 we talked about, basically we introduced this problem, like measurement induced entanglement phase transition. We have a random unitary circuit and randomly perform projective measurements and it's associated class of measurements considered individual trajectory separately. Each trajectory calculate half chain entanglemental P or whatever is sufficiently large subsystems and average over different trajectory and average over different unitaries and then the claim is that there's a phase transition. Okay, so we introduce a phenomenology and we introduce a subtle thing. You know, if you talk about competitions between entanglement productions by unitary and entanglement destruction by measurements, this naive picture does not give us really desired competition that leads to the phase transition because in this naive calculation and naive intuitions, the measurement always wins in a sense that measurement happens over everywhere but entanglement generation happens only on the boundary. That's the recap. In my second lecture, I wanted to become a little more technical so that we learn some real technical tools that you can potentially utilize once you go back home and then do your own research project. So that's my aim. The objective is a second objective that I wanted to talk about, that I mentioned in the beginning of the first lecture. So that's a recap. It's naive model, very naive model. Let me just clarify one particular aspect. We previously said entanglement change will be entanglement change due to random unitary circuit and entanglement reduction by measurements and this is order one for one D because it only happens on the boundary. This is order, maybe some entanglement density, if you like, defined and this is what I'm going to say wrong but let me just write it anyway and the number of measurements on average that's performed on the subsystem and because they are completely disentangled, this is entanglement reduction. So setting this equal to zero implies that entanglement density has to scale as one over NA. Therefore, in the sufficiently large system size entanglement density has to equal to zero and we do not have any phase transition in this picture and I explained using a particular circuit, namely we start from a plus state and then consider applying control the phase gauge here and there and then perform measurements in X basis and I showed you that before the measurement this is maximally entangled between this qubit and the rest but even after the measurement still it's maximally entangled because even though the middle one is disentangled now the first one is entangled with the third one. What's happening is when you perform measurement here this qubit used to be entangled with maybe second one or some combination of a second or third one but when you disentangle the second one this information is actually teleported the exact quantum teleportation mechanism such a way that now it's entangled with the third one and that's generically what's going to happen in the complex chaotic dynamics. We can engineer the dynamics so that this happens exactly and that's the notion of quantum metal correction but what's interesting is in generic chaotic dynamics similar phenomena happens. Therefore, what's wrong about this argument is this connection. Even if you perform measurement you shouldn't think of it as killing the local observables or local measure, local parameters rather this is a non-local behavior that we should be very careful and the lesson of this example is okay basically we need new theoretical framework to understand this type of behavior especially if you want to study the critical behavior or the behavior of the criticalities and understand the universalities or extract some universal behavior, large-scale behavior or even write down the field theory description maybe Sebastian will be able to elaborate that more later we do need new framework that's different from the conventional one and that's going to be the topic of today's lecture. So I'm going to first provide a broad overview but of what I'm going to do for the rest of the time. So the overview is the following we are going to map our own problem into a classical step-mang model so that the solving the step-mang model is basically solving our problem that we will introduce this mapping and correspondence and this mapping will be accomplished through four steps. The step number one is going to be that initially we talked about this average entanglement so for subsystem A the average entanglement is defined by averaging over unitary and then averaging over different measurement outcome for a specific choice of unitary of entanglement entropy of subsystem obtained by this wave function. But instead of considering this especially we talked about the von Neumann entanglement entropy instead of considering this we'll introduce new quantity. New quantity. So I'm going to just write it as S, A superscript and bar. It's just a name, I didn't say anything about this but I'm going to design this object such a way that A it's easier to calculate because otherwise meaning this and B it has a property so that if I take limit so I'm going to have a series of quantities N enumerated by N equals two, three, four all the way to infinity and well defined for the integer numbers of N larger than one and we define this quantity such a way that if I take the limit that N goes to one and then