So the lecture today is going to be about the tables and the views of what they do today. So I'm going to start by reminding you how part of the tables work in quantum mechanics. So suppose we had, this is elementary stuff, most of you know it. We just look at this, at the segmentation. Okay, suppose we had a consensus to put an amplitude in which first is squared by 2m, as bmx. The canonical first quantum mechanical system anyone studies. It's the quantum mechanics of a particle moving in one dimension with canonical kinetic down the sub-potential.  The evolution operator in quantum mechanics is even from my minus i h bar h times t. Okay, this is something called u of t. And it's the operator that takes a state at time t1 to a state at time t1 plus t. So suppose you know your state, you know the state is psi at t1. And you want to know what is the state psi, what is psi at t2? The answer to that question, as all of you know, is e to the power minus i. So this is a Schrodinger equation, h of t2 by minus t1. The evolution operator is clearly a useful equation in quantum mechanics. It's the operator that allows you to translate things in time. It's the operator that allows you to ask the basic question of old physics. That is given a state at any time. Given a state at any time, what is the state at t? Okay, as all of you know, 70 years ago now, Feynman came up with a very nice partial temporal representation of this evolution operator, which we will now discuss. Okay, so this is an operator. The operator has two indices. We can directly represent it by a number. The only way we can do that is to look at matrix elements of the operator. So that's suppose we are interested in matrix elements of the operator e to the power minus i h t starting at x1 and ending at x2. The basic idea, the key idea, the key point was this. So this is the operation that we will sometimes, you know, in words, say it's the amplitude for propagation of the molecule from x1 at time, let's say 0, the x2 at time t. Of course, I could really equally say that x1 at time t1 and x2 at time t1 plus t. That's the structure that we invented. I'm not going to show you the matrix. Now the key observation here is that this operator can be built out of breaking it up into operators for evolution over smaller times. And the idea today is the completeness of states upon mechanics always is just a true statement of decomposition of the identity of the basis. So at any point of time, we would rewrite this operator e to the power minus i h, let's say t1 t minus t1 e to the power minus i h. This is trivial. I'll use the fact that these two operators commute because h is the same h. And then I can insert the identity into here. Now once I insert the identity into here, so let's call this object g of x1 to x2 at time t. Clearly, what we have is the identity that g of x1 to x2 at time t by inserting the identity is equal to the integral of g of x1 to x at time t1 times g of x to x2 at time t1. So we can base this property that if you know it for smaller times, you can build it up for bigger times by compounding and integrating. Since this was the case, we could try to repeat this procedure many, many times and build it up out of compounding in infinitesimal dimensions. So what's the idea? Let's suppose we had what we call this, suppose we define various time steps. So let's say that this is at time 0. So we have x1 at time 0. Now we break this up. You know many, many ways it calls time steps. That's called the first time step. I shouldn't have called this t1 because it can't live with my notation. Let me go ahead. u. That's t1. Let's say that this, let me also call this x initial and x minor. So suppose it's at x0 at time t0, which is 0, and we'll call this x1 at time t1, x2 at time t2, up to xf, xn at time tn, x1, xf at time tn. Let's break up the resolution operator into, and we'll use the notation that delta t1 and delta tnm is equal to tm. What we're doing is breaking up this entire integral from 0 to t into n different subparts. So the first subpart is at t1. The second subpart is between 0 and t1. The second one is t1 to t2, up to n minus 1. Then we take the resolution operator, break it up into those products of those various resolution operators, and between each of them insert the complete set of states that we discussed. We're breaking the integral, but it's actually a product. Let's write the formula. So the formula is the g of x initial to x, final a times t. So what is it? So now we're going to have a product over all intermediate times. So m is equal to 1 to n minus 1, n minus 1 to xn, in time tn minus, what do you call this? Yes, I've got n minus 1 guys, this here. So I've taken this from 0 to 1, and if this is only up to n minus 1, let me write this in less fancy. Ex1 dx2 up to dx n minus 1. And then these g's, there is a product like this. So what we have is a product of n factor of resolution because we've broken up the time limit of n to the sum into it. But because we've broken up n to the sum into it, there are n minus 1 intermediate positions. I'm going to explain all of those in a minute. But obviously I did. And Feynman's main insight to following up by an observation, by Dirac, was that when this thing is infinitesimal, I mean each of these infinitesimal guys are not easy to compute. So let's go ahead and compute. Just to compute this evolution operator, we want to compute, let's say, e to the power minus i h delta p. And then when delta p is now some infinitesimal object, we want to work to first order it. And then we build a combination of those follows. The first thing to do, I dropped out this h, but you remember, it's p squared by 2m plus 3 of x. And the first thing to do is to remember that in general, it's not true that e to the power p squared by 2m plus p of x in general times of t is not equal to e to the power minus i at p squared by 2m t times e to the power minus i at p of x. So the p squared and p of x commute. However, just what makes the finite evolution difficult to evaluate that much in the non-convenient nature of this operator. Obviously, it was an infinitesimal. The lack of commutation is expressed by the Campbell-Baechler-Hausdorff identity. And it was a commutator of this operator p squared by 2m times t with v of x times t, the first connection in both sides. And that's about a t squared. So if we're moving down to order delta t, we can ignore the fact that these two operators move together. So you can write this as a problem to two different languages. So that's what we're going to do. We're going to write this as follows. And you write the any order, the difference again is proportional to commutator to delta t squared. So let's write this as v e to the power minus i of v of x times delta t by h plus p squared by 2m. But this is really easy because as you know, it's the property of any function of the operator x acting on a eigenstate of x. You just replace the function with the value of x. So this is included v to the power minus