 the discussion that we're going through as possible. From next semester onwards, I hope to get more calculation of stuff, to calculate scattering amplitudes and try to get more of the feelings. OK, so fine. OK, so let's return to the discussion we were having last time. You remember that at the end of the last class, we wrote down this scattering amplitude for string theory. And so basically what we wanted to do was to compute the path integral overall world sheets with certain insertions, insertions and insertions. We converted that path integral over all world sheets to a path integral over the x fields and some closed fields plus some moduli. Some moduli. So the final answer was an integral over moduli, a path integral over x fields, and a path integral over the closed fields, dc and dc. Then the things that we integrated over had a world sheet action and had some insertions. So the world sheet action was exponential of minus the formula called action. So that was just the action we started our discussions with in this course. But then to that demand is the dc action, plus b, then c, e to the power c. Again, the b and c came from the closest engagement, as we discussed in detail in the previous class. And that was it. Pulczynski finds convenient with some normalization, which we will follow him with. I'm not going to write that down now, but you can do it. Of course, it makes no difference because you can change the normalization by rescaling the fields you can see. So it's physically different. And then this was the action. We also had some insertions. The insertions were the following. For every factor of the, for every modular, modularity, we had an insertion of del G by del dm, del G by del dm. So what was the semi-like that we did? Remember that b started its life as a 2-pacer. So the insertion that we actually had was b alpha beta squared G integral del G alpha beta by del dm. And then we had product over it. One such insertion for every field. And then we got a particular conformal basis, right? So you should write out this insertion in that conformal basis. Essentially, this becomes a factor of million, 8. And then we also had an insertion, which was the product over c times c tilde times the vertex operators. And the number of these guys was the number of the formal healing vectors. The number of diffior morphisms that wasn't fixed by a positive and formal location. So you have product over these guys. And then you also had a product over integration vertex operator. Yeah, I don't know. This is down in the schematic here. But this is a final form of the answer for these S matrix for the S matrix element that we're interested in structuring. Now, this class in the next, in this class in the next, we'll analyze aspects of this formula, aspects elements of this formula in various ways. We would want, by the end, to show this formula, both by answers, for example, that's for formulaing arguments. This formula has a sensible formula defining sensible scattering processes in an extension of what we're doing. We don't want to show many things about this formula. Even before we can start talking about some of these properties, there's some formal discussion that I need to deal with. And this formal discussion is a discussion of the BRSD symmetry of the action of this action. So we're going to get back to the discussion of various aspects of this formula in a bit. But in order to start talking about aspects of this formula, and especially ask the question of what vertex operators here you are allowed to put into this path in paper, it would be very useful for us to have a better understanding of some of the symmetry properties of this action of what we start with. The symmetry property that we're interested in has a name that's called the BRSD symmetry. My discussion of the BRSD symmetry of the transfer of these properties follows. It's basically very close to what we're putting into this. It's a really, really beautiful discussion of the BRSD symmetry. It's so much nicer than the discussion we find in a standard theory next to a formula. So we're going to put this action on a moment for a moment. And I'll ask a slightly more abstract formula question that's relevant to any gauge theory and any gauge fix. And the question that we're going to ask is, suppose we take the gauge theory and quantize it using the Kadi-Pokop method, OK? Is there an x? We ask the question because the answer is in the denominator. But the question we ask is, is there an extended symmetry? An extended symmetry of the gauge fixed action that involves the coast as well as the matter fix. And the reason we ask this question now is the answer is yes. And this is the thing that I want to describe to you, mainly in this lecture, and to show you how it applies to the string. So the first part of the discussion in this lecture will be, formula that abstract can apply to any gauge theory. But that's not strictly the string. So let's add up some notation. Suppose we've got a theory whose various fields I denote by phi i. And maybe it's completely, for the equations, maybe it's completely abstract and compressed notation. So this i index would be a flavor index of the field. But it could also be a space index, where the field is. In general, in field theory, phi i would be phi i index. But this i use both position of the field as well as any other quantum assumption. The gauge theory has a gauge symmetry. So let's say that the transformation properties of these fields phi i under the gauge transformation. So firstly, I'll use the principle alpha to denote the various inequality transformations. Once again, alpha could be labeled by space time point. Gauge symmetries are generally local space. So alpha could be brought over and continues under the gauge. I'll use the simple delta alpha to denote the variation of any quantity under the gauge transformation parameter, respect alpha. And in particular, delta phi i will represent the variation of phi i under the gauge transformation parameter, respect alpha. And phi alpha is a parameter that labels infinitesimal transformations. So alpha is labeled in the reagent of the gauge transformation. But remember, as a gauge transformation, it's labeled as a group of gauge transformations. There's an alpha for every point in space, as well as whatever it is. So if we were just interested in gauge transformations within the one structure, we were discussing gauge fix actions. So I also had to tell you what gauge fixing conditions I used. So let the gauge fixing conditions in my theory be denoted by f