MIT 量子场论·8.323·第04讲：自由标量场论的正则量子化（原声文稿）

Transcript of MIT 量子场论·8.323·第04讲：自由标量场论的正则量子化 （中文完整文稿）。

So last time, we quantized this theory. So let me just write down the theory again.

Okay, so we quantized this theory.

And so, so let me also just write down its canonical momentum, density, and then the Hamiltonian density would be, uh -huh, square, okay? and the classical equation motion.

So we'll not repeat the quantization procedure, and we wrote down the most general solution at the quantum level.

We're written as the following.

This factor is the convention, and then we have a k, Okay, okay, plus, okay, so from now on I will suppress the hat, okay, so I will suppress hat, you should always now view it as a quantum field, quantum operators, and the plus, okay. Stop.

UK is the complete side of solution, which is given by potential minus i omega t.

So this is just the basis of solution.

So this is the complete solution to the operator equation for this operator phi.

And then you can also find its conjugate momentum, density, so the pi, you can just take the derivative straight forward. I will not write it explicitly.

And then we discussed that you can impose a canonical commutation relation, which is that the pi gx, gx prime, prime, you could choose pi kx and pi kx prime should be 0.

And then pi kx, x prime, then given by the delta.

So that's what we did at the end of last lecture.

And then you can just plug it in.

plug this, plug those expression into here, then you can find the commutation relations between those integration.

So, ak and ak dagger are integration constants, they're of your operator equations, they're constant operators. Okay.

And so, when you plug them in, then you find the canonical, then you find the commutation relation between ak and a dagger, and then you find that ak and ak prime, prime equal to a dagger k and a dagger k prime is equal to 0.

And a k prime dagger is equal to the delta function in the wave space, OK?

So now, if you look at those expressions, So they essentially look, we essentially find the infinite number of harmonic oscillators, and each harmonic oscillator labeled by a set of continuous number k, and then the commutation relation is just like a continuum generation of standard a dagger equal to what?

OK, and they are independent because of this delta function.

And if k is not equal to k prime, then you get 0 as the commute.

So you can see the story a little bit sharper.

That indeed, we just get infinite number of harmonic oscillators. So these are just commutation relations.

This still does not tell you we have a harmonic oscillator.

To see harmonic oscillators, we actually we need to calculate the Hamiltonian.

So now, let's try to calculate the Hamiltonian.

Now, we can try to calculate the Hamiltonian.

So this is just, you can just do it straightforwardly. Just plug that in.

Just plug that expression in. We draw the above.

And then you just plug in the expression for the phi pi and the pi into that equation, and then you just straight forward to calculate it. And then we'll let go through the details.

Yeah, it's a straightforward calculation.

So then you find, yeah, maybe after some minutes, then you find the following answer.

So now when you plug them in, then you find you can actually do the spatial integral because you just have the plane wave.

You can just do the spatial integral. And when you do the spatial integral, then that gives you a delta function, momentum space, et cetera.

And then you can reduce everything into a single k integral.

And then what you find is you get omega k.

and then you find ak dagger, ak, and then plus, say, ak, ak dagger. Okay? So let me see.

Yeah, actually, yeah, I think one half.

Okay, so this is the standard expression assume for harmonic oscillator now, OK?

Actually, it's exactly identical to you just sum, yeah, integrate like sum, just sum over continuum harmonic oscillators.

And each with frequency omega k. And each harmonic oscillator is labeled by this number k, OK?

And so we can do from the standard trick to write it actually as the, to write it as a commutator, we can just write it as k, and then we just have this.

So we combine these two terms together and introduce a commutator.

You just have a dagger a, and then plot the commutator between them.

Okay, let's get that.

So in the standard story, this is equal to one, Then you get the standard answer that you get omega a dagger a plus one -half, okay?

But here, so we can again, yeah, let me just write one more step.

So this is k omega k, OK, dagger, OK, then plus E0.

And now E0 is the sum of all the zero point energy of all the harmonic oscillators.

OK, so 1 half omega k.

Except that the tricky thing is the following.

OK, except now we have this commutative of a k with a k dagger.

