So, in the last class, we will discuss in this funny book of my third book, I'm making the gauge theory passing table, better defined in the continuum limit. And we come to the conclusion that the passing table, the area over the volume of the gauge group, this thing that we call the tecton d omega, could equally be written as d m m d v d c, where v and c are both at joint value, both at joint value and t. Okay? The exponential of i s, okay? plus v delta f, okay? times delta h. At joint value, we mean there is one, there is one b or one c. You know, they have these indices, they are the indices. And these indices are in one-to-one correspondence with the indices of the reaction. Okay? That's really... Okay, I shouldn't have said i joint value here, right? That's misleading. Just that there are as many f a functions, and so there were elements of the group of the reaction. Okay? So guys, you remember what the purpose of this f a was? f a was an object that was going to fix the kj values. f a to the zero gave you one point up in the orbit of every k group. Okay? Now, how many kj transformations could we have? kj transformations were labeled by group elements, or infinitesimal elements, by liangular elements. So if we're going to fix these kj transformations, we need as many conditions as there are group elements. So we need all functions of spacetime to be set to zero, where all is the i complete. Okay? Number of generators. Is this clear? So the u are theory. We need one condition, because the i is one. And SU2 theory, we need three conditions. If the i is three, if the i is three functions. And these indices here are in f. Okay? This index here was a... Which condition is next? This index here was a kj transformation. Because the number of conditions was equal to the number of kj transformations, this was a square difference. That makes sense to talk about this. The determinant. And this determinant, yeah. So what I really meant is that we are just indices, labeling the generators of the generator. Is this clear? Yes. In what? In electromagnets. We have the visibility go, go, go, go, go, go. Yes. But then we still have... Yes. But you know the rest of your gauge transformation is a function of fewer variables. Okay? Okay? So on this thing, you know this counting of how many things is a bit imprecise. Because what you do is counting functions. Okay? So in that counting, if you have a function of four variables, compared to that any function of three variables is counted as zero. Now, why should we think that way? For instance, take any function of four variables as a data expandant in the fourth variable. There are infinite number of coefficients. Only one of them is the zero. When I'm doing this counting, I mean counting as a function of four variables. It's a bit imprecise here, because you have to regulate them. But roughly so. Is this clear? Any other questions about this? Any questions about the procedure used again? Was it clear to everyone last class? Let's move on. So let's... Let's study... Let's move on by studying one example of such a gauge transformation. The example we studied is... is the one that was just revealed. But it's to be studied again. For instance, let's study the example. Where f e is equal to b mu, a mu. Again, why this condition? Well, this condition is sort of nice because it's Lorentz variant. You can imagine many other conditions. You could say it's at a zero. A zero, a equals to... It's sometimes a useful condition. It's called temperament. Or in a way, it's some things you can do. These other things are sometimes useful. In my own research, I've often found it more useful when you've got a non-trivial problem. If you use a non-Lorentz variant gauge, then it's Lorentz variant. But for the purpose of generating perturbation theory, this is clearly a variant of the equation. Is it preserved? This matrix is brought from the Lorentz variant gauge. All right. Now, let us now compute... And another question that you might have is, well, why not put, you know, d mu mu, whatever that means. So what that might mean is the following. You might think, perhaps it would be more than natural to put a gauge condition. Which is d mu minus i, a mu, commutative. Suppose we define d mu of this. This would be the covariant derivative when acting on an asterisk. Because the gauge field is not quite asterisk. It's the inhomogeneous structure of the matrix. But this is a part of a structural thing, an asterisk. You might think, perhaps this is a slightly more natural gauge condition. But why not put that? Now, the answer to this question is what? Suppose we did put this condition, how would it be working? The condition I just put. Is this the addition or clearance? Well, it's a variant, probably. Let me see something. Fine. Let me take... So, let me regard a mu as we put an a, a, c, a as we always do. And now, let me work out d mu a mu. Okay?  Just because the commutative in any matrix that says it's zero, doesn't make a trick of course. That trick lets us use either plus or minus. Plus or minus. Okay? So, that's no difference. So let's work on this condition. Okay? Now, let us try to be a bit concrete here. Let's try to be a bit concrete here. And to try to compute this, we have a closed section. By the way, these fields here are called closed. They're b-field and c-field. And sometimes conventional call being c bar. That's the answer. Why do we say that commutative in the matrix equals c matrix? Because the commutative in any matrix with itself finishes. So what you're looking at is the example of the fact that you can all... You can take an obvious fact and make it look confusing. But it's still true. So let's see. What you're saying is that no. Suppose I got some matrix. This is equal to bb minus bb, so it's zero. Now, b might happen to be sum over acad, da. But these are some other matrices. But you won't stop this from being zero. Okay? Okay. Other questions about... By the way, you could, of course... I mean, it was a good question. It was not my intention. It was a serious comment. You should always try to understand something. Look for the simplest way about it. You could work out that it's zero by expecting this one. And you'd find this one. But there's no meter. Okay? So, now let's look at this condition. Let's look at this condition and what are the closing actions for this condition. Okay? So we've arrived. Now, what was del Fabc? It was given by the condition that under-against transformation by epsilon b, delta Fb equal to this del Fab... del Fab... let's say I have a x is equal to del Fab x minus y, x plus b of y. This was our definition. This was our definition of our delta Fb. The statement is playing English. Delta Fabc is the coefficient of the transformation of Fb under-against transformation. No, no, no. Right, right, right, right. I just wrote that. X comma. Whatever. Whatever happens to it. Of course, we choose a translation there. Yeah. Yeah, an independent condition. What to do is to understand how age changes under-against transformation. Okay? But we know how age changes under-against transformation. Delta Amie is equal to it takes the covariant derivative under-against transformation. Of the gauge transformation term. Okay? So delta Amie is equal to Amie of epsilon. When I'm using the notation where I look at the whole matrix here, rather than its components. Now, what was our... By the way, look. This is at some position. This is at the same position. Nonetheless, this kernel matrix has a slightly non-trivial dependence on x and y, because there's a derivative.  