a. There are some functions of phi i. And my gauge fixing conditions are chosen to be f a of phi i. In implementing the Fali-Pobov method, that we described for string theory last time, you just take the discussion we had two lectures ago and apply it to any gauge theory. The whole discussion was true. So you remember finding the conclusion first. Finding the conclusion was that we put in a delta function that enforced that the gauge fields were on-shell and could make this equation. In analyzing this path, the path integral of the delta function, we found that it continued to exponentiate this function. So the final action was I've got a S, Y, and the angles action, the gaussian, the original action, which is function of v of phi i minus i times, I've got a b a. This is the path of the action, such that the integral of a b a in force of delta amount to the vector a is equal to 0. F a of phi i. And then blast up the Gaussian at the cos part, which came from the path integral of determinant. Now remember how we got the determinant. We did a small variation of F a with respect to a gauge transformation. We were the determinant of that, rather than the inverse determinant. So we did the gaussian integral over the small variation with respect to two anti-commuting fields. So the small variation I've made is delta alpha a. And then in order to get the determinant, we multiply it by some field C alpha and some field B a, both of which are taken at the end. We have the structure. The gauge fixed action has the original action as a path that enforces the gauge condition by doing an integral over the Lagrangian line. And as they fucking go off now. In order to derive this, I'll ask you to rerun the derivation we went through in the gauge spring theory in your head. But in the abstract case, it's easier in the abstract case. You don't have to figure out what the variation is. It's the variation. It's a hybrid. We don't go through it. It's just the same logic. But I'm going to ask you to read it anyway. Maybe we should at some point have a discussion session in class where I'll ask you questions. You ask me questions. I ask you questions and we discuss things. But maybe that would be useful at some point. Let's move on. Now, this is the action. Now, I want to point out that in the complete general, I think this action has a global symmetry. So the symmetry is the same. So the VRS is the symmetry. And variations under the symmetry are known by delta B where B stands for VRS. So in order to define the symmetry, we have to take the variation of all fields under the symmetries and reject that symmetry. So delta B of pi i is equal to minus i times C alpha delta alpha. Delta B of C alpha is equal to minus i by 2 minus i by 2 f alpha beta gamma C beta C. So let me get your answers. The answer is the structure of this case. So delta alpha of A is equal to alpha beta gamma It is the meaning of everything which is written down in the book. I should check the answers. The meaning of the equation for the book here. We will check whether they work or not. What does this mean? Consumption. Yes. So if you want to really make a symmetry, I mean, in your theory of variation, you multiply it into delta alpha. Yes. Multiply and decommute it into delta alpha. You should multiply everything on the right-hand side here, right? By an x along which side? I'm going to use this simple delta for the variation without the epsilon. So you should think of variation as epsilon times delta A. Okay. I'm going to use it just as a matter. Okay. Yes. What is the... What is the... The PRST charge? I mean, the thing which is anti-competitive in these fields gives you the stuff. Yes. So this will be the opposite of the corresponding to the delta. Corresponding to delta. The anti-competitive of phi i will be delta alpha. Yeah, that's the reason I'm at it. Delta will be delta alpha. Okay. Good. So, now the first thing we're going to notice, the first thing we're going to notice is that the angles that we actually go down on the board can, in terms of this operation delta that I've defined, can be in an interesting form. Again, that we can rewrite S and we put the S angles plus there's some i which we'll get straight in a moment. But delta B of B A. And with this sign here. Okay. Let's check. So S angles is S check. So this part is this part. So this is true. This is what I think some of these two terms. Okay. Let's check. Delta B acting on B A. Delta B acting on B A. Okay, wait. Now these things are anti-competitive. I should get the order of this one. Okay. So delta B acting on B A is equal to capital B A. And so that means this one here. On the other hand, delta B acting on F A is what? Delta B acting on F A is... Okay. So firstly I become minus sign because it goes through an anti-competitive which itself is an anti-competitive operation. Goes through an anti-competitive field. So I get plus i times B A. Okay. Times the variation of F A. And how does F A vary? It varies because phi varies. So the first thing we want, but the first thing I should have said is that notice that the action of the transformation on the field of phi i has a gauge transformation on the field of phi i with gauge parameter cf. So formally the variation of any function of phi i under a BRST transformation is the same as the variation of that function of phi i under the gauge transformation generated by cf whose infinitesimal parameter is cf. But this thing is the same as c alpha and delta alpha because my definition of delta alpha was the change of the quantity alpha under the gauge transformation. Do you like to change the, the change where you know the both fields, the both fields are local fields, right? Yes. So I mean the, despite of that the other system is a global system. It's a global system. The epsilon is constant. Yeah, epsilon is constant, exactly. You see, because, right, you, exactly. You see, this is not a new field, a new parameter that you're going to get from. It's a field of theory. There's only one BRST, there are many different BRSTs that depend on some functional parameters. There's one. And we need a minus i. Because the gauge transformation is minus i, you see. Okay, so we put that together, we get minus i times i plus, we get plus ba. First thing we've done is we check that the action is equal to s y n plus delta b of minus i times delta b of ba, that's the thing that we're going to, that we're going to, this variation squares to zero. So I want to get at delta b and delta b. How do I do that? Let's just check this for you guys. Okay, first let's come