MIT 量子场论·8.323·第05讲：复标量场理论与反粒子（原声文稿）

Transcript of MIT 量子场论·8.323·第05讲：复标量场理论与反粒子 （中文完整文稿）。

So, last time, we more or less finished discussing quantization of the free -scale field theory, and then we showed that the excitations of that theory essentially give rise to relativistic particles.

And so, we also discussed various other things.

say, here with structure of the Hilbert space, conserved quantities, et cetera.

So, so, so at the end, we talked about the, how to understand the conserved charge at the quantum level, okay?

So classically, so we have, say, symmetric transformation means that if you take phi A to some phi A prime, okay, phi A prime, normally we write it in the infinitesimal way. Introduce some parameter A, some parameter epsilon.

And I can also have multiple symmetries.

So for each transformation, say I have an epsilon, so alpha labels different transformations.

And then, so this is the transformation which can depend on phi A, also depend on its derivatives, et cetera, and then so if you have a such a symmetric transformation we discussed before, then that generates a conserved current, and it's labeled by alpha, alpha just delots different conserved currents, and then yeah, so if it's conserved, okay, so this is the classical statement, oh yeah, also lots of theorem tells us how to write down on such a current explicitly.

So we can just write it as the current that can be written explicitly.

So k mu is the transformation of the Lagrangian density.

Suppose under this transformation, the Lagrangian density transform as, say, alpha partial mu, k mu alpha.

So alpha doesn't matter.

Alpha just label different symmetries. is the index doesn't matter whether it's up or down, okay?

And then the zero's component of the charge, the zero's component of the current when you integrate of all space then give rise to a conserved charge.

Conserved charge, okay?

And the, yeah. Now let's discuss.

So when you go to quantum level, so when you quantize the theory and this just become a quantum operator and yeah, the whole current become a quantum operator and then Q is also a quantum operator.

Okay, at the quantum level, at the quantum level and then now Q alpha, because it's conserved, so this become time independent operator.

We see the explicitly in the case of the, say, Hamiltonian and the momentum, okay, so we showed explicitly that they actually don't depend on time.

But at the quantum mechanical level, actually, this Q can play a more important role, can play a more important role.

In fact, it can be viewed as a generator of symmetries.

Okay, so we can show, we will show the following statement that if you consider the epsilon, so consider such a quantity, yeah, consider the commutate of q with phi a, okay, so now let's continue over here.

here, let me write it here.

And that generate minus epsilon alpha, alpha A. OK?

So the commutator of this q alpha, epsilon alpha vis phi A, essentially generate this transformation.

Essentially generate this infinitesimal transformation.

So that's why we call the q alpha also a generator temperature of symmetries, okay?

So this is the statement that I'm going to show very soon, okay? So before I do that, do you have any questions?

Sir, current's equation, how can f have an alpha in the mean?

Yeah, yeah, just alpha, don't worry about the location of alpha index.

Alpha just label a different current and just one, yeah, here we don't worry about this location, yeah. I just put it up and down just for convenience.

Yes? In its position, is that a sum of all alphas? Sorry.

In this closed RL, that's one alpha partial between alpha, is that a sum of all alphas? Yeah. Yeah, that's sum of all alphas.

So any repeated indices is summed.

So if you only have one, say, for example, if you have one epsilon, of course, there's no sum. But if you have two different charges, and then, yeah, this is sum.

So, we are going to show this statement.

And so this statement is very easy to show in the general case.

So let me call this equation star.

So let's first consider the case.

First of all, it's considered a sub -case that let's first suppose f a alpha, just this transformation of phi a, does not contain the time derivative of phi a, okay?

And also the zero's component of the k is equal to zero, okay?

Okay, let's consider first a more limited case.

Okay, so this more limited case actually covers many, many examples, okay, actually covers a majority of examples.

For example, if you look at the translation symmetry, which give rise to the stress tensor.

So this applies to the case i mu, t mu i, so this corresponding to the spatial, so this is the conserved current corresponding to the spatial, so this is the current for spatial translations.

When we do spatial translations, then of course this F gets given by the spatial derivatives and then does not involve time derivative and in that case the K also does not involve K zero components zero yeah just I try to remind myself which you worked in your p -set and so so for this example I is the counterpart is the alpha here okay so for each direction you have a translation and so this so the alpha of course, as i is the alpha here and the mu is the j mu here, okay?

And also, this applies to the case which for all internal symmetries.