In the standard case, this is equal to 1.

But in model case, when you set k prime equal to k, we get the delta 0. So we get delta 0.

So then we get that 2 pi cubed delta 3 0.

And remember, 0 is in momentum space.

So this is in k space. We have two kinds of data functions.

is the data function in k space, data function in coordinate space.

So here I put the subscript k here, means that this is the data function 0 in k space, OK?

And so this is a zero point energy.

So this zero point energy is divergent, but we will comment on that a little bit later, OK? Do you have any questions on this?

Let me just make a couple of remarks.

First is that something interesting happened.

So when you look at the expression for phi and the pi, both phi and the pi depend on time explicitly. OK, so there's a time dependent here.

And here, we only integrate over spatial direction, okay?

We only integrate over spatial direction.

So in terms of integrating, we don't do anything with time. But what do you notice here?

What do you notice there?

Yes, it's not that we draw time dependence. This time dependence disappeared.

No, the Hamiltonian, phi is time dependent.

And pi is time dependent.

If you plug those expressions into here, certainly this inside there is dependent on time explicitly. This guy is dependent on time explicitly.

But in the end, after you do the integration of spatial directions, then you find in the end of the time dependence actually cancel during the calculation.

So do you have a guess why somehow this should cancel? Yes? Like energy concentration or something? Yeah, exactly.

So you have done in your piece that the time translation, so this system is time translation symmetric, and means that you should have the H. It's a conserved number. It's a conserved quantity.

So indeed, you see that this is independent of time.

This is a conserved quantity.

Okay, we see it explicitly, okay, so there's no time dependence since H is a conserved quantity.

So this integration is absolutely crucial, because if you don't do integration, the Hamiltonian density is not, okay, it's only the total Hamiltonian.

Okay, so this is a comment, and the second, yeah, an image.

So, yeah, since I only have two comments, let me just call it one, rather than call it zero.

So the second comment is that then we've, yeah, just repeat, we already said, we just create t of phi.

We see this create t of phi reduces to, essentially, a system of a continuum of harmonic oscillators because labeled by k is frequency omega k. So we call this omega k defined to be k squared plus m squared.

So as we mentioned last time, the fact that we actually see infinite number of harmonic oscillator is actually not surprising from the perspective that this field theory can be written as a continuum limit of a chain, say of atoms connected by sprints, okay?

So when you have a chain, yeah, we draw down a one -dimensional system. You can easily generate two three -dimensional.

Yeah, so you have those chains of atoms connected by spring and they're just coupled oscillators And then you can just diagonalize them and each will find the normal spectrum and each of them just a harmonic oscillator.

And so when we find those solutions by doing free transform, and essentially we are just diagonalize those interactions between those springs, okay?

And yeah, so that's why we get infinite harmonic oscillator.

So from that kind of way to think about it, it's totally unsurprising we get the the infinite number of harmonic oscillators.

But it is surprising if you think about it from the point of view of a field theory.

So when you quantize a field theory, and then you find in the end you get a bunch of harmonic oscillators.

And before doing that, it's hard to, before you actually carry out the quantization, it's actually hard to anticipate that if you don't have this intuition from this discrete system.

Do you have any questions on this? Yes?

So when you say the infinite amount of harmonic oscillator, are you saying that the infinite amount at any given point in space, or that each given point in like a continuum of locations is one harmonic oscillator?

Yeah, yeah, yeah, that's a very good question.

And so it's at each point you have a harmonic oscillator.

Yeah, at each point, essentially at each point you have a harmonic oscillator, and then they connected in space.

Yes? Is there any reason why you didn't compute E0? You have a delt function in there. Sorry, say it again? So that the integral E0 equals why didn't you just?

Oh, well, this we cannot do with this integral. No, up there. The zero point. Oh, right, right.

Yeah, we will talk about this later because the data we are going to elaborate on this a little bit later because we are going to try to give interpretation of this.

In homework one we found that a hat was time dependent. Is a hat not time dependent here? And is that why the Hamiltonian is not?

Yeah, yeah, a hat is a constant operator.