i. That's insert. Let's insert a complete set of states using the momentum basis. So this is v e to the power minus i by h plus p squared by 2m delta t p p a. Integral dv. The momentum basis, this is trivial to compute because here we just act as operator p and replace it by number p. So that is equal to power minus i p squared delta p by h cross 2 h cross m times this inner product of b with p. The inner product of b with a. So this is equal to the inner power i h power b of a 19. Because the inner product of b with h is equal to the power h by b times a. That's equal to the power i h by b a. This is the weight function. This is to represent a eigenstate property. This is just the weight function of the momentum eigenstate in position basis. If by h by b. This is complex. We have e to the power p squared delta t h by m times i plus i h by b. h by should be in the inner product. Here. Thank you for leaving the integral over p. Let's note that this thing carried a pretty nice form. Because we divide another product by delta t. We are going to carry this e to the power i b p of b by h cross. Let's carry it. This minus i v b by h bar delta t. This is the rest of the form. This is simply equal to the power i or e to the power minus i e to the power i times b x dot by h bar minus h of b and b. That's that. When i x dot by definition, it can be minus a. What I've written on this form is that it might remind you of something. It's one that reminds you of something. As you know from your study of classical mechanics, that one that if we use canonical detection, if we use Lagrangian transform, substitute for b in terms of x. This is what is simply the action. The fact of life. Actually, there's more general thing to say, but I won't say at the moment. So it's a fact of life that a quadratic path in time. So what's a Lagrangian transform? Lagrangian transform asks you to solve for b by thinking of this as a function of both b and x, writing down the equation of motion for b and plugging in the solution of the equation of b. The series of steps has this fancy name, majority of charge. Imagine that it's actually a function of both x and b. Use the equation of motion for b. That gives you an equation. The equation will be x dot is equal to del h by del b. Use the equation of motion, solve for b.  So that's the thing that we do with the action. It's a fact of life about quadratic integrals, quadratic in the exponential, Gaussian integrals. But after a deterministic factor, they implement that procedure. You know what I mean? So what I'm saying is that in our particular case, our h was quadratic in b. So you just did the Gaussian integrals. The answer to that involves a major factor, a determinant. A determinant of this h bar m by delta t. In this case, the determinant is just the determinant of a number. So 1 over square root of h bar m by delta t. It is equal to square root of h bar m by delta t. Times of pi business coming from e to the power minus. Some numbers. Times just solving for the equation of motion for b. And turning that back in. Is this for many of you? So I wouldn't bother to evaluate the Gaussian integrals. I just use the fact that I know what the answer will be after normalization. Which we will probably never need in this course. So I'm going to make one statement. So I'm going to make normal. Why is b appearing here rather than i? Why is b appearing here? You see, that's because I broke symmetry by inserting the complete set of states here rather than here. And we have to remember that all of these expressions are approximate. That is, they battle only the first order and delta t. So to this first order and delta t, as we will soon see, b and a will always be forced to be infinitesimal in your mind. So it won't make a difference. You see, what we are going to do is to get, what we're doing is breaking up this path integral into steps. And whether you use the potential here or the potential here or the steps are taken to be very small, doesn't really make a difference. Because it's all going to be some parts of which these steps are. Finites are pretty much going to be like this. It's going to be very surprising. So higher orders, this won't be correct, right? Higher orders and delta t? Definitely. This won't be correct. But you see, we never need higher orders. What we're doing is like an integral. If you think about how you sum an area on that curve, you've got this. And you say, well, what's the answer? It's the height here that exists. Now, you could ask, no, no, no. Why not the height here that exists? And you're right. This is an approximation. The height here that exists. So the height here that exists. But there are approximations. But the point is that the error you make is like delta t squared, delta x squared. And then you take the limit delta x goes to zero. So it doesn't matter. It's the same kind of thing. That's not correct. If there's an error, delta x goes to 0. But in the limit that we'll be interested in, we don't care about that. I guess the usual thing. OK, correct. So now just using what we know, what we're going to care is that this is simply e to the power i by h times the action, which it was simply the action we started with, which was x dot mx dot squared by 2. x dot, remember, is code for b minus a. Yeah, b minus a. OK? Minus v of b, the whole thing. And it's times a constant normalization factor, which will be of no importance to much of what I'm going to say. Is it? Is this correct? OK, so we've got, OK. And now we take this formula together with the formula we designed earlier in this lecture about how you take the evolution operator and decompose it up into a product. And remember that the product of exponentials is upstairs. OK, derived in the formula, the g of xi to xf is times t. d equal to power integral e to the power i by h plus s dt m of x dot squared by 2 minus v of x. And we also have precise meanings for this dx. And this action. What does this dx mean? This dx just means this dx is just brought up because it came from the product popular. This dx1 to dx n minus 1. Sorry? Now everything is a number. You see, this is some number. Numbers are the same. OK, so all the operator stuff was dealt with here. Now we're going to matrix everything. Matrix everything is a number. And now we just process numbers and numbers are processed. What does this mean? This is product of this times the factors we got by doing this integral here. I'm not even trying to find. Times normalization. But you can easily do that. You see, it's just some pi's and factors of i a, square root of h by m by m. That's what this dx means. What does this action mean? This action just means the action for each integral added up. This is just this object added up over a unit. OK? We've got n intervals, but n minus 1 integration variables. The two n points supply the data need to complete evaluation of the action. But I think the argument of extra action is taking the data that everything has to do with it. Right. But your