This is not an ultra-local equation. It's not like the change in A and x depends only on epsilon and x. It depends only on epsilon and the root of x. But there's some logic at it. Okay? So from this, of course, we can deduce what this delta F will be. It will involve some covariant derivative or some delta function. Okay? But we don't really even care about what delta F is. What we really want is this action. How do we get this action? We get this action in the following way. What we do is replace epsilon by c. How do we have to do the following? So I'll give you the answer and then you see. All we have to do is is to do trace b d mu of c. That's the answer. I'm saying that from the fact that you know that delta A mu is d mu of epsilon, if all of the means you take of the course action is trace b d mu of c. Okay? Let's understand this. The only confusing thing about this is that this is written in with indices up here and I've written the final answer in some kind of mathematical explanation. So first, you don't understand this. Let's work this out in the indices. Okay? This of course is just delta A mu of A is equal to d mu of epsilon A. I'm saying nothing. I'm looking at the coefficients of t here. What's that? Epsilon here is some epsilon A t A group. So we just replace that by c A t A group. That's clear, right? Okay? And then of course we have to multiply by a p. b t A, I'm sorry. Suppose I call epsilon epsilon b t b. Then we just replace that epsilon by c b. Why? Because that's the rule here. This is exactly exactly, you see, what was the rule? The rule was that f, the change in f under a gauge transformation has this epsilon here. This operator has c in place of epsilon. So c here was just precisely in the infinite explanation. Okay? But what we're also supposed to do is to put a B A with this A here. But because it's easy we're doing that matrix rotation. Because in matrix rotation delta A A was equal to delta A A t A. Now, if I want to get B A delta A, the easy way of doing this is to take B, delta A twice. Because where we use the fact that we've chosen a basis in which the price of B A A t t is equal to delta A. Just substitute it. This is price of B A t A. This is delta A c t c. And then I use a place for it. About the epsilon zero, why are you putting here delta A in here? Why is this formula to be asking? No, that's fine. But then I would I would think that you found delta A and then put extra machine in that. Oh, I'm very sorry. Yes. You see I skipped earth, death, I missed the scale. Thank you. Sorry for that. So what you have to do is now find delta A. Right? But delta A was just a derivative of this. Yeah. Okay? So this is being you I'm sorry. You're absolutely right. I still have let me do it. Let me start again. Let me do it. Let's do some abstract stuff. Suppose I know some delta because F e is the definition for every attribute. Instead we can think of it as F is equal to 0 where F this is the same thing as F e equals 0 for all F because the coefficient of every T a must match. Suppose it's true that under a gauge transformation Lf something we call this something times X Suppose that any that the I think of determinant is simply B that's something that is C this is the first abstract is firstly you understand the claim you understand what I mean by secondly do you understand why it's true?  In order to find delta F I did a side step and then I forgot actually it's a compute delta F so I completed the delta emu was equal to d mu was equal to d mu at epsilon this we know to be true therefore delta F is simply d mu d mu of epsilon is this clear? So now what's our general rule? Suppose we have a gauge transformation like this and the fadi pop up to 10 we replace the epsilon by a C we put a B behind it and take the bridge that's what this formula is is this clear? and so what we have to do is we replace d d mu d mu of C in this particular case that's the fadi pop up is this totally clear? any questions about is the fact that this rule that this to this the same as this is that in the particular case of the range gauge now I'm clearing this out in the particular case of the range gauge pop and take rule becomes dA mu dP dC exponential of high ASA whatever the action happened to be in this case it's exactly what we do doesn't matter okay plus BAA plus B where we've got this delta outside the exponential of FAA of C FAA was what a generate perturbation theory it is useful not to have this delta function explicitly developed in this particular by the way if you were able to A0 equals 0 each then we would just use the delta function to send A0 equal to 0 okay but here since this is a non-local delta function in order to solve this this condition we've got to solve an equation in terms of some data that's not convenient for formal purposes okay so it's conventional to take this delta function and rewrite it in a in an integral representation so this delta function of course is the same thing as exponential D I think it's called if you belong here B capital B I think if you belong here you've got the capital E to the power of rA trace now we've got a path integral but at least it looks completely traditional it's like the usual kind of path integral the usual kind of path integral the kind we have for a scale of B local action after this only with lots of fields the price we've paid by the way is that we've introduced many new fields into it okay we've introduced the B field the little b field the z field and the capital all this was just a mathematical tip the original path integral was in terms of these a's the Hilbert space is entirely greater than the a's as we discussed it was as a mathematical tip to compute that original path integral which had a clear Hilbert space interpretation as we discussed and sometimes it's usually introduced more fields okay there is an interesting okay and let me say one more thing by the way you know there are more manipulations you can do here at least alpha we'll get to that very soon maybe we'll get to that very soon the essential ideas the Hilbert space of this you know there is an interesting way of regarding the Hilbert space okay now if you look at this path integral suppose somebody had to do a very deep just gave you the path integral and asked you to analyze the Hilbert space of this path integral sir all these fields which we have introduced obviously follow this algebra like the commutator algebra which we started off with so like can we somehow introduce some field which doesn't follow this I didn't understand what was the commutator algebra I mean for example like all of these fields can be written as B it can be written as D A T A so that T A T B delta A B algebra is common right so my question is like can we introduce some field which does not follow this algorithm that may be something important well you know we were we were introducing fields for at least wasn't just I have of course you know in this field I won't say no to that but I've just said that we want to do things just for fun right I mean we were introducing things for the reason that we were led to this now given any integral that means the number of manipulations you can do that will not change anything integral so there may be other interesting things to do I won't say no but now look suppose you just looked at this path integral okay you looked at this path integral and you tried to give a Hilbert space interpretation you would at first say okay maybe you would not give you would see of course that this B here has no kind of integral and so it somehow some sort of a large part of that field but you would be tempted to associate a Hilbert space to this problem you would be tempted to associate a Hilbert space to this problem in which all the A fields have promoted operators okay all the B little B and C fields have promoted operators so you might be tempted to take one thing the whole problem was was the Hilbert space was very functional of all four categories little B little C subject to one constraint that it promotes a B field and that of course is totally wrong okay because that was not the original Hilbert space the problem was not what we stopped what we do now that was totally wrong you would have seen many things okay you would see for instance that this problem if you try to give such an interpretation the Hilbert space would not would not get an interpretation with with you you would not be able to get an interpretation that was both where the physics was both unitary and energy boundary problems okay and there were many sources of this problem first is the A0 thing that we talked about okay the second source is the fact that we've introduced anti-community field is the problem even though this anti-community field is transformed under the lens transformations that's dangerous and if we will discuss in more