 There is another dependence on sigma 1, that comes from the determinant of the seahorses. In general, there is additional source of dependence on sigma 1, that comes from the determinant of the seahorses. Can you tell me what, you know, competitors should take an average of this? What? Comparatives, not some comparison. We could have just taken an average. We could have just put it on a fixed position. Okay. Is this the same answer? That would be the same, right? Whether we go to a wrong position or not, we get the same answer. So, either way, I have a joke about it. And the point is that the average... Well, the point is that the average, does not have an extra weight in this way to sigma 1, from the determinant. So, the reason why we're doing this is to put these four operator benefits in the same way. In this case, we can. Oh, yeah. Okay, fine. Well, yeah. Fine. So, now we've got a nice formal expression for coronation function of the femoris, which is supposed to integrate over the fundamental domain of the torus, and it's supposed to put in these insertions of C tilde, B tilde, at some point of the torus, and we put in the vertex operators as we want, all matter vertex operators are integrated, and we're very happy. Okay. Now, the next thing that we have to do is to evaluate what the matter vertex operator correlation is. Okay. So, the first thing that we're going to try to do, and in the case of the torus, this will already have physics in it, actually, whether we evaluate no non-trivial correlation, but no scattering amplitudes of the torus. However, what we are going to do is to evaluate the partition function of the torus. Okay. Now, see, what does that mean physically? You see, what we're doing on the torus is a one-loop vector. So, when we evaluate the partition function of the torus, what are we doing? What is a one-loop diagram with no insertions? What's a one-loop diagram with no insertions in free theory? Free theory? What is a compute? It computes a constant path to the E. Okay. So, what we're going to be doing is evaluating the vacuum energy of strain theory. Actually, we'll do better. We'll take strain theory in Euclidean space for the time circle on a finite circle. We'll take the time circle, the x-knock circle, put it on a finite circle, and then do this one-loop diagram. Then what we'll be evaluating? Yes, we'll be evaluating the partition function, the free partition function. So, you see, unlike in this free-level case, when we go to this one-loop order, there's already physics. See, there's already physics just in the vacuum energy. In particular, what we get is the partition function of strain theory by evaluating this vacuum energy. Once we've taken the time circle on a finite space and made it compact. Okay. So, the focus of this lecture, what remains of this lecture is not very much. Soon, we'll stop soon. And the next lecture will be to perform just a vacuum amplitude. Just the amplitude on the shoulders with no insertions. Okay. And that already has physics in it. It corresponds to vacuum energy in one limit or more generally. The partition function of strain theory is fine. Okay. So, this is the thing that we will be evaluating directly. I'm trying to see the general lesson. But it's five minutes more. But if it's five minutes, what I want to do is quickly start setting up what we need to compute. Quickly start setting up what we need to compute. Hence, we have no vertex operators. All we want is the path integral of the matter sector and the path integral of the busy system on the topics. Okay. Now, in general, it's pretty subtle, pretty difficult to keep track of this path integral without insertions. So, ratios I keep track of. The path integral itself is pretty subtle and difficult. It's possible to track it certainly. It's sort of difficult. However, the dollars, there's a very easy way to keep track of the answers by using the Hamiltonian interpretation. Easy. Because when... All right. Let's first look at the rectangular dollars. Suppose we have a rectangular dollars like this. Then, we have this formula here in space of size 2 pi. And what is this dollars function function here? Suppose this was 2 pi i a. What would the dollars function be calculated? Give me an Hamiltonian interpretation of the partition function of the formula here in the dollars of this aspect ratio. Exactly. It's trace of e to the power minus 2 pi a times Hamilton of the theory. Trace of e to the power minus 2 pi a times the Hamilton of the theory. Okay. When the theory is put on a circular line. Okay. Now, you see, more generally speaking, suppose the aspect ratio of this dollars was 0. Okay. In propagating over... You would be propagating over a certain distance at the time of fiction. That fixes your temperature. The distance is the imaginary part of the hour. Okay. But after doing this propagation, you identify your space with a translation. Right. So you take, say, you go up here and translate. That's the identification theory. Now, remember you were a certain... So this translation was really a rotation. So, if you were computing the partition function on a dollars like this, what would you be computing Hamilton? You would be computing trace of e to the power minus 2 pi... and in the hour times the Hamiltonian. Okay. Times the operator that translates you by 2 pi times the real one. L naught minus L naught. So, the thing that generated was L naught plus L naught. e to the power minus 2 pi. And then, let's take the L naught. Okay. So, that's L naught. L naught into minus the imaginary part of the hour. So, let's take an i out here. i. Times e to the power 2 pi i L naught power into what's changed is this thing. So, that's... We get a minus i here. And we have this down. Minus i to the power 10. Supposedly, if I find q at e to the power i q bar is equal to even power minus i to pi times that power. Then, what we found is that z of tau, the partition function, on a dollars of... of... whose module is this tau, is the same thing as trace of q to the pi of 0, q bar of pi. Because this thing is simply tau of these things. This is a interpretation of the... of the partition function of a... It's simply the trace. The trace of q to the pi of 0 times q bar to the pi of 