normalization factor will go. Will go. So how do you go around that? How do you go around that? Well, as you can see, what we have always been interested in is the ratios in which the normalization will be. OK? However, this is a clip out. Because it's not totally satisfied. Often in physics, you don't feel totally satisfied unless you can control it. And that's a good insight. Now, there's a better answer to the question. Maybe I'll take that as the first assignment. I'll ask you guys to look at it. The better answer is that if you want to account for this normalization. OK? You can do it by starting with a renormalization. You know, it's often said that in quantum mechanics, there's no renormalization. Well, it's true for a busy non-trivial, but there is a sense in which the action you have to use here is not quite the classic action, but includes what's called a counter. And the inclusion of this counter term cancels this internal factor. I'll try to explain it here, but I'll give you an assignment if you want to answer. Actually, fortunately, the canonical way to answer this question is to just work out the path integral for a harmonic constant integral. OK? And to work it out including these extra terms. OK, I'll... OK. So two answers. The first one is that we never need really to answer the question. OK? The second one is we can, but I won't take the time to answer it. OK. You know, you were both having the same touch, OK, that you're a Hamiltonian, but I think that's a very good question. Yes, nothing... Nothing, I said, would have changed v of x to the function of v of x of t. Nothing. Except that, you know, we would have to be careful in the notation. It wouldn't be the product of the same operator all the time. You know, what people... Yeah, you might be wondering here, you might be wondering by saying, look, you're starting off saying that you can break this up because the operators commute and then v of x is unfavorable and they don't allow it to commute. You see, that is a great help because what is the right evolution operator? You have to start with the right object. What is the right evolution operator and it's going to be the function of time, is that what you're telling me? Path order, prime order, right, exactly. It can be formally written as this object. Path order e to the power minus i h of t. OK. Now, this is a formal object. Exactly the same. This is a formal object, p here is very important. What is it? What it means is exactly what we're doing in the path. That is, break up this evolution operator into evolutions over little time steps and multiply the corresponding operators. You see, where does that come from? Not from the fact that over any infinitesimal time no operator's time limit. The general principle by inserting a complete set of states works infinitely. The general formula that you can break up about infinitesimal evolutions doesn't care about anything else. OK. And the object that, you know, the object that we wanted was this path order object that was the product of the infinitesimal evolution operator by definition. So, that's the object we started. And then, just between these infinitesimal operators we insert complete sets of states, everything works. No more differences. Yes. You said that we'll cancel the infinity in the end by adding counter terms and all that. Right. But that should be wrong, right? Because I would assume that the infinity in the end cancels the fact that we have infinite integration going on. We have infinite integration going on. No, no, no. So, the infinity in the end you'd have an infinity in your end and that's what we're going to do. Even if there's a limiting process where you have two factors, it's going to cancel. Yeah. Then the limit is probably, yeah. No, no, no. What are you saying is correct? What are you saying is correct? See, there are basically two ways of doing this. If you are, if, of course, actually what are you saying? What are you saying is correct? See, if you just do exactly what I did, that I started with a well-defined object, that I end up with a well-defined object and the infinitesimal cancels the values. Okay. So, you should say, right, then I would draw my statement. Sorry, that was wrong. Now, let me tell you what I managed to say. Okay. What I managed to say was the model. You know, in order to keep track of these normalization constants exactly the way I permit, okay, is all the way I permit. Okay. It's often very big. Okay. Often, what do you do? What do you end up? What do you actually do? See, suppose I wanted some evolution from initial days, right? Suppose I wanted this evolution from a existential days, right? What you can do is to look at the sum over all the paths as a classical path plus fluctuations and integrate over the fluctuations. Okay. The normal approximation. Okay. Now, the big difference is that between the, since the classical path already has the right boundary conditions, the fluctuations are a problem with boundary conditions 0 and 0 at the initial time. Let's suppose we were, let's say, the arm on a constant. So, what would we be doing? We would be looking at paths. We would be doing some positive integral over fluctuations. Okay. We would be doing a positive integral over fluctuations. Starting out at time 0 ending up at time t with the boundary condition that x is 0 at time 0 and 0 at time t. Now, there is a convenient basis for such functions. The basis is sign functions with the right real estate. Okay. And what you can show is that this measure that I do, that's my former Jacobian time, okay, is the same thing as all. Suppose I take x of t and write it as sum over a n times the sign n times y. Then this measure, dx1 then dxn, n minus 1. Okay. Is the same thing as the problem as measure a n for r to y? No, without a precision. Because I've ignored a normed decision. Now it's no longer true. Okay. Now it's no longer true that the positive integral that I started with is exactly what I want to do. Okay. It's true after a normed decision. Okay. So if I define the positive integral in this manner, okay, then I would need to add a quantity to it. And this is often the operational definition of the positive integral. Just because that object is a bit tricky to get to work. Okay. So I'm sorry what I said was wrong initially. This is the correct version of all. This is the problem I had in mind. I don't know all the ways where it's very big. Yeah. This will get basically a value-managed income. That's a trivial problem. A quantum field theory might be to understand these things that everything will be less trivial. It will be a problem. Then it won't come just from these trivial things about changing a positive integral into a... It'll come from more interest. It will be a problem. Okay. Thank you. Any other questions or comments? So we've been able to take this evolution operator and represent it as a positive integral. And the positive integral took this nice familiar form. There's this product, the sum of all