 So let me remind you what we saw in the last case. We performed this quantization with our extreme. This quantization, we performed this quantization in a elaborate manner using this phase space quantization technique. But let's now remember the end result. The end result was as follows. The input space of our theory was made up of 20 infinity time oscillators times functions in 26. Functions of x or functions of p in d variables, where d is the length of the input space. That was our input space. Now, this input space was the input space of an, was too large input space, as we see from the passive player functions of d variables. The correct input space that I found was d minus 1 by 1. And the fact that this input space was too large was told to us by constraint that we had to impose on this one of theory. The constraint took the form of an operator equation that was obeyed on states. Another operator equation, well, there were two operator equations, one for less, one for right, and one for left. And the operator equations were, so let's call it a sense, tilde, and also alpha prime by 4 into square, which is alpha prime by 2, 2n minus. Just to remind you where all these terms came from, this term here was just the constraint if actually it was simply the Hamiltonian on the worksheet of strictly 0. More precisely, the Hamiltonian could be written as a sum of a Hamiltonian for left-movers and Hamiltonian for right-movers. Independently, the left-moving Hamiltonian and the right-moving Hamiltonian have to be 0. So these are the two conditions, Hamilton and left-moving Hamiltonian and right-moving Hamiltonian, 0. So let's remember what all these contributions were. This mass squared term, so this mass squared was minus p squared. And it was simply the contribution to the Hamiltonian from the kinetic term for the 0's. It's to say, this is the kind of Hamiltonian we got in part, the mechanics. But the reason that we have the same Lagrangian to the 0's is that we have the problem. OK? The factor of 4 is fact that we divide the Hamiltonian up into a left-moving Hamiltonian for the part that there is a factor of 2. Fine. The sum to the right-hand side is new. OK? The sum to the right-hand side is new. And it corresponds to the contribution of the Hamiltonian from the oscillators. Remember that in both for the left-movers and the right-movers, we have a sequence of oscillators of frequency 1, 2, 3, 4, up to infinity. n and n tilde are here a short-hand notation, which denotes the weighted occupation number of the oscillators. So let's write that down. n is equal to the sum over little n, nn. And tilde does equal to the sum over n, n, n tilde, where nn is the occupation number of the nth left-mover. Similarly, nn tilde is the occupation number of the nth row. Right. It's a little bit of a system of harmonic oscillators. So omega, the frequency, and also positive. OK? So what we get in the end was a bunch of oscillators, times functions, 26 variables, and a constraint composed of the system with a constraint being Hamiltonian 0, which gave us this equation. But this is new physically. This part, this p mu v mu, is implemented on wave functions by differential operators. So if you write down a wave function, a wave equation, you write down this constraint as a wave equation on states, you will get an if, for a moment, be concentrated on states with particular eigenstates under the occupation of this. What you would get from here, which would be the same thing here and here, by subtracting these two equations, we come to the conclusion that the only allowed states are those for which the left-moving occupation given this left-moving level number, n, is equal to the right-moving coefficient. OK? So that's our first conclusion. And does this put an n-pilter on another state? And the second conclusion is that when this constraint is satisfied, the wave function of our system, the wave function of the system in any particular state with a particular value n and n-pilter is the Klein-Gordon equation for a particle of mass and square n is equal to, now let's get everything to this side, so 2 by alpha prime into 2n minus e minus 2n, 12, which is the same thing as 2 by alpha prime into 2n. Because this equation simply becomes p squared plus n squared is equal to 0 with this particular value for the mass, and that is the Klein-Gordon equation for a mass of particles. Look at that. So what's the physical interpretation of this system? The physical interpretation of this system is clear. What we have is a bunch of many different particles. How many different particles? Well, the particles are one-to-one correspondents with the occupation numbers of this infinite number of harmonics. We have occupation numbers of left-movers and occupation numbers of right-movers subject to this one constraint, only these are the only allowable units, that's it. Minus e minus 2 by 12 times 2 by 2 by alpha prime, and that particle moves freely in space. It remains the Klein-Gordon equation with that mass. Occupation numbers give you different particles, which could be clear different mass. What is the largest mass of the system? At this level, there's no largest mass. It's not as large as you want. Occupation numbers are only constant. It doesn't all come from trombones. So we've got an infinite number of particles of arbitrary high mass, or two-legged around with each other. Well, at the level that we've discussed things, these particles are free. They're not interacting with each other. But of course, string theory is interesting because we can free ourselves of this frequency, and things will switch out. So the quantization that we perform in this intuitive manner has given us already an interesting result. Why do we require this system for the system of particles, of free particles, an infinite number of them, of arbitrarily high mass, all of which are made of free Klein-Gordon equation? Is this your question of what the highest mass is in the system? Now let's tell them the question of what is the lowest mass of the system? We want the smallest value of n squared. Yeah, the smallest value of n squared because it is equal to n tilde, that's equal to 0. None of the hominons in this R are occupied. In that case, we get m squared is equal to minus 2 by alpha prime d minus 2 by 2. This is