detail in this course it's a big order there are very few general theories here about quantum field theory that not specifically about quantum field theory in particular but there is one simple general theory called the spin statistics theory it says that if you want to get a good theory you are going to raise certain conditions there is a connection between the spin and the statistics of all fields of physics now if you have anti-community fields they have to be they have to transform in spin or in representation let's go in commuting fields they have to transform in vector by which I mean a representation that can be built out of the entire products of vectors okay so this thing here while it's the spin statistics theory in real life and it's going to give us a little bit of spin control okay but there is an elegant way of restricting my Hilbert space that you get by quantizing this bigger path in general to a consistent sub-sector okay and that sub-sector matches with the Hilbert space that we were originally interested in and an interesting way is called PRSD quantization and we describe this perhaps by thinking with this class more like the beginning of the next class but in preparation for that we want to go away from these very slightly formal discussions so two other things more general things about the path of dependence and how that will spread okay so this would just be an appetizer we're going to move away from these formal considerations to slightly more concrete also formal anyway do you see anything about the nature of these de-feels, capital de-feels yeah, these capital de-feels are clearly important with that because they don't have kinetic okay and of course one could okay if one were naively doing this quantization okay and in a naivus procedure what you would do is to make a Hilbert space for all the other guys and then try to implement the constraint that you get out of as a constraint very much like a zero in a correct interpretation but you know that for many reasons but there is a correct subject which we would discuss today in a way of thinking very useful when we try to understand many things called huge states for normalization so we come before we get there preparation of a few other things that you want to discuss this BRC fields if all the chief fields were I mean get the operator status then in the Hilbert space interpretation I mean the states corresponding to BRC fields can we show that these are somehow unphysical or after we have to know that these we don't we shouldn't contain these fields in the case now firstly of course a priori we know but as we discuss some more they just gave me this thought I said try to make a physics theory that respects all the usual rules this unitarity is energy value from below by doing the 90th thing you have trouble as we discussed so you would have to do something in addition to this knife but this path of tetral is no good for business because if you work on the path of tetral you would get a Hilbert space that included B and C excitations but since we buy it it splits the statistics there and something will go wrong I think with a t-bar and a t-bar I am about to give you some or maybe both okay similarly with AC something is going to go wrong okay and the thing that goes wrong you know something is going to go wrong okay now way of thinking of this property I will describe this to basically by the end of this class but there is a way of thinking of this problem that identifies some sub-sector in this 90th Hilbert space which is isomorphic in every sense in every physics sense to the actual Hilbert space of the body namely the square and tetral by the way of functions of the eyes so if you are in the 90th thing you would see trouble but there is a way of fixing that even on it's own terms okay without referring to what we knew the answer to be we hang on with your questions after we have discussed the other things BRST quantization is the one setting question okay okay other questions to come on let's move on so so we are acting with this but the proper thing and this basically ends our first brush with the path in there we have discussed in some very formal way we have discussed in some actually a very formal way how every field theory we are evaluating down there in this course can be written in terms of a party also in semi-practical terms but before we proceed with further formal analysis you shouldn't already know the answer to this question which is definitely if you do some of you are wondering some of you would be wondering what is all of this you know this is a lot of formal stuff you pronounce them mathematics of course you said that the mathematicians were not accepted because it's not true with this so you pronounce them no books I mean so in this play around with these entanglers what can you do to use these entanglers what physical quantity of interest can be computed in part of it and of course the answer is maybe this is why but let's start with the discussion of this okay so now for the purposes of this our restarting the discussion of the part of the techniques let's go back to our original problem one particle okay and you remember we had this lovely formula we have dx exponential of i with the right boundary conditions a starting point in our understanding of part of the techniques the following suppose instead of this part of the table here I have a modified part of the table where I compute a part of the table weighted by the values of fields at various times about the table of dx because what is the left-hand side corresponding to this right-hand side you understand we have this right-hand side here which is an integral which computes the matrix elements of this illusion of that okay already useful okay now I'm going to just take this right-hand side and deform it in a way that from the point of view integrals are quite interesting we think of the action as a measure we're giving some sort of measure in some space and now instead of just computing the volume of that space subject to boundary dimensions we will compute average years of quantities weighted by that measure not the point of view of integrals is quite an actual procedure quite an actual thing to do question, do we compute anything interesting on the left-hand side do we compute anything interesting on the left-hand side that is procedure okay so by the way of course you know when I start on this question what are these quantities good for somebody could have said they go for computing transition amplitudes and that's of course true but he has even seen in our study quantum field in this transition amplitudes complicated and rarely of interesting themselves we are only interested in computing simpler things than transition amplitudes which is where this the transition amplitudes depend on some wave function of your wave function over there and don't die we have to compute that the wave just computes okay so now does anyone have a suggestion for what this would be of course all of you know the answers suggestion let's let's build this up you see it's very simple because you go back to the derivation of the part you will see that what's going on is the following okay firstly let me call this df-d just to make the formula of the function and then as you know this is the starting point of this we are thinking of part elections df-d i can be written as df-df minus df times df-d i dn df2 df-d i d1-d is generated by a part of the table of the sanctions of you know if I could complete set of states in between here we saw these different part of the tables so what we have done here is break up the part of the table into a part of the table over a segment from di-d1 d1-d2 each of these is a part and then in the full part of the table we have to do that in the tables x of d1, x of d2 and so on now to find the derivation of this table with these insertions there's a way to do it suppose in between these things I put these f's so suppose here I put f1 of x of f1 of x and in the next place I put f2 of x and then finally here I put fn then if you see that we repeated our path integral there which we will get the right hand side as the answer you see because in each of these places we