Yeah, because a hat is the integration constant of your operating equation.

Yes, yeah, yeah, yeah, that's analog of coherence state.

Yeah, yeah, you can, yeah, we will come, you will see, yeah, this question will become clearer when we talk a little bit about the Hilbert space, yeah.

OK, good, good, good.

Now, before we talk about the Hilbert space, we need to talk about two things here.

And the second thing, so Hamiltonian is an important quantity.

So this is one of the conserved quantities.

So there are the conserved quantities.

So there are other conserved quantities.

And one of them is the conserved quantity corresponding to spatial translation.

So the spatial translation gives you conserved momentum, momentum conservation.

So from spatial translation, then you get this conserved charge, Pi.

Again, you should have done this in your homework.

to have the following form, this 3x pi, OK? So this is the loss of charge for the conserved quantity associated with the spatial translation, OK? And we interpret it as a spacetime momentum.

This should be interpreted as spacetime momentum.

So this is, so here we have two momentum here.

You don't confuse of them.

So this pi is the canonical momentum conjugate to this canonical variable pi is the canonical momentum conjugate to this field variable phi.

And this is the general physical spacetime momentum. It's the momentum of your full physical system.

And again, we can just plug it in.

We can just plug in the explicit expression of pi and the phi. We already know it's time evolution.

We already know they are time evolution. And then we can plug them in.

And then you find that the answer is given by, again, after some smaller number of minutes calculation. transformation, and then you find that the answer is given by this.

In this case, there's no zero point energy, you just have this expression, okay?

And furthermore, you also have a Lorentz transformation, Lorentz symmetry.

And then that will lead to, say, the conserved charge associated with Lorentz transformation.

And again, you can find it explicitly, okay?

And so in your p -set, too, you will find the expression of this n mu mu, say, in terms of phi and the pi.

And then you can, again, express this guy in terms of a and a dagger.

But I didn't have the guts to assign it as the part of the p -set.

So I used it as a bonus problem. Because that calculation involves a little bit extra calculation.

Yeah, involves a little bit slightly more tedious calculation.

So as I use that, it's a bonus problem for those people who like to have some more fun.

Anyway, so this you can, we'll see it in your p -sets.

Good, so then again, p -i, you see it explicitly that this is time -independent.

It's time -independent.

Good. And then the next thing is let's talk about this zero point energy.

So this zero point energy we can write it as a following.

So we can, yeah, so this is a constant.

So we can take it outside of this k integral.

Okay, this does not depend on k because, yeah.

So this is just 2 pi cube delta 3 0.

So remember, this is a case space delta function. And then you have these 3k.

So that's the expression we get.

So now let's try to understand what's the meaning of this term.

Let's try to understand the meaning of this term.

To understand the meaning of that term, we need to do a little bit of a mathematical trick. We need to do a little bit of mathematical trick.

So remember the definition of delta function.

We have 2 pi cubed delta 3k equal to the Fourier transform of the exponential i k x.

So this is the definition of the, essentially, the definition of the delta function.

And now here we want to set k equal to 0. So let's set k equal to 0 here.

And then we find that this 2 pi q theta 0.

So let's remember this again in the k space.

And then we just set k equal to 0 here. And then what is this? If you set k equal to 0, you get what this? Yes?

The volume. Yeah, that's right. You just get the volume of the full space.

The volume of the full space.

So let's just imagine you, in order to make sense of this quantity, you can put the whole universe in the big box.

Imagine the whole universe is in the big box, and then this is the total volume, OK?

And then now, now since this is the volume, now we have very good interpretation of this quantity.

And now we can write E0 as real volume times epsilon 0.

And then this quantity, epsilon 0, then have the interpretation of the energy density.

So now, this is energy density.

So this energy density, we can write it in one more step.

So continue to here. So we can write it one more step, a plug -in the explicit value of omega k.

So this is just 1 half.

Sorry, I forgot to pi to dk.

k, and then you have k square, yeah, so that's what you get, okay?

So how do you like this integral?

Yeah, sorry, say it again, what's two picture?