one thing about this formula is the minus sign. It's just a fact that the bosonic string, this shouldn't just be quantized, it includes a spectrum, a particle of negative mass, at least one particle. Particles of negative mass have a negative because of that. Negative mass sometimes brings to mind the n problem, especially the word tachyons. Sometimes it brings to mind the exotic phenomenon like fastness like a potion. This is all wrong. You see, negative mass, in our last class, what it means for a molecule to have negative mass, you should think of the mass as the second derivative of the potential for the particle, for the high field described in the particle, about the value of field equal to 0. Generally speaking, I mean, if you have a stable theory, if you're doing perturbation theory, you have a stable minimum, okay? Then the second derivative is either 0 or, by most cases, positive. So as you have to move your field a little bit away, it wants to roll back to 5 equals 0, okay? The second derivative is the mass squared, as I said, and the positivity of that second derivative is the positivity of mass. However, suppose you're doing an expansion, you're trying to do an expansion perturbation theory about a potential which curve the other way, which is negative, negative second derivative. If you draw down the Lagrangian of the particle, truncated quadratic order, you would write down the Lagrangian of the particle with formally negative mass squared. Negative mass squared tell you that it's not. But it's telling you that the point on which you're perturbating, if you're writing down a perturbation theory in your Lagrangian, it's not a stable derivative. There's an unstable direction in which you move. You can see in this theory, if you're trying to solve the equations of motion for a client-ordinate system with negative mass squared, you'll find that at low enough frequencies, low enough wave numbers, at wave numbers smaller than mass. Our wave, whatever, inverse wave that's smaller than mass, you'll find that the solutions don't oscillate in time, but they'll exponentially grow and decay. This is a characteristic of an understatement system. Okay, so while there's nothing pathological, there's no faster than like, prouder than nothing, theoretically pathological about negative mass squared, it doesn't indicate that if you're trying to do a perturbation expansion of some sort of theory, you're starting about a particularly unfortunate point. It tells you that you've started about that point. If you want to deal with this, you should find the true minimum of the system if it exists. Move there, and then do your expansion about it. Then you have all positive masses and you're able to do a usual perturbation theory with no internal divergence, okay? So, we're clearly part of an unfortunate and undesirable feature, immediately, not that we have to become extra. Now, namely, in fact, we discovered that we've got one ounce, at least one ounce, in that direction, and not in other directions, okay? This theory, we've been unfortunate enough to expand this theory wherever it is, you know where it is, about an unstable point. Now, this problem, well, the question of whether the Bosonic string of theory here describing here does have a good stable point about which to expand, or just some signal, you know, or the maximum amount of water that's potential is a fascinating question, which has not yet been resolved in my opinion, clearly, one day of the year. Okay? However, so, this theory that we've understood how to quantize in class so far has not proved useful for physics so far. It's just some forward thing. However, that is another system. There is very closely related to this. A theory, a world-sheet theory's quantization is very similar to the one we performed, which includes in relation to Bosonic theories and the world-sheet of the string. So, fermion. Okay? In that system, this problem goes away. Okay? So, all the physics that is done with string theory and in the Beethoven string theory is done with the fermion string. The string theory that does not have to be. Okay? However, dealing with that string is technically complicated by the fact that you have to deal with fermion to the fact that you have to introduce notions of supersedentry, because the string only on the well, the actual world-sheet string is supersymmetric and sort of un-matching to talk about the irreducible introduction to supersedentry. Okay? So, in any course of string theory, you're faced with these crossroads and different times of teaching this course have chosen to go different ways. But this time we're going to go along the route of bunches, okay? That's the way to say, look, believe me that when you grow up a couple of months from now or three months from now, we will tell you about the string that doesn't have this problem. If you can't do real physics with this system, allow me to forget about that fact and go on with the system, because you will learn in a simple, a technically simpler setting, many lessons that will be useful that you will be able to bodily carry over into the more complicated setting, where you have a string with fermion to the well-sheet and you have no, no technicals. All right? Actually, I've never thought of the course this way. Partly because my name is a student I can't really answer this, I think. But we want you to experiment, right? So, this is the way we're going to teach. Now, we're going to go this way. Again? Sir, regarding whether we have chosen the route vacuum in the bisonic string theory. Yes. Can we really answer it without going into string theory? I mean, from remaining within the range of string quantum mechanics? Here's one way to try to answer this, might be using renormalization group techniques for the well-sheet group. You know, the most, why string theory is very useful in the study of open string, maybe. It's a tack-tackinom compensation in the open string. It's not very useful so far, with the study of a tack-tackinom compensation for closed string groups. The techniques are useful there for different reasons. The techniques that try to go off-shell by using renormalization group for techniques for the well-sheet in the string, you're purely first one, I say, techniques. Okay? So, I would say that that is, you know, for