 It's actually yeah, it's not going to be the whole lecture on this is the algebra, but this is this is This is where we got to last time Okay, so last time what we'd done is we had worked out supersymmetry transformations in free field theory and then worked out the neuter current the neuter the neuter charges and we had got to this point here where we had Where we would we had found the Famous n equals one supersymmetry Algebra, okay, and as I commented this is this shows again really shows that this is a spacetime transformation because the essentially the square of a Susie generator is a spacetime translation Okay So I'm going to stop writing. We've been using hats for operators. I'm going to stop doing that now So now let's look at some general Consequences of this supersymmetry algebra, so not assuming anything about the theory other than the fact that it has this Symmetry, okay, so now you can sort of invert the supersymmetry algebra by projecting out The momentum generator guess I didn't drop the hats yet. Anyway, I'll drop them soon enough And so basically the zero p zero which is the Hamiltonian can actually be written Essentially as a sum of squares and so we can see that the energy of any state in a supersymmetric theory must be greater than or equal to zero and in fact the If the vacuum is supersymmetry invariant, right We expect in general that the vacuum state should be invariant under all spacetime symmetries Then what we see is that the actually for momentum of the whole state vanishes in particular the energy vanishes Okay, so unlike non-supersymmetric theories because the Hamiltonian is the square of some symmetry generator, right then the invariance actually Makes it natural for the vacuum energy to vanish and this seems like this could be a very important clue for the cosmological constant problem Although to make that precise we would have to discuss supergravity Okay So the other consequence general consequence I want to talk about before we talk about real theories is Just what kinds of particle representations are allowed by supersymmetry when we have We know that when we have Lorentz invariance or Poincare invariance the one particle states are described by momentum in a spin Right, so we have a three momentum in a spin and that describes all single particle states now for massless States the spin is just a helicity the projection of the spin along the momentum and the way We analyze these in supersymmetry is the same way that we do in field theory We chew for each particle massless particle We can choose a frame where the momentum takes this form and then in that frame the Susie algebra looks like this And so what we see is that q1 is Acts trivially and q2 act like raising and lowering operators like fermionic raising and lowering operators okay, and So what we can do is we can see that actually by looking at the commutation relations with spin We can see that q2 and q2 dagger raise and lower the spin Okay, but because they are fermionic operators You can only act once with the raising operator before you get zero and so what you find is that the massless the massless single particle representations There's one where the ground state of the lowest spin state has has spin zero and then there's Backed in with q2 we can create a spin one half state or we can start with a spin one half state as the lowest Spin state and get a spin one state Okay, of course we can write down fermion multiples with any spin But these are the only ones that we can write we normalizable theories for Okay, so in a very simple and direct way the supersymmetry algebra tells us what kinds of particle multiplets we should expect Okay Questions on this Okay, so now we're going to go into superspace right this is a part in the science fiction film We're like this okay, and I Actually debated with myself some time about whether to Introduce this sub topic because it is a little bit technical, but I think it's worth it after all as We saw from severe's talk really We human beings and we physicists in particular are explorers of space and time and the subject is just Too beautiful and gives too much insight. I think into the whole subject of supersymmetry To omit it so we're going to go into it so things are going to get a bit more technical now but That's it. This is a technical subject So I'm going to be going over things Well a good reference for what I'm going to be talking about here is is given here It uses almost the same conventions as me some very simple changes. So that's that's good Okay, and I think mine are more standard by the way, so this doesn't violate my my rule Okay, so let's let's remember what You know what let's remind ourselves. What's the role of space time in quantum field theory? Well Obviously Lorentz invariance acts on space time right it acts linearly on space time and so that you know, that's That's that's that's important, right? Okay, now the claim is that this there's a space that supersymmetry acts on naturally Which is a generalization of space and time which is called super space and this is in addition of those space time Coordinates we have one vile fermion and it's complex conjugate. Okay, which are additional fermionic coordinates and The fact that they're anti commuting coordinates It just means they all anti commute with each other and they commute with x mu like this Okay, and just as for a quantum field theory the natural variables are functions of space time Those are the dynamical variables the dynamical variables in supersymmetry the natural ones are going to be functions of this super space Now these are called super fields and what do they mean? What does it mean to be a function of an anti commuting? Quantity like these theta coordinates Well one way you can define a function is just by Taylor expanding it so we can define functions of theta and theta bar by Taylor expansion and Taylor expansion is very simple because these coordinates anti commute so their square is zero so the Expansion terminates after a finite number of terms so if we had a single theta There would just be two terms in the Taylor expansion of a function in the case of the super fields that we have We have two Thetas and two theta bars So the highest component we can have has four Thetas. Okay, so we have only up to fourth order an expansion and because of the Symmetries here terms with two Thetas can always be written in terms of contractions of the Thetas right with our with our trusty vile spinner indices That we learned about last time and so if you look at a general super field Okay, it It has a lowest component an order theta term It has theta squared terms which we can write like this it has theta cubed terms and theta to the fourth terms That's what it is Okay So one way we can think about theta and theta bar is that they're just some sort of place holders We can think of this super field s as a kind of vector that contains all these different components But notice that very non-trivially these different components have different spins, right? We have let's see Oops, I forgot a very important term Who wrote these slides? There's a term that I forgot which should be up there in the middle which looks like this Okay, that's very important. Okay, and this a mu should also be in that list of components Okay, so obviously we're going to want okay, so that should be it's missing up up in here Okay, so this thing had this thing contains spinner fields scalar fields and vector fields all mixed together Okay, so there you can okay. All right, so now You might not like anti-commuting coordinates, right just like Mathematicians of the 19th century a lot of them did not like the square root of minus one I what is that? what kind of number is that and If you want to be if you don't if you're a conservative and you don't like I You can always say that a complex number is just a pair of real numbers And you can define addition and and and multiplication of these things in this way But nobody in their right mind would actually do complex Analysis in this way. It's very useful to have I as an organizing principle in the same way Theta it's a fictitious mathematical object if you like but the more you get used to it the more real it becomes the more You know actual it becomes Okay, and so the same way Just like I is a great organizing principle to understand for example, how where is this weird multiplication rule come from? You can just multiply these you can add and just multiply these superfields together And just expand the result and you get a new superfield So there's a way of combining superfields and analogy the way there is a combining complex numbers Okay, so that's the basic idea Now I haven't the whole point of this is that super symmetry is going to act on these super field operators So let's remember house Translations, which is part of the super symmetry algebra acts on ordinary functions of spacetime Oh, that's slide guy screwed up again There's supposed to be a phi of x here this e to the ip a mu p mu is supposed to act on a phi of x right here Okay, so spacetime translations act by guess what translations and those are generated by a Derivative operator p mu which is i d mu right okay, and in the same way What we're going to show is that super