Yeah. Just that delta function is one for all k? No, no, because this is a constant, right? Just when you have an integral, you have a constant, you can always pull it out.

Because that delta zero does not depend on k anymore.

Good. So how do you like this integral? Can you do it?

Good. So it's fruitless to do this integral because this is divergent.

And there's very good reason for this divergence, physically.

So this is just energy density.

So essentially, this is energy per unit volume.

Say in some unit volume. So let's imagine you have a unit volume here.

Remember this, remember we said before, this is corresponding to you take a discrete system with some lattice spacing a, and then you take the lattice, under each lattice point you have a harmonic oscillator, and then you take a lattice spacing a equal to 0.

So that means for any unit volume when you take a equal to 0, you have infinite number of oscillators inside, okay?

So essentially when you, yeah, you have the lattice of oscillators at each point, yeah, anyway, I will not try to draw it.

And then when you take A0 to 0, you have any unit, any volume unit, you will have the number of oscillators go to infinity, and that's where these divergence come from, okay?

It just comes from in field theory, theory, you have a continuum of degrees of freedom.

So at each point, you have a degrees of freedom.

And within any volume, then you definitely have infinitely normal degrees of freedom.

So this is just from continuum of degrees of freedom.

So this is the first time you see a divergent quantity in quantum field theory.

But you will soon find that this is the normal. normal.

It will be a fact of life. You will see divergence very commonly.

Because you have continuum of these freedom.

So the whole thing about quantum field theory is to find a way to deal with those divergences.

And one of the key difference between quantum field theory and just finite quantum mechanics or finite number of degrees of freedom is because of all those divergences, and the big part of quantum field theory is to understand how to treat those divergences.

And they actually don't affect your physics, but you do have to develop sometimes sophisticated tricks to treat them.

And this infinite answer is also closely connected to a very famous problem you may have heard. It's called the cosmological constant problem.

Because this tells you that any quantum field theory have an infinite, say, 0 point energy.

And so you may say, OK, so infinity, whenever we see something infinite, there's one thing you always do to treat it.

Can you guess what is the thing you always do when you see infinities?

yes yeah that's one idea that's one idea but you subtract infinity is very hard you know in your calculus class when you subtract infinity infinity you can get infinity right so you have to be very careful yes that's a lot of very Indeed, but to do all those things, you have to do one thing first. Yes.

Yeah, it's also very close.

You need to do, yeah, exactly you need to find a way to make it finite first, and then you can subtract it, and then you can, yeah, just like when you sum one plus one half plus one third, et cetera, you get infinite series, series, but if you want to estimate, yeah, you get divergence, but you always try to cut it off the series, and then approximate by a finite answer.

And here is the similar, OK, so here we always put some momentum cut off, OK, say imagine the momentum is smaller than some momentum, say some value lambda, and lambda corresponding to maybe to the scale which this quantum field should no longer apply, because nobody told you that this quantum field theory should apply at all the scale because, for example, in this lattice model, this scale will be one over the lattice spacing, okay, will be one over the lattice spacing.

Anyway, so once you cut it off, still you get a pretty big number and you cannot ignore it because this is a physical zero point energy, in principle you can measure it, but in real life, we don't see it.

So we have quantum fields flying around all the time, but we don't see this big vacuum energy.

So this is called the cosmological constant problem.

Actually, there was just a colloquium last week about this cosmological relative to this cosmological constant problem.

Now we can talk about Hilbert space. this.

So as a harmonic oscillator, so we can first define the vacuum state.

So now what we will do is indeed in quantum field theory itself, we can just ignore this E0, okay?

From now on we will just ignore this E0 because this is just give you a constant which does not do anything, okay?

It's like the potential energy in E and M, yeah, just like the unit, yeah, yeah, just like the, anyway. And so from now on we will ignore this term, okay?

So often I will just write Hamiltonian, I just write this term.

But later we will actually see examples, okay?

Later we will see examples actually this E0 can actually have physical implications, okay?

But just for our current purpose we ignore it, okay?

So now the lowest energy state, then it's clear from here because this is just a constant.