instance, the Alex Kulchinsky-Silverstein paper that described the decay of a cone still. Let's face it, use that technique. Now, I would say that the techniques are not bad here. There's no conceptual barrier of techniques. I think first one, I say theory might well be enough. It's just that nobody's come up with a good way of addressing these questions. It's very possible that the was on-stream, it's just itself, you know, just by itself, I say theory, okay? On the other hand, you know, as people say, we're going to go for months studying this theory, and apart from that, we find no other problem. Somehow we find it could be that it plays into the game in some way, but basically, you know, nothing's not about it. All right, good question, now the question is on it. Okay? So I'm going to ask your lead to discomfort with the fact that we're working about an unstable minimum, and let me continue for the next two or three months, and then we come back in here also. So, we've got a lowest mass particle where it just indicates an instability. Okay, the problem is also with the net. So let's say n is equal to n tilde dot equals 1. This we can do, remember, because we've got it, how many times we've got an equal frequency, what? So we can accept one unit of that by half. Okay, so firstly, let's count how many such objects we have. Well, we had d minus two options. Remember what we could do. The what state is this, this is alpha i, and we call this minus one on zero. Alpha tilde of j minus one. Remember our notation? This minus one was the lead of, was we, minuses were creation operators. One, two, three, four, they were the frequency of the oscillators. Okay, so I'm acting on the vacuum with the creation of one creation operator of frequency minus one of the left, of frequency one in the left, and one creation operator of frequency one in the right. Clearly this sets n equals one and n tilde dot. Clearly this is also the only way to get n equals one and n tilde dot, which one? Because you either have two more excitations of the harmonic oscillator frequency one, or one excitation of the harmonic oscillator and you have a frequency, you would go about it, which one? Okay, so if we go to the next mass level, that's what is that mass, that means 12 is equal to two minus d minus two over 12. We find that the number of such particles, the number of particles that we have here is 24 into 24, because each of i and j run over 24. d minus two into d minus two, sorry, d minus two into d minus two, z over, each of i and j run over d minus two values. Okay, we've got so many components of the particle here, or so many fields for a particle to discuss. Now, there seems to be a more, if you look at this answer, there seems to be a more serious problem with this answer than the problem of the instability that happened. And this problem seems, this problem, this problem that we are going to look at it, is the following. This particle, wherever it is, here it transforms under spacetime rotations because it has a vector and then it spans another vector. It's supposed to be one of the non-zero mass. So we come back to that in a moment. And now, the problem is the following, you see. It's a general property from group theory, from the study of the representation of the Poincaré group, that massive particles transform in representations of SO d minus one. Is this familiar to you? Is this familiar to you? You know, basically what I'm saying is that when you get the wave functions of a particle, you have both its momentum as well as its spin direction, so the momentum is zero. And in that frame, the part of the Lorentz group that's unbroken by this choice of frame is rotations. So rotations must be expressed. You know, this particle must transform in some representation of this remaining unbroken rotation group, which is SO d minus one. Now, how many of you are familiar with this regional theory of representation of the Poincaré group? So only a little bit. Oh, a little bit. Okay, this is something I'm not going to teach you. The teacher can't even know. But it's something I strongly recommend to you about. Okay, representations of the Poincaré group, okay. Which tell you that, in general, for a massive particle in these dimensions, it must transform in representations of SO d minus one. Okay, this is in four dimensions that's probably the analog, the massive particles, transform in representations of SO d. This seems like a problem, because these states clearly do not transform in representations of SO d. Okay, so let me work the problem by suppressing the fact that there are both left povers and right povers. Okay, let me think of just the left povers of our world. Okay, you've got all jets. You've got 24 different states, d minus two different states, each of which clearly, and these states clearly transform in a vectorial representation of SO d minus two. Is there any d minus two-dimensional representation of SO d minus one, such that it includes all of the states transform as a vector of SO d minus two, and there are no other states, and clearly the answer's no. The nearest thing to this kind of representation is a vector of SO d minus one, which, in addition to these d minus two states which transform as a vector of SO d minus two, include a state which would be similar to SO d minus two, but is needed to complete the model that is a full vector of SO d minus one. Is this clear? Good. That's some big SO, okay? And that breaks us up into one lower volume, and so this is one, and this is d minus two, d minus two, and this is one. Okay, we've got, we've looked at, let's look at a familiar representation of SO d minus one, namely the vectorial representation. How many elements does this have? That's d minus one. Okay, this d minus one, I'll just give you a follow-up on this problem. All these d minus one elements with this choice of breaking SO d minus one and the output of SO d minus two of the state. These guys transform as a vector of SO d minus two, but there's one more element in the representation. Now under SO d minus two is a state. Is this clear? So there is another representation of SO d minus one that includes that when you, when you break it up into representation of SO d minus two, includes the vectorial representation. But, you know, but this representation has