to the support. Let's do that ba. Well, since one variation is zero, two variations are certainly zero. Let's do it a little bit. The first variation is ba, the second variation is ba, which is zero, so that's good. So there are a couple of ones of these. Okay, so let's first write this file. Okay, so what we want to check is that delta b of the variation of ba. So if it's zero, zero up to constant, this is where it's going to be minus i. So delta b of c alpha delta alpha phi i. I want to calculate this and check whether it's zero. Okay, so let's get it. So this is two terms. The first term here is the company of the changes c alpha. And the changes c alpha is i by 2 f alpha beta, gamma, c beta, c gamma. And then you want to calculate delta. The second term where we pick up a minus sign because we go through an anti-communic object. And then we perform the variation, we perform the brsc variation of, so we get minus delta alpha of minus delta alpha of c beta. Let's call it delta beta minus delta alpha of c beta delta beta of i. And this is c alpha. I'm sorry, minus i. And the minus sign. Thank you. This is because this, because this I don't make a sign. Thank you. So let's make it a plus sign. It is the c with itself, you know, anti-communic with itself. So this delta alpha delta beta, we can rewrite as a sum of a commutator plus anti-communic. A is but 2. Commutator by 2 plus anti-communic by 2. And it's only the anti-communic in the piece that survives because c is anti-symmetrical with itself. Because this c is anti-symmetrical. Is this correct? So we've got, so this delta c is anti-symmetrical. Right? Symmetric under the, let me alpha and c become equal to 0. Yeah, that's right. I'm just wondering whether I should really have the c inside this delta. Just, just. No, what we have to do, I think what we should have done is that the c beta del beta, the whole thing would have come up of your delta alpha. Delta alpha i is a function. And we are looking at the gauge transformation of the function. So you just get minus i c meter, the meter of delta alpha 5. Yeah, exactly. I think you're right. I think you're right about exactly what I said. You know, it's an infinitesimal variation. It's a particular, you know, linear function of 5's. So we could write this in matrix form. You know, this is some matrix. Then alpha pi is some m alpha ij of 5j. And the second del acts only on a 5j. On the whole, then alpha pi itself is a whole function. And we want the gauge transformation. Exactly. This m is independent of 3. So at the end of it, it's a linear function. So it acts only on that 5j. So it's ready. Okay. Okay. Now the next step that I want to do is to, now I'm getting... Okay, this goes through. Okay, good. The next step that I do is to replace this by anti-commutator of delta beta over delta alpha. Okay. And then I get a factor of by 2. So I can replace this by by 2 and anti-commutator of delta beta over delta. Okay. And then at the next step I use that the anti-commutator of delta beta and delta alpha is fb. So this then becomes f beta alpha gamma delta gamma 5 which is minus to this. Change it up. Okay. Then as you see, when you think things through straight, the only difficult part is your calculations to keep track of minus sign. So often, difficult to get confused about your minus sign. Okay. Whenever you're alive, you think that through carefully. Exercise, the checking of the double variation of c being 0. Let me just give you the main steps. Remember that the change in c itself, delta b of c alpha was proportionate to f alpha beta gamma c beta c gamma. Okay. So what you want to do is to check that the variation of this is 0. Now the variation of this is two columns. One with the delta acting of the cp and the second with the delta acting of this c gamma. But each of which will be the same up to a factor of 2. Okay. Give you the minus sign. Okay. Then one of these two columns will be the following column. See how it's got beta gamma. And then let's do the variation of beta. See beta, let's say some theta phi. And you'll get c theta c phi c. So anyway, when you take the variation of both columns add them up, you get something proportionate. And the two columns will be the same. Okay. Now this thing here is an empty, because c is a completely anti-symmetric. This is a completely anti-symmetric product of two elements. Okay. Which is 0 whether you're going to hand. This is the main idea. I want you to convince yourself that in detail that all this is correct. Okay. So we're not going to let you spend time. We checked that the BRST charge squares c. It's important to set the reasons. The first reason it's important is the point. You remember we started out with this action by saying I want to provide a new symmetry of this action. So I'm fine. I'm not talking about a symmetry of this action. I'm not sure of you if the action is embedded under the structure equation. But now, everything we've seen is obvious. Because remember that the action was equal to s by m plus delta b of BAF. Check how the action varies under BAF. The variation of this action is because the BRST charge is equal to 0. Take two variations where you can see. What about the variation of this term? This term is equal to s by m plus a gauge of varying function of phi i's. And the BRST transformation on phi i's is a gauge transformation labeled by a parameter c. Both these terms of the action manifestly have 0 BRST. You know, if you need practice in dealing with anti-commuting fields, algebraic anti-commuting fields, I advise you to just expand the version of this part and explicitly check that when you do the variation, the BRST variation, you get 0. Again, the only difficult thing in that calculation was not difficult. The only painful thing in that calculation is keeping track of minus and minus. Okay, great. Okay, so that's the first thing. We've seen that the symmetry, that our action, the action that we've written along with those fields, has a new symmetry. Okay, it's a symmetry. First, in a total sense, it's an anti-commuting symmetry. It's a symmetry whose parameter, epsilon, would have been an anti-commuting field. Why is that? It's because, for instance, you see that the statistic, the commuting, anti-commuting nature of each field has been changed in the symmetric transformation. It's changed in five to the portion of the c, which is that. So you need an anti-commuting parameter on the right. Again. Now, okay, so we've seen that we've got new symmetry and that this symmetry has twice the c. So the generator of the symmetry, whatever