So the lowest energy state, the ground state, which is we often call the vacuum state, means there's nothing there, is given by ak0 equal to 0 defined to be for any k.

So the ground state satisfy, which at the load by 0, then satisfy or are related by this a k.

And then the general states, OK, you can just. So general states have the following form.

Say I can write nk1. So the k1 oscillator excited n times nk2.

And the k2 oscillator excited nk2 times terms, et cetera, which is given by, which is proportional to, say, ak1 dagger to the power nk1, ak2 dagger nk2, et cetera, acting on zero, okay?

Just like we have a large number of harmonic oscillators, OK? Yeah.

Questions on this? Yes?

Yeah, you postulate, but you can actually write it down.

You can actually write down its wave function, just as in the standard harmonic oscillator case, you can start in from A dagger, say, allotted by A.

You can start from this equation to write down its wave function, and here you can write down the wave function for the vacuum states too in terms of phi.

Here I'm just saying, depending on which k are excited, yeah, that's a very good question.

I will comment on related issues very soon. But here, here I'm just write down some states.

Yeah, I excite k1, k2. Yes, I want.

I'm not saying that this is the, yeah, you can have as many as you want.

That's right, it's true.

When I write it this way, it implies I only excite countable number of them.

But in principle, we will soon touch your point which related to your question.

And so for example, the simplest excited state would be just excited one of them.

So let me denote this by k.

And then the last simplest one is just excited two of them.

So this k1 and k2 can be the same. If this k1 and k2 are the same, and then just like the square.

And here also, I will not be very careful about the normalization.

And now we ask, what are the physical interpretations of those states?

So now let's ask, what are the physical interpretations?

So for this purpose, we can just look at the quantum numbers under, say, the Hamiltonian and under the spacetime momentum. quantum.

So for example, H0, so if you act on H on 0, then of course you get 0, assuming we throw E0 away.

Say, ignore E0 so far.

And then so for the ground state, and the Pi acting on 0, of course it's 0.

So you can see it just because they have a k here.

And so e also, when we throw this away, you have this.

And now, let's look at the excited state.

Now, look at this state.

So this answer is obvious because all different oscillators, they are independent of each other. And this just give you over the k.

OK, we can just use our result for harmonic oscillator, OK? And now you can look at the momentum.

So this is a spacetime momentum operator acting on here.

And then it has Ki, OK?

If you look at this expression, OK, when you do the act this on that, OK, OK, you ask this on that.

And then you can just use the standard trick when you do the commutator. And then the particular k for this is chosen.

And then the eigenvalue will be just given by that particular k i.

So there's one step of calculation there. And you should do it yourself. One minute calculation.

There's a data function you will be generated.

And then we'll get rid of that integral. And then you will pick up this Ki.

So now this equation now has an obvious physical interpretation.

So that means that this state K has spacetime momentum.

So now it's right down to 4 momentum omega K and K.

So this is exactly the momentum of a relativistic particle on shell. OK?

So this is of a relativistic, so this is a momentum of a relativistic particle because omega k is equal to k squared plus m squared of mass m, OK?

Yeah, so this means that the p squared, so this is four vector squared is equal to m. Yeah, so this satisfies the p squared equal to minus m squared.

So we can just, so it's very natural just to interpret this as a particle of mass m.

Because yeah, you can just interpret as a particle of mass m.

Good. So now let's look at this one.

So again, the calculation is very simple.

So you find for that one, for k1, k2, you find the energy just defined.

Again, you'll find that this is the energy eigenstate of the H.

It's the eigenstate of H with energy eigenvalue omega k1 plus omega k2.

And it's a momentum eigenstate of Pi with eigenvalue given by k1 plus k2. OK?

So e k are eigenvalues of h and p, OK?

Yeah? Yeah, so could that be?

So here, so this mass is not 0, right?

Because omega k is defined to be this. That's defined by my theory, defined by my Lagrangian.

Yeah, defined by my Lagrangian.

So this is the parameter of your action. Yes?

So that happened the same way?

Yeah, for here, it's uniform.

Here, just for each k, it's uniform.