at least had exactly one more element. Okay, and of course if you think about it, you will convince yourself that there's no representation of SO d minus one that has 24 elements, so that has three minus two elements, or all of which transform in a, in a vectorial representation of SO d minus one, and d minus two. The symmetric between the d minus two dimensions and the remaining one dimension. Okay? So this confusion is not changing when you can insert this thing with something else from the right-hand side. So we can prove that the states that we've got at level one do not transform in representation of SO d minus one. Now this is a direct contradiction with some general theorem about how particles in d minus one dimensions in a Lorentz and that is clearly d minus one. You know, in this space time, that dimension should behave. And so it looks like a contradiction that's more serious than incident. Looks just true. Like you're getting something wrong. And you might ask, is there any way around it? What I said about particles transforming, representing d minus one, is true for massive particles. But it's not true for massless particles. You see, the intuition is that why is it always possible to get to the rest frame of a massive particle? That's not possible for a massless particle. The best you can do is use the red transformations to make sure that the momentum of the massless particle is in one particular direction. Let's say that d minus one is the direction. Okay? Now once you get this choice of frame, the rotational part of the symmetry that you have left is x to the d minus two rather than x to the d. Is this clear? Because time is special and the x direction is special. The other direction is all you got. So, and this incubation is if you're trying to be correct, we know as gender analysis tells you that massless particles fields for massless particles transform and representations of x of d minus two rather than x. Okay? So, what's going on is that this particle at level one is massless. If that were true, all of this would be consistent. In fact, how would that be true? Well, it would only be true if 24 was equal to d minus two. So it would only be true if d was effectively move on here. We say, well, let us before thinking it doesn't even make any sense unless we have massless particles. Perhaps we can be, you know, we can respect your own invariance and be in us there. Okay? But you might think now we're very simply so. Because you might think, well, the way we've done things we always have things that are manifesting representations of x of d minus two but you're going to have a hard time building things up in representations of x of d minus one. You've got out of it, you've wheeled it out a bit at level one by saying this guy is massless but unfortunately if you're massless at level one you can't be massless at level two. So what are you going to learn in the next? So let's see. Is there a question here? So let's work with level two. So now we're working with n equals to n plus n. Let's list all the states that we have. Okay? I'm going to just list the left movers and we're going to answer this whatever we have with the same thing for the right movers. All the groups here, everything works sector by sector. So let's just list them. But firstly, we have alpha minus one i alpha minus one j. We also have alpha i minus two, thank you. We've got everything in representations of x of d minus two. We've got a miracle of the transform representation of x of d minus one. Let's see if this miracle is in view. Is it true? So let's look at a particular representation of x of d minus one. Namely, the symmetric traceless representation of x of d minus one. It's elements are matrices are d minus one and d minus one are dimensional matrices that are symmetric and traceless. Because unlike the representation we looked at before it's a bigger matrix representation and breaking out into representations of x of d minus one. So d minus two. Now, let's see. Bigger matrix here. So once again we take this matrix and break it out into one and d minus two and one and d minus two. So this is the d minus two times d minus two matrix. Is this correct? In the full space remains symmetric in this space. So this thing here is a symmetric matrix of s of d minus two. This is not traceless, not necessarily traceless. Traceless representation in the full space tells you that this element plus the traceless is zero. Okay? By agreeing to take, you know, so the trace is not zero but it's related to what's happening here. It's not an independent degree of freedom. In counting representations we can remove the trace if we agree to count this element. Okay? So the question I'm asking now is if you have the symmetric traceless representation of s of d minus one what are the representations of s of d minus two? Does it break out? So what do we have? Well, we've got the symmetric n star of s of d minus two. Let's remove its trace. So that's the trace, the subtlety of the trace on this side plus we've got these rows in these columns. Because that's going to make it so symmetric, they equal each other to become the equivalent. So the net conclusion is that what we've got is a symmetric traceless representation, a scalar and a net. We've concluded that the traceless of d minus one is equal to sin traceless and the string traceless of d minus two plus vector d minus two plus scalar of d minus two. If you want to, let's say that the counting component set the component counting matches the formula you're going to have to verify is that d minus one into d by two minus one so that symmetric traceless of d minus one is equal to d minus two times d minus one plus d plus one which I can't do in real time but it's a reasonable case. And this is plus d minus two and then maybe I can. You can take this out of the common but let's get this straight. That's the tricks. This is d minus two d minus one by two plus a vector so that's d minus two okay. Okay. Let's just get this right so since we're starting here. Ah, thank you. Thank you. There's also minus one here. And that is it right? Let's just verify this. So what is this? So this is d squared minus two d minus two over two and what is this? This is d squared minus three d plus one plus two d minus four oh, this is plus two four over two