parts weighted by e to the power definition of the action. All of you know this, of course. This was a review. Okay. Any questions about this before we... We need to start applying these ideas now to quantum field theory. But any questions? So let's start. Now... The part is just an evolution in time. You can consider other evolutions. Right. But it's the path order explanation. So what is a notation here is what you mean is product of infinitesimal equations. And that would make me look bad. Excellent. So let's keep going. Now everything I said here would apply... would work for two particles. Everything I said here would work for any particles moving in any number of dimensions. Okay. I just worked one particle for no takes 12 seconds. Nothing depends on it. So any any action of this form with arbitrary numbers any action of the general form of course you start with the classical action of the general form a i j x i dot x j dot okay. The V of... I think we said we'd go through without basically without. Okay. We just I wouldn't need more notation in the complete set of states but tables for each of the different variables. Okay. One thing with theory we often end up studying such actions. So let's start with the simplest the simplest action one that you're familiar with one that we studied in your earlier courses. Suppose we started with an action which was a model. The action for us is canals. So okay let's set conventions. Let's choose in this course the mostly positive convention for the action. The eta is minus one. One more time. Okay. The advantage is to write the convention but we try our best to the extent that we can get conventions right. We'll try our best to stick to this. Okay. So now what is the kinetic term for the scalar field? Well because the time part is I'll write it down it's db5 db5 that kinetic term because if you work it out in terms of phi dot this will have a positive phi dot space. That's the rule for action. Actions are going to give rise to energy once you get this. Always come with positive kinetic like x dot space. Okay. Moreover now the position space. So this is equal to phi dot square by 2 minus li5 both things are satisfying. First the kinetic term is positive. Second that the term that is beyond the potential is negative or more positive than negative. So the thing that we added is v of x is li5 squared by 2 which is positive. Let's say that we include here to start with some minus v of minus v of 5 v of 5 will most of what we do will be 5 to the 4. But at the formal level that we're going to work here it won't matter. Intermediate steps.  This is more the symmetry of the problem by working with a particular slicing. I choose something with a choice of time whatever that is and I slice up my path at 10 according to that slicing time. That's slicing time. With that choice of time the action is what we added. It's plus 5 i 5 of x squared by 2 minus li5 of this way. 2 minus v of 5. We're working in some d-dimensional space time. It looks a little confusing because we've got a continuous number of degrees between them. In order to make sense of this theory we can imagine regulating it by a sort of lattice type of an x. So imagine that x is not continuous but discrete. We'll eventually take the spacing to zero. That's basically what we mean by this thing. Okay? What we've got and imagine also for a moment that we put the whole theory on spatial norms. This action here is exactly this one. With the x i's, the road to the x i's being played by 5 of the way this is discretized is the scale. Since our derivation of the particle integral went through for any action outside of this form I'd put another torus just so that a finite number of degrees would do it. We'll eventually take the torus to zero. Since it's true part any action of this form is true out of this action. So without further ado that the evolution operated for quantity, isn't it? Just like in quantum mechanics. Okay? It's given by into the power i times minus d2 of 5 d2 of 5. You see what do we mean by this? What do we mean by all these quantities? This quantity is pretty clear. It's the action. What we mean by this quantity here is a path integral over 5 at various time positions and at every spaces. You see? Because space plays the role of the label. So you have to do the path integral over each of the label variables. So each spatial point and then space time is discretized according to the velocity of the particle. So the way we get sense of this is imagine if you've got a grid in space time the spatial positions are this i index the time positions are what we do before. Okay? We have to take a product of integrals over each time but for every label. So that means a product of integrals over every point on this grid in space time. Okay? So that's what we know formally by this path integral. The integral over all functions of space and time fully by beautiful so we started we do this nice beautiful classical Lorentz invariant and in order to try to set up some evolution operator. Okay? In order to set up some evolution operator we broke the Lorentz invariant of the problem. We're choosing a time space. The final arms are we going to the Lorentz invariant. You see the measure of the product over all fields and all parts of space not distinguishing between space time. Okay? And the action was the action which unlike the Hamiltonian for instance. It's a matter of evolution operator. The evolution operator is precisely as as we've discussed in quantum field theory as in quantum mechanics we write a Hilbert space. I should have said this. I'll tell you let me say I'll tell you. In quantum mechanics the problem we discussed we had a simple Hilbert space the Hilbert space of square integral functions of space. Okay? If we generalize the quantum mechanics to this problem we have where i is equal to square integral functions of the m variables x of i x1 is to the power xn. Okay? Now once again in quantum field theory at least put this discretized once we discretized we've got a Hilbert space. The Hilbert space of 3 is the we will see different ways of looking at it. Since you've got a lot of Fourier analysis in quantum field theory 1 you already know many different ways of looking at it. But one conceptual of looking at it is just square integral functions of phi at each point in space. Square integral functionals of phi. What's a functional? A functional is an object that takes a function and outputs it. So from every value of this function you have some number. For instance an integral is a function. You've given a function you associate a number with an integral. Okay? So square integral so showing away functions of quantum field theory are functionals of phi of spatial axis. That have some square integral properties. This is the Hilbert space of the theory just starting to view it canonical definition of canonical framework after you discretize it. Alright? And now just like you had position eigenstates position eigenstates x we have the