But of course, when you add them together, you get something very complicated.

But for each k, it's just uniform.

So the most natural way to interpret this is just So you have two particles of momentum of four momenta omega k1, k1, and omega k2, k2.

And similarly, you can do this for any state like this.

There are all eigenvectors of H and P.

So nk1, nk2, you catch it, et cetera.

So this corresponding to nk1 particles of momentum, they're all on -shell, OK?

On -shell particle of momentum, k1.

And nk2 has nk2 particles of momentum, k2. OK?

Okay, so this tells you one thing, okay, so now let me just make some remarks.

Any questions on this before I make my remarks?

So the first point is that now you can see this can describe any number of particles.

So mathematically, in our description, they are harmonic oscillators, but each excitation of the harmonic oscillators from the spacetime point of view corresponded to a particle.

OK, so this is the beautiful thing of this theory.

And then due to, because of the commutation relation, ak1 dagger, ak2 dagger equal to 0 for any k1k2.

So you have four symmetries.

So when you construct the state, you can just commute them as you want.

OK, so that means the four symmetry in permuting all these different particles in the general state.

So this tells us these are bosons. OK, these are bosons.

And two is that all particles have positive energy.

OK, so even though that e squared equal to k squared plus m squared, this equation have two solutions plus minus omega k.

But when you look at the physical state, when you look at your state and look at the agglomaniac states, so all particles have physical energies, okay, so you don't have this negative energy problem associated with taking the square root, okay.

Also you have total energy of a state is equal to sum of energies of all the particles.

So this tells you there's no interactions between them.

Because you have potential energies between particles, and then that will change their energy.

When you put the two particles together, they will no longer be the same of the sum of the individual energy for each particle. And so that tells you there's no interactions.

So this is a series of free particles. particles.

This is a good starting point.

At least now we have particles. Okay.

Questions on this? Yes?

So now, is there a way to change this if you apply again, like in this picture, it's like you have another particle.

No, no, there's no way to change the momentum of a particle.

So once you created that particle, it just goes straight.

Yeah, it just does not change any more momentum, yeah, for that particle is conserved.

Is there, if you wanted to change it by way, or is it just?

So after dealing with this free theory, and then we will consider the next simply theory, and then that way we introduce interactions.

And when you have interactions, then the particle momentum can change, yeah, yeah. Yes?

So the thing of this? Good, good, yeah.

These are the momentum, yeah, it's like the plane wave in the non -rativistic, yeah.

And if you work like, try to mobilize it. Yeah.

So the commutes things don't change because everything is filled up by AK and they always commute and so that won't change.

But indeed, we will talk about the way packet to localize in space. Other questions?

So the last, we will talk a little bit about the technical point.

is that so far we haven't talked about the normalization of such a state. So now let's look at the normalization of states.

So let's just look at these states.

This single particle state, let's look at this normalization.

Let's look at k with k prime.

So now if you take the overlap of this with the k prime and then you can reduce this to the commutator of ak k and ak prime, ak diagram, ak prime.

So you will get just this.

So again, this is a five second calculation.

So you find the d. If you do the overlap.

But you already did in your p set, this thing is actually not the Lorentz invariant.

This is not a good thing. This is not a good object on the Lorentz transformation.

But when we construct state, we would like to have our state to have good properties on the Lorentz transformation.

So we will choose a society of different normalizations.

So instead of this state, we will define the following states.

We will define a k now without the vector, to be the square root 2 omega k and this k.

And so this is the 2 square root of k, a k dagger acting on zero.

And now, if you compute the overlap with this k and the k prime, So now, you have square root k, square root omega k for one of them, and then you have square root k for two of them, and then you have two omega k, and then you have two pi cubed, delta three, k, k minus five.

So now, you recognize this object from your p -side.

The omega k, multiply this guy, actually, have good Lorentz transformation properties.

So this guy actually transformed nicely on the Lorentz transformations.

So this will be the normalization we will use from now on.

So these states have very, so indeed you can show, so that will be, I think, in your P set three. You should look forward to it.