this is d squared minus three thank you. So that's the answer. Okay, sorry, sorry. Why am I saying this combination? You see under s o d minus two we have precisely a vector a tensor which can be thought of as a symmetric trace instead sometimes the trace. So the representation content that we have under s o d minus two is precisely at the source basically compiles into an s o d minus one. Since this works for left movers it also works for right movers and therefore it works for the tensor product. Every two, there's a potential problem which actually is not a problem. The group theory is okay. But if you guys want to take a video of assignments for this purpose that's a way to formally check. Okay, so certainly I would advise every one of you to check for the levels. Check for the levels three. Check for the levels three. There are ways to formally check but this continues to happen at every level. Other than that. There's one cheat. The one cheat is not actually a cheat because but it fixes something you didn't know. It fixes what I mentioned before. That is that we're working with this consistent with the variance in diagonals and the group theory. But only if we were working in the equals matrix. And the reason we need that equals matrix from this one clue was for this one little thing there is states of one particular level in the first level would bother the random variance unless it was true that they were asking. Okay, now you know you might ask where in our procedure where in our quantization procedure where in our quantization procedure you know what's logically necessary to ask for equals 26. See it's a little unsatisfying to follow a procedure which you think is okay and then finding parameters in that procedure not on the basis of logic but on the basis of the answer being sense. It's a little unsatisfying. We will answer this question in the next four or five lectures but in order to answer that question we have to do at least a slightly more serious job with the quantization. If you very love to answer that question immediately. You remember that in this procedure of quantizing the string the place we got this minus 1,12 included a physical assumption that was the assumption of conformal invariance of the worksheet question. That was the thing that uniquely fixed that was the thing that uniquely fixed 1 plus 2 plus 3 plus 4 equals minus 1 okay now the conformal invariance that we have wanted was a consequence of symmetries that we wanted to have wanted to keep true in our quantization. These symmetries were re-triometrization invariants to watch the string and winding invariance on the worksheet question. So why these two symmetries are only maintained provided that provided that 1 plus 2 plus 3 plus 4 is equal to minus 1 over 12 these terms are not to be the only condition. If you do a serious job analyzing the path index of this theory and asking the violin invariance and the detromophist invariance and the worksheet of the string remains an entry of the theory. You require that but that's about what we will see in the next few lectures after we understand the performance. Okay, but I just wanted to assure you that there is a systematic procedure that if you do it more seriously you will see where the procedure thinks great about unless it is equal to 26. These those 26 has another more satisfying derivation which we will get through in our previous optimization of the string. Okay, and now let's immediately look at and look to see what the consequence what the result is. So we have got this one unstable particle which we decided not to bother with but here we should think that's a master's card. Master's card is always interesting always a really interesting physics. So let's understand exactly what the master's card is. So the master's card remember were in the 24 times 24 of the vector and the vector is a 24. Now we want to take this vector time vector and decompose it in irreducible representations. In some way you are asking it's how we decompose vector times vector and it reduces the representation of any SO group. Very good, very good. Symmetric tensor traceless symmetric tensor antisymmetric tensor and the trips. Okay, we have got a traceless symmetric tensor but the antisymmetric tensor has a name which is called between union and we will encounter it in many many devices as we go along master's table string theory. The trace also has a name it's an important field you would know as you go along master's table string theory but the thing I want to focus on at the moment is traceless symmetric tensor. The traceless symmetric tensor is very interesting it's very interesting because if you take understanding gravity and linearize linearize different small fluctuations and you ask in what representation physical excitation physical excitation so the chronograph stops forming in the dimensions. The answer to that question is in the traceless symmetric tensor representation of SO group minus 2. Okay, so the fact that we've got a symmetric tensor floating around the place the fact that we've got a symmetric tensor floating around the place is a tantalizing indication that we maybe a master's symmetric tensor that's very crucial is a tantalizing indication that we may be dealing with a theory of gravity maybe more than an indication there are various claims levels of how much you should believe these claims out to the reader but there are various claims the most systematic of them being by Feynman in this book by the lecture of gravitation there are various claims by various people in most systematically by Feynman that if you want to study that any theory that includes master's symmetric fields and makes a consistent unitary quantum field theory must be a theory of gravity in what do I mean by theory of gravity in the sense that it must implement if you want to do that let us for our generalized purposes any theory that implements a few more ways of invariance as theory of gravity the various claims floating around that say that if you have they only by the way of consistent quantum theory of master's fields in master's tensor fields is by making it a theory of gravity let me give you a one minute introduction to the idea behind these claims the idea behind these claims is very similar to the idea behind a similar likely to be