symmetry transformations are generated by derivative operators acting on super fields Okay, now the transformation parameter as we've actually already seen the transformation parameter of Super symmetry has got to be an anti commuting number as well Okay, stop me if there are questions Otherwise Keep going Okay, so what are these derivative operators these derivative operators now act on super space So in addition to the ordinary spacetime derivative I can have derivatives with respect to the anti commuting coordinates, okay, and you can you can understand the rules for For for taking derivatives using anti commuting quantities using these basic relations that just say ah This slide guy So there should be some the appropriate bar thingy here, okay, so I don't know what this is supposed to be But this one's right These are the only two that are right, okay, I'll fix these so let's look at the ones that are right So they are they just tell me that if I take the derivative of the coordinate I just get one if I use the same coordinate or zero if I use a different coordinates That's what I'm trying to write here. Okay. This was made while I was still jet lag I'm no longer jet lag so no more excuses. I'll actually try to answer your questions Okay, all right, so and then you have to be careful about signs You have to be careful about the fact that any two fermions that you move past each other There's always a minus sign and so that gives you some funny minus signs like if you're taking this ordinary derivative This theta hits this theta But because you have to move it to the left first you actually get a minus sign like this with those two rules You can work out any any derivative Any theta derivative, okay, and because the metric this will quote unquote metric epsilon alpha beta's anti symmetric There's some additional signs coming from that and you can work out for example this thing right here Which is which is a little bit. There's an extra minus sign compared to the way we usually raise and lower spinners Okay, now I can just with that technology Okay, obviously by the way if you want to learn any of this you have to practice it yourself I'm not going to do calculations in front of you. There's no time for that, but I am hopefully going to give you enough Information that you can do them start doing this yourself and really make the subject your own so Now I'll tell you what the Susie generators are. They are a linear combination of theta derivatives and Ordinary derivatives, okay, and now that you with this definition You can just work out what the algebra is and you find that it obeys this algebra right here Which is the Susie algebra if I remember that id mu is p mu acting on Fields or super fields Okay, okay, so super fields do give you a simple representation of super symmetry Okay now Just like whenever you have a symmetry Right and in field theory you Often find that you need to redefine your derivative operators to be co-variant under that symmetry for example If you have a gauge symmetry you have to define co-variant derivatives to get so that the derivatives of things transform Simply under gauge transformations the same thing happens in super symmetry if if I have a super field I can ask do derivatives of that super field also transform in the same way Do they transform according to q and q bar under super symmetry and for the ordinary derivative that is the case Okay, d mu of a super field is a super field that is it transforms like a super field And to see that we just check that if I take the transformation I have to take d mu of the transformed thing then the q is inside the derivative But I can just this derivative doesn't act on q or xi because they're independent of x And so I can just pull the derivative through right and so d mu of a super field is a super field and Formally what this is is this is saying that the derivative the ordinary derivative which is a derivative operator commutes with q Okay, and that's actually obvious That's because d mu is basically just the momentum and the momentum is supposed to compute with commuting with you So yeah, it's it's all all consistent, right on the other hand if I look like if I look at d by d theta of Acting on a super field that thing does not Anticommute with q okay, and that's simply because the definition of q or rather q bar Involves theta right so the definition of q involves theta and theta bar, so they don't anticommute Okay, but we can define covariant derivatives Okay, and here they are here the Susie covariant derivatives notice that the formulas look very similar to the queues But there's an important change of sign here So they're not the same as the queues right but they're they're a little bit different And then these things are constructed so that they anticommute with all the queues and q bars Among themselves they obey actually the Susie algebra again, okay sort of conservation of ideas The only the algebra only algebra in town is the is the Susie algebra Okay questions, okay So this means that if I can take covariant derivatives of a super field and get a super field Okay, so a covariant because again I have the same sort of computation if I want to find the very the how the d alpha of s transforms It's this but now d alpha Anticommutes with q so I get a plus sign here That's not a typo the the the slide guy was paying attention here because D alpha actually also anticommutes with xi right. It's also an anticommuting thing all right So it actually just like this but anyway And I just find that Oops, okay here. I tried to explain it. So the point is is that this thing transforms like a super field, okay? We're good All right any questions Okay, now an unusual feature of Superfields is that the the superfield that I wrote down for you the most general Superfield that I wrote down for you is not actually an irreducible Representation they're smaller representations just like when we started with the direct representation. We could reduce it to the vile representation This general superfield Contains within its simpler superfields that transform into themselves under supersymmetry and The simplest one is called the chiral superfield and it is defined by imposing the condition that the anti Susie covariant derivative d bar acting on the field vanishes Okay, now if this covariant derivative were an ordinary derivative this would say that this Superfield phi is independent of something right if I say the d mu of phi is zero it means it's independent of x So does this mean something like that? In fact, it does okay, and you but you have to because this is a covariant derivative and not an ordinary derivative You can you have to change variables to find what variable it's independent of so we can change variables in superspace by just shifting the x Coordinated by this theta dependent thing okay, and Something I didn't I should have I should have mentioned before this theta these Thetas have units of the square root of a length Okay, and I could actually see that if I go back to the Susie algebra here If I go back to the Susie algebra I have to assign the units here consistently right d by this is d by dx that has units minus one so the units that I have to assign to To to theta are the square root of a length okay, so that's not that's not something I get to make up Right, so if I come back to here, where are we? Okay, come back to here this shift has units of length so everything's good Okay, I'm not introducing any new dimensionful parameters Okay, and now with this shift I can work out what d and d bar are in terms of the new variables And I did it here now that you've absorbed that I will tell you the answer as I said I'm not going to spend time on Doing calculations, but they're provided so that you can look on the notes will be posted And you can use these to sort of get some practice doing these calculations, okay, but the point is is that? Sorry, I'll reveal it just for a second the basic idea is just the chain rule, okay There's nothing fancy going on here. I'm just for example for d bar I have to use the chain rule and I find that things cancel for d bar, but they add for d Okay, that's that's that's the basic idea Okay, and so what you find is that d bar in these coordinates is just d by d theta bar Okay, and d is got this extra term in it so this condition that d bar of phi is zero tells me that it is it is Independent of theta bar in these shift it in these new coordinates Okay, and that means that I have a lot when I x Taylor do my Taylor expansion in theta in those coordinates I only have three terms which I write like this Okay, so this chiral super field has a much smaller set of fields Okay, and if you don't know that do we know what this thing looks like these are called this this y by the way It's called the chiral representation these coordinates y theta theta bar are the chiral representation of super space If I don't like that I can always expand these functions of y in terms of x and theta bar Right, I can just expand this out to figure out with a relation between this and what I wrote down before questions, I'm either Pounding you into submission or I'm much clearer than I think I am Okay, so And now some very basic things about chiral super fields if I any power