And internal symmetries, we mean the transformations don't involve space -time coordinates, okay?

For example, you did in your piece at the complex scalar field, which you can just rotate it by a face.

And that's called internal transformation because that transformation does not involve space -time coordinates.

In contrast, when we find the stress tensor, we will do a space -time translation, and that does involve space -time transformations.

Yeah, so this corresponds to the space -time transformations.

Space -time symmetries, and this is the face, is the internal symmetries.

okay so for all internal symmetries you don't the transformation will not contain the time derivative so so in this case then then it's easy because in this case then we just have the one then the zeros component of j mu j just given by the first term by definition is just the canonical momentum conjugate to the, to the A, okay?

So the, the, the first factor partial L, partial partial mu phi A, when you set mu equal to zero.

So that's just the derivative with respect to the, a time derivative of phi A.

And that by definition is just the, the canonical momentum density conjugate to phi A.

And then, then you have f, f, alpha, a, okay, f, a, alpha, okay, good.

So let me just make one comment on this expression.

So this expression, of course, classically, you can write whatever way you want.

But quantum mechanically, there's a very important subtlety, because this typically will involve phi, okay?

And then this is momentum and they don't necessarily commute.

So this is the field three version of so -called the operate ordering ambiguity, which you're already in quantum mechanics.

Okay, in quantum mechanics, when you go from classical, even it's already in the long right to basic quantum mechanics.

When you go to classical mechanics to quantum mechanics and then there's an issue of how you order operators when you have X and the P at the same time.

So here, you also have potential, operate ordering ambiguity. You have to pay attention.

Often, such kind of ambiguities can be resolved by physical considerations.

So we will see some examples in a little bit.

So now, let's just imagine there's some ordering we have chosen. And now, let's look at this commutator.

So we can just write down the definition.

So this is just integrating over all spatial direction, epsilon alpha, and then we have pi a, f a alpha phi a.

Yeah, let me just, sorry, here let me call phi b, because phi a, now they are different. So these two indices are summed, OK?

So let me call them b, and then you have phi a here, OK?

Okay, so now you can use commutation relation between the momentum density and the phi phi.

So we discussed in general that the, so we discussed the scalar case, but in general you have this commutation relation, which is delta 3, say if this is the x, this is x prime, and then this is x minus x prime, okay, and delta AB, okay.

So each one, it only has canonical conjugate momentum with its own momentum, okay, if they have different fields then Of course, yeah, so you have that.

And now if you use this, OK, so here is x.

Yeah, let me call this prime. And this could be x.

Sorry, I should label more explicitly x here. And then here would be x.

And so here is x. And then here would be x prime.

Sorry, I didn't say enough space. Let me just rewrite it.

So this gives me d3 x prime epsilon alpha, then I will have this pi a, then x prime f Let's alpha a x prime, and then we is commutative with, sorry, alpha is down upstairs, b, and then phi x. OK? OK.

And yeah, so we can evaluate this, say for example, of OIT equal to zero, so I suppress the T.

And then, yeah, and now you can just trivially use this commutation relation, and then this just give you minus i, epsilon alpha, f alpha a, f a alpha x.

Okay, evaluate at x, okay.

So that just confirms that expression, okay? Okay? Confirmic secretion.

So this is not satisfied when you have time translation. Okay?

So for time translation, so first any questions on this?

In the simple case, yeah, the idea is very simple.

Essentially just the first term proponent to the momentum and the momentum conjugate to this and then you just take that, okay? Yes? Yes?

Is it a term with the, like, come together and multiply because of our restriction? That's right. That's right. Yeah.

So if there's no partial 0, there's no time derivative here. It means there's no momentum here.

If there's no momentum here, then this will commute with this one. Yeah, no matter it's in here, it's always commute.

Yeah, a very good question. Yeah, I forgot to mention this point. Other questions?

So, so now the special case is that if you have time translation, okay, when you have time translation, for example, one example is the, the, yeah, the, the, the, the, the So the L here for the time translation L is the analog of A and the partial T phi A is the f here and then here is the pi A. So here you also have time derivative.

Here you also have time derivative.

And so this by itself is also a pi. And then L may also contain in time derivative.

So in this case, the k0 is non -zero, and which is equal to L.

And so in this case, the conserved charge is just h.

So in this case, our general argument here don't apply because now you have pi here also, and you may have a time derivative here also. So the story is more complicated. it.