true state in a lot of factor theories a quantum field theory about vector fields a theory involving a theory economically quantized when you use the quantum field quantization of the vector level in fashion you get a new data you get and you define a new dagger as the object that I had it's vacuum so that's great you get a canonical computation relation a new dagger equals e to the mu it looks good when mu are spatial indices this is a standard computation relation when mu and mu is done these are the wrong computation relations it takes creation is equal to 1 rather than creation with that addition and that leads to all kinds of problems basically conflict with unitarity and boundness of the energy spectrum you might think that it's really hard to build because of this problem you might think that it's really hard to build a consistent quantum theory with master's electromagnetic fields with master's vector fields and the fix to this turns out to be the fix to this turns out to be and it is hard to build a theory which is based on the granted del A mu the whole thing squared is the sense but there is fix to this and that is introduce a data connection to the problem which allows you to eliminate the kind of like the degree of freedom this is the use of maximization you eliminate the kind of like the degree of freedom then you have only states of positive norm and you have no trouble when you look at a theory based on symmetric tensors because if you look at a theory based on symmetric tensors you've got t less a mu nu by eta mu alpha eta mu beta plus symmetric these are perfectly sensible in combination relations but everything is spaced or actually in this case both are done because there are two minus side states but when you have one space in one time you have trouble with the wrong side of it in combination relations and in order to cure that you need to gauge invariance once again something that will allow you to eliminate the bad degrees of freedom these arguments are trying to show that the only way to make sense of a massless la la la theory try to argue that the only consistent gauge invariance in this theory is certainty of offices which is an actually symmetric now I don't want to go on with these general arguments very much because who knows whether they are true general arguments or make lots of assumptions assumptions might be violated however the basic idea seems flawed seems hard to make certainly there are no examples of theories interaction theories interesting theories with symmetric fields that are massless and that don't help if you are office men therefore not taking this crap the fact that the Bosonic string has massless symmetric fields is a stronger indication than you were in first class that it is a theory this is much more detailed than these words we will see as we go along in our study that we will report by building this theory in practice supposing this is probably support without supposing it tells us that whatever we see as quantum mechanics these theories are mostly theory does that violate belzing qualities I mean I think that's I mean I don't know that much about it either it's not easy to reduce quantum mechanics by any classical theory right I mean there are these sharp inequalities that protested experiment in any classical theory you get something less than this the quantum theory violates the bound it's not I'm asking this because we are most oriented for the actions they don't know if we read classical yes yes yes what seems to work what seems to work I mean the classical well any quantum system which is weakly enough which cross more than half is approximately less so that's true but it's not spoiled the fact that it is quantum you know quantum physics is fantastic in quantum view it's really well emphasized by these lectures of Feynman gave Feynman lectures of physics in the first three chapters in which he tried to track down which form the bullet is it's maybe we can make this weird but at least quantum mechanics is very different rather important way also as Feynman those lectures emphasize if never consistently in the quantum theory can have half of the world quantum that's never consistent can somebody tell we'll take two minute break to answer this question can somebody give me an argument a simple physical argument tell me that's not world would be consistent if there was anything classical let's be clear if gravity was classical anything else was one could this be consistent it would make sense I mean at all length scales there should be you put it in certain length scales but we want to shop around I mean if it has to be classical then it has to be classical we answer the difference in the refinement good can you sharpen that basically we can make that that's exactly let me just say that let me just say that more in the context of given experiment experiment Feynman told you he told us about those lectures if you remember he was dealing with double slit experiment he was dealing with double slit experiment and he was throwing I don't know what electrons and this slit and it was funny that if he had one slit he got a curve like this another slit he got a curve like this should do the same thing but if we had both of them open he got a curve like this like of interference he was very puzzled presumably the electrons either go through that slit or through that slit presumably the electrons go through this slit and this one is open at the moment so how can it be that what you get with both slits are open is different from the sum of what you get where each one is open and then he tried to track it down with an experiment let me do the following let me do both slits open but see which one the electrons go through if I do that it must be that I get some of those that go through the first plus those that go through the second okay so how do you try to do the same try to do that with the photon and shooting shooting these photons at the electron and then imagine the example the difference is directly between the photons can you see which one the electron went through but then he had a problem he had a problem because if he chose the photons to be too small at the moment then the kick they carried was so strong that it could take an electron that one headed there and make the head go but strong enough to