of a chiral super field It's it's d bar on that vanishes. So that's a chiral super field Right in fact any function of phi any function of phi at all if I take d bar of it It's a chiral super field But notice that it's very important that this function what I'm calling f of phi is a function of phi and not a phi dagger Right, that is f has to be a holomorphic function, right? So for a physicist holomorphic is the same as analytic I always forget what the distinction is for mathematicians that has to do with singularities, but basically It's a function of a phi and not phi dagger. Okay. Now. This is actually extremely important We'll see that this has a lot of far-reaching implications the fact that chiral Super multiplets have a holomorphic structure is at the root of many of the beautiful and important properties of supersymmetry So phi dagger It does not is not a chiral multiplet as you can easily see from their components But it is an anti chiral multiplet just by complex conjugation the alpha of phi dagger is zero And that's called an anti chiral multiplet So obviously we can learn all about anti chiral multiplets by just complex conjugating the results for chiral Multiplets chiral super fields Now Here comes another somewhat technical point, but it's it's it's it's useful. Okay And that is that we now would like to define the components, okay of A super field now one way to do that is just the way that we've done it so far just do the Taylor expansion In terms of theta and theta bar, right the Taylor expansion in terms of theta and theta bar is Is is is how it is the equivalent to doing a Taylor expansion using the derivative operator d by d theta, right? But we've seen the d by d theta is not a covariant object The d alphas and d alpha bars are covariant objects So it turns out to be useful to do the Taylor expansion in terms of the capital d's and capital d bars, okay? so basically we define the lowest component theta to be the Just the setting theta and theta the lowest component phi of the supermultiply of the chiral supermultiply big phi By just setting theta and theta bar equal to zero We denote that by just putting a slash at the end of it a vertical slash Okay, and now the the the fermion component is Is equal to to d by d theta of this which is just Actually in this case just equal to the covariant derivative and the same thing for the f component So we can we can write so this actually agrees with the previous definition for chiral super fields It's actually Generalized for other kinds of super fields We'll see that it actually is a little different than the d by d theta expansion here The two things coincide because the extra terms you can easily see vanish when theta equals theta bar equals zero And now the point is but the point about this formulation is it makes it very easy to work out the Susie transformations of these guys Because we can just use the fact that the d's anti commute with the cues So for example, we want to know what is q acting on phi what we can do is we can use the fact that The the cues are equal to the d's up to the additional theta terms and then we just get this for the lowest component Okay, that's that's pretty easy. So the lowest component phi transforms into zeta times psi Notice that differs from what we had before by but for when we did it before by a factor of 1 over root 2 And that's because there's this factor of square root 2 that we we wrote down in the definition of the the super super Q okay, so it's all consistent So you can get a better feeling for for for what happens if we look at the fermion case We write q of d alpha phi evaluated at theta equals zero now We can anti commute without any cost the d through the cues, right? Which gives me a plus sign because of the size now I can replace the d's by the cues by the d's because the extra term vanished when theta equals zero and then I can Just do some a little bit of algebra and I can get this Okay, now if you go back to what we had before there's again these silly these factors of root 2 which are because of Standard conventions. There's this additional term involving f f wasn't there before we'll understand the role of f in a bit Right now. It's just there and we'll understand what it's doing there in a little bit and in a similar way You can work out the transformation rule for f so you can use these super fields to find out how the component fields transform okay Questions Yeah, please Yeah, the question is please remind. What are these components? What am I talking about? Okay, so For a general super field Here it is a general super field is a function of theta and theta bar Okay, and if I expand it out I get all these different components including the one that I forgot Okay, and these fields a of x Psi of x and so on these are called the component fields So the analogy is that you can think of this the super field has as its component fields all these fields here Does that answer the question? Well, we're working that out. We're in the process of working that out. We're like this is like when you first learned about Complex analysis. I'm just teaching you what the complex numbers are. We're going to be working out the properties of these things We're going to be writing quantum field theories in terms of them. That's what we're going to be doing Okay All right, so I mean in my opinion you can do you can do Super symmetry just by writing out lists of these fields and writing down the Susie transformation properties like the ones that we've just derived By the way, you'll notice that I now have I have numbered slides now So I'm trying to make it easier for you guys to ask me questions So here are the definitions of the component fields in terms of the super field Okay, that I'm using and here are their super symmetry transformation properties now We could just work directly with these and check in variance under supersymmetry But that would be sort of like the way Maxwell did electro dynamics He didn't use vector notation and he had you know, he had e1 e2 and e3 B1 b2 and b3, you know, that's a horrible mess, right? It's much better to use vector notation in the same way. It's much better to organize these super these fields Phi psi and f into a super field. That's the idea Are there questions? Yeah, please That's correct. So I the the question is do I do not assign? reality conditions to the anti-commuting coordinates X of course is real x mu is the same x mu that you grew up with in kindergarten Thetas are vile fermions. They are complex objects Okay, the super field when I wrote down the general super field I did not impose any reality condition on it either So the component fields are complex in general for that as well Okay, now this phi here. I also did not my capital phi. I did not impose any reality conditions So this phi is a complex scalar This psi is a complex vile fermion and this f is a complex scalar Other questions now I have a theory with two scalars and a fermion Okay, and I now have the technology to write down interacting in Suzy invariant quantum field theories This is a very very big deal, right? Okay, so I want to understand if I have a super field like phi. How do I write down? Invariance for that right just like with ordinary field theory. I know how to write down in variance I want to write down an invariant action. I integrate overall space time and I contract all the indices, right? That's how you do it. I want to know what the procedure is for Super fields and it's actually not so different. It turns out, okay So the claim is there are two kinds of supersymmetry invariance that I can write down. The first kind is like this I can use I'm again using these covariant derivatives. I'm going to be using them a lot, okay? I can take I can take as many covariant derivatives as possible of a super field K And I claim when I do that I get something that is invariant under supersymmetry or more precisely Transforms under supersymmetry to a total derivative. So it's a good enough to be a and it gives me an invariant action, right? Okay, so here we're using the shorthand when I square something like D I mean contracting two D's together, okay, and Since I want this Lagrangian right here to be her mission again up to total derivatives I need this K the super field to be real So here I am imposing a reality condition on this super field So K is a general super field except for it needs to satisfy this reality condition, okay? Now in in in supersymmetry as you will see about half of the symbols in Super symmetry are some form of the letter D, right? There's D mu there's capital D And there's even more to come. Okay, and here's an example Which is that this thing right here is called a D term not as you might expect because it has D derivatives but because this Thing is equal to the highest component of the super field K, which is also called D Okay, and I'm just not going to apologize for this because you got to learn this everybody in supersymmetry uses this So you have to learn this horrible notation if you're going to look at anything in the subject Look at any papers in the subject. Okay. All right, but anyway, this is is it clear what this thing is That's the most important thing Yes, okay, so now let's understand. Why is this thing? How