evaluate this quantity this becomes equals to S, A bar. We'll introduce quantity that follows this. So the argument is even though this is well defined only for the integer values of N larger than one maybe our physical behavior can be understood by analytic continuation of this function by extending this definition of N even for the regimes that's not explicitly defined. Then we are going to hopefully solve STEMAC problem associated with HN each N will give you different classical spin model and we solve all models and extract the critical points for all every N and then again take the analytic continuation of N. The disclaimer, we won't be able to do that. Turns out in the most desired situations we have a qubit systems random unit choice when N equals two we can carry out almost everything almost exactly. Actually exactly. But when N is larger than two calculation seems to become trickier and harder. So we won't be able to extract all the desired properties. But if you take a certain limit for example instead of qubit since the qubit where the local Hibas dimension is not two but the D dimensions and then take the limit D equals infinity turns out we can actually carry out this calculation exactly and then extract the critical point for every single N and then do the analytic continuation that's doable. However, that's not exactly what we want because local Hibas dimension is too large. That's not why we want that because it often neglects important contribution from the quantum fluctuations. So that's what it is. But our approach is to introduce this one and that's going to the stem number one. And this idea is called the replica trick. And stem number two is going to be 10 star network. Now that our goal is to evaluate this new quantity. Turns out we can write down a tensor network diagrams that represents some important quantities that let you evaluate this at. So that quantity is Z. I'm going to just call it Z of N and bar. So if I know this, I can also calculate this. That's going to be a relation that I will explain that. And we will evaluate this as contraction of some tensor network. So there are many lines and tensors at the intersection of the lines and some boundary tensors, such a way that if I evaluate the contraction of these many tensors, then I'll evaluate this value. Once I evaluate this value, I can compute this. That's going to be the case, okay? This tensor network is nothing but diagrammatic representations of contraction of many different matrices, but we will find it useful. What I'm going to do is two steps within this stem number two. First, I'm going to draw a tensor network diagram that represents the desired quantity for individual particular instance of U and the particular measurement outcomes M. And then what I'm going to do is I'm going to average over all unitary measurement outcome without evaluating the diagram. And that's the beauty of this approach. If you're given a particular set of unitary, it's randomly chosen unitary and contracting this large-scale tensor network is computationally very heavy and it's not possible. In fact, there's a computer science results where contracting this type of large-scale tensor network is prohibitively large, it's exponentially large, not only in memory, but also in computation time. But the beauty is, once you have a good structure and average over unitary, we can average it out the diagrammatic level itself and get the new tensor network diagrams on another tensor network diagram different from the previous one. And now these guys, now they're averaged over unitary and summed over M, this measurement outcome, weighted by PM. This is insensitive to the particular measurement outcome or particular trace of unitary because it has been already averaged out. We are talking only on the average behavior. So this tensor network, we'll get a new tensor network that's kind of universal. It only depends on the geometry of the circuit or the parameters such as Q, which is a measurement rate. So we get a new tensor network diagram. And then we'll see that this diagram actually corresponds to the diagram that you obtain if you want to write down the partition function diagram for the classical spin model. And it will show this is actually partition function of classical spin model. For every N, for different trace of N, we'll get different classical spin models. And this diagram is nothing but evaluating the Boltzmann weight and the summing over all different configurations. And that's step number three, actually. So averaging over unitary. And then finally, step number four, you just provide the interpretation. We solve the problem and provide, get the results, interpret the data, interpret the results from the classical spin model. So to do that, we need to understand what quantities in the classical spin model corresponds to what quantities in our random unitary circuit and measurements. Any questions so far? Why? Because that's what we do in order to solve the problem. So it's not, this mapping, we have to remember that this mapping is something not somebody asked me to do, but I do it so the convenience for me. So I make it classical, I'll try hard until it's classical. So that's why it's classical. There's no fundamental reasons, yeah. There's different fundamental reasons. Here we are averaging over unitary first and we talked about dynamics of how correlation propagate in a particular