So you can show, if you act the Lorentz transformation, the lambda is a Lorentz transformation, and u lambda is the quantum operator to generate that Lorentz transformation on such a state k, and then you just get lambda k. And lambda k is the Lorentz transformation acting on that k. It's the Lorentz transformed k.

Okay, okay, and but yeah, you should see your maybe p3, okay, so so if I remember it good any questions on this Yes, yes Okay, you have a data function here Yeah, because omega k only depend on k only depend on the spatial part other questions. Yes Yeah, is that your problem?

No, just this guy, when you, it's not about whether it's Lorentz invariant, it's not Lorentz covariant.

Just when you transform it, just this object, when you transform it, it's very awkward.

That's not a very, you can transform it, you can write down a transformation for this object.

Just does not transform nicely.

Then that means if this does not transform nicely, means each of them don't transform nicely and then they're just not convenient.

So this one have the properties that when I act you on this scene and then actually you can just directly corresponding to the state with Lorentz transform and momentum. Yeah and then it's very easy.

This property does not apply to this one, does not apply to this k.

So you can We can also talk about a wave packet. So this is a plane wave.

So this is like a plane wave. This is the momentum eigenstate.

So we can also consider a general single particle state. We have the following form.

So we have the following form. We can write it as psi just as a superposition of this k.

Again, you only need to integrate over spatial momentum, momentum, because only the spatial momentum are independent.

And then some arbitrary function of k, and then on this k.

So any single particle state, you should be able to write this way.

So in particular, by choosing appropriate fk, you can construct a local wave packet. it, okay?

So you can, so choose k, you can choose fk to construct a localized wave packet in space, okay?

So again, we will have some exercise like this in your, in your, in your When I call this to be a two -particle state, because each is a plane wave, so essentially, they're not localized, they cover all space, OK?

And in one sense, they correspond to two particles.

So now you can solve this problem.

Say by constructing, say, using, say, dk1, 2 pi cubed, dk2.

So by choosing a proper function k1, k2, this k1, k2, you can construct two widely separated wave packets, okay, widely separated, again, it's in space.

So you can really talk about, so these are the generally two particle states, okay?

And so that confirms, actually, Maybe this can, it should be interpreted as the plane wave version of the two particle states. Okay.

Good. Good. That's a very good question.

Indeed, there will be a problem in your PS3 which asks you to show there's no eigenstate, there's no perfected look. Yeah. So in non -rativistic quantum mechanics, you have this state.

You have this wave function, say, you can localize it at the point.

But there's no analog of such kind of state in the relativistic case, precisely because there's no position operator anymore.

So you will explore this a little bit more later yourself.

Yeah, just with this formalism, you can explore all these questions yourself.

That's the key. Actually, you already have the power to do that.

So now let's say a little bit more on the structure of the Hilbert space.

So here, the structure of the Hilbert space is a little bit special because there's no interaction.

So you can just separate it into the full Hilbert space which I write as H, script H. Then you are separated into the vacuum state.

And then you have the one particle field of space.

And then you have the two particle field of space.

They all, they don't have anything to do with each other because there's no interaction, OK?

So in principle, you can have infinite number of them. You can have infinite number of them.

So if you restrict to always a finite number of particles, OK?

So the sets of states.

So this is one particle, this is two particle, et cetera.

So that's of states with a finite number of particles.

It's called a folk space.

In the folk space, we only have states of finite number of particles.

But we can have arbitrary number of them as large as you want.

10 billion, 100 billion, but it will be finite.

So finally, those k, either this k or this k, they are just plain wave normalized because they're normalized by delta function. So strictly speaking, they are not normalizable states.

And so strictly speaking, they are not in the physical here space.

So strictly speaking, k or any of those states rates of k, or k1, k2, are not normalizable. They're only plane wave normalizable.

So just as, say, psi is equal to extremely i kx, in non -right -based quantum mechanics, a lot of normalizable, a lot of normalizable.

So they are convenient for various mathematical operations, but they don't correspond to general physical states. states.

So it's not generally in physical state.

So physical state, we have to consider, for example, this kind of state, and then you choose fk, so that is normalizable.