have put it completely that integral in terms of full part integral was the integral of x at d1 but because we are going to complete this somewhere this guy acting on the basis vector will give us f1 of f x of d1 so maybe here f2 of x of d2 that's all so if we were interested in computing so this integral computes this object from here that's what it is any questions or comments the one who is computing takes the initial thread evolve to di-d1 so put this operator f1 of x then evolve to d2 put the next operator that's all questions or comments pretty ugly formula ok and ok how we need to change this to some size in this some time sir on the left hand side this extra operator on the right hand side yes what is that kind of eigen states not eigen states these are matrix free because some products are more critical expected no mistake so much is it there allow me to change this x to x to xf to size and die what we now have to do is of course add the extra combinations so this is not the integral in the right we have to do a positive integral also over the initial and final axis as we discussed ok just to make the discussion clear ok now so far we've been using what people call the Schrodinger equation when we discuss the left hand side then there's also this other famous picture called Heisenberg which is much less natural if you think about what we are studying quantum mechanics quantum mechanics because it's like quantum mechanics the most convenient way of thinking of quantum physics ok so let me remind you of the Heisenberg picture quantum mechanics first let me remind you of Schrodinger pictures Schrodinger picture we've got some states side as they are and then side t they're not functions of time x is the operator x it's an helbert space and that matrix for it this is the Schrodinger picture because it's conceptually clear as pretty much however Heisenberg noted the fact that what Heisenberg is doing is equivalent to what Schrodinger was doing because of the operating fact that what are you doing with the operator and the operator here you can be bothered by you're clear as much of time here right but what I meant is like these operators x in the basic the basis of it yes you can define operators but the basis operator 5 is 70 we have to define this kind of combination ok ok ok so there are some operators let's just talk about x x and b i if we're interested only in computing if we're interested only in computing matrix elements of operators then what the Heisenberg picture defines Psi Heisenberg as we put it by h times Psi Schrodinger and this definition is done to undo the time so Psi Heisenberg is Psi Schrodinger at t equals 0 and it's not a function of time the Schrodinger operator you know the same function of x is in this times by my Psi h this is done in such a way so that obviously i is 1 let's let's say that we suppose we want to compute matrix elements suppose we want to compute matrix elements of some product or operators between some states at some time we get the same answer whether we're working with Schrodinger states and Schrodinger operators or Heisenberg states and Heisenberg operators just because the difference between these two all cancel out and because he can state which has this additional factor the first insertion of the operator will have this additional factor to the left and so on so in computing matrix elements in computing matrix elements you're going to equally well work with these objects that is the reason for the definition of these objects now that we have the definition we take the definition of the terms until so the first thing we're going to ask is what is the left hand side for this object in Heisenberg so let's say we knew the Schrodinger picture what it was in the Schrodinger picture this object was e to the power i h minus i t f minus d i right? by the way I started dropping h bar I noticed I feel excuse that in the first lecture you saw lots of h bars this was the Schrodinger representation of the left hand side e to the power plus i t i of this guy minus i t h remember with an active to the left that involves a complex conjugation so can you say that this is the same as guy Heisenberg and side hand it's quite easy you're hard to get your head around you're going to obviously understand that so in the Heisenberg picture in the Heisenberg picture the farther technical just completes the overlap of the two Heisenberg pictures so this if we compute the s next is computing the expectation that that's going to be heavy so now the question is why is it computed as was commented why does it compute when we build these x so now let's work this out again you see we knew what it was computed in Schrodinger picture once again we can teach the graphs t i into psi and make it sideways once again we take e to the power minus i h t i of the guy and make it guy Heisenberg so what does that mean why does that leave behind each of these operators dressed by one such factor from this guy in the other guy which exactly tells it into the Schrodinger operator the Heisenberg insertion of that so it should be level and then the guy inside should have the time to I should have the time information so what what this guy s is is the operator sorry what this guy h is is the operator it is the Heisenberg wave function for the Schrodinger wave function which at time t was guy which at time t too was guy so you're right no one of course knows about t21 how does that know about t21 it is these two guys what are they this is the Heisenberg wave function for the state that is Schrodinger representation was psi at time t 1 or this is the Heisenberg wave function for the state which in Schrodinger representation at time t 2 let's be a little clear let's be a little clear you see we have to convolute the path integral with some psi and some guy that psi and that guy will not be the wave functions that appear because we are asked to convolute by the Schrodinger wave functions at that time so the time dependence comes in the translation and what these states the translation is given by the formulas of this translation and then of course the translation of introduction so this notation here has both time dependencies implicitly so psi h what was it in terms of Schrodinger language was psi times t to the power i so clearly depends how much you want psi h is this let's call it that's it yes yeah so we answer this in Schrodinger representation did you understand that answer now we will answer it Heisenberg that's the question what does this correspond to the original translation this is the answer you see the original translation happened to you was just inner product between two states because these f still we go back to the original what corresponds to the original translation is breaking this up and putting f f is it clear I am asking you before before you are reading the steps what is this possible no it goes through nothing I mean the question you should be asking is what is the Hilbert space integration and what is the Hilbert space integration there must always be an answer and Hilbert space integration is all for a number it's always inner product between two states of sub-operated sessions what I have to identify for you is what operating sessions there is no further answer any other questions or comments what we conclude here if you follow carefully our discussion is that this object the guy that involves this Pry Heisenberg p2 Pry Heisenberg p1 f1 of x of p1 because the Heisenberg operators are functions of f f2 of x of p2 fn of x of p2 you can write this as f2 of x comma p2 what is the function of x of p2 because there is no explicit time dependence on the operator these are the same so that the x field at p1 function of p2 okay so this is the Hilbert space interpretation this is the Hilbert space interpretation of this part of the instruction is this clear yes this is time order if you because the way we do it look in this t1 was smaller than t2 was smaller than tn because we had to make up the evolution operators first to t1 then to t2 t1 came here t2 came here t3 came here also there is no explicit time is this clear that was going to be actually my next comment the comment here is that you will often