why does this thing transform into a total derivative? Well, I just do what I said before for the components I just were now that I've this technology. I'm just using the same trick over and over again I I write the transformation I can pull the cues out from the D's because of the anti commutator I can replace the cues by D's and then I can use the D algebra to show that this thing right here is a Total derivative, okay, and the basic reason is that if I look at if I look at D Acting on this thing here There are three D's next to each other and three D's next to each other have to vanish because D anti commutes with itself a D bar next is not right next to the two D bars But I can put it next to each other with a commutator and that commutator is a derivative So consequences of the algebra for D's are relations like this that give me a commutation relations for the D's and the D bars Okay, any questions on this Okay, so that's one kind of term we can write Okay, if I have any super field K as long as it is real I can write an invariant There's another kind of invariant I can write which is a similar idea But if I have a super field which is called W, okay, which I call W That's the conventional name Then I can just take two D derivatives of it and get a supersymmetry invariant Okay, and since W is chiral, it's not real So I have to add the Hermitian conjugate if I want to get a real Lagrangian and the The proof is the same kind of thing that I talked about before Okay All right, and I guess something I should have should mention it before too If I want this Lagrangian to have dimension for this W This super field has to have dimension 3 if I go back here that I mentioned it here Yeah, I didn't say it but this K right here has to have dimension 2 if I want the Lagrangian to have dimension 4 Okay, is there another typo? Oh, oh, you're right. I'm sorry. Ah, thank you. Thank you Yeah, there shouldn't be a theta bar squared here I'm sorry because the highest component of the chiral super field is just the theta squared term in the chiral representation Okay, so sorry. Yeah That's slide guy again. Okay But but I really appreciate you pointing out these typos I will fix them all in the posted version the post the posted version for the first part of the notes I tried to fix all the typos. So if you find some more, please send me an email. Okay, or just talk to me Okay Now in the literature we don't you don't Usually find the notation that I've been using You don't find that you write things in terms of d's and d bar squareds instead The d term is written as an integral over all of super space and the f terms are written in terms of integral over half of super space Now you might think what could be more different than integration and differentiation They're usually inverses of each other, but actually for anti commuting coordinates. They're actually the same thing Okay, if I have a single anti commuting coordinate the integral is is the function just has a simple Taylor expansion here An integral is be and that's the same thing as the derivative okay, and So this is why that this notation is not crazy Okay, so but rather than developing the theory of integration over super fields I'm just going to define d4 theta to be this and d2 theta to be that there there They are the same thing after you do the whole theory. So let's just Okay, let's just use that but I will use this notation because I like it and it's what people normally use in this subject I'll use the the integration notation Okay, questions. Okay Okay Now we're ready. Okay, we're ready to start being field theorists instead of mathematicians So we have let's say we have one of these chiral super fields Okay, it has these these component fields. Okay, and let's say that remember before we Wrote a free theory involving a complex scalar and a vile fermion Let's suppose we want to do something an interacting theory like that try to make an interacting theory like that out of this chiral Super field then we want the scalar field to have dimension one, right? Well that works because now the fermion field has dimension three halves and F has dimension two We still don't know what F is but we'll get there Okay, but at least the scalar field in the fermion field have the right dimensions And so that means that the whole super field has dimension one right as the same dimension as its lowest component okay, and Now we can just use these tools that we have to write down the most general Susie invariant Lagrangian and for simplicity, I'm going to take the case where we have only dimensionless couplings Let's suppose we're interested in that case Then we can write d4 theta which was really d squared d bar squared remember And we need to make something that is a real super field and has dimension two Well the unique thing we can write down is phi dagger phi right okay, and For the the super field we need something that has dimension three So if we don't want to introduce any dimensionful couplings, we can just write phi cubed and I've chosen the coefficient of phi dagger phi to be one just by rescaling the fields So there's only one parameter in this theory lambda so you can remember these so anyway The dimensions of these integrals are two and one which looks wrong because theta has dimension minus one But remember integration is the same as differentiation Okay all right Now okay, so here's the Lagrangian great okay, but To do physics with this what we need is we need to know what this looks like in terms of the component fields phi psi and f Right, that's what we need to know so but we have the technology to do that now We just Expose this integral for what it is. It's really differentiation and then we use guess what the chain rule right? We just have a whole bunch of derivatives acting on this thing. We just have to use the chain rule Okay, we have to be careful to use the right Susie algebra Okay, and let's see. Do I want to say anything about this well? Yeah, you know you get what do you get you get terms with four derivatives acting on this with Three and one and this and you use some algebra and you get there's what you get Now this is pretty cool. This is in fact very cool because what I see is I have an ordinary kinetic term for phi Okay, there should be a dagger here Okay There's an ordinary kinetic term for psi the fermion field and there's a mass term for f Which we still don't understand but we'll get there Okay Questions all right Now we can do the same thing for the The the this w I don't think I said the words I think it was on a slide somewhere this w is called the super potential Okay, so this w for the super potential which is just five cubed in our case for the super potential What do I have to do? I again have to do two derivatives. It's again the chain rule So it looks like this. I just worked it out for a general super potential not for the phi cubed case All right, but it's just the first and derivative first derivative of w times f second derivative times psi bar psi So the second derivative, okay? Oh So For our theory I have the sum of these terms. I have kinetic terms for this guy I put the dagger there. Thank you and then I have this super potential term here Okay So the thing that you recognize is you recognize these kinetic terms those are good And then you recognize this this is a yukawa coupling, right? It's a coupling of phi to two vile fermion fields So we're starting to see so we see some interactions. This is really great In fact, we see ingredients that we need in the standard model, which is really great, right? What about this f? Okay, we could finally have to face the f well the point is is that this is our Lagrangian We didn't have any choice, right? This is what supersymmetry gave us. We have to deal with what we have What we have is a field that has a linear term and what we would normally call a mass term, right? but f has dimension 2 and so this mass term is dimensionless, okay and f is a field like every other field we're just supposed to for using a pathological quantization For example, we're just supposed to integrate over it. Okay. There's no special rule for f f doesn't get to be anything special But the point is is that the Lagrangian is quadratic in f and there are no derivatives acting on f So f does not have any dynamics Right in any normal sense if I looked at the equations of motion for f, they would just be algebraic equations Okay, the fact that f is quadratic in this way means that we can integrate it out exactly in the path integral And the point is that we can write the because f is quadratic plus linear We can complete the square so we can write this as The square of something Minus the extra term. This is completing the square, right? And now I want to claim that this term here does exactly nothing. You can just erase it Okay, and the argument is a simple path integral argument The point is that if I look at the the the full path integral is integral over all of the fields Including f if I look at the path integral, right once I've completed the square The only dependence on f is in this perfect square But I can just shift f by this amount and then I have an integral here, which doesn't depend on any of the fields Okay, and so it just goes away doesn't it