geometry. And that dynamics is not necessarily experiencing interference. And that's why you can map to the classical dynamics. Any other questions? Turns out it'll be okay in this other case because we will draw the diagram such a way that when we perform the averaging, yeah, so they do not affect each other. So that's another non-trivial thing. So when you perform the averaging, we basically average individual components of the electromagnetic diagram separately without contracting. So we'll never, at any point, we are going to evaluate this number PM. That's not what we are going to do. Just average over the diagrams. That's a very good point. So if I get a solution for classical as PM model, is the universality class of this vague transition to classical as PM model basically universality class of this random entanglement entropy model? No. The reason being, if you want to compute precisely this particular quantity, we need to take the replica limit that n goes to one. However, for each n, the critical point is different and the universality point is different. So what we want to do is we want to figure out the universality class of the limit that n goes to one. And that's not always easy. And actually, it's very hard, it seems. So we are trying, I mean, that's an open problem at the moment. Would you speak up loud a bit? Why should? No, these quantities are not random entropy. It'll look similar. I'm going to present the equation. It'll look similar. It's not random entropy. Maybe I can explain the later. Individual n is not exactly the same as this guy. And these are not average random entropy. It's slightly different. And that's all to actually make some big difference. That's right, that's right. The reason why we can do the averaging is because it's a random unitary circuit. And that's what random unitary circuit gives us. We can characterize random unitary circuit ensemble very well. We know their average, we know their variance, we know all the statistical moments at the tensile network diagram level. That's what allows us to evaluate this example. If we have not chosen how random unitary circuit, but some other random unitary circuit or some ensemble of circuits, we could evaluate in principle do this. It's not clear whether the final outcome is going to be useful or not. Because here, we ended up getting classical spin model with a very physical intuition that we can play with. But that's not going to be the case if you average over a different ensemble. So choosing a good ensemble so that we get a good, interpretable classical spin model is going to be, that's going to be the art. Did I answer your question? It also depends on the geometry, but not really for depending on the value of N. Yeah, I mean, I think I'm just providing the overview so maybe we can talk about the details. So you had another question? I'll talk about it later. So this is what overview strategy I think is clear. Okay, okay, cool. And then, I'll be honest, I'll be worried that I won't be able to finish this whole lecture within time, so I'm going to tell you the results first, okay? So if you do this whole calculation procedures, you will get classical spin model, and for the special case, so let's just go to the results. So let's take the special case of N equal to. We'll get a particular spin model, and that's going to be ising spin model. We will have different sites on a triangular lattice, and then these particles will be interacting. They'll be interacting for everyone's like that, and then they will in fact have ising symmetry. They have ising symmetry, so you can understand this whole thing as a partition, you can talk about the partition function Z of the spin model, which is going to be summing over all classical variables. So they have a classical variable associated with them. I'm going to call them sigma. A sigma could take a value plus minus one. Every dot here can be plus minus one. So we are going to sum over all configurations of sigma, and for a given configuration, this whole thing will give you a Boltzmann factor, and that's going to be the product of the Boltzmann factors coming from individual Boltzmann factors, okay? Because it's going to be the summation of the terms in the Hamiltonian in the exponent, but here it's some Boltzmann factor, but it's easier to write down interaction as a three-body interaction, even though I know it'll be factorized to the pairwise interaction, because it's ising symmetric. Product over all triangles, Boltzmann weight associated with particular triangle for sigma one, sigma two, and sigma three. So consider this particular triangle, and we have a classical variable, sigma one, sigma two, sigma three, and depending on their relative configurations, we'll have some Boltzmann factor, okay? And you multiply all those Boltzmann factors for every upside-down triangle, and then there will be global Boltzmann factor, and then we sum over all different configuration of sigma, and that's the partition function, and this partition function is going to be the object that we are going to talk about, and then we will see that this object, that you want to evaluate, is going to be the free energy difference over the following two situations. We consider two cases. Case one, where you have this spin model, spins, and on the bottom, the spins has open boundary conditions. At the top, we have a fixed boundary condition, where all