it is, we call it QB and this thing's twice the c. Okay. Now, there's something that I want I want you to remember. And the thing is this, you remember that c alphas, you remember, there's something with the other three properties of c and b that I want to remember. Okay. The thing that I want to remind you of is that this busy action came from looking at a determinant of the gauge-fixing the gauge-fixing term, the gauge-fixing term under an infinitesimal gauge transformation. The gauge transformation was real. Okay. And each of these gauge-fixing parameter terms are also real. It's not like we've got a complex value in the set of gauge-fixing terms. Okay. So in our detonation of the behind-the-action, where we had an integral representation for the inverse determinant that we were interested in, we were doing an integral over commuting realness to this. In order to get the representation of the determinant, we have to transform that into a positive integral over anti-commuting for realness. You know, this C parameter, the C parameter, the C field and these D fields that we're talking about here are realness. So because of that, you know, the BRST charges itself as a realness of it. Complex conjugated, you get the same thing. For instance, we've got the transformation of phi. Phi is C times delta. If you're complex conjugated, both things go back to that set. Okay. So the quantizer of that statement that you expect, that will be that this operator QB is a Hermitian object. Okay. In explicit examples, we actually work out, you know, or the commutation relations of all fields are right on an object expression Q. You can check this explicit. For instance, string theory can see this very explicit. But what you always expect is that this is actually a QB, a QB diagonally. This nice symmetry, along with this commission property, consistency properties of correlation function calculations in gauge theory. Remember that the gauge-fixing function that we fixed in order to calculate in order to facilitate gauge calculations was completely arbitrary. Which was actually completely arbitrary. If any calculation we do reflects properties of the theory, and not an arbitrary choice of gauge-fixing parameter, then there will be that any calculation we do is invariable under the choice under, let's say, small detonation of this F-A gauge-fixing function. Any calculation that depends on F-A is a calculation that talks about what the way we set up the calculation rather than the theory. Now let's see. Remember that S was equal to S-Y-M plus delta B-B-F. Suppose we change F-A a little bit. How does the action change? So the action, under a small change of F-A, the action changes by delta S is equal to delta B of B-A delta F-A. But this delta has nothing to do with B-I-S-D. It's just a smaller change in F-A. We can write as the anticommutator of Q-B-I-S-D with B-A times delta F-A. Definitely, the B-I-S-D operator is the operator such that its anticommutator is the smaller reaction under B-I-S-D. So now let's see. Now let's see. You see, support before trying to compute some amplitude between an initial state and a final state. And I want to see whether the amplitude changes if I perform a small B-I-S-D on the B-I-S-D plane. If I change my n-sync condition a little bit. So suppose my final state was side-screen and I'm doing some e to the power minus the propagator calculation and I have some side-screen. Suppose I change my B-R-S-D charge a little bit. I change my n-sync condition a little bit. I've changed the action and therefore the Hamiltonian of my system by something proportional to an anticommutator on the B-I-S-D plane, but something. So what was the change in this amplitude? The change in this amplitude would be proportional to side-screen times QB B8 delta F. But B is zero for arbitrary delta F. It can't be because delta F is completely arbitrary. Okay? The only way that can be is the way that how do we do that? But this must be some condition on the states. It won't be true for every state. For sure. It must be some condition that you impose on states. And the only way this can work because delta F is completely arbitrary is that it must be a condition involved with QB on the states. Now, since this delta F is completely arbitrary, we don't want a condition of QB times this time state. So it must be a condition of QB acting this way. Okay, so let's write this out in 20 days. This must be this is this QB VA delta F A plus VA delta F AQB There's a condition. We can write this in the first term. So we can write this in some two terms. We can write this as QB on site fin A delta F A site in in class plus the other way site fin VA delta F on site. Now more or less the only way which this can be which this can be zero for arbitrary delta F is site fin and site involved to separately satisfy the condition that they are annihilated by QB. That is true this will be zero arbitrary delta F I'm not sure you know if this is really completely arbitrary it's probably possible to prove that this is the only way. Alright. Okay, but anyways with this action we're going to make the following cut. We're going to say that look states we set up some calculation okay this calculation can then be interpreted in some new words based on you know whether we've got a path container you take a path container slice it and constant tax license and give a Hilbert space interpretation for that path container or at least if the path container is really good enough okay but not every state in this Hilbert space that you obtained by slicing this path container is going to be treated as physical for instance you know that's safe and that's clear physical states in the Hilbert space should be safe that they oscillate only at the five islands the C and A are the districts we put those fields in to help our path integral calculation clearly they if you have an oscillated adjustment the C of the B is in the tax on the state clearly that's something physical it was some calculation it was not something physical at all so not every state that you get by taking this path integral and slicing it will be treated as a physical state only those states that are and highlighted by QP is on the silence a guarantee that something very good will happen and that something good that will happen is that any calculation we do will be invariant under choice of issues and therefore you know as a calculation of the theory of procedure of calculation similarly if we were interested in an operator acting in these states we will also require that all correlation functions