just add it's just a constant in the path integral It's like rescaling your path integral measure, which does nothing Okay, and so integrating out f just means we throw away the term from completing the square But we have to remember that there was this term left over. This is the scalar potential So this Lagrangian is a Lagrangian that you recognize, right? This is a Lagrangian of a complex scalar a vile fermion with a yukawa coupling and a quartic self interaction Okay, notice that there's only one there's two couplings here We would normally say but they this one is lambda this one is lambda squared the couplings are related by supersymmetry because we constructed this Lagrangian here to be invariant under supersymmetry So this is just what I said supersymmetry relates the yukawa coupling and the quartic scalar coupling So if you remember back to the introduction This is exactly what we claimed was the structure for the stop the top and the scalar partner of the top in The in a supersymmetric version of the standard model, which we need to have naturalness Okay, so we'll of course return to this subject But not we won't get there in this lecture Question. Yeah Yeah Yes, okay good the question is and The previous slide when I was integrating showing how you integrate out f, okay Is this equivalent to imposing the equations of motion and the answer is yes? Okay, this is precisely equivalent to using the equations of motion to eliminate f Okay, however, you may wonder that is a sounds like a classical procedure Right, how is that justified in an interacting theory say to all orders in perturbation theory or perhaps even beyond perturbation theory? And that's the pathological argument that I gave Okay, but I'll come back and comment on that and Additionally in just a little bit. Okay Another question. Can the super charge be measured? That's the question Not the answer is no. It's a fermion. It's a fermion field So you can just as well ask can the Dirac? Kind of fermion feel like a electron field to be measured and the answer is no More more more more, you know to say a little more about that You know in in quantum mechanics and in quantum field theory a very important concept is the concept of an observable Right an observable is supposed to be something that an experimentalist at least in an ideal setting can measure the eigenvalues of Okay, so momentum operator in quantum mechanics the position of a particle and so on these are the operators that are accessible to an experiment Now you can just show that fermionic operators like field operators. They actually do not have eigenvalues Okay, they just do not have eigenvalues But however squares of these things, you know are operators that have eigenvalues So those are well-defined operators. You can an experimentalist could assign values to them. Okay, so in the same way You know Q is a fermionic thing. So no Okay, so as I said in the last lecture when we always of course We're using the language of invariance when we talk about supersymmetry But you can't picture supersymmetry as some sort of a transformation You know you can't measure the charge But the mathematics of the invariance is very parallel is exactly parallel to the invariance for things like rotations that we can picture Okay Other questions Okay, so let's just generalize this to a general theory with n chiral super fields, okay, and So we assume that the the d4 theta term is just the sum of the phi dagger phi's that gives us canonical kinetic terms for all of the component fields and then we have a general super potential Okay, and you can work out again. This the generalization the second derivative of the super potential multiplies the fermion so this can generalize the yukawa couplings and then the square of this the first Derivative of the super potential gives us the potential that we get by integrating out f Okay, the most general renormalizable super potential is cubic the couplings Now this is the question that was asked earlier Integrating out f is precisely equivalent to imposing its equations of motion So its operator equations of motion are simply that f is the first derivative of w. That's what the equations of motion are So that means that the potential that I get which is just the square of this thing. I can think of it as the square of f Okay, all right, and that is a useful way of thinking about the potential as we'll see and Notice that consistent with the general properties of the Susie algebra that I talked about this potential that we found is always Positive, it's the square of something right it's f dagger f. So it's always positive Okay Also remembering supersymmetry is broken if and only if the vacuum energy is positive Right, it's unbroken if the vacuum energy is zero So from this we see that a good order parameter for Susie breaking because we will eventually want to break supersymmetry is f F itself is a great order parameter for supersymmetry. That's one of the main reasons. It's so incredibly useful Okay, so you might not like this f because we have it and it went away It's there to first of all to make supersymmetry manifest It also gives us a very good handle on understanding how super what how supersymmetry is broken What breaks supersymmetry? What vet? What's the Higgs? What what what gets an expectation value when Susie's broke? Okay, so now I'm going to explain something that's a bit technical, but I think it's very beautiful It's also very important and it also illustrates the power of these super field techniques This is the famous non renormalization theorem for n equals 1 supersymmetric theories The point is this is this is really the tip of an iceberg of a vast subject The UV divergences of a supersymmetric theory are highly constrained Okay, usually what we're used to and renormalizable in In quantum field theory is that every term that we can write down in the Lagrangian in general has UV divergences It needs to be renormalized in supersymmetry. We'll find that that's not the case. I'm going to give you a complete Modern discussion of this which is based entirely on symmetries You're never going to see an actual Feynman diagram, so my view is a good thing, but you may not agree Okay, so let's remember what the right language is the language the right language is that we look at the one PI effective action Okay, remember the one PI effective action just summarizes the correlation functions of the theory So it's a highly non-local object But the UV divergences the UV divergent terms in the one PI effective action are always local That's because they come from short distances, right? If they come from short distances, they are going to be local things. That's the basic idea Okay, so that means that that already implies this is That already implies that the the most general UV divergent terms that we have in the one PI effective action have the same form as the Lagrangian that we started with Okay, that's the basic Fundamental theorem of renormalization theory in general non relativist general relativistic quantum field theory Okay, now let's suppose we have some UV cut off lambda then by power counting z is dimensionless by the way Sorry, Delta Z just or just or just a lot of here Delta Z is the coefficient of Phi dagger Phi right these other things these deltas are the Things that multiply the terms in the super potential so Delta Z is dimensionless so it can depend logarithmically on the cutoff Kappa has dimension to it can be quadratic M can be linear and little lambda can be log lambda just by power count by Dimensional analysis these are the most general divergences I can have okay now these UV divergent terms must respect the symmetries of the original theory right you're used to this for example in Q Ed you have the word identities that guarantee that the UV divergent structure as the same structure as the terms in the Avigil Lagrangian It's they are gauge invariant in the same way these counter terms have to respect all of the symmetries of the theory Okay, now this is true as long as we believe that this regulator doesn't break the symmetries Okay, and in supersymmetry we believe that The nature for various reasons for example from string theory We actually believe nature is is supersymmetric at arbitrary short distances So we we better hope that we can preserve supersymmetry with some sort of regulator and in this theory We certainly can we can use very specific regulators like poly VR's or higher derivatives So this is a bit technical But we assume that we can keep all the symmetries of the problem and I claim that in this case you really can Okay, but then You might say well, we've already done that we've already kept track of all the symmetries because we've Required that these divergent terms are supersymmetric Okay, but it turns out there are more symmetries to be had here and to see what these symmetries are an Extremely powerful and general tool that we use very often in particle physics And you should learn even if you don't care about supersymmetric theories is the idea of Spurrions and the idea of spurrions is that you take a coupling constant, which is a constant Which is a number you can experimentalist can measure it But as a theorist you have the freedom to promote it to a field Because