But this case below trivial is true because it's by definition h acting on any field is minus i partial t phi a.

And this is exactly the time transformation of a field.

So in In this special case, this equation is also satisfied.

Good any questions on this?

Okay, so this is just followed from the Heisenberg equation itself.

Okay, and now you should just understand this equation as, so H now can be considered as a symmetry generator for time translation symmetry.

And when you act on the field and generate the time translation.

Okay, time translation just infinitesimally just a time derivative on phi.

So these are infinitesimal transformations, but now starting from here, you can actually obtain finite transformations on the field also using Q.

Okay, so we can generate finite at transformation.

So let's consider, say u lambda alpha, so lambda alpha just some parameters, let's consider this quantity, okay, now I, now I explain shade this q.

So q is the operator and lambda is just some numbers, okay, lambda is some constant parameters, okay, and again the alpha is summed, yes?

It's like the Heisenberg equation of motion, but here how would you evaluate this commentator not knowing the form of element, but because you said it can't depend on the time derivative of this as well. Right. Yeah. Yeah. Yeah.

Here, just this general argument doesn't apply anymore.

So in this case, you really have to, you can check this explicitly.

You can just write down the explicit value, the explicit form of h, and then work out explicitly, but you don't have to do it because we know this has to be true but just by self -consistency.

Because when we quantize it, essentially we are solving the Heisenberg equation.

Yeah, but you can check it explicitly yourself that this is true.

So here, so lambda alpha are here just some collection of finite transformation parameters.

So these can be used to generate the finite transformations.

So if you act u lambda alpha phi a and u lambda alpha dagger, and then you just get phi a prime.

And now this is a finite transformation or finite semi -transformation of phi, okay?

And when you expand, when lambda alpha is infinitesimal, and then you can just expand this exponential, then the leading term becomes this one, okay?

And the leading term becomes this one.

And yeah, so when lambda alpha equal to epsilon alpha goes to zero, and this equation, this equation star and the star star reduces to star at the leading non -trivial order when you expand the exponential, okay.

So, this can be done in general just, so the reason this is true, without even doing any calculation, you You can just imagine you can build this u, this finite transformation by infinite number, infinitesimal transformations.

And so when you build them up, and then you will just generate the finite transformation.

So you already worked out the example in your p -set, so you already worked out the example point in your p -side that if you can see the ht minus i pi xi acting on phi is 0, minus i ht plus i pi xi.

And then this just give you phi tx. OK? OK? Just give you t by dx.

And so this is, so here, ht minus pi xi is our lambda alpha q alpha here.

So alpha here runs 0 and i direction.

So in the 0 direction, the t now is just some parameter.

and the lambda alpha equal to t and the minus xi, okay?

And then q alpha is just the, and the q alpha here is equal to, yeah, t xi. Let me just do t xi.

And the q alpha here is just h minus p.

Good. So this is just an example of that. Okay, this is just an example of that.

And also you can consider Lorentz transformation. So now the example is the Lorentz transformation.

So let's consider, so in your piece that you already, so in the Lorentz transformation, the conserved charges, actually there are six of them.

Okay, and you knew going from one, zero, three, okay, Okay, so they all together six conserved charges associated with the Lorentz transformation, three rotations and three boosts, okay?

And then, so when you write the finite transformation, and then you will have six parameters, okay? You have six parameters.

And omega mu mu, because mu mu is anti -symmetric, which you worked out in your p -set. So this parameter should also be anti -symmetric, okay?

And so they are finite angles, finite angles, sorry, not finite angles, finite angles and boost.

Okay, so essentially they give the parameters for the finite rotation angles and the boost.

And so that generates a low -range transformation. Okay, that generates a low -range transformation.

And so for example, you say if you act on u lambda on phi x, and you can check dagger on phi x, what you get is phi prime x, and phi prime x is also the same as phi lambda 1 on x, OK?

So here lambda 1x is the Lorentz transformation of x mu with parameter determined by, yeah, just omega mu.

So this capital lambda is just the corresponding Lorentz transformation.

So you assumed that you can represent, I guess, the operators that perform these transformations like this, but don't you have to think about if it follows the algebra such that you can represent it like this, or like the group of transformations? Yeah, yeah, yeah, indeed.

So here is not, here is not say I assume. Here I'm just directly telling you the result.

It's just when you build, yeah, so if you want to generate the, yeah, I'm just directly already, directly tell you the answer. It's the composition of infinitesimal transformation. Exactly.