property that the operator x actually x is number x actually x. You have position eigenstates states phi of x which have the property that the operator phi of x acting like this is equal to phi of x. Notice that the Hilbert spaces have distinct points and position are distinct Hilbert spaces and that the operator phi of x positions are given time slice all commute to the time because the operators act on different Hilbert spaces. Okay? So it is possible to simultaneously diagonalize the Hilbert space because they will commute when simultaneously it is being at all points in x. And so there is one with one basis state such that phi of x and any x on that phi of x is the property that the function of x is this clear or with my notation to the cruiser. If this is I am sorry I said this in America if you use it the way I know it is. Okay? It was answered to anyone please. So do you also need to put in the the phi of x acting like this? Yes. You would need to do that? You would need to do that. So this for a wave this wave functional is a wave functional on the Hilbert spaces that are square integrable square integrable wave functional. Okay? And we will do this automatically to the way we actually use this by going the way we actually do. Yes. Any other questions? Is what I am saying clear? Do you understand what I mean by this state in quantity here? Requiring that the wave functional is square integrable is it some sort of restriction in my English in configurations in hand? Yeah, yeah. Right. It is very much like in quantum mechanics. Basically you will not you see what does it mean roughly roughly speaking in quantum mechanics? What does it mean that the wave functional is square integrable? More or less it means that it dies out at infinity. That is the same kind of thing that is going to happen. That the functional has no support when a file is taken away to infinity at any case. Roughly speaking. Yes? You have to do this more carefully as a major issue to give yourself that problem. But roughly this is whether it doesn't blow up if the field goes to infinity anywhere. This dies off every infinity. I don't feel that way. Okay. Now of course you know if there is I sometimes even causes to be in places where mathematicians also take it. And then you get real trouble to throw at this because it stays at the head much places square integrable and then you use these x's which as you know the quantum mechanics are not square integrable because you square and then how do you particularly get it? Okay and okay so let's let's put it on that side. Sure. Okay. You know what I mean? Like you know most of these things cancel out places where you don't but they don't cancel out usually it's the physics that helps you understand why it doesn't get good. But at each point you said there is an inverse distance or an over distance. Yes and each point that I mean is number of inverters how do I find out if you find a certain inverse distance at each point? At each point the inverse space is simply claimed by the harmonic oscillator. You see at each point we've got an inverse space which is square integrable functions of five of that value x's x's x's x's x's x's x's x's x's x's x's x's x's x's x's x's x's x's x's x's x's x's x's x's x's x's x's x's x's x's x's x's x's x's x's x's x's I will find out the number of states, number of states is infinite, I mean how do I say what are my states? What are your states? Suppose we take this thing and on the lap we take space and break it up into the x, let's say space is 1, let's say 1, let's say 1, let's say 1, let's say 1, let's say 1, let's say x1, let's say x2, let's say x2 up to xn, let's say we're on top of xn, let's say xn, let's say xn, okay? So what is our Hilbert space? Our Hilbert space is just square and variable and wave functions of 5x1, 5x2 to 5xn, okay? Now what's the Hilbert space? Now if you want to specify, one way of specifying a state, okay? It's just to specify only some function of 5x1, 5x2, 5xn. Now suppose we're doing three things. You want to know what is the specification for that thing? In fact, it is a function of this form that you can easily write down. Any basis for specifying that humanly reasonable state. Because while the Hilbert spaces are distinct, the Hamilton is not a product of Hamilton. It's not a sum of Hamilton, it's indifferent to Hamilton spaces, right? Because we've got this Li by Li by 2. This thing is phi of xn minus phi of xn minus 1. That's two double things, right? So while the Hilbert spaces are completely distinct, dynamics doesn't factor into dynamics. This way of thinking is not usually the one, at least to be complete. It's not usually the most convenient way of representing states. It's more convenient to go into a basis that diagonalizes the Hamiltonian, which as you know is a Fourier space. What about the issue of Hilbert spaces? When can I have the Hamiltonian space? At each point I have a Hilbert space, same dimension. Same dimension. So the dimension is infinity. Yes. But I can find out time by the procedure you just said. I just have to find out the harmonic oscillator states. I mean I don't use harmonic oscillator, but maybe I square the integral of a wave function. Square integral. Any other question? Okay, so why did I start saying this? I started saying this because I wanted to tell you, what is this object? So, I was a little cavalier in the mountain. So let me. So this object, let's say I integrate from time t1 to time t2. Now, in order to make this well defined, I need bounded conditions of the same type. Just like you want to make ethics, you want to make positive type of well defined, I need bounded conditions of initial and final types. I need to say why x was an initial and final type. Because this action, once you discretize it, is not well defined. Unless you know what's happening about this. Okay. So this object is only well defined if I specify that at t1, I know what the function is. Some phi of x. What do you do? I know what this is. Some phi, two of x. That was a clear number. Now, the interpretation of that is phi 2 of x, e to the power minus i h d, phi 1 of x. Well, these states we defined two minutes ago. The eigenstates of the object. Yes, the boundary conditions of the object. Yes, the boundary conditions of the object. Yes, the boundary conditions of the object. We are asking a question that breaks the answer. We are asking a question, what is the amplitude for a state defined on this slice? To evolve to another state defined on the other slice. Since that question breaks the answer, breaks the answer, the boundary conditions have to. The basic object of which we are computing. The fact that there is a basic lurence invariant object of bearing, as you will soon see, will give us a lot of, will be very nice. Easily, it enables us to derive many lurence invariant elements. It is called spatial count. I will see. Again. Yes. So, we have defined, as we said, the inverse space on a slice. Yes. But that will be sufficient only