see this new written in the following way there is a starry operation name it means put the earliest operator to the right the next earliest operator just next to it and so a fancy way of rewriting in this time order of course this is in any order because time order tells you what order to put the product to the right so I was careful about writing f1 to the right is this clear sir if there is some way to write this cyan kind in terms of the back then maybe we can compute those correlation constants okay cyan kind at this point are completely average maybe you will think of some operator but hang on let's go to iteration theory very good question very good question you see firstly firstly this most general weighted expression is this so from the positive angle there is nothing more left to do okay so now if you got functions of p it must be that there is some translation formula functions of p thing functions of p in the integral side because the integral is not there are no p's in the integral however on the operator side of course you can have functions of p you can have operators that have functions of p so you could ask suppose instead of putting these time orders expressions here suppose I had operators that were also functions of p then what would happen and the answer is that in doing the positive angle to over p you would effectively generate some function of x let me let me give you a simple example if you see you can have derivatives with respect to time in the integral but there is no sense in which you can have d by d just an integral let me give you an example let's look at the insertion of the operator p now can you tell me what you would expect the operator p to become suppose our action was what would you expect the integral p to become p to become that's what you should expect it to become that's what you should expect it to become as soon as we have got p inserted at some time okay what we have is let's say we are interested in computing you know we do this one step of the positive x is becoming plus one as I am doing it in showing up that one step of the positive x n plus one e to the power minus i in each delta c and x n suppose in this I put an insertion of the operator then let me now work out then redo the derivation of the positive integral with this insertion as you remember because the delta c was very small we could put p h plus p squared by two n plus v of x we could write that equally as e to the power v of x on the left p squared plus two n on the right and we could insert h times plus one v of x delta c times v squared by two n this is this so now this character becomes v of x n plus one then we get the inner product of v of x this guy both these p's just add to this then we get the inner product so as you can see this becomes what it becomes e to the power minus i v of x n plus one delta c e to the power and there is an i here minus i minus i minus i minus i p squared delta p by two n times v in the denominator that was these two inner products so e to the power i p times x n plus one minus x this was the term that was the i divide this and multiply by delta k this is the i p x dot the rest was the Hamiltonian you remember this now previously we did the same thing previously and then we integrated out the p and that just generated into the country now we are going to integrate out the p again with the insertion of p in the denominator and downstairs what is this going to do well the following is true right so if you got a housing e to the power minus a x squared by e x and you put some x down here the answer to this in general then we will get zero why don't we get zero we don't get zero because we can shift but two we write this as a power minus a x minus b squared plus whatever we write it like that then we got the x down here and then we write this as x minus b plus b x minus b has zero expectation value of b has expectation value of b and then much nicer way of saying it the nicer way of saying it if you got a housing in general and you got a linear term you complete the expectation value of the linear term then that linear term is just replaced by the solution to the equation of motion of the gas in general because the solution to the equation of motion of the gas in general was x is equal to b so we have this x down here we just replace this x by the solution to the equation of motion that is the minimum of the experiment it's a general rule for gas in general I have worked it out to remind you of one variable but it's already true so here and now we can apply the same rule we got this b what is the solution to the equation of motion what is the equation of motion that comes from the numerator well it is the b by 2m by m delta with a minus i plus i x dot delta is equal to zero and now the b is equal to x dot as you would have expected remember we will be so simple b squared because in a quantum theory b squared is not the square of b I mean the expectation value of matrix elements of b squared I am not the square of the matrix elements of b so you shouldn't expect that b squared will be mx dot will result in the replacement of mx dot squared of x squared that won't be the case in comparators it's open beast but it will always be the case that some insertion of momentum is an effective insertion of x and x dot so if actually you still avoid answering this question that I need to answer the question before and I could have asked the question what are time-ordered expectation values of the quantum technology and then I knew this question would come would you force me but I remember how actually it doesn't matter it doesn't matter because if we ask a question from a path in time it's an integral of the electron over x there's no p but if you want to work out time-ordered expressions of p's by themselves written that way you have to do more work and find out what the electron insertion is and then we say I tell you that in quantum theory we will never want to do this but when you do quantum field theory we work as you would see more from this Lagrangian than Hamilton so we will never need more than this insertion I don't know if you'll ever remember this but most things that you see in textbooks you don't need this so you won't even get into this issue okay, good after the means after the indicator of the key you can always put what we get in the numerator as some right up there at some potential so we can always find some other action for which in these free insertions work is a that's correct in this particular case that other action would be low over x times if you want to put it up on the exponent okay you can think of it that way so what's natural not true if the insertion is downstairs let me say suppose you had generating function for insertions suppose you want to put it up naturally then that's a very natural thing because if the thing is downstairs you can still think of that way because that's natural okay, so I agree with you but maybe I am is there something else you want to put you know the point I was trying to make was like when you said that we can forget about the P insertion means I have got a strength to put no you are trying to explain that then we can always have some other Lagrangian with the same problem absolutely so you see that from there not the individual what we did here just any other questions because what on the integral side we can have arbitrary functions of x of t but x dot is a function of x of t because x dot with the x of t plus x of r minus x of t so when I say that you can have an arbitrary function of x at every time in particular of course is this clear in fact this is what it means this is what x dot means inside the problem but x dot always other questions okay excellent so now we have understood that the path of technical is pretty good at computing one set of observables not from transition it's pretty good at computing amplitudes you know amplitudes between two different states of time or order operator insertions actually the path of technical is not very good what has to be modified if you want to compute other operators suppose I have the insertion of three operators that's d1, d2 and d3 and I want to put those operators not in time order