If you do the composition of the infinitesimal transformation, you will get that. Yeah. Yeah.

The reason I write down this explicitly, immediately because you should have already seen this in quantum mechanics because that's how you do the angular momentum.

In quantum mechanics, when you do the angular momentum, the way is the same thing.

Here we just generalize it to general symmetries.

So it was kind of a similar question, but just like, so like this is like a general procedure if you have a symmetry to get the representation on your open space. Yeah. But it's theorem like this. Yeah.

But like in QM, we usually just postulate that there's a unitary operator and it demanded the group law goals, right? Yeah, yeah.

So like, don't we need to show that those things get this, that this preserves the group law or is that not? Yeah, yeah, so this is guaranteed.

So, so, yeah, I didn't go into that, but this is guaranteed.

Essentially, that's how you, yeah, in the, this is how you build the Lie algebra.

Because Lie algebra is also the, the finite transformation is also to build up by infinitesimal transformations.

And so essentially guaranteed, yes.

So there's a series of regroups in the algebra behind this. Yes, I didn't go into that.

So some of you may already seen a little bit of these when you talk about angular momentum, say in quantum mechanics. So it's exactly the same story there. Yes.

I know that rotations correspond to angular momentum being preserved. Yeah.

But what is the intuition for the thing that's in boosts? Yeah, yeah, indeed.

There's no very intuitive way to think about it. Also, it's not that useful. You can define a boost charge.

So if you can see all the spatial components, and if you can see all the ij here, then these are the angular momentum operator.

And this is your standard angular momentum operator, which you have seen.

And if you look at zero i and then that was sometimes we call boost charge it generate the boost But in terms of the conserved number Yeah, we don't have We don't use it in the in the essential way say say say in other context.

Yeah Yeah, yeah, it just does not give you something additional So, normally when you deal with physical problems, say angular momentum conservation, momentum conservation, energy conservation is right enough, yeah, you normally don't need to use a boost charge conservation to solve your problem, yeah, so that's why we don't see this object very often, say when you discuss special relativity or other things, but here plays It's a very important role that in quantum field theory, this is the operator which generates the boost symmetry.

Okay, if you want to do a boost in your Hilbert space, that's the operator to do it, okay.

Yeah, so in your P -set, yeah, another example is the phase rotation. Let me just very quickly wrote here.

And so another example is the phase rotation if you have a complex scalar field.

So there, we show there's a loss of charge corresponding to the phase rotation.

There's a loss of current corresponding to you can rotate a phase to the field.

That's a symmetry. And so that will generate the charge Q.

It's going to do the first state.

And then if you have alpha q, and then when you act on phi, and then you can just check explicitly that we generate the phase rotation in phi.

And we generate phase rotation in phi.

Any questions on this?

So let's conclude our discussion of the real scalar a field, now we can move to a new part.

So before we move to new part of your last chance, do you have questions about this part?

So now let's very quickly talk about, let's very quickly talk about complex scalar field.

And when you can see the complex scalar field, we get something new.

We get the concept of antiparticle.

Okay, so now let's can see that the Fourier Lagrangian density, again, we only can see the quadratic, the simplest, only quadratic Lagrangian.

But now the phi is complex, by its complex.

So complex is also the same as true real.

So you can decompose phi into its real part and imaginary part.

And if you plug them in here, and then you find you just get two separate p's. And the one is the real scalar field for the real part.

And the one is the free scalar field for the imaginary part.

Because each one, if you write phi equal to a plus i b, and here it just becomes a squared plus b squared.

Here it's also a plus b squared.

But this reason we write in, yeah.

So essentially, we just get two copies of our previous theory. We just get two copies of our previous theory.

So you say, oh, then it's trivial. Why should we just consider this? We just get two copies of our previous theory.

But there's some advantage, but there's something conceptually new when we write in this form.

There's an advantage to writing in this form rather than write as two separate identical scalar field theory.

It's precisely in this form, this phase symmetry is manifest.

If I just write that as a two separate real scalar field, and then this phase symmetry is not manifest.

And so here in this form, there's a phase symmetry manifest.

So phi goes to measure i of phi, and so you immediately see that symmetry.

Okay, so when you write it in terms of real and imaginary part, this corresponding to a rotation between the real and imaginary part, and yeah.

And equation motion is the same because you can easily find equation motion is the same, okay?

And yeah, because just two copies of your previous theory, of course, equation motion should be the same.

