the sum of a 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, there's some interaction potential between these particles. They interact in pairs and this, in this, in this, in this picture.
So this would be the Hamiltonian of these particles which are interacting in some arbitrary way with each other, flows that interact in pairs. What we can say is that is that h is its 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, long 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. Of this. Of this transformation is going to be a conserve quantity. So this so this transformation. Well, it implies. Conservation. Of the generator. Which is going to be the the total momentum being the sum of the momenta of the individual particles.
Which of course we recognise as is the total momentum of this system is going to be conserved because action and reaction or equal and opposite back to back to what? Isaac Newton. These points you haven't seen probably made in this way before. But it is. I would like to make the point that. They are actually very fundamental points of physics. They are not peculiar. They're 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're 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 something to what I skipped, which was.
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 in an awful lot of devices. And people also now stick their crystals in magnetic fields to see what happens.
So it's still a it's still an important way of probing systems when you're trying to understand systems which whose physics is 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. Right. 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 two M 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 Lorenz force, the cross b is perpendicular to v so v dot v cross be identically vanishes, and the Lorenz force never does any work. So it can never contribute to the potential energy of your system. And therefore you can't look for you can't look for magnetic contributions in here. So it turns out that. And you need to do it because magnetism is a relativistic correction to electric statics, right? Fundamentally, that's what it is and I think it's one of them.
I'm always amazed and I don't 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, your. I mean, we make the electricity, in fact, using a relativistic correction to electrostatic.
We do almost nothing with electrostatic. There are a few electric a few scientific instruments use like electrostatic drives but it's it's it's almost an unused you know Coulomb so 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. So, so but it is a relativistic correction to electric statics.
And so in order to, 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 Q. What is the charge on your particle?
And of course, B is the curl of A, so a is the magnetic back to potential that generates the magnetic field. So I'm not really able to justify this because to, to, to explain why this should be so we need to do some relativity, which is way out of scope for, for the quantum mechanics. But what we 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 the magnetic field and ultimately.
You know, only experiment can tell you whether this is right or wrong. So let's let's let's use Aaron first. To recover classical physics out of this. So what we have is i h bar dvd t oops dvd t of the expectation value of x i. What's that? That's equal to the expectation value of x, comma h. We're going to will we'll drop this because we're not really interested for the moment.
We're not interested in 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 one over two m expectation value of x i comma p minus q a squared comitato. ABC. Now we know how to take combat cases.
Why does x i not commute with this? It doesn't compute with this because it contains the momentum operators while in particular it contains the ith momentum operator. And this is going to be one over two m if we're pedantic, we could do it more quickly than this. This will commute. This is there are two of these couple together. It'll commute with the first one.
We should do the commentator with the first one. Now X is going to commute with a. We're going to have that x comma a equals nought because the this vector potential is a function of x write it to 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 it's going to hit with this bracket is because it contains p. So we're going to have x x i comma p, the vector p dotted in to p minus q a.
Close brackets. That's one of the two terms. And then, unfortunately, there's there's another term which will in fact be identical. But just to be pedantic, let's get it. Let's keep it right. There's going to be P minus Q A. Dot x i. Com up. All right. So what we've done is regarded this as P minus times P dotted into P minus Q A, which is a product we've used the rule for doing the commentator on a product to the commentator on the first term,
leave the second alone, that's that. Then leave the first time alone and do this. Commentator On the second one and the commentators on these brackets reduced media p because of that this so this product could be written as a sum over K of PK dot PK minus Q K Right. And this commentator is going to be nothing except when K equals I when this will be an h bar which will cancel that h bar and we'll discover that dvt t of x ii is equal to this h bar.
Is is is a mere number. It'll commute with this. So we don't need to worry this this term is going to generate the same as that. So this gives me a one over M of P minus P-I minus q i. So this what this is telling us is that the classical velocity, because the rate of change of the expectation value, the position is what we would call the classical velocity.
The expectation by the velocity, if you like, is not equal to the momentum over M It's equal to the momentum minus Q a over m. Or alternatively, it's telling you that pie is equal to m v. I always write these what their expectation values here this is. This is their 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 all this is the momentum. Of the IMAG field. Point is that if you move a charge 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 electro magnetic field. And there's a calculation which makes it look as if I think it probably is broadly true that that that. Yep. The removing charge to get a charge moving, you not only have to give it momentum, but you have to give the field 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 anyway. So 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 A7 may have encountered this formula. I'm not sure whether it goes quite that far. Let's have a look at the other of the other equation of motion. We should have a look at which is harder duty of the expectation value of pi. So. Is going to be a sigh. P comma h but what's H its peak? I'm going to write it out now. P.K. minus Q. A K squared some over k commentator sticking another upside.
So this is the commentator of the momentum with the Hamiltonian, where I've now written out the Hamiltonian in Gorey in reasonably gory detail. Over to him. Over to him. I'm missing a lot of it to him. I not one of it to him. In all that. That's one over two more times. This is the Hamiltonian. Why does PI not give you this? Well, obviously, P.E.I. views with Peak, that's a problem.
P.E.I. doesn't commute with this because sitting inside, because this is a function of X. So when we work this out, we get a one over two M CI big bracket. We're going to have this thing commuting with that. So we'll have a minus q p P.E.I. comma a k. This is going to be some do over K of maybe it'll be better if we put a some over K here. Got to have one somewhere. Times peak minus Q a k. So that's that's p commuting with the first of these two of these two brackets and then we will have.
We will have P.K. minus q ak of pi ak. Commentator And the factor of minus Q from here, close big brackets sticking out. Another sign. So this is a disgusting mess that we have. And now we have to address the question of, so what is this comitato? What is pi, comma, ak? We need it. We need it in two slots. We need it here and we need it here.