0. Okay. Now, we should probably... then we should work out this thing. This time to the free boson. Two minutes. Two minutes. We work out the free boson, and ask you to work out the VC system. That's what we want to do, is for a single free boson to work out this trace. This is a very easy job, because we know the input space of the free boson perfectly. So what's the input space for single free boson? Well, it consists of harmonic oscillators of energies 1, 2, 3, 4, and then the zero. Let's first work out what we get from the harmonic oscillators. Now, if you've got a harmonic oscillator of energy of n dot energy 1, what is the function? What do we get? What's the function of the vacuum? q to the power of... second exercise today. This gave a real question. A harmonic oscillator of energy 1, a harmonic oscillator of energy 2, what's the function of energy 2 then? And you can square it. And also how much is the function of energy 3? So what do we get again? Full of 1 in infinity, 1 minus... So we've dealt with the harmonic oscillators. Wasn't that simple? We've not dealt with it so far. We've dealt with it so far is the zero point energy. This would not literally be true of harmonic oscillators. It's only true once... I mean, there's always zero point energy. Which would generally be divergent. But we've understood how to deal with these energies. And then we understood what the right answer is. We understood that the right... the right answer, harmonic oscillator, a factor of q to the power minus 1 by... a factor of q to the power minus 1 by 12. It wasn't 1 by 24. There was fully energy in the power from q 1 to q 1. That's q to the power of 1 by 12. So what we've got here is including zero point energy from zero point energy. We've got q to the power of 1 by 24. In strength theory, we don't have one harmonic oscillator. We have 26. So what's the right... function when we account for the factor? We have 26. We've got the power of 26. But not at this stage. You see the... What are our rules? We're doing the hardest of the answer. Now, ideally, what remains? What remains is the zero points. Okay? So what remains is the zero points. So... Now, the zero points, we have q to the power times h. And what does the hand turn into the zero points? Okay? It was... The strength theory action was 1 over 4 pi over prime times h. You know, one of that. Dead alpha x mu, dead alpha x mu. But we restricted it to zero points. So this thing... The d26 might technically just do. That gives us back to 4 pi. So we get 1 over 2 alpha prime times h mu dot squared. This is exactly the Hamiltonian of a non-relativistic particle of mass 1 over 1 over alpha prime. So we have the same spectrum, characteristic particle of mass 1 over alpha prime. So what's the spectrum? No, this is a branch. m by 2 x dot squared. Right? So it is p squared by 2. p squared by 2. We have q, q bar. We have q, q bar, that's all. q, q bar. q, q bar means the imaginary part of that. The minus sign? 4 pi times... Yeah. It was minus... So, e... q, q bar was q, q bar. That is equal to power minus 4 pi times h down. So this thing becomes... So the contribution from zero over it is simply k exponential and m tau times k squared. The partition function from zero over it is simply this quantity. What is the Gaussian integral which we can do? Constants, which we keep back of the original answer at this stage. After constants, it's simply 1 over m tau, the whole to the power 26 by 23. Okay? So what's the final answer to the partition function of... What's the final answer to the partition function of... What's final answer to the partition function of the free boson? It's 1 over m tau to the power 13. Then we had q, q bar to the power 26 by 24. 13 by 12. Then product n is equal to 1 to infinity. 1 minus q to the power n to the power 26. Times product... Times 1 minus q bar. I think we'll stop here for now. Okay. So that's the answer. Actually, we should keep the problem to keep track of the factors. And we have a 1 by 4 pi squared to the power 13. Good? Well, then I'll talk about the next one. Let's see if we have the other one. No. We have 4 pi squared. Thank you. So the power of alpha prime is easy to keep track of. Same as the power of m tau. But 2 pi's are probably harder to keep track of. Because yeah, 2 pi's are the power. We trust both of us. Okay. Fine. Now, as an exercise that we've asked you guys, use the same technique to evaluate the bc, the partition function of the bc system. Okay. We will put the class in the next slide. And then we will try to evaluate the full amount of partition function of string theory, of string theory on the tolerance at zero temperature and finite temperature. I don't understand the physics. Okay. If you can see it on camera, I have the right answer. So I'm going to say this again. Sorry, so we messed up the factors of 2. Okay. So this qq, when we do, this is qq bar to the power, I don't know if you should have been qq bar to the power. It's not the second term. It's not power by 2. So qq bar, I haven't looked it by 2. And so that the right formula has an extra matter of 2 here. These two cancels. I mean, then all of these factors are very easy to understand. The other problem to the path where q is because the Gaussian integral from here, the Gaussian integral produces square root of pi to the power 26. Then you get 1 over pi to the power 26. 1 over square root of pi to the power 26 from here. That goes cancel. And all the remaining 4 pi is out of 2 pi to the power 26 in the integral. Yeah. Sir, please continue. Did this computation in the right code or where you don't have a code? You just put where we put. Yes, you can do the calculation in the right code. We knew just what we were doing. We bring it in the school over here. So we should form the rules. The rules say put 26. And the rules say also equal to BCC. Okay. Which you match dimensioning with the... Where you get from the right code. But of course, you know, this is what was written in the sense that we know that what made her to integrate over the modulus of the daughter. All these things would be much harder to guess from if you wanted to be consistent. Yes. Yes. So in the end, we should get an answer that is that in the end, our sum's only over.