And the Lagrangian density, the momentum density for phi, then it's become partial 0 phi star, and the partial phi star become partial 0 phi.

So we can treat phi and the phi star as two independent fields.

Okay, and then and then the canonical momentum say say if you take it really to be partial 0 5 Then you get the partial 0 5 star and the vice -versa.

Okay, so so they're complex Conju momentum and like that Okay, so we can now write down again.

We can just write down the most general solutions So the basis of solutions is the same same set of solutions as before we just have u k, x, you get your minus i omega k, t plus i k, x.

And then you have u k, x star. OK, because it's the same equation.

So the basis of equation, of course, is also the same as before. Yes?

So if the field is an operator, it's an observable. It doesn't have to be permission.

No, the field, it depends.

So, yeah, indeed if it's something directly observable, it has to be a permission.

But I can have two observables, for example, I can combine them into a real variable.

Yeah, yeah, just not necessarily the find itself having to be directly observable. It's just a field.

Yeah, the field itself don't have to always be observable by itself.

Yeah, yeah, yeah, this is not the momentum.

This is just the canonical momentum for the canonical quantization, right?

Yeah, yeah, this is not the spacetime momentum.

And is the adjoint of the field operator the same thing as the start?

Yeah, here we are talking about the classical theory.

And then when we go to the quantum indeed, we will replace the star by dagger.

OK, so now we can just write down the most general solutions to this equation.

Now we can just write down the most general solution to this equation.

So there's one slight difference from before, almost the same identical as before, except with one difference.

But here, previously, we have AK dagger.

We have AK star, say classically. Let me just first write classically. So previously, we have this.

Previously, why we have AK star here? When we have a real scalar field.

It's because the field has to be real.

And then we have always add to its complex conjugate. But now the difference is that now phi is complex.

And the phi is complex.

And then we no longer have the real condition. That means here I can choose another arbitrary constant here. So this is arbitrary constant. So by convention, let me just put a star here. It doesn't matter.

It's just the name you call it. So now the most general solution.

Now you have two sets. You have A, which is complex number. Then you have another complex number, which is B. Now you have two sets of complex numbers, OK?

And it makes sense, because when you have a complex variable, you double the degree of freedom. And your integration constant also doubled, OK?

So BK, independent of A.

So now you have two sets of variables, OK?

So you can get the phi star just by taking the complex conjugate of it. Just by taking complex conjugate of it.

And then the full set of integration constant, just you have a k, you have a k star, you have b k, and then you have b k star.

OK, so this is the classical story.

So now I'll go to quantum.

We just do exactly as we did before.

We just promote all those into operating equations.

So you just call this dagger, just call this dagger. Now you just view every one of them as operators. And now these just become constant operators.

They become constant operators.

And now here, you also just call them to be dagger.

So now equal time correlation.

Still, we have to impose the equal time commutation relation.

So now we have to impose the equal time commutation relations.

Again, so these are should be simple. So we should have five dagger.

They can see that the independent operators say they should commute with each other. They all field variables.

And the five should commute with self.

So this is all evaluated at different spatial.

Yeah, just to save time, I will not write this explicitly, but all this should be evaluated at the same time by different spatial locations, OK?

Such as all this equal to 0, OK?

Except the only thing which is non -zero is the phi with its own canonical momentum, OK?

So everything else will be 0.

The only thing non -zero is phi, which is canonical momentum.

again should be given by i, that has three x minus x prime, and then then the similar thing for the phi dagger, okay?

Okay, so again I save effort, okay?

So so so that's what you should impose.

So now if you plug those things in, then then you can find the commutation relation between a and a dagger.

Again you find and similarly with B, you just get now two set of, two infinite family of harmonic oscillators.

Okay, previously we have one family and now you get two families. Okay, and all other commutates vanish.

So this part is boring. You can almost guess everything without doing any calculation.

You essentially guess everything without doing any calculation.

So now again, you can write the Hamiltonian And in terms of ak and bk, just will be as you would guess them to be.

Just the harmonic oscillator that corresponds to ak, harmonic oscillator corresponding to bk.

So and then the ground state, again, is the state allotted by both of them.

So this is the vacuum state.

And now you can act ak dagger on 0. 0 and the bk dagger on the arrow, et cetera.

So now we have two kind of particles.

So now we have two kind of particles generated by ak dagger and the bk dagger now. Now you have two kind of respectively.