when we have an evolution equation of the functional, like a Schrodinger equation. So, what, what guarantees that to be a base? Well, we do quantum mechanics. Quantum mechanics always, actually. You see, all that quantum in theory has is suit up quantum mechanics. You take everything in discrete lines. And then you've got quantum mechanics of a number of, large number of degrees per unit. Quantum mechanics is coming back to Schrodinger. I h bar, del psi by del t, you put a psi. Okay? The differences arise in what inverse space you're working. What h is? You're working in a pretty complicated inverse space. Inverse space is a square and integral wave function of this. And h acting on this inverse space will be a little complicated. But we're always doing, your quantum field theory is not an extension of that quantum mechanics. It's a specialization. So, we have the Schrodinger equation. This is here? What? I mean, in whatever case, we're interested in a last term. There will be some very complicated superposition of these five states. Yeah, we'll write this down. Yeah. What if you wanted to write an attitude for going from an active state to a one-parted state? This would be really bad. Well... When you have to, your basic object are these three magnetic elements. Are these matrix elements? But you see, now suppose, let's ask this question of quantum mechanics first. Okay? Even in quantum mechanics, let's say for the harmonic oscillator, the eigenstates of the harmonic oscillator, or any state you might reason with, really want to consider. It's pretty far from position. Yeah. For the path 10. So, how do you... Suppose you... Sir, if you could ask, suppose I was actually interested inside you to power minus iHd and psi. Not an x1, x2. Then what should I do? Well, you just use the same thing that we always do. That is, put an xf. xf is the x. Yeah. Okay? And even for minus iHd, xi is, as the object, a five-minus-doctor study category. What you have to do is one million of it against the issue of it. And the final issue. Those would be extremely urgent. Yeah. Honestly, those would be extremely complicated. Those would be extremely complicated, right? Okay. The harmonic oscillator, these will be how I thought it would be. Yeah. Okay? In quantum field theory, if you were really interested in computing some finite time evolution, this should be a bit harder. It's not that hard, now, it's not like you ever want to do this. It's very new to me. Okay? But it's not that hard. You see, we write this down at some point. We write down the vacuum wave function of scalar fields in this language at some point.  It's not too dead. But let me, sorry for one more thing. Part of the integral is in quantum field theory. In physics, I have used not so much because the integral is easy to compute. Okay? In fact, you almost never exact, but you almost never, you almost never exactly compute a part of it. Except when it comes to it. Okay? The one exception, actually, is the superspecial speed values, where you can use the tricks of supersymmetric exactly to compute part of the spectrum. It's very yellow, because of the numbers. One exception. But you almost never use the path integral. You're not actually evaluating the integral. Okay? However, the representation of various quantities has a path integral. It's a device that allows you to transparently see formal connections between the constants. That is the main use of the path integral. Okay? So it's a very useful object to have in your head so that you can understand the properties of the constants. Okay? The actual computations. Of course, you'll see. Yeah, you'll see. The actual path integral is your actual computer. Basically, you'll see. And then you compound the representation there. You do all the things. Okay? Sir, run. Sir, run. Sir, please. We are spacing up. And we're assuming that at every point there exists an inverse space. Yes, too. What guarantees that there exists an inverse space? No, no, no. There exists an inverse space. At every point of space. There exists an inverse space which lives for all time. Okay? So, again, assuming that there exists a state basis out where finance is diagonal. Are we assuming that? No, no. We're not assuming that. You see, what we're doing is just canonical quantization of this Lagrangian. Suppose I take you to this Lagrangian and I ask, what is the quantum mechanics according to the usual procedure that the classical Lagrangian gives you? Well, you would tell me as well what I have to do, I've got m different variables. The variables are in size. Then I set up the classical mechanics in canonical formulation. Okay? And I do the usual correspondence. That will give me a Hilbert space which is square integral function of x1, x2, x3, xn. Okay? Now, this field-faring Lagrangian, once we have described it, is exactly of this form. Well, the rule of xi is played by phi's at different positions. And therefore, the usual procedure of derived correspondence, going from a classical theory to one. Okay? Tell us what the Hilbert space is. The Hilbert space is this. Hilbert space is a square integral function of x of the various points of space. In the example we set up in one direction, that is x0, x1, xn. Phi of this, phi of this, phi of this. Just to stand and prove it, it leads us to this conclusion. Correct. Now, what is the operator phi? The operator phi acts purely in this Hilbert space. Phi of x0 acts purely in this Hilbert space. Phi of x1 acts purely in this Hilbert space. Is that correct? So that means operators commute with each other. Is that correct? It's from the usual rules. If by assuming you mean how are we assuming that the usual rules apply? Who are we assuming? Who are we assuming nothing? That means using the usual rules to set up a quantum system. Okay? Now, how could that go wrong? It could go wrong in one of two ways. A, it somehow is not internally well defined. B, there's some ambiguity. Okay? The second thing actually often does happen. You know, it's something that one should keep in mind that quantum mechanics is a bigger period of classicality. Okay? So it's usually, I mean, it's often not the case that you can uniquely answer the question what quantum system has this classical system as a classicality? There may be many answers to the question often now. So if you're wondering about the ambiguity, that's a fact of life. What we're doing here is producing one sensible quantum difference. That will have the right classicality. And just this question of ambiguity for value. Then there's also a question that you're asking about. You could implicitly be asking, is this theory well defined? Now, once you discretize it, what we're doing is clearly wrong. The issue is taking place in discretization. I think you'll continue with it. This may or may not be well defined. In fact, it's a very tricky thing to get it straight. That's a program of renormalization. If you're going to discretize it, then everything will be done as a question. The