kick the electrons enough to square out the individuality on the other hand the photons too large because of the big waveguide of the photons you couldn't detect which the direction was not sensible enough to say where the photon was coming from from there if you try to make the photons large enough so that you didn't kick the electrons enough to square out the individuality you made them so large that it couldn't detect which electron which electron came through so as a function of photon size as you took the photons very small you would be able to see the electrons you would be able to see which electron came through and you would find then you make the photons larger and larger and larger your resolution of where they came through becomes lower and lower and lower and the pattern becomes the other way around and it lives on the age of not making sense that's very easy finding a cure in theory now suppose these photons have been classical then you see you can make the photons very small and yet very small energy make them very small to detect which still the electron came through maybe a very small energy you would kick the electron almost not at all so on the interacting with classical photons it clearly wouldn't have made sense at least it would have been greatly improved everything I showed the photons to a good level it just wouldn't work you had some particles that were classical enough precisely because you were invited to be on set so that was a philosophical side better good other needs for quantum math any other questions or comments before just a mini round one we completed a little quantization of the bosonic strength and we've seen the result we've seen that we had master's and two particles so presumed we have the answer now we're going to go back and do things more systematically through a more detailed study of quantum theory in terms of the world-sharing strength that we can form in quantum math and maths free lectures relatively faster study of quantum okay so let's go with the quantum theory and much of what we do in this course which is quantum mechanical we'll be done following questions from now on we're following questions in chapter 2 so perhaps not at the order of the questions in chapter 2 which are kind of very cognitive okay okay so the the study of quantum theory and much of the other quantum mechanics we'll be using the path detector for quantum mechanics okay so partly it's warm up in using path detectors but largely because we need the results and now we need to use path detectors to derive two or three results in quantum they have a lot of what classical theory are related to symmetries the first thing we need to do is to derive the novel theorem okay and using the path detector for quantum mechanics oh what a waste of space okay so first of all you know what's the path detector formulation but you know there's some partition function defined by an integral over fields by interaction in quantum mechanics those fields are functions of time only in quantum field theory those fields are functions of both time and certain numbers space dimensions we're working in one plus one dimension so our fields will generally be functions of time and one space dimension interesting objects that you compute in quantum field array and will be of great interest to us as we move on in the course as you can see correlation functions that is objects of the correct form firstly let's define terms x1 and x1 and x2 are positions on the field theory space which in this case is two dimensions I often use the symbol x to denote both the space and time it's x me okay now what are these O's these O's are local functions of the fields okay so over on operators local operators for instance the fields of the theory as will be the case in the both of our experiments that we studied are x nu of sigma1 and sigma2 of sigma2 and you can make many local operators out of x nu for instance you can make the operator where x nu itself is an operator you can make operators like there in other form x nu acting as a given point in space or you can be given by ik at x at that same point in space functions of the x's that depends only on the value of the x at the particular point of interest and is immediately on the fields and there because of the fields of that in case of correlation functions for the action of the stream will be very important to us okay so please go define it we are going to try now to something that I will often do in this course is work interchangeably with the Lorentzian and Euclidean formulations okay you might find me switching this is to minus s in which case I am working with the Euclidean action when we actually move from Lorentz to the when we actually need it for physics when we actually use Euclidean action for the worksheet of the string to an order of calculation we justify this very carefully why the physical calculation that we want to do which is a Lorentzian calculation can equally be well done in a Euclidean setting but until we get there formal purposes the manipulations I am going to do will sometimes be Lorentzian sometimes it won't matter so for manipulation purposes for formal purposes I will often work in Euclidean theory in order to answer physical questions we must be very careful about what we are doing we always do that next few lectures and next five lectures will be about so in formal theory we make no string theory we look at the application of this to the study of string theory only five lectures from now so we will be hard in the next five lectures to answer questions like where will this go in the string theory or why is this correct thing to do on the point of view of string theory because we will get that out of the equation so this is just formal exercise for studying the kind of quantum theory of course the reason we are studying this is the kind of quantum theory that appears in the equation so we will be interested in these objects until derive some well known but anyway in case you haven't seen this way of dividing properties of these correlation functions imply by the existence of symmetry imply by the existence of symmetry in the theory ok first we are going to have to define what is symmetry that takes this form now suppose you can find a variable change as the sub function of x prime such that