but in some crazy then what I will have to do is to evolve up to the time of the insertion then evolve back in time to get the next insertion and then evolve forward in time again to get the third insertion to go on so I could still do this by path of technical but a path of technical that loops back in time do you understand this it's a funny contour actually it's something useful but I want to swing up the thermal context context of equivalent non-equivalent statistical mechanics I want to swing up a keldish kind of a path keldish path of technical okay for thermal physics we may have the time to discuss such issues however the most natural path of technical is the most natural path of technical the most natural operator expectation values particular path of technical language are these time orders okay so any quantity that is easily expressed from time order insertion to operators is very naturally to produce in path of technical in particular we want to do these funny guys we would have to choose a different integration contour for every observer forward and back maybe to zigzag any time time order guys is just one one contour of this okay great so now we understand what you can compute so nicely using this path of technical we are going to examine some properties of these insertions inside path of technical and the suffers of my things we are going to see is how the equation of motion works inside a path of technical you know I should have said this but it is obvious to you the path of technical is one particular way actually a very good way of making connection between quantum and classical okay because the structures that appear inside the path of technical the structures of classical and with the action as we have seen as I mentioned before the stationary phase approximation in the limit of h bar goes to 0 the stationary phase approximation which tend to tell you that you have that they are the path that dominates is the classical trajectory the path that exposes the action okay and from this point of view all of the structures of classical physics should somehow find some sort of role not sort of modification in the quantum theory that is of course very natural quantity in classical physics okay and that is well very natural structure that is the equation of motion think that you get by very much okay so we are going to ask so by the way everything that we said for quantum mechanics generalizes without any adieu to the field theory path of technical study is this clear these operator insertions time order no issues okay by the way another thing that we might worry about how could it be how could this possibly work for one of these because time ordering depends on how you slice the path of it but by different time order I can get different path by different slices I can get different time order let me say this again what is the generalization of our let me work with our first scale what is the generalization of what we just compute the scale of it so the scale of field theory we've got isomal operators pi of x even shorting operators the lingual mechanics but now these are functions of 15 these are the functions of x this is by the way why it lies in perpetrator so in action in quantum field theory because it removes this distinction between space and time operators are labeled not just by position but also by time actually so we've got these guys and the generalization of what we found is that state time order of let's say this pi of x it looks like a great formula until you start thinking about it because when you start thinking about it it makes no sense because in order to get these operators in what slicing is this statement true we've got some operators here let's say in one dimension I've got some insertion that's here and here this guy is before this guy is before this guy but by boosting I can go to a frame with this guy is before this guy before this guy for instance and you change the order of the time order of events in special relativity by boosting okay how could this possibly be true because it's not in your an unambiguous order what is the answer to this question exactly this is consistent because as we were analysing more detail you've already studied this before but we will analyse it from our point of view because operators that are space life separated necessarily come true what is the simpler causality causality what I mean you know this is true you make the theory from the way you construct the way you construct quantum field theory is true because if two operators are space life separate you can always go to a frame whether at the same time and your construction of quantum field theory introduces a separate Hilbert space for each point in space on a given time space operators that act in distinct Hilbert spaces obviously come out so from the construction the basic construction in quantum field it's clear the operators are space life separated points completely true so this ambiguity is not there so this generalization of quantum field theory is not clear because of this nice additional structure quantum field theory but now what we can do is to try to understand as we said the equation of motion so what we're going to see is that we can write down some equation that acts on these operators we write down an equation these objects have a lot of called co-relators time-modeled products of operator insertions we'll tell you about that okay we write down an insertion an equation that acts on co-relators in the following suppose I have a path integral like this like what I've written down here and I modify it in the following way let's see it with respect to of x of if what I'm going to do is care this path integral here is a separate path integral this path integral here is a separate path integral for phi at each space time function of phi at some space time point x this integral everything here is a function of phi at every space time point in particular phi at x I think the total derivative of that object with respect to phi path integrals are convergent I will discuss this a little more when we discuss analytic continuation of this path if all the integrals are convergent you might expect that the integral vanishes it will only get boundary contributions the boundary contributions add infinity and those things add infinity contribute so without path integral way of thinking it will only make sense of things and infinity don't matter so this thing had every zero okay so we're going to start with the assumption that any second where we're giving sense to this path integral making sense to this path integral that's our test but now let's break up what this object is this is integral d phi of now this can act now this derivative is just an ordinary derivative that's the ordinary function of that it obeys the language so what are the various terms here we have minus i s and then there are various terms this is actually s plus okay so i times del s by del phi of x what is the other name for this object equation of equation of the action with respect to a field is equation of x1 mu phi of xn that's one term but this hits all of these well of course you know phi of x and x prime are distinctive are distinctive variables so whether you have rules of functional derivatives right this will become delta of x minus x1 delta d of x minus x1 that's phi of x2 i'm not putting on the use one delta x minus xn plus n such this path integral with appropriate equation of final states whatever they are I'll just denote by this expectation of whatever insertions I have so what do we conclude that i times del s by del phi of x phi of x1 phi of xn let me take the i out and take the other stuff to the right hand side now to the right hand side minus i plus i times n is equal to 1 m is equal to 1 to n delta of x minus xm phi of x1 phi of xn with the one removed being the phi n minus 1 phi phi of xn xn it tells you that we got an insertion of the equation of motion and many ok let's look at what this tells you fixing a constraint when we we are putting a constraint on one of the that comes to the delta function with the sum more on the I got one from I got one from I got one from and then let's look at the case with no