So if we take this, and then so with psi given by that, then the normalization of psi, you You can calculate it.

And then this is given by, so you can easily guess the answer.

So when you do the overlapping texture, you find that it's given by the modulus of fk squared.

So if you do the calculation to calculate the norm of that state, you find it given by that.

And then we can choose this to be finite. And then this will be a, then psi will be a normalizable state.

So for this reason, sometimes we don't do it very often.

Sometimes you can choose just a basis of fx, say, of fi.

So i run some number.

And then this alpha, then you can define alpha i, okay?

And then, and then this will provide, then alpha y will, will provide a basis of all, basis of the single particle, or, or, or, or, a single particle state.

So here's a fun fact related to the earlier question, uh, uh, was asked.

So, reliably, here, you have, if you look at the k, so k form a basis, OK, let's just look at the single particle states.

So, reliably, this k form a continuum, uncountable continuum of basis of states, OK? Uncountable continuum.

But once we impose this normalizable condition, means you need to choose the normalizable f.

And then you can show the basis of this normalizable f actually is countable.

So in the end, the Hilbert space, the one dimensional, the one particle Hilbert space is actually generated by countable basis. Just like in your ordinary quantum mechanics.

It's actually rather than uncountable basis like in k.

Once you restrict the normalizability. So this is a far mathematical fact.

Great. Questions on this?

OK, so the last few minutes, we talk about conserved charges, OK?

That's rho of loss of charges.

So classically, we say if the action is invariant under some symmetry, some transformation, some continuous transformation, say, say, of the infinitesimal form.

So the alpha are the infinitesimal parameters.

So the alpha label different transformations, and A label different fields. Alpha label different transformations.

And then you will get the conserved currents, j mu labeled by this alpha.

So alpha label different symmetries for each symmetry, And then you have a conserved current, which is satisfied the j mu alpha equal to 0. OK?

Alpha equal to 1, 2, et cetera, label different symmetries.

And then you can write down your Lothar current.

So we derive a general formula for the Lothar current is given by the form, say partial L, partial mu phi by a, and then f a alpha. So this is essentially the same as before.

Essentially, I just add the index alpha. Now to label, you can have more than one symmetries.

And then minus k mu alpha. So k mu alpha is the derivative.

Total derivative, suppose, data L is given by epsilon alpha partial mu k mu alpha.

So the k mu alpha is the total derivative which would correspond to that transformation.

And then the q alpha, that would be the conserved charge for that current, the 0's component.

So this is a classical story.

Now I'm using a little bit more general rotation for each alpha.

So now the key thing is that the quantum level, So at the quantum level, first as we already seen in the case of the Hamiltonian and the momentum operator, so q alpha is time independent.

It's a constant operator.

So this is a reflect that this is conserved.

And also, another important property of q is that when you act q alpha on your field, If you look at the commutator between this Q alpha on your field, it actually generates such a transformation.

So it turns out, yeah, so if I put a parameter here epsilon alpha, yeah, so it doesn't matter for Q, say it's up stairs, down stairs, and then you get, okay?

So you just generate that transformation for f, for phi, okay? OK. So this can be shown explicitly.

So you can show this explicitly. We're running out of time now.

So you can show explicitly just using this expression.

So let me just quickly outline the way to do it.

So if you say, so let's look at the 0's component of this J. j.

So the del's component of the j just equal to partial l, partial partial 0, phi, a, and f, a, alpha.

So let's consider the, yeah, I think today we will not have time to finish Anyway, yeah, let's just leave this to the next time.

Yeah, let me just say that the quantum level, this will generate the infinitesimal transformation. And then you can also express this q.

OK, you can introduce, say, exponential i lambda q.

So if q is the conserved charge, and then you can express q like this, And then this operator then generates, when it acts on phi, then generate finite transformations.

So this takes you to phi a prime, the finite transformations.

Anyway, so for Psi 2, you will actually check yourself a couple of simple examples. And you will see this works. works.

And in the next lecture, we will give you a simple derivation to show this actually works. Okay. It works in general.