that are meaningful in this theory operators that commute to anti-community development statistics with the BI state charge for exactly the same place provided it commutes anti-community with the BI state charge then we can take this up you put some operator here you can run the skew through that operator to add in the state if it didn't you couldn't and the calculation would depend on the choice of the same function this is very plausible guess for how the Hilbert space so I'm not emphasizing here you know generally when you're dealing with the gauge theory you can often fix gauge or you know do some thinking before you do some careful thinking in your problem to understand what the Hilbert space is without all these hosts so the question of what the Hilbert space of the theory is is rather physically it's a clear rise what we want to do is something different what we want to do now is the following you see tell us this calculation procedure that's very convenient it's a calculation and changing our action if you take the action for the calculation procedure slice it on a on a take your integral slice it and dice it and associate a Hamiltonian to the dimension generator of that path integral a larger Hilbert space it's clear that's a larger Hilbert space okay now we want to identify the physical the original physical Hilbert space some some sector of the Hilbert space of this this larger auxiliary problem and the claim we're making is that it's a very natural basically it could have been some subset of this but you'd have to find a natural subset in any given example you can check with us whether it is actually the same thing as we would do carefully for string theory as we will say basically what's the problem that we're supposed to do we can't we can't yeah there is an additional recognition between this constant in a moment but the condition the claim yes, so there is a subset the claim is that the physical Hilbert space of interest is given only by those states that are annihilated by Q okay, now we come to low time okay, so is this clear the motivation for this this statement now we come to low time let us now look at the following suppose we put the form Q times consider the inner product cyclist is equal to Q on cyclist any state in the world any physical state in the world including car itself what have we done you see we were interested in states we said physical states are states that are annihilated by Q remember Q realize this question because one very simple way of solving this condition that states are annihilated by Q namely let's choose the state to be Q times any other state because Q equals to 0 that will automatically be added so we've got a large number of states which are BRST okay, so the terminology something that's annihilated by Q is called BRST closed something that is Q times another state is called BRST exact okay, so any state that's BRST exact is also BRST closed this is the reverse we've done now states that are BRST exact are BRST closed are in some sense physical and are physical by definition so far however they have zeroing a product with any other physical state two states one states psi and the other states psi plus Q times that should come okay, any Q times up now products will get the inner product of these two states with every other physical state the theory is like that physical measurements are concerned just as far as physical measurements are concerned and this is true also not just inner products but correlation functions with operators all of which are BRST closed okay, so as far as physical measurements are concerned these two states are physically identical are physical according by definition okay, they must clearly represent the same physical configuration through physical inverse states now in one state for the physical configuration the true physical inverse space is more subtle than just the set of all BRST closed states so the physical inverse space is Q is BRST the physical inverse space is the set of equivalence classes of BRST closed states where any two states BRST closed states are treated as equivalent if their difference is BRST so by module I mean this equivalence class and I'm just a statement for operators remember that we wanted that an operator commute with the BRST charge this is a very simple way of arranging that and we choose the operator the commutator of some operator with the BRST charge but two operators that differ by such a commutator give identical matrix elements because the Q on one side kills the first state and Q on the other side so any two operators that are equivalent with commutator with commutator and commutator with the BRST charge are physically equivalent they give identical matrix elements so once again the set of physically evaluative operators should be thought of as a members of BRST co-op by the operation of BRST closed objects and modeled by the BRST exams that has a mathematical name it's called co-op so just summarise any gauge theory much with gauge fix we've got a path integral which displays just the level of path integral an interesting symmetry that symmetry is familiar and expresses it the action of our theory is invariant under that symmetry but more importantly the change in the action of the theory under a change in gauge friction condition is BRST exact otherwise BRST exact once we identify the correct physical input space namely the input space of BRST closed states of BRST closed operators because it's BRST exact the two different hander tokens that you get by making choosing one choice of gauge fixing bandwidth of the other of the second hander because they develop a BRST which applies to every gauge theory well when gauge fix is part of co-op and later part of it using part of co-op also applies to the action of the word of co-op so now that we want to understand and in more detail the BRST the BRST transformation properties the BRST symmetry and the word particle but just so that the algebra is simpler we first do it for the point particle where it's a little more transparent and then we'll get out of it for a second ok so let me look at ok so let's look at what the action for the point particle s was equal to d x mu dot by e plus m squared when we did in the first two lectures of this course we wrote down the action for the point particle which when we put a little glue in the space takes this fault but this point particle example it's due to quantization of this point particle and instead of fifth way remember we had one lecture we did four different ways of quantization of this point