And both of them have the same on -shell condition. OK, both particles have the same mass.

They satisfy the p squared equal to m squared minus m squared.

OK, so because they all come from the same equation.

OK, so now the question is how do we tell them apart?

OK, now we have two type of particles.

They have the same mass.

And we, yeah, just by definition, they don't have any, they have the same math, they all have the same spin because they all spin zero, scalar particles, there's no directions, okay?

And yeah, the question is how do we tell them apart, okay?

So here is this U1 symmetry become useful, okay?

Here is this U1 symmetry become useful.

So this u1, so this phase rotation, this phase rotation is a symmetry.

And mathematically, this is called the u1 transformation.

So this is normally called the u1 symmetry.

U1 is just a mathematical term for phase rotation. Mathematically, it's a u1 group.

It's a mathematical term for, so the way maybe take them apart is by looking at the conserved numbers, look at the quantum numbers. If we look at two particles, how do we tell them apart? We look at the quantum numbers.

And the quantum numbers, so far, they look almost the same because their mass is the same. And by definition, they are scalar particles.

And so they have 0, they are spin 0, they don't have spin.

And so the only other conserved number, yeah, if you want to talk about quantum number, the quantum number has got to be conserved, okay?

Because otherwise, if it changes with time, it doesn't make sense to use that number to label a particle.

But if we get one more conserved number corresponding to this symmetry, okay?

So this U1 symmetry, the Lothar theorem, yeah, from the Lothar theorem, so this U1 symmetry, then tell us that there's a conserved charge.

So if you work it out, you work out the Lothar theorem, then you find it's the corresponding conserved charge which you should have already worked out. So let me just write it classically.

So as I just write them as 5 star.

So classically they have the following form. Okay.

So at the quantum level this become operators.

But here there's an operator.

But now here there's an operator.

When we go to the quantum level, Now, let's operate ordering ambiguity now.

You see, here, the phi don't commute with its own conjugate momentum, okay?

And so here, we have phi multiplied conjugate momentum.

And if we change the order, and then we will result a data function, okay? We will result a data function, okay? And that will give rise to infinite, yeah.

So, so at the quantum level, different ordering, differ by some infinite constants, okay?

So you can take five, you can, yeah, if you change ordering them, then you get a delta function, but delta function again is evaluated, because these two are evaluated at the same x, and then you get the delta function you value it at zero. So you essentially get the infinite constants.

But we can fix this ambiguity by requiring your vacuum state, your lowest energy state, should have charged zero.

Because by definition, when you add the lowest energy, there's nothing there. It's a vacuum.

By requiring the lowest energy state have charged zero.

OK, so that uniquely fix whatever infinite the ordering.

So by applying this equal to 0, and then you can show that the Q, so now you can essentially, the Q have the following form, OK?

So you can already, you may be able to guess the answer.

of a k dagger a k, the minus b k dagger e k. That's what you get, OK? So you plug those expression in.

You plug those expression and it's time derivative, OK?

Time derivative, you pi. You just plug those expression in.

And again, you find because this is conserved, you find all this time dependence cancel, et cetera.

and then you get these two expressions up to some constant, okay, and then this condition requires that that constant must be zero, okay, yeah, when you write in this form, okay.

You see this annihilates the cube because the A is on the right -hand side of a dagger, so this automatically annihilates the vacuum.

So essentially this is just occupation number for A, and this occupation number for BK Okay, so we essentially get q, essentially the occupation number for ak and the occupation number for bk.

Okay, so the occupation number I just write as n, okay?

Yes? I still don't see how this fixes anything, because if you switch them, you still get this issue. What?

If you permit them, you still get the infinite constant, so I'm confused why this is considered effective.

No, when you permute them, then you no longer evaluate the vacuum, right? So the constant will cancel.

So only this form, with A sitting on the right -hand side of a dagger, this will you're not allowed to do the vacuum.

I guess, like, because earlier, the line of argument was, if you pre -meet them, then we'd get it out to push, because you would get the same thing here, right? If you pre -meet them, you'd also get it out to push. No, no, no, no, no. Here, here, there's ambiguity.

Here, I don't know how I should order them.

Whether I should put phi to the right of pi or put the pi to the left to the right of phi, OK, I'm not sure how should I do. Here, there's no prescription.

And here, there's no prescription.

But if I impose this condition, OK, then that tells you when I impose this condition, then Q must be of this form, OK, must be of this form.