insertions what we get is the object that is 0 by the equation you know what I am going to say del s by del phi del squared plus m squared phi that's it so that's what we got this is exactly like in classical physics that the insertions obey insertions inside the pattern obey the equations of motion on the other hand let's now look at the case where you have one insertion ok and in order to first let's look at how are you placed with 10 or 15 or 16th extension and ok so now let's look at the case where we have one insertion so we got expected value of del s by del phi of x then let's look at phi of y is equal to i times delta of x minus y then you remove the phi of x minus y this is a prediction of this pattern and I want to ask you um how doesn't this worry you del x by del phi of x is not just zero, in respect to which insertion, why is this happening why do we get the stage option why should you expect the del s by del phi of x zero you might expect it because as you probably know in the Heisenberg picture the equation of motion is an operator image at least nine this you know you know right all right ok why is this an operator equation why is it always true why are these anomalies if you can't operate out of zero so this is less by zero inside any so we didn't pass the tables this was a question it is what it is but we had a Hilbert space interpretation of this object how could it be that this in that Hilbert space interpretation this was not just zero because in the Hilbert space interpretation it involved an expectation of an operator that now is least eleven zero sedation but in the Hilbert space formulation we had an exact operator exact operator suppose we are taking quantum mechanics sir we I learned this picture the Hilbert operator this you can prove please sir in the Heisenberg picture I think at a particular time it is zero like so the delta function basically implies you know like if we go at that time then I understood you see we are supposed to complete the time of some operator so let's be specific let's suppose that our equation of motion let's write down our action s was in quantum m to pi delta mu pi minus sign because we use the mostly positive plus sorry by 2 plus m squared minus m squared minus m squared by 2 so let's be concrete then this will help to complete what is del S by del S so what is del S by del 5 well here I get something that I integrate by parts to kill the minus sign so it will give me del squared 5 no minus del squared minus m squared 5 what the term means is that integral d5 is equal to is equal to delta function del S that's the concrete term now what does it mean to have this del squared minus m squared inside that part it means something very simple we got an insertion at 5x we got an insertion at 5x this is the del S with respect to x so what this means is you know without this this is some number and we differentiate some function of x and y and we differentiate that function okay so it's meaning is clear that if we take this value that's this function of x and y this is already is a definite okay now this is really clearly stated that this is equal to this is the correct now let's look at the inverse phase interpretation of the left hand side let's write it down carefully so what was it it was expectation value of the order of 5x 5y with the del squared let's say the inverse phase can this operator just kills 5 it's true let's write it down as a formula so time order means the guy that's earlier goes to the right so theta of x0 minus 5x 5y now y0 minus y5 this is clear is this clear now we see that it's far from clear that the left hand side vanishes you see it's far from clear that the left hand side vanishes because why this does manage to act on 5x it doesn't manage to act on the theta function in particular the time derivatives here acting on these theta functions the time derivatives here acting on these theta functions give rise to del function exercise exercise compute use the equation of how to provide x so what that gives you is that these guys acting on these two are just here what it leaves behind is either one time derivative acting on the deltas and one time derivative acting on 5x or both time derivatives acting on the deltas show that the term with both time derivatives acting on the deltas cancels show that the term with one time derivative acting on the deltas and one time derivative acting on 5x does not cancel but instead but instead proportional to 5x dot communicate use the canonical commutation relations see all times see use the canonical commutation relations to turn this into a full four times more or less x star column works for you and then ask you to complete this is it clear is the sequence of steps you have to take anyone unclear anyone again from the Hilbert space point of view this equation is true but when infertilously this was of course a simple example so you can argue it quite easily when there are some equations that are much easier to use pathodactylism this by the way generalizes completely to many more incisions because there are many more incisions exactly the same logic from this from this Hilbert space point of view can be used to derive the full formula what's going on is that the observable derivative changes discontinuously when the times are true because the time that discontinuous change happens at a particular time so we need to translate instead of time so basically is this clear equation is like we can interpret this exactly like that just one minute but is the Hilbert space interpretation I am going to leave this as an exercise for you people by the way I think I will send you an email sometime in the middle of this week asking you to do a few exercises for summation I think we will make this one can I request someone to make note of these exercises when I ask you I can ask you just ask me there is just one thing I want to do and then we will I think extra time extra time no worries what you try to do is what you are asking is to try to use this equation in a particular case and the equation that we have let's use it for insertion of the one so without the equation del x squared minus x squared on pi of y is often this equation can be followed as a differential equation for the 2-point function pi by 5 y is often called 2-point equation that is the differential equation whose solution turns out to be 2-point equation notice by the way notice by the way we are working in the free theory the abstract analysis that I talked about where this was del x by del 5 was true for anything but when I replaced del x by del 5 by del x squared minus m squared and now we are in a free theory I haven't included the possibility of let's say 5 to the power 2 if I didn't include the possibility of 5 to the power 2 what would I get? del x by minus m squared on the 2-point function is equal to some expression acting on a 4-point because we would have 5q that's fine that's the reason we can solve the reason we can get this equation that we can just solve is that we work in the free theory it's usual okay now that this equation is here in order to process this equation let's make the following Fourier transform whenever we take the Fourier transform the momentum actually will always be accompanied by f2 pi to the power d in the denominator when positioning terminals will not have a there is one one resistive convention something to reverse an action you want to be there okay this is just a definition of the Fourier transform of 5 pi by what was let's say an operator searching side about the table 5p is Fourier transform operator and Fourier transform searching side about the table definition okay now what does this equation become this equation for if we insert this up here what do we get well what we get that's what we have 5p dot x then we will get plus e to the power i p prime dot prime okay pi of p pi of p and two of these sorry p by 2 pi d p prime by 2 pi this is this expression here but this expression just becomes minus p squared minus x squared is equal to whatever it was in the right we will leave the right okay let's look at the left hand side so on the left hand side now on the left hand side now what do we have so this is equal to i times now I will also write that as a Fourier transform space p squared by the same squared must also be inside the p squared by the same so this guy is I will write an integral representation for this data function so I will write this as first I will write it in two steps first I will write it by p dot x minus y p by 2 pi this is you know the integral representation data function and then make this as similar as possible in that object and write this as is equal to i is pi p dot x plus i p prime dot y p by 2 pi by d p prime by 2 pi by d prime into 2 pi by d and delta of p plus d prime I know it's a perverse thing to do but it's true right the reason I've done this is that now I can I can completely compare these two expressions the integral representation Fourier transform functions I need a bracket to find some functions so since the two functions are equal Fourier transform are equal Fourier transform functions in this language what we are concluding is that minus p square plus x minus p square plus x and thank you the bracket pi is the power d and delta of p p plus p and this does make minus anything is equal to i i I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I