question of taking a continuum limit is an excellent question which we'll come to soon. But not very much. So if your question was about those subtleties there, deep subtleties, we'll leave. But if the question was about a more elementary thing, we're not making any other problem. It's just following the usual. Otherwise, it's not. So you said that Hilbert space, for a given time, when Hilbert space attacks CO, commutes the Hilbert space at X1. Yes, these are spatial equations. Actually, even the word commutes is wrong. The full Hilbert space is the product of Hilbert space. It's one at each location. You know, operators commute. Obviously, if you put a product of Hilbert space, one operator that adds only one of these Hilbert spaces and an entity in them, they obviously commute to the other. Something that's hard to see, it's obvious. Those of you who have been interested in entanglement activity, this is the basis for that model idea. It's obvious in quantum field theory that the Hilbert space, these are regarding subtleties of Hilbert. The Hilbert space is a product of Hilbert space. It's one at each point in space. See, you can divide it in the title of an entity associated with the region of space. Because that region of space is associated with the sub-Hilbert space. Maybe the Hilbert space of all degrees of freedom in that region. Any other questions? We're just looking at a transition amplitude within one unit. These two x's? No. You see, what we're asking for is a matrix element. So let's forget these two. Your question can be addressed just above this one. So what we're asking for is a matrix element. You see, we've got a basis of states of Hilbert space labelled by this number x. For any operator a, we can ask this matrix element. But x and y are not even. That's how they do it. We're looking at the matrix element with x label on one side by label. Is this clear? The matrix element enables us to compute the more general matrix element. Any operator? We don't need to work on a basis. We can compute y and y on one side and y on the other side. But general basis matrix element is related to the special one that's easy to compute. That goes by combination. Is this clear? Okay, excellent. Okay. So that particular architecture is actually a transition from 5, 1 to find all x's and t1 up to 5 to find all x's and t2. Exactly. There's 5, 1 tells you that operator 5 acting on that state is 5, 1 or x at each x. Now of course in general we're looking at this. 5, 2 times u is an even operator like that. Knowing this you can go to this 5th convolution. 5, 2 times 5, 2 of x 5, 2 of x u, 5, 1, x, 5, 1. This is some wave function. This is your Schrodinger wave function. Sometimes. This is your Schrodinger wave function. So take that Schrodinger wave function and do the integral of d pi 1 of x space d pi 2 of x space. Then use your transition amplitude between any two states. Is this clear? Other questions about this? Okay, excellent. Let's move on to something a little less straightforward. As we will see maybe in the next lecture depending on when they come in. As we will see this path integral representing wave is very many common properties. It's a very useful thing to have in mind. So we work on this path integral representation with a scalar. We find it very useful. So we're sort of determined to write down some path integral representation for it. And then everything here is done in this way. Oh, we've tried. Now if you try to think about the theories that are qualitatively you know if you have to do a field with two scalar fields and ten scalar fields and the same thing. Then you have qualitative differences. Well you could have qualitative differences if there's something funny in the magnetic field. Or you could have qualitative differences if there's something funny in the magnetic field. So these are the two problems with an address. So we don't want to spend a lot of time dealing with some artificial theories. So you can ask is there some theory with volumic fields that for some natural reason has different structure of kinetic terms. That we have for let's say five scalar fields or four scalar fields. And if you think about it this is the theory of electromagnets. So since much of what I say just goes directly about the non-illegional theory. Since you cannot discuss the non-illegional theory we will be doing that. Let's stick to the ability theory. Five minutes. So the theory with this address is the theory based on the action s is equal to let's start with pure 0. We'll take it at the same time. Let's define it at mu nu is equal to del mu and mu times del mu and mu mu and mu in the space of time dimensions. And we have arbitrary dimension. In order to see for the purpose of what I say in order to see what I mean by this let's do this time slice. So we'll single out 0 especially. If we single out 0 especially we can rewrite this action as 0 i f 0 i minus this plus 1 by 4 f i j just allowing this mu next to time. Taking out the terms in which time appears especially. Let's see here immediately f i j is del i a j minus del j a i it involves subatial derivatives of a i. So this is the analog of the wheel of fire. This roughly is the analog of the kind I mean. This is del 0 a i minus del i e 0 plus because I write everything below del 0 a i minus del i this next grid. Why is it kind of because it is 2 standard? Why didn't we have a half year scheme? It's because in this contraction between mu and mu either mu could be 0 or mu could be 0. That is an additional factor. It's in order to get standard half for kind i 0 that we put the one fourth convention of mu. Notice here as a del 0 a i in the whole thing is squared plus half. So the a i is understand with n. But a 0 has no n. There's no term that involves time derivatives of a c. It's obvious if you think about what n is. x is completely anti-symmetric, anti-symmetrical derivative of a and del 0 a 0 is not anti-symmetric. Okay? So now we've got something interesting. Okay? So the question we're going to ask is about suppose we write down the path in X but you know this is such a beautiful action. Such a beautiful action. And we just learned that one way of defining modern mechanics is in terms of path in X. Okay? So what we're going to ask is the path in X. Suppose I you know you might have been tempted originally you might have been tempted originally to say well if you didn't know about this first class constraints of now you might have been tempted to say that the canonical procedure suggests that this is a very strange action for what to do with X. It's not a good deal with it, of course. You might have been your first reaction. However, from the point of your classical action it's beautiful. And therefore needs to be a beautiful path in X. So what we can try to do is the path what we can try to do is to say let us take this beautiful path in X but that's the definition of what's here and reverse about it. Let's try to understand what Hilbert space interpretation this path in