Of course, when you permute, then it's the same thing. When you permute, then you don't change the operator.

You don't change the operator.

But if you say, let's do this order, if you do this order, what you get is this expression with the infinite constant.

And then this condition tells you somehow you have to change your order here, so that constant is 0.

When you write it in this form, when you write it in a and b, and that constant has to be 0. Yes?

Yeah, yeah, it's, yeah, this is a good question, which it's hard to say, so most of the time you can fix them by some physical environment, as here you can fix it by some physical requirement and then you don't have to worry about it anymore.

But the ordinary ambiguity requires you try to find such a physical requirement.

Yeah, yeah, because otherwise what ordinary should you use?

And the process of finding that requirement by itself is understanding the physics.

When you got the infinite energy on the previous lecture, Krishna required the brown state to have like zero energy and then get some different order. Right. No, in that case, we don't have ordering ambiguity.

So in that case, because we just have pi square itself plus phi square and plus phi square, there's no ordering ambiguity here. And so whatever you get is whatever you get.

And if you get infinite constant, then you get infinite constant. And you don't have freedom.

Other questions? Yes?

So I don't also see how like phi and phi dagger is clear that they can be like how do you know that there's no coupling between the fields or how do you represent it and all this is a definition you say where this come from yeah yeah this is just a definition of the this is the same this now have nothing to do with field series only this equation have to do with field series this have nothing to do with field series This is just quantum mechanics that different degrees different degrees freedom they commit with each other Yeah, just x1 commit with x2 in quantum mechanics Yeah, fine here. Just fine fire dagger.

Just it's analog of x1 x2 other questions No, it's different from that here Here it just means that the here you should view them as the variables in the configuration space and And in quantum mechanics, the variables in a configuration space always come with each other.

Good. So now the key thing is that here is the minus sign.

So that means when you look at the commutator between Q and A, you get the A.

And if you look at the commutator A with BK, you actually get the minus BK. Okay.

Yeah, let me just do the diagram because this is used to create a particle.

So if you just calculate the commutator, you just find that.

So that means if you look at the states defined by one particle state corresponding to A and one particle state corresponding to B, OK, then both are eigenvectors of Q, but here with eigenvalue one, but this one with eigenvalue minus one.

OK, so it means they actually have opposite charge, OK, so k, and oh, by the way, so let me explain a little bit of this notation.

This k does not mean the magnitude of k. So this k means the bigots.

Normally, our convention is that the four vector, say, the x mu, we always just write as x.

The p mu, we just write as p. So here, you should view this as a four vector, as a four vector.

It's just a short -time notation.

So k and the k bar, then they have the same mass, the same spin 0, but the opposite charge.

So by using this q, we cannot distinguish these two particles.

So normally, we call such a particle, we call in this case, which they are all quantum numbers are the same, except they have opposite charge. We call them particle and antiparticle. So this we call a created particle.

But here we say they create antiparticle.

So here let me just make a remark without proving it.

So in the relativistic quantum field theory, you can show that any particle has antiparticle Okay, in any quantum field, in any relativistic quantum field theory, any particle has anti -particle, okay?

In the case of the real scalar field, you say we only have one particle, in that case the anti -particle is itself, okay? And, yeah.

Good. Do you have any questions? Yes.

Um, yeah, the conservation of q, it follows from the symmetry.

Yeah, it's, um, no, actually the, the, the number of A particle and number of B particle because they're not separately conserved. They can actually, in principle, they can allow it.

Yeah, here in free theory, they cannot allow it, but if I consider slightly more complicated theory, say if they allow it to interact, and then they can actually allow it, and then disappear, and so in that case, but this formula still works, this symmetry is still there.

Yeah, so that's why the symmetry is powerful, also applies to the case -based interactions.

Yeah, indeed, what you said is correct in the free theory.

Yes? So does this mean like a different?

Yeah, yeah, essentially their property, it means their property almost identical, except they have over the charge.

Yeah, yeah, they have essentially, Yeah, just like real world, we have electron, we have anti -matter, we have anti -matter. They have essentially identical properties.

And if the world is made of anti -matter, everything still will behave the same. And yeah, yeah. But except just the charge is different.

In the number of antiparticles and particles.

Yeah, it's conserved, yeah, right. So I guess this is like the electric charge or is this some other charge? At least some other charge.

So in the case, we will go to, when we talk about electrons, say electrons involving more complicated formulas, we have to introduce spin, particle with spin.