particle we'll do one more ok and the contact with others we are by the particle of the BRST method so we're going to we're going to do a positive method of quantization of this point particle but by choosing the terms of gauge what gauge will we choose well it's we choose the obvious one we choose easy to do going to the discussion we had today what should we be doing we should be enlarging this action we should be enlarging this action to include a i and b e minus one that's the part that enforces this data function ok and then we should in addition to that the particle is the particle of the determinant so what's the particle of the determinant here well you see what we have to do is to see how e changes under a gauge transformation now under a gauge transformation a gauge transformation is a diffusion ok so how does e change e changes in two different ways under an infinite average transformation it changes the first thing which it changes is that e is a function of magnitude so there is a change which is of the form so t is equal to t plus t is equal to t to the dot but that's from t to the dot we will have a change in e proportional to f times del e just from the fact that e is a function but then e also transforms into weight so there will be another transformation that will be proportional to del f by del del f by del d times e so e is the square root of the metric of the factor like this the change in e will be some two terms of the appropriate relative coefficient minus that now there is a variation of this quantity and multiplied by e to the power c the cp equals this f the infinitesimal change the infinitesimal change in the infinitesimal that label sign so these two terms will give us what they give us terms like b times del del e times c and we get plus v times e times del c sorry c is one time there is no contraction here there is just one pattern one time if you do this I am going to integrate by plus v dot e v c dot I will get back to the next class what I my logic is leading me to is that this would be after integration by parts v times e times c v dot times e times c I want to warn you that the same thing with the dot of the c okay let's just continue we see if we are in trouble with this and definitely I think it would be something very similar to this question this is what I mentioned I must be messy now okay good so now what we have here it could be the next class and I will comment that this is b v c dot sorry okay so this is the so now the next question was let's try to write let's try to write on the brs in the as you said okay good now what is the brs variation of all fields okay that's quite simple that's quite simple let's look at delta x mu delta x mu is just an infinitesimal variable changing x mu under a small homomorphism that just changes as with some sort of time dimension i times c times x mu dot delta of b 0 delta of little b was equal to b delta of c okay this is quite interesting because what you have to do is now look at we have to identify what f r is what we cut down nice okay in order to identify what f r is what we cut down nice what you have to do is to perform two different diffeomorphisms in different orders and see what you get so suppose you perform a diffeomorphism generated by a function v and then perform a diffeomorphism generated by a function g okay now so what is the diffeomorphism question suppose you say that t is equal to tau 3 tau is equal to tau plus let's call it f of tau delta and then you say tau tilde is equal to tau tilde tau tilde plus g of tau tau okay all you do the other way around you interchange the functions of f and g you want to see what difference you get what difference you get in this one three definition okay so the difference is very common how we put the definition here or you do the other way around so in one case you get a change in tau that is g times derivative of f but in the other case you get minus f times derivative of g you do this two diffeomorphisms in different orders and you generate effective difference diffeomorphism of this one you want to put that in f alpha beta gamma c beta c gamma what this is simply so now you have to put both g and f you put the same function c okay so that's c times c dot but two such terms and the effect of minus sign goes away with this c times c dot minus c dot times c and you you put it through and you get something that looks like c times c dot okay so why don't you do this more carefully so on but at least in structure we are also going to be able to perform i times c times c these are the brsv transformation these are the suggested brsv transformation transformation properties for this point model okay very good very good so perhaps we should do a now I am suggesting and check to see whether this action so we should check to see whether we should say this one right b and v minus 1 so one term of course is just b times v minus 1 into this term and the second term is dcc dot so sorry is y is b times brsv transformation the variation of b is the variation of b under the gauge transformation b exactly like x not exactly like x b is like square root of x okay if it was exactly like x it wouldn't be worth the difference but I think it's a second term like what we have so same calculation you see let's do it suppose variation of b under the realistic transformation was c times del e plus del ec the root is after the del of c times e and integrate by cons and it goes to this but I think it is not this far right because let's work this out the minus sign always are on the diagonal but let's see this carefully so how does e change under our coordinate let's do it really carefully suppose we have td and this is equal to t prime plus okay so we have that now e transforms like a del because there is one index there exactly exactly let's do that t should be equal to e prime of d prime and so we find that d prime d t that's right d t prime plus del f d f so we get that del so e prime is equal to e plus del f of t e prime theta and therefore e prime e of t and this is e of t now we put e of t prime plus e of t and so we get the other term exactly the other term and the other term is exactly del e times x which we can write as del of f and c so that's what I was writing here and then integrating by part I think it's probably just enough I should have checked fine so let's do that now we I'll check this more carefully we have many others we should check do you know what website I've written do you know what model it contains in which it lists typos and potency books it's always useful when you think