X has. So that's what Hilbert space interpretation is. And if so, what is it? Okay? That's the question we're going to address now. So we have e to power i times minus x mu right. Or this is the this is the path in X that we wish to consider. And the question we're going to try to address is what Hilbert space is the path in X that defines the work mechanism? If so, what is the space? Okay? Not completely straightforward because a zero does not have this time in X. So this path in X here is D, A, I, D, Z. Hilbert space is very interesting. We're slicing time. So time and E zero are coordinates. It's the same zero in X. So this path in X in this way because I don't know what to do with a zero I will do the path in X over a zero last. Think of it as integral over a zero. Time is integral D, A, I. For an action of integral I'm going to drop each path. We'll restore the dimension. This is the traditional thing to do in quantitative. Just because it's traditional time. No. Okay. Record it. Take some comments about text boost and future. Okay? Um, but okay. So, if you know this guy anytime, so integral I, D zero, A, I this is summed over I and this summed over French. This is our action. It's confusing from point of view on top of a Hilbert space. If we look at this part that part could stop any segment because each A.I. has a standard time indicator. So if we just want a new Hilbert space interpretation, this part of the path would be okay? We will follow the usual procedure. What would the Hilbert space be? The Hilbert space would be wave, wave functionals. It's where integral to wave functions. Over A.I. is at each point of space. Each of the D minus one A.I. is. Or three A.I. is that we were dealing with. We were dealing with four-dimensional but physical. Four-dimensional. Okay. Next. I mean, first time we were dealing with here also I know we are dealing with a term in days Q and A. We also have a linear term in days Q and A.I. But that is not a problem. As long as the action was quadratic in time derivatives that whole derivation went through. For instance, in quantum mechanics in the presence of background electromagnetic. A.I. you have a linear term in this kinetic in x dot. Okay? I did deal with it but it's totally trivial too. What was important was that it was quadratic. Zero here is an arbitrary function of space because we are going to do a path for all these zeros. Okay? But this goes back to your question. Reeta. Let's go back to Reeta's question if we have x was a function of time could we do the same thing? Of course we could. So this object here is going to give us a standard path integral and it's going to give us a standard inverse space for a system that breaks free symmetry because of some background field. Okay? So what is this quantity computing? This quantity of course is computing the evolution of it. Okay? So the evolution of it as always is e to the power minus i. Let's compute what the Hamiltonian for this object is. The Hamiltonian of this space of square integral functions of A.I. at each point inside. So what is the Hamiltonian for this object? Well we just do a standard. What's the Hamiltonian? The Hamiltonian is pi. So what's the momentum conjugate to A.I.? It's clear right? Momentum conjugate to A.I. That's zero A.I. More precisely the momentum conjugate to A.I. at x is this object at x. Okay? And we will not, we will not So let's suppose I find this. This is pi.I.of x and we will have the commutation relations A.I.of y space times pi. pi.j of x is equal to delta i j times it's because the action has little delta x's which we don't take at all. Is this okay with everyone? Yeah? Now let's let's what Hamilton is. Where is it? A.I. pi.j minus the Lagrangian. It's supposed to be written in terms of coordinates and momentum. A.I. of momentum. And that's easily fixed. H is equal to this. H is equal to this. That's right, the I dot. Oops, whatever. The I dot has pi.I minus pi.I plus delta i just from this relation integral. Now when you work out this pi.I squared pi.I squared and then the Lagrangian has plus pi.I squared by 2. So this is now the first whatever is the standard. You get pi.I squared and pi.I j and pi. So I'm going to come into the first but in addition there's a second term. It's plus delta i is 0. e to the power e to the power i h e to the power i times this where a 0 enters the physics of this auxiliary the quantum system we get from this auxiliary is in this term. That's a totally standard kind of kind of system. It's a path integral for three scalar fields. a1, a2, a3. With a particular potential namely f ij, f ij totally standard. The system depends on this one additional a0 thing which in auxiliary problem was back on. So this might be some arbitrary as far as I'm concerned with AI path integral. This might make the Hamiltonian non-positive. This might make the Hamiltonian non-positive definite. I don't think so because you see I need not positive definite but it won't make it non-positive because this term is quadratic. It will be completely bounded by this term. Sorry but pi i is though they are conjugate to a n. But pi i contain a 0, right? So how do I think about that? No but you see that's not how we should think. That's not how we deal with it. Pi i is independent of it. Pi i is the operator that obeys this combination. While writing path integral these are just numbers. So we have a path integral which was this. Now we are trying to do is to understand what that path integral is. So the case is that this part of the path integral is computing the evolution operator generated by this Hamilton. The Hilbert space built out of three fields. Okay? Straight forward. No confusion here. But now what are we supposed to do? Now what we are supposed to do is the following. Now what we are supposed to do is to implement the path integral over a zero. Provided I am allowed to integrate by paths and well after boundary, such it will tell you a lot about. Okay? Provided I am allowed to integrate by paths. Okay? Um I am making I don't even want it. So let me give it to you. Okay? You see now if I integrate over all a zero something quite interesting. Okay? Let me explain the interesting thing for the analogy first. Suppose I have e to the power i times p t where e is a number. And I act with this of psi of x uh on psi. Some wave functions. How is my wave function psi under showing the representation psi of x? Then as all of you know this object is a translation. Okay? This object is a translation operator. The reason the translation operator is that this object is conjugate um Yeah. So So this this object here is psi of x plus a I took this object and integrated over all a what I would get is psi of x plus a integrated over all a out the most general wave function in Fourier space. Let's say we are on a torus. If I wrote out the most general integral in Fourier space then this is a project onto a particular state. Namely the state that is translation in Fourier. The state that has zero formula.