it appears as a symbol in your action, right? And if you put it inside the d4x integral you can just pretend it's a field while you're doing your calculation Okay, you can always set it to the experimental value at the end when you want to tell the experimentalist What the answer is but you keep it as a field now in the case of supersymmetry we can take our couplings kappa m and lambda and Now we can promote them not just to be fields, but to be super fields So this is super spurrion analysis, right? And if we want this to be but if we want this to be super symmetric We have to promote them to chiral super fields not just any old super field, right? Because we want to keep them inside the d2 theta integral and we want to make them super symmetric, okay? So let's do that. Okay. It's not obvious right now why that's a good idea, but we'll see why it's a good idea Furthermore once I do that I Notice that these kappa zems and lambdas have all these indices, right? These indices are just because I have n chiral super fields, right? But now I can imagine that there's a UN symmetry that rotates these things into each other Okay, and normally you would say you can't have a UN symmetry Why because these kappas m's and lambdas break all the UN symmetries in general, right? However, now that these kappas m's and lambdas are spurrion fields ta-da by magic I just say no it's perfectly invariant. I just have to transform those guys according to their indices as well And now this Lagrangian is perfectly UN symmetric Okay All right, if it's perfectly UN symmetric then all the UV divergences have to be UN symmetric Right and UN symmetry just means that I need to contract all these indices in a way to make an UN invariant Okay questions on this So this is what's used for example in in chiral perturbation theory and in everything Okay, so now what we see is these these these things delta z kappa delta kappa delta m delta lambda These have to be functions of my original Couplings and they have to respect this UN symmetry now, right? So what can they be? Okay, well, here's the thing and this is now the crux of the matter this delta lambda delta m and delta kappa have to be chiral Superfields that means they have to be Holographic functions of these kappa m's and lambdas they cannot depend on the dagger guys Okay But now there is no way to contract the indices basically because look right these guys all have lower indices Right these fies all have upper indices Okay, so if I want to make a kappa Out of well, let's look at I don't know that's let's look at this let's look at this guy Okay, so if I if I want to make a lambda of course delta lambda of course it can be proportional to lambda duh, right? But I claim there's nothing else I can write down You might try to write down kappa times m, but that doesn't work by dimensional analysis because kappa and m have positive mass dimension Okay Now you might be a little bit suspicious because you might say well wait a minute What if I only have one field that I have no indices, but then I have charges There's a you one symmetry and you can check that the charges also don't allow you to write anything else Okay, so this is true even for n equals one Okay all right, but now All I have to do is check what what's going on with these with these C's okay Well, it's it's it's clear that Okay, so this is what I just said. Okay, so because these things delta kappa delta Delta kappa delta m and delta lambda. Okay the slide guy screwed up again Are linear in the couplings I can now just compute these things in perturbation theory I don't write and and I can just easily see that there are no corrections at loop There are no loop diagrams that are that are That are That are that all the loop diagrams have at least two couplings in them Okay, so these C's have to be zero and so all of these super potential couplings are not renormalized at all There are no UV divergences in these guys at all Okay, that's the argument questions This was originally proved this this non renormalization theorem is a classic result It was originally proved using the full machinery of super field perturbation theory super Diagrams and all of that this proof was understood by Natty Cyberg in the 90s Okay, it's fairly recent Now on the other hand the coefficient of Z Z was the one guy didn't talk about Z is a totally different beast because delta Z is a real super field remember it multiple it's under a d4 theta integral it multiplies Phi dagger Phi it is a real super field and There are definitely Things that I can write down there because there are just no problem in using the Phi daggers In fact, the leading one root loop result is proportional to Phi dagger Phi I see that I can very well contract the indices. Thank you and I can have a log divergence Okay, and in fact I do has this coefficient Okay Now this of course It can't depend on kappa or m by dimensional analysis Okay, and now we can treat this UV divergence in the standard way. Okay. What is the standard way the standard way? Here I'm going to explain it in terms of a single chiral super field to make so the notation doesn't get to to bog down Okay, so I have a single chiral super field now what I do is I add a counter to I have I Add a counter term to the Lagrangian that cancels the dependence on lambda the UV cutoff But it introduces Dependence on a renormalization scale mu Okay, the their renormalized Lagrangian is defined by sort of taking out the counter term That has now a wave function coefficient Z that depends on mu in the way. I've done things Okay Now usually what you do in and and this this z of mu satisfies this Renormalization group equation where the z here is the Cz is just that constant that came from the one loop calculation All right, it's minus 1 over 4 pi squared Okay Now this is not normally the way that you do perturbation theory normally you set the coefficient of the kinetic term to 1 at each scale Okay, so we would like to define a new field phi hat by just scaling out this wave function Renormalization now one thing to note here is that this step we can no longer treat z as a super field anymore, right? But that's okay. We know we have to set it to be a number eventually So now is the time to do that, okay? And so now we have to z is back to being an ordinary number and it's just equal to this and now we have a We have a canonical kinetic term, but now we have to redefine the super field couplings Okay, they've been rescaled by these z's and so what we find is that they in fact the the If we canonically normalize the kinetic term the physically normalized Couplings do have non-trivial logarithmic renormalization, but only logarithmic, right and They're in fact given by the same function Cz, right? They're all determined by wave function renormalization, okay? So this is this is the general structure of renormalization in these in these theories, right and theories of chiral supermultiplits So one implication of this it's often used in model building is that if we no matter if there's a symmetry or not if We set some super field coupling to zero it stays zero to all orders in perturbation theory because it's just purely multiplicatively renormalized questions Yes The question is what if the background symmetries are anomalous? That's a beautiful question Okay in the West Seminole a model that I'm looking at there are no gauge fields And so there are no anomalies in that case. Okay, so in this case, it's totally legit when there are gauge theories There are interesting anomalies. They do have very interesting effects. I don't have time to talk about them I Can give you references for that. Okay, but yes, there there are very interesting effects that precisely arise from the anomalies Symmetries like this UN when I have gauge fields present, which I haven't talked about yet Other questions So gauge theories you asked for gauge theories. We want gauge theories, right? This is all great we certainly like like scalars and Fermions, but we also need gauge fields So we need to understand how to formulate gauge fields in super space And it should not come as a shock that the guiding principle is going to be gauge invariance, right? That's how we construct gauge theories as field theories So now we have to have gauge invariance in super space. So we'll consider the case of a U1 gauge group so if I have a a Complex scalar and a vial fermion And I want them to be charged under a U1 gauge group I should I need to give them the same charge since in the same supermultiplet. I give them some charge Q Okay, and I want to I want to write down a Lagrangian that's invariant under this Okay, but these guys are contained in a chiral super field capital phi and I need some kind of The transformation parameter theta, which I'm sorry has nothing to do with the other things, but that's life This thing is a function of x, right? That's the whole point of gauge invariance is the transformation parameters or functions of x But now I need the transformation parameter to itself be a super field Right in order to preserve in order that this thing here be a super field i.e. Transforms like a super field under super symmetry and it has to be a chiral super field So suddenly the gauge invariance has been blown up to a huge