Hello and welcome to this session. In this session we will learn about construction of a right triangle when its height of the news and one side are given. Now suppose we have to construct a right triangle pqr the size pq and put in news qr are given. For this we will start with the steps of construction. In the first step draw a line segment pq. So here we have drawn a line segment pq. Now in the second step at p draw an angle xpq is equal to 90 degrees as this triangle pqr is the right triangle. So we have drawn angle xpq is equal to 90 degrees. Now in the next step with center q and radius is equal to the length of the hyper news which is qr draw an arc. So with q in the center and radius is equal to qr we have drawn an arc to intersect the ray xp. Now in the next step drawing so we have drawn qr and thus the triangle is the required triangle. Now let us see an example for this. How to construct a right triangle is equal to 4 centimeters and the hyper news qr is equal to 5.5 centimeters. So in this example we have to construct a right triangle pqr in which its one side and hyper news are given and now we will start with the steps of construction. In the first step draw q is equal to 4 centimeters. So we have drawn pq is equal to 4 centimeters. Now in the second step at p draw an angle xpq is equal to 90 degrees. So we have drawn at p angle xpq is equal to 90 degrees. Now in the next step with center q and radius 5 centimeters which is the length of the hyper news qr draw an arc to intersect the ray xp. So with q of the center and radius is equal to 5.5 centimeters which is the length of the hyper news qr we have drawn the arc to intersect the ray xp at r. Now in the next step drawing qr. So we have joined and thus we are getting the required right triangle pqr in which angle p is 90 degrees. The side pq is 4 centimeters and the hyper news qr is equal to 5.5 centimeters. Thus the triangle pqr is the required triangle. So in this session we have learnt about construction of a right triangle when its hyper news and one side are given and this concludes the session. Hope you all have enjoyed the session.


Hi, everyone. Joe for JaspeysCaseBreaks.com coming at you with 2020-2021 Panini Prism Basketball, Salo Pack number 84. And we're giving away six spots in that hip parade preferred basketball mixer. It's two cases of hip parade basketball in a box of some old 2012, I don't know where we dug that up, a 2012-2013 preferred basketball, which is very interesting. So really great break there. Now we're going to do the break itself first, and then we're just going to do the normal thing, as if it were just a normal break. And then we'll re-randomize our names at the end and see who's going to win those spots. Big thanks to everybody here for gathering together and making it happen. I appreciate the late night folks here, too, who polished off this break and closed it out. So I appreciate the closers. I appreciate the starters, the middle relievers, the closers. I appreciate everybody. All teams are in. Let's do it. Roll it, randomize it. Six and a three. Nine times are names and teams. This is just for the break. Two. And ninth and final time. After nine times, Sean down to John. Six and a three. Nine times for the teams. Two. Three. And ninth and final time. So a new dice roll for the giveaway at the end of the break. Heat down to Pelicans. All right, here's the first half of the list right there. Daniel Katz hornets. So it could be some surprises in here. Here's the second half of the list here. Tom has the timber wolves. There's kings right here too. Warriors over here for Bennett, so on and so forth. No trade windows. Let's just get on with it. Let's sort by column alphabetically by team. And let's just print and rip. Good luck. Let's see what we've got. Here's the final list. Hot off the presses. Thanks for staying up late on a Sunday with me. Let's see what we've got here. Kemba Walker silver. Marcosol green. Any big rookie names? Terrell Terry Christian Wood. Nothing too crazy here. What about red, white, and blue? John Collins. Daniel Tice. And Langston Galloway. Nothing too crazy in this cello pack. But let's get crazy with those spots that were given away. Daniel down to Daniel. And everyone in between. New dice, new list. And once again, it's going to be the top six after. Top six after six. Four and a two. One, two, three, four, five, and sixth and final time after six. So from seven on down, that means sad times. Very close, Jason. Thank you, everyone. Very close. No cigar. But I appreciate everybody from top to bottom for taking the risk, for taking the chance at it. And for the late night rally. Top six. Happy times for you. Daniel Katz is happy. So is Steve. So is Tom. Barry. Ben. And Mark. Congrats to the top six.  Preferred basketball mixer. I'm Joe Jaspy. I'll see you next time for the next one. We've got to do, if you're watching live or if you're watching this tomorrow morning on the 30th, you've got to do one more of these filler packs. Sell out a few spots straight up. By the time you look at Jaspy's casebreaks.com, you'll be pretty close to it already. So check out the website, Jaspy'scasebreaks.com. Go to the basketball section. And I'll see you next time for the next break.


I'm Judy Darcy I'm the new Minister for Mental Health and Addictions and I'm here in Vancouver today in the downtown Eastside visiting sites on the front line and talking to the frontline workers who are dealing with this massive overdose crisis that continues. Four lives are being lost every day on average and my question to them really is what's working and what more do we need to do? I visited Insight a little bit earlier Canada's first safe injection site and right now we're at the Powell Street getaway. It offers food and spaces for people to relax and engage in activities and today what's been added is a safe injection site. From staff's perspective it's so much easier to respond to a situation in a room like this in a controlled environment where you've got trained resources around. This is very very exciting because they're able to bring all of those services together and also surround the people who come to the safe injection site with other services more quickly including health services. I really want to get the message out there that this affects potentially any family in the province and we need to reduce the stigma and we need to make sure that the supports and services are there for people. We're the first clinic in North America and all of the Americas to provide prescription heroin for treatment of severe opioid use disorder. I went to school for my professional cook, William, in 2014. I don't think I would have done that without this program. There are wonderful people working on the front lines in mental health services and addictions but there are not enough of them. There are not enough of those services. The wait lists are often too long and sometimes finances are a barrier but the bottom line is ask ones get home fast. That's our goal.