you have a mistaken potency it's a model you may or may not be right okay so which is the website this is the address half of which have been pointed out so okay so let's move on so we've got some nice actually okay so now the next thing we want to do enhanced action in addition to the added we've also added this because we were to do the following and then of course that's the most important thing so suppose we were to do the following suppose we were to integrate out of the ground and it would get to the ground and it would say e to 1 under that process first side it seems that it makes no sense to talk about the realistic transformations after we've done that procedure the variation of b little b has a big b yet and b is no longer a field in the action but you do how do you define this? you see in integrating on b we lose two fields we lose e and we get a motion we get out we can integrate out e and the equation of motion for e which is sort of what we're losing from the problem it determines b in the following way so let's write down the actions the action is this plus and we'll stick to which is the science so this plus i times b into e to the minus 1 plus e to the b times c that's it i times b into e to the minus 1 plus e times say b dot c if we integrate out this b field we lose the equation of motion that we've got with respect to e what is that equation of motion? the equation of motion is just differentiating with respect to e and equaling e to 1 so it's just x mu dot x mu minus 2 b s right? 2 then 2 plus m squared by 2 plus i times b plus b dot c is equal to the following way if you do the integration out procedure you do the integration out procedure but replace this b here by one word from these equations of motion it sounds as if you're doing something sense actually to derive this in a more satisfactory way what you would want to do is a problem the path integral with this thing and then see because we want to keep track of what happens to a variation in b once we integrate out first we take this path integral alpha times b integrate out b and then differentiate that result of the path integral with respect to alpha what we have is the answer to what an insertion of b in the path integral would have been even though it's been integrated out suppose you have a path integral and you want to insert a factor of b one way of doing it is to take this action add alpha times b to the action and differentiate with respect to alpha differentiate the answer with respect to alpha that gives you a variation with respect to b so now suppose you want to know what the insertion of a factor of b would have been even in the procedure after you've integrated it out this is an invariant way to do it so add this insertion into the path integral then integrate b out and differentiate the final result with respect to alpha then say alpha is 0 this gives you an insertion of b in the path integral originally would have by definition and I'll leave this for you as an exercise but you can check that this procedure gives you exactly an insertion of this quantity an insertion of the quantity that you get that you get by replacing b with x mu dot x mu dot by 2 minus n squared by 2 plus b dot c all with the minus alpha let's do the scheme action so what we're doing here is taking this path integral here and adding in an alpha times b plus alpha times b the next thing we do is to do the integral over b you see when alpha appears in an additive shifting so instead of setting e to 1 you set e to 1 minus alpha with respect then you want to differentiate the answer with respect to alpha which is the same thing as taking the derivative of the action with respect to e at e equals 1 which is the equation of motion free is it clear? this is the most satisfactory way of seeing that equation of motion from the point of view of the path integral is the same thing that is to be used for an insertion and this insertion of this operator is the same as the insertion of this operator once you integrate the operator in this mu reduced theory which has a bc cause otherwise that will leave the same cause is delta x mu is equal to i c x mu dot delta b is equal to minus i times x mu dot x mu dot by 2 minus n squared by 2 plus 3 dot c and delta c is equal to i c c now these vlsc transformations here these vlsc transformations are again the algebra is more or less the same I want you to check that these vlsc symmetries directly are symmetries of the action but only on using the equations of motion okay this is now not an on-shell this is now not an off-shell symmetry of the action anymore it's an on-shell symmetry of the action essentially because you've solved some field using the equation of motion I want you to check just explicitly that this thing is a symmetry of the action that squares to minus that squares to 0 only once you impose the equations of motion which of course is fine because we are interested in some operator you know operator relations inside a path integral and equations of motion are true there's no problem but so so what have you got we've got these vlsc transformation properties in this reduced area this is the analog of what we're going to be doing in string theory but there will be many more fields and it will be less transparent what we would want in string theory is to deal just with the xmu fields the b and c courses not the main trick so deal just with the xmu fields and the b and c course that you have to deal with blsc symmetry so we're doing the analog of this for the problem okay great now that we have this nice explicit blsc transformation properties and nice explicit action for our theory we can try to write down the formula we can try to write down the formula for the vlsc challenge okay so first let's look at the let's look at the action so the action has remember e has been set to 1 so let's pre scale this okay and it has a bc system now bc system is a quadratic of anticomputing fields of the type b.c okay the canonical interpretation of this is that of a system of c with b we see that where the canonical momentum conjugates to b okay in momentum conjugate to c anticomunitator c is equal to 1 okay okay when we start dealing with the fermionic stream I'll have a lecture on fermionic path that goes in the Hilbert space interpretation in which we'll discuss things like this in a more satisfying in a more satisfying framework but at the moment roughly you see it from the usual rules of canonical organization and you just momentum with operator commutator okay and when that is true you know how to solve