gauge invariance that depends on an entire chiral super field Okay Just now we're really just following the script of ordinary gauge theory an ordinary gauge theory We say okay great We want invariance under this and then we discover that the kinetic term is not invariant under gauge Transformation and that will lead us to introduce a gauge field. We do exactly the same thing here Here's the kinetic term. Here's how it transforms under gauge transformations Okay, it is not invariant why because omega is a full chiral super field. It's a complex object Okay, I cannot impose phi dagger equals minus phi without violating super symmetry And so this thing is the kinetic term is not invariant okay Now the way that I make it invariant is actually the stupidest possible way you could imagine Turns out to work which is that I introduce a real super field V that just shifts under this this Omega combination omega plus omega dagger Okay, and I get this This is sorry this thing is gauge invariant. I've just defined V to shift oppositely to cancel this factor. That's all Okay, that seems really dumb like the sort of thing that would never actually work, but it turns out that it works Okay, so just as with ordinary gauge theories We had to introduce some new degrees of freedom to make the kinetic term gauge invariant And now we need to define the components of this thing V. What are the components? Okay? Well, here they are in all of their glory. There's a lowest component fermion Two theta two theta derivatives a theta and a theta bar derivatives three guys four guys Okay, and there's some factors and some eyes and they're there to basically make everything real Okay, and I need to tell you what the components of omega are omega is this chiral super field Which is now my transformation parameter and that has a complex. That's a complex thing It has its lowest component. I've written it like this because here theta is the theta of x the the gauge Transformation parameter. I really should call that something else. I'll call it something else I'll call it alpha in my when I fix the notes. Okay, so anyway, this thing right here is the is the gauge transformation parameter the ordinary one as well Well, no not as we'll see as we as we know and then we have a fermion and then we have this e Which is like the the like the auxiliary field of omega, okay? And now I can just work out what the gauge transformations are of all of these components Okay, and it's again the same kind of algebra over and over again that I keep doing so the lowest component of the gauge field just shifts by omega The fermion shifts by Ada, which was the next component of the gauge field the a mu transforms That's good That's really good, right? Because we see that this a mu transforms the way a gauge field should Okay, so just by following our nose We found that one component of this v does transform like a gauge field now We know we're on the road to success, right? Okay, and then lambda doesn't shift at all it's invariant and d doesn't shift Okay, this is I'm not going to go over this, but there's some Help for you if you want to people who want to compute the algebra. So here. I've just summarized it So what happens is that the lowest components of the vector super field C? Chi and B. They just shift. They're just pure shifts Okay, and then a has an interesting transformation and transforms like a gauge transformation these other guys are invariant That's the result Okay So now notice that because these lowest fields here. They just shift by some remember these Omega's Adas and ease are just general functions of x so we can just set these fields to zero by using the using up this gauge freedom That still leaves the gauge freedom in theta, which I'll rename alpha. I promise, okay Okay, and this gauge choice is called West Amino gauge So in that gauge things simplify a lot because the lowest components of V are all zero Okay, and now I can just work out using the same kinds of Susie algebra that I did before I can work out what the gauge invariant kinetic term is and guess what Okay, I got of course covariant derivatives of the scalar covariant derivatives of the fermion. I still have my f dagger f There are no derivatives no need for conformal for for covariant derivatives So all of this structure is just dictated by gauge invariance It couldn't be anything else right because the the amu transforms as a gauge field Okay, the only thing that is that is not dictated by ordinary gauge invariance is this term right here there's a term phi dagger phi d we can understand where that comes from because D is the highest component of V. That's a theta squared theta bar squared term and so we have a term like this Okay Questions all right, so we see in fact this silly idea of just putting in this V really does work It makes the kinetic term gauge invariant in just the way that it should Okay Now we also have to write a gauge kinetic term Right, we have to write not only the kinetic term for the scalars, but also the gauge Kinetic term, okay, and we can take a page out of if we go back and look at how we define the components the fermion lambda Okay, the fermion component lambda was defined by taking three derivatives of V in this way and then taking the lowest component and we found that that was gauge invariant It turns out that it's also gauge invariant if you don't take the lowest component Okay, so exactly the following our nose I just want to give the care the idea that what we're doing here It's somewhat formal, but once you have a few basic ideas. It really is just following your nose Anyway, this thing right here this combination is gauge invariant Okay, so this is a super field that is gauge invariant. It's also a chiral super field You can see that d bar acting on it is zero because if I take d bar and act on this I have 3d bars right in a row which is zero because all the d is anti-commute and so it is a chiral super field It has nothing to do with the super potential which is also called w. That's a standard These are very standard notations. So I'm not changing this one. I'm sorry. Okay and Now that allows us to write and this thing has dimension three halves if you work out the dimension here V was dimension less remember it went up into low Lagrangian. So up into the Exponential so it's dimension less and then I work out what this thing is. What is d2 theta and I get? Guess what I get a gauge kinetic term. I get I times FF dual and I get a kinetic term for lambda and I get a d squared so if I want to give a complex Coefficient there's no reason why this coefficient in the d squared theta integral cannot be complex and if I give it This complex coefficient. I work it out. I get a kinetic term Okay, I get a theta term I get a Kinetic term for the for lambda and I get this d squared term so you can see now this d is a real Skater field it's going to be something like an auxiliary field kind of like f except this guy is real Okay, and I have these five funny one over G's because I put a one over G here Okay, but so these one over G's are not the way that you're probably used to seeing the gauge theory written But I can rescale V goes to G V that means I rescale all the component fields by a factor of G And then this thing looks more familiar Okay, so this is it's actually a useful thing to go back and forth between these two Definitions one where there's a one over G squared in front of the kinetic term and No G in the covariant derivative then if you rescale you have a G in the covariant derivative But no G in the kinetic term. It's useful to go back and forth between those Okay So let's see. I want to sort of wrap up. Yeah, this is a good place to wrap up Okay, so now I've kind of made a little bit of a thing about this theta angle in a U1 gauge field The theta term is Doesn't do anything. It's a total derivative In in any gauge theory, but in an abelian gauge theory. It actually doesn't do anything in a non abelian gauge theory It does something So I'm just trying to illus just trying to illustrate that you that a theta term is completely compatible with supersymmetry, but in this theory here we could have just left it off okay, and the As I said, there's this D term here, which is going to be an auxiliary field very much like F It can be integrated out using its equation of motion We'll talk about that next time. Okay, so I'm going to stop here. I think I have is that right? I have a 15 minutes for questions. Yeah, I just want to make sure so I'll stop here and then I'll just take some questions Okay, so I start with questions. I'm not taking this as a good sign Can you say something about this lambda field? This is something like a ghost. Oh, sorry. Sorry Sorry, sorry. No. No, it's not a ghost at all. Sorry. So So Remember that all of these fields are just following my nose all these fields are fields that are in the path integral I have to integrate over them. This is a field Sorry, this is a a fermion field with a good well-to-do ordinary kinetic term Okay, so this is a physical fermion. It is exactly the super partner of the gauge field Remember at the beginning of the lecture I said that the one particle rep the massless one particle representations of the supersymmetry algebra included one One with