under the variable change the part of the integral takes the same functional form when expressed in terms of x prime and it took when expressed in terms of x that is then if you just make this variable change into this integral it turns into the integral x prime at exponential of minus s what I am talking about here is functional form because all I have done is a variable change a variable change can't change the value of an integral of course whether you make a variable change by symmetry or by anything else the value of the integral hasn't changed but most variable changes will change the form of the integral if you can find symmetry in terms of integral changes you have changed the form of the integral it is the same function the integral doesn't change such a variable change is called a symmetry of a variable symmetry is interesting symmetries are interesting because they are associated with conservations and related to water ok so let's try to understand we have been interested mainly in continuous symmetries symmetries depend on one parameter but when this parameter is taking the zero then it takes you to the identity so that's why we can take the value of this parameter to be very small so infinite as milling our variable change changes the form and our x is equal to x prime plus epsilon times some g so in particular it will make this variable change this is the when you go to z as z equals to x prime where what? there is a symmetry if there exists a variable change such that if a change makes a variable change the integral it does not change that's what it means to make a symmetry a symmetry given an arbitrary path integral that may or may not exist symmetry because variable changes are identity something is equal to itself always exists, that's true so that's the value to change the example that takes that form okay, now one consequence can be drawn from path integral using the path so if you have same path integral you could be able to absorb the integral so no, this is symmetry after you have taken care of the jupovian this result is true so once you have reassured that you have taken care of the x prime what you have here should be equal to a path integral there are many ways in which you could have could be the jupovian one and the action is the same could be that the jupovian is not true the action is not true, changes but the two can see both forms are interesting again see, important mechanics all that is important is the path integral the actions that between measures separately are not distinguished okay, good now let's move on so let's make an infinitesimal variable change of this form if our action was if our therapy was if we have a symmetry then making the cerebral change we get the same we get the same function in terms of of x prime okay, that hasn't drawn us very much yet but now we will do the following let's take this epsilon which was a constant in order for this thing to be a symmetric maybe in a function of inter-equivalence okay, so let's for all the variable change x is equal to x prime plus epsilon of let's call it sigma alpha g of it now what happens to the path integral under this variable if the epsilon is constant making the variable change simply gives us the x prime exponential of i and s of x prime okay, let me do the argument always be pleased with this that there are fewer i's okay, e to the power minus s of x prime okay this would be true if epsilon was constant where epsilon is not a constant because that is not necessarily the symmetry of the action we will get the change but that change must be proportional to a derivative of x to reduce the fact that when epsilon was a constant we got no change so let's say there are a function of epsilon and some function j of alpha what if it has two derivatives a function that has two derivatives of epsilon and the answer is well, if it has two derivatives you can divide the derivative by parts for one of the derivatives instead must have a he's one there you might also ask well, why is this exponentially divided why not the numerator how it says the answer is the first order in epsilon which is where we are working is no difference you have to put things wherever you want so this is what we have got upon making the derivative a change in the action in the partition function but of course since it's a derivative the change is a value of the partition function has to change so this is the same thing as integral to the x exponential of minus s of x the original function for the partition function these two are as an expression let me change the change to dummy variables and call it x prime equals x and then subtract these two expressions so we get the equation we get the equation dx exponential of minus x minus s of x we have a one let's expand this leading order of epsilon so we have one plus this business so the one term there is one answer that that business survives the integral of L at alpha epsilon J alpha of x is equal to zero let me choose for epsilon and therefore we conclude that the expectation value now that's integrated with my parts so we get epsilon here at alpha here with a minus sign here we can choose the epsilon to be anything we want delta alpha that tells us the expectation value of del alpha J alpha is equal to zero so we have a first conclusion that everything we have a symmetry inside the part that implies the existence of the observed charge we've not fully proved that we've only seen that it implies the existence of a charge that as it comes the expectation value vanishes expectation value of divergence vanishes because we want more but is this all the reason just one question or is it kind of confused? no, it's just one question we'll see exactly how in the quantum theorem the classical angle of the statement certainly is not valid if you violate the equations we'll see exactly how on-shadow the statement we'll see the statement but as we said the thing that we were often really interested in is expectation values of various operators so the thing that would be more useful to have is a statement about what then are the five J alpha that's four one four N so what is this? that would be the more interesting statement let me answer this question well, let's try the same manipulation as we did above but by inserting some operators into the problem