this problem so this is a you have to find the representation of this algebra and the representation of this this algebra is clear it's given by the operator 0 1 0 0 1 0 this is like one is the raising operator the other is the lowering operator okay so what's the Hilbert space we've got Parthen-Temple what's the Hilbert space interpretation of three Parthen-Temple that we have the Hilbert space interpretation of Parthen-Temple that we have is there are three bosons for the x field remember 26 of them as many as there were both spacing time coordinates plus the two states system corresponding to the b and c system now the next question is what is the formula what is the formula for the for the b and c challenges okay now with the traveling yes they do because you see now when e is in set you watch the action is just x dot squared plus b c okay not the x that's two completely completely so just the two state system plus the x okay so what's the Hilbert space of the system it's functions of d variables times up and down times vector in the two state system up and down yes and it really like strings here it's the same thing happens yeah completely there's no coupling between and of course yes completely yeah by the way so let's move on so a state that is up down some state in the two columns here times some function of x mu in the momentum space of action of k actually we could ask is now in this Hilbert space can you write down the canonical expression can you write down the canonical expression for the Hamiltonian process for the we are actually charged we are actually transformation charged okay so what do we want we want something that when commuting with x's gives us c times x to what and then commuting with b gives all this stuff okay so the thing that works the thing that works is and commuting with c gives us all of this so the thing that works is what it is c times h where h is equal to where there was a plus here I think no okay okay okay so this is minus sign somewhere that's what the c is actually the thing that would work would be something is actually of this form okay since c for instance we take the property commutator c times h with x the momentum part you get with x is x dot so this we should think of as b mu squared by 2 so 1x will eat up 1p you will be left behind with a b which is this guy here and we have the c if you if you take the commutator the commutator with a b you will hit this c and get everything else and if you take the commutator with a c you will hit this b and get everything else even a motion of the theory you see the term of the action has b dot c or c dot b it doesn't matter now if you say it equals 1 you use the equation of motion of the theory and you get b dot equals 0 or c dot equals 0 so on shell you can just draw this it's important for generating it's important for generating commutations but if you want to evaluate it on shell it's on shell action this term is not there so for many purposes many purposes you can think of this brsd charge that's just been c termization where it inches this one now if you want to complete the brsd quantization of the point particle what would you say well we have to find the brsd coho to share this point particle so we want to find the brsd coho washed what we do the brsd charge the qb is c times h and we're going to make this on shell replacement where this thing is just so that c times p squared by 2 will outcome minus n squared you say there is a minus n error okay I'm going to ask you to forgive minus n errors the result will be an equation of motion which will be the correct equation okay fine so qb is this this is the brsd charge so now let's make some notation let's say plus the upstate is the state of the system that is annihilated by d such that b on up is equal to c and so the downstate is the state of the Lagrangian such that c on down is c and b on up b is down b on down gives up c on up gives down okay so if we want to specify a state in the full elimination of theory if we want to specify a state full elimination of theory we have to specify when it's up or down let's say so the basis for the inverse state is up k mu and down so look at the action of this brsd charge this brsd charge on this basis okay so we look at qb acting on let's start by saying up to b now qb has an explicit factor of c c on up gives you down so this is the same thing as down that is k mu and then you get whatever you get by acting h on k mu which is k squared plus m squared that's where I'm correcting the equation of motions okay now we should be able to figure out what it is because when you clean it in space it should be bad and if you continue it should get minus k0 so this is correct so it's k squared plus m squared whatever that so this is the actual qb on down on up and down on down that's k mu now this thing here is missing because the action of qb on the basis of of our analysis so the next thing we want to do is to list all states that are brsd but that are brsd closed so brsd closed that are states that are annihilated by brsd that's what I lost well it's qb down to k mu that are the states that are up k mu when it's a brsd z that is of the form down k mu that doesn't have k squared plus m squared equal to 0 is qb on something down k mu k squared plus m squared not equal to 0 is brsd this is brsd go homology and we can find the basis you know some equivalence class you know some typical elements of the equivalence class of brsd go homology this is just generated by up up k mu with k squared plus m squared because this is closed and nothing that's of the form up is brsd exact and also down k mu where is k squared plus m squared okay great first the answer this is almost the answer that we got in the first couple of lectures about this remember the answer was all the functions of k such that this equation of motion was away it's almost the answer because it's twice the answer you see we've got one set of functions with up and the outside functions with down okay so that's not quite what we had then that's not quite what we had in our discussion in class okay this is an example of you know this phenomenon that in the standard of brsd quantization we had a guess for what the right hand of the space was and it's a very reasonable guess so what was wrong is a super super space of the correct answer then maybe a little more as we've seen by this calculation now the correct answer I want to tell you that the correct answer in this case is which one which was the one that was the one that was exact so I'm clarifying that the correct answer is