a massless spin one particle Plus a massless spin one-half particle. That's exactly what I have here I have a massless spin one particle and plus a massless spin one-half particle. This is beautiful This is just like in the scalar case. I had a single Single complex scalar and a single vial fermion here. I have a Vector of massless spin one and a massless spin one-half okay, so This is a gauge boson. This one is typically called the gauge. You know, so when we want if we want to have Bose Fermi Symmetry we need to have one fermion for every boson and vice versa And so we should expect we should have we knew if we thought about it ahead of time We would know we have to get a fermion And we do You come due to this ww term there and then you multiply it with a complex number Yes, and I don't get why this vacuum angle appears only in the f f tilde term So that in the other terms, yeah, so I mean of course out somewhere Sorry, they must be they must be cancelled out somewhere What must be cancelled out somewhere this vacuum angle terms in this normal f f term and in this d square term Is your question? How did how does this lead to that? Yeah, yeah, is that your question? So i'm not gonna I mean there's a detailed calculation to be done, but it it basically comes from this Okay, so when you do this calculation Without any Hermitian conjugate or prefactor or anything if you just do this calculation You find that there's this this is a complex quantity Okay And it uh, it comes it has its lowest component as a real part which is f squared and an imaginary part which is f f dual So you really once you accept this then the the other thing follows Okay, because if I have multiplied two complex numbers together the real part is this real times real minus imaginary times imaginary that's all So it comes back to the fact that this is a complex number That i'm that i define the kinetic term not as you might expect as the square of something which is manifestly real But as a holomorphic plus anti-holomorphic thing Does that answer your question? That you don't look that way, but why don't we we can I can I can try I can try again after the lecture But I certainly didn't explain where this comes from so if that's your question. It's it's a bunch of algebra This is okay, but then look Look, I have i times the imaginary part of this. I have real times the real part of this Because I have this hermitian conjugate Okay, yeah So can you also build a spin two? Supermultiply and somehow play with the transformations and get rid of a couple of the components Yes, so spin two multiplets are of course of great interest The massless spin two field is the graviton So your question is exactly equivalent to saying can I construct a supersymmetric theory of gravity? Okay And the answer is yes with a lot lot lot more work Okay gravity is always more complicated than ordinary field theory and You have to because you have to we in the end, you know, you have to gauge Lorentz transformations essentially when you just construct ordinary gravity to construct a supersymmetric gravity you have to You have to gauge supersymmetry And that is that's that that can be pretty complicated One comment I do want to add is that it's complicated But we have to do it if we believe that supersymmetry is a fundamental symmetry is I keep saying it's a spacetime symmetry And we know that the spacetime symmetry is gauged because we know general relativity exists So we actually know that supergravity must exist. It even has phenomenological consequences It's one of the topics. I'm not going to be able to cover in this course Could you go back to the spurion slide where you first promoted the parameters to fields? Yeah Yes, so I I wanted to understand the logic because in the beginning you have a Lagrangian which does not have this un symmetry Here's the here's the Lagrangian. I mean there are the kinetic terms the kinetic terms The kinetic terms are invariant under the u1 Right because the kinetic terms just look like phi dagger a phi a Right, so the kinetic terms are invariant under the un un right The interaction terms are not because the couplings carry un indices and they are numbers Right if this described the real world the particle data book would tell you the value of kappa 1 kappa 2 kappa 3 and so on right So before you promote them to fields there is no un symmetry only afterwards So but I can write it like this And once I write it like this if I as a theorist, right Imagine a universe where these are just fields where kappa m and lambda are fields I can say well gee look they transform under a un symmetry And then you renormalize the field values, correct? Instead of renormalizing the couplings you renormalize the fields lambda of x essentially That's right. The way I can think about it is if these were actually dynamical fields You would be say well if they're dynamical fields you're you're you have to put a coupling constant in front of these fields, right However, if you do that I can say well knee on you because I'm just going to absorb that coupling into the fields But you also need to add a kinetic term for all those fields or not? Yes, if they had kinetic terms I would now exactly if they had kinetic terms I would now have get some non-trivial factor in front of the kinetic terms of those fields But I never have to look at those kinetic terms. So I'm fine with those Ugly factors being there. So I can really think of this as just some fields Coupled some new fields kappa m and lambda coupled to these old fields. And if you'd like I'm just keeping their vegs Okay, in fact, this is how we If this is exactly what we do in the standard model for for much of the standard model until the Higgs was discovered in 2012 The the only appearance of the Higgs in the Lagrangian that was tested was in terms of its vev Right, you could take the standard model Lagrangian Replace the Higgs everywhere by its vev and for all intents and purposes. That was the Lagrangian right That was the Lagrangian for every process that doesn't involve actually creating a Higgs boson So in exactly the same way if you think of these as fields as long as I never look at processes Where I'm creating these fields if they in fact are fields I don't know what mass I'm supposed to give them right But if I just look at processes where I don't create those fields and they just I think of them as all just having vebs That's exactly the Lagrangian that I have So just as the standard model is not the the standard model with Higgs vebs put everywhere You could possibly imagine putting a number v They're put in in a very correlated way because why because the Higgs is a doublet under electroweak symmetry Right here. It's the same kind of idea these kappa's m's and lambdas don't appear in a completely arbitrary way They appear in a way that is dictated by this un symmetry So I don't know if that helps I know that there is these Description of prst symmetry in terms of superspace. Yes, how is this connected to this description of the gauge field? Does this depend on this gauge you choose? well, it's very important first of all that This supersymmetry is conceptually completely different from BRS symmetry Okay, so coming back to because I sort of suspect that this problem I sort of suspect this problem this this this connection can arise most easily in your mind in gauge theories So let's talk about gauge theories now this this Lagrangian here is invariant under these supersymmetries Okay, now that is a real symmetry that enforces It relates the properties of this physical spin one half field to the properties of this spin one field It's a physical symmetry that relates different things BRS symmetry is instead is a principle that we can use to understand the Correct counting of degrees of freedom in a gauge theory. So physically conceptually they are completely different things Okay Formally they have a lot of the same properties because they're both fermionic symmetries. So the mathematics gets recycled Okay, but physically they are completely different things Giovanni, there's Is there any advantage of the other gauge than the best immuno gauge? Yes, actually there is You know Just as in gauge theories one gauge is enough, right? I mean, it doesn't matter what gauge you use So he's never anything wrong with west immuno gauge, but sometimes certain properties are more obvious in some gauges than others It turns out that one that I can think of there's probably more But one that I can think of is just like there's remember there's the spurious un symmetry It turns out it's also a very useful idea to gauge that symmetry Okay, so you introduce some fictitious background gauge fields That's also always a good idea if you have a global symmetry in your theory No matter what quantum field theory you have gauge it and you will probably learn something Okay, um, and it turns out that the lowest components of these things Make it make certain things manifest like for example the structure of a vacua and other other things So the quick answer is yes, but certainly for doing practical Calculations in particle physics. I think you should always use west immuno gauge people should be allowed to go to lunch, maybe Okay, so let's go to lunch and thank the speaker again