Well, this is we now use our rule for doing the commentator of a function of X. We use previously almost the rules for a function of P. The rule was that this is equal to the a k by the x. High Times. The Communist commentator p comma exi. Right. The reason this isn't. This is a function of exi. That's why this computer fails to vanish. And we derived this rule quite early on that you can you can buy Taylor series expanding your function.
You can convince yourself that this is true, that we just have a derivative times. The commentator with respect to whatever it is we taking the derivative with respect to this is minus h bar. So. By the I. So we're going to get some ice balls, which we can cancel over there. So we're going to have the DVD t of PI, the rate of change of momentum, which should be equal to false all being. Well, this is turning out to be.
Oh, yeah, sorry. Yeah. So we're going to have a one over two now we can take out a factor of Q of A to M This minus sorry is going to cancel that minus sign that Q I've taken outside the sum over K has not collapsed. No, it's the left. Yep. So we're going to have a summer of okay, what of we're going to have a K by TXI for this one here. We've got all the factors, peak minus Q arc and we're going to have essentially the same thing.
But in the reverse order, peak minus q a K of the K by the x, i close the brick bracket and stick a matching cat upside 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 compute with that p. Similarly this one. So so this thing is trapped on the left side of P and that one trapped on the right side of P and I can't combine them.
And then in quantum mechanics, this is as far as I can as I can go. I now have to. So this is a this is a respectable, totally aboveboard quantum mechanical calculation to go further, I, 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 in this is predicting the correct classical physics. If I'm predicting the correct classical physics, I can.
If I talk 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 we don't worry about these fluctuations. We assume that they all they all average wage zero like the interference pattern associated with the bullets. I mean, 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 a product of the expectation values.
And then these become in 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. I can stop fussing about about this so we can we now say in the classical limit. We've specialising now to the classic limit when we can neglect fluctuations, we can write this is Q over.
M Because I'm going to combine these two terms, the sum over K of the expectation value of decay k by the Z times the expectation value of peak minus q a k. Now we can simplify again because this expectation value p, k minus Q. Remember we showed above. We honestly showed above without any fudging was equal to the mass times. The expectation of value of the velocity. Well. Well, that's. That's unjust.
Yeah, right. So this thing here can be replaced by M the K. And the EMS cancel while we're about it. Why don't we replace this p I with M? Well, with what we get from up there, I've lost it. Here we go. It's in the VII minus. Sorry. Plus Kua. So this now comes down to DVD T of M VII plus q expectation of a I 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 D a K by the X II times. What did we say was weak? And we ought to put an expectation value on 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 if. What are we trying to get? I'm trying to get that mass times acceleration is equal to V.
Crosby. And it may look as if I'm still some way from that, but it's not so bad, actually. Why? So what? On the left, what we have on the left here is the rate of change, of obviously the velocity and the and the vector potential evaluate is 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 the if this thing were nonzero, it would generate an electric field that would be quick. And then you know right of time varying magnetic field creates by Faraday's law creates a curly field and we that that leads to more complicated equation of motion. So we got we're just 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, and sensing the impact the, the vector potential at different locations. So what we have is that d by d t of a i is equal to what is it equal to? It's equal to the by the chain rule. It's equal to the x k by d t sorry.
That should be a total total derivative d by t times d i by the x k. 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 change is with location. So what we now have is the m d vii by e t e mass times acceleration is equal to Q Q is going to be a factor Q on the right. I write down the terms. I've already got some of a k k by the x i. VK and then I'm transferring from the left side.
This time's Q Right. So I have and it's going to come minus this is v k and this is the I by the x k. Street speaking. This bracket should be here because that summation sign is over. Is over. Both these both these signings have. Now this is actually equal to. Q The cross B ice component. Uh, well, because it's V cross. Ask yourself what what this would.
I mean, just I claim that this is true. Let us see whether it is true, what we in order to expand this vector triple product, we would say it was this thing does it 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 a VK VK K direction of this thing gives me a nebula. I because I'm trying to calculate the ice component and then minus this thing dotted with this thing, direction of that thing.
So that's a VK Nebula K. A I. 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 does it, 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 with. If you move that around in back. Right. These these two terms are the same. So we have indeed recovered mass times. Acceleration is equal to Lorentz force. In the classical limit. Well, I think that's really all that I want to do. Yeah, that's all I want to do. That justifies provisionally the use of. So this Hamiltonian. Where was it? P minus. P minus A all squared over to him being the Hamill.
That change in the Hamiltonian introduces a magnetic field into the physics, and we will use that when discussing atoms. Down, down the track. And if you look at the back end of chapter three, you can see there are some quite entertaining things you can do with the with the motion of a particle in a uniform magnetic field. When it turns out that you can recycle, you can recycle the physics well, you can recycle the formalism in the mathematics that we did for the harmonic harmonic oscillator.
You can recycle it for this uniform magnetic field case. The basic principle is that if you 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. It goes round. You know, we have we have empty square of all.
Is equal to Q the b b. So we have that that v. Over R is equal to cube of m is equal to than the more frequency. So the the the the the angular frequency of which the particle goes on its orbit is depends on the strength of the magnetic field and the charge in the mass, but not on the energy. It doesn't depend on how fast you're going. So fast particles go in 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 or it allows allows us to reset that. The 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 mean, I hope some people will have some fun looking at that in the vacation.
It is very good quantum mechanics. It's it's very important physics. But unfortunately, we are not going to have time to cover it in the in the lectures. But magnetic field will be important in in the context of atomic physics. Okay, so that's it until next time.