And in that case, we also have electron and anti -electron. And in that case, the story is parallel. well. And yeah.

So this does not necessarily involve BNM or like Young's repellent? No. No. Yeah. But this is the analog of that.

It's the analog of that. Yeah.

So the charge here, they can be some other kind of charge. It does not have to be yeah. Yeah.

Can you get like something that's physical that's conserved from a symmetry that's not really physical?

like this is not like you can translate in space or like translate time, this is, I don't understand, I don't know how to think about that. Yeah, yeah, that's why we call it internal symmetry.

So for things we don't have good intuition about, you just invent the new name for it.

And once you get used to this new name, you say, oh, I understand it.

Yeah, yeah, it is a more or less to getting used to it.

And indeed, at first sight, it's not very intuitive because it's not something you can, yeah, from your normal experiences, et cetera, but yeah, but this is the, yeah, but this is as, at the fundamental level, it's not that different from energy conservation, it's just some symmetry and then this to some conserved quantities.

Yeah, yeah, that's right, they can, they can in principle have multiple charges, but the particle and other particle, they're always opposite.

Yeah, yeah, in terms of charges, they're always opposite. Yeah, whatever charges you have.

Two different charges that each one can have instead of four things in a different language. Yeah, yeah, I think, yeah.

Yeah, you can have them, but for particle and antiparticle, they're always opposite.

They can have multiple charges, but they're always opposite.

So, this concludes our discussion of the complex field.

So, we have a few minutes, then let's do the next topic.

So, hopefully, now let's talk about the concept called the propagators.

So again, let's recall in non -right basic quantum mechanics, so this position vector plays a key role.

So this is the eigenvector of this x, which is the position operator.

So again, let me just put a hat here to emphasize. This is the operator.

So this is the eigenvector of x.

Okay, and because this play a key role because we need to use this to define the wave function and then the amplitude of the wave function the square Give us the probability of a particle at location x.

Okay, so so this is a crucial quantity So so this language Normally we talk about the the shorting of picture you started with psi x x at t equal to 0.

And then you try to find, yeah, or maybe say, you start to say x, say t prime at some t prime, you want to find psi x at some t.

So this is the D question of quantum mechanics.

So if you know the wave function at the initial time, and you want to find that the wave function at some future time.

So now let's imagine we consider this question in Heisenberg picture.

So in Schrodinger picture, there's just one eigenvector, yeah, just one set of eigenvectors for x for different eigenvalues.

But for the Heisenberg picture now, your position operator now depend on time.

Okay, so the position operate at different times.

They're different So now if you want to talk about the position eigenvector now you need the label Okay, so now the position eigenvector Need the t label.

Okay This t does not mean that this actually will this time. It just this is the label.

Okay, this means that this is the eigenvector of Position vector at time t t.

It just tells you this is eigenmarked at time t.

And now this question of this z question of quantum mechanics then can be reformulated into a different question.

So in this story, the z question of quantum mechanics becomes the following. What is the value of this object?

Suppose you are in the eigenstate, you are the eigenvector, you are in the position eigenvector, you are in the position eigenstate at time t prime with the value x prime, and what's the amplitude for you to go to location x at time t? Okay?

Okay, I remember this, This does not involve time evolution.

It's just, of course, one is laboring different eigenvalues at a different time, OK?

So if you load this object, this is called, this will normally be loaded G.

This is called the propagator.

This is called the propagator.

So if you load the propagator of a quantum system, then you have full knowledge of that system. So you have to solve that system. Why?

Because, say, let's imagine we want to find psi tx, wave function at time t, and then by definition, this is given by this.

So this is the Heisenberg picture definition. It's the psi overlap with the position vector at time t.

Now we can insert the complete side of states.

I just insert the complete side here. just this is identity, OK?

And then this object is just g.

So you just get g x t, x prime, t prime, then psi, t prime, x prime, OK?

So if you load, so yeah, so this just gave the wave function at t prime.

So now if you load this, given the initial wave function, If you load a propagator, you just need to do an integral, okay, then you find the wave function here.

So the full knowledge of your quantum mechanical system essentially is encoded in this propagator, okay, it's encoded in this propagator.

So, yeah, so now we want to talk about what is the analog of this object, say, you write we see quantum field theory, okay, in quantum field theory. How do we define a propagator?

And what is the, what is the analog of the position eigen states say in quantum field theory, okay? And we will discuss next time.