the classical spins are pinned to all the positive configurations. So consider putting extra layer of classical spin variables, and those are not variable, that's a constant, and that's a pin to be positive. That's case number one, and we can evaluate the Boltzmann factor, or the partition function of them, and that's z, okay? I just define z to be the partition function of this case, and case two, I basically have the same spin model, and then the same open boundary. However, we have a different boundary conditions. On the top, we divide into the two parts, a and a bar, that's a way that on the a bar, we have a pinned boundary condition where everything is pinned to be positive configurations, but everything within a is pinned to negative configuration, and that one defines a partition function, z, a. It depends on the choice of subsystem a. The size of the width of the spin model is going to be the number of qubits, and the height of the spin model, this is two of the spin model, is going to the depth of the circuit, and this a is obviously the choice of the subsystem for our circuit model for which we want to compute the entanglement, okay? Once we get the partition function, we can evaluate the free energy associated with them. That's going to be a minus log of partition function, and we also have a free energy for a minus log of the partition function, and then we evaluate the difference in the free energy, delta f, which is f a minus f, and that's going to be our quantity s a of two average. So this is a special case for n equal two. We believe basically the same thing or very similar thing will happen for the large n, but there are technical subtleties and difficulties, but we are not going to talk about too much, but we believe similar things will happen. Quantitative agreement is difficult to achieve because as I said, we eventually need to take n goes to one, however, we'll learn some intuitive picture of what's going on already with this picture. Further, now we tuned the Q, which is a frequency, basically likelihood of performing measurements, right? In some sense, that's the only remaining parameter, right? Because we have average over unit. Actually, there are exactly two parameters that control this model, aside from locality and geometry. One is Q, the other one is local free bus based dimension which assumed to be two because we're assuming Q bit model instead of Q dit model, okay? So Q and D are the two parameters that control this model, but let's consider the fixed case in D equal two. So we can talk about what happens as you tune Q and the outcome is going to be Q much larger than QC is going to be T larger than TC for ferromagnetic to the parameter, the disorder phase transition in the spin model. Oh, sorry, I forgot to mention this classical spin model with ferromagnetically interacting spin model with z two ising symmetry. So it has a finite temperature phase transition, it's 2D, and that phase transition temperature is TC. So when measurement rate is sufficiently large, larger than the critical point, then our temperature is going to be larger than the critical point. And the measurement rate is much less than QC or just less than, it's just, you see, and then that corresponds to T less than TC. And in the first case, we have a disorder phase or paramagnetic phase, and in the case, we have an ordered phase, ferromagnetically ordered phase, okay? So what happens when we are in the ordered phase, because we are pinning the boundary, basically that breaks the symmetry and the entire box system will pin to be positive configuration. Everything will be rigidly, you know, ferromagnetic. It could be all positive or all negative, but due to the boundary, it will be all positive, okay? In the case two, it's a little subtle because this part is all positive, this part is all negative, and it's a ferromagnetically ordered. So there's a frustration. There's a domain wall here. So what's going to happen is basically, everything is minus here and everything is positive here. And this domain wall has to move around and maybe choose some path like that or maybe to the right, whatever the path. And different path will give you different free energy coast because in the ordered phase, as long as you have a domain wall, this domain wall costs free energy. Therefore, the difference in the free energy will be basically the length of the domain wall in the leading order, okay? If the system size is very deep and subsystem size is potentially large, this length will be basically proportional to the length of the subsystem. Therefore, free energy difference will be proportional to the volume of A in the ferromagnetic phase. So therefore, this order phase in the RUC model where it corresponds to volume line kangal phase. On the other hand, if the measurement rate is so frequent, we just don't take them very much, corresponding spin model has a high temperature and the spins are disordered. So even though you are pinning the top boundary to be the all positive, effect of the top boundary will deteriorate and then you know, attenuate as you go into the bulk and majority of the bulk will be 50-50 random plus and minus. It'll fluctuate a lot due to thermal fluctuations. Therefore, if you have different boundary condition, it doesn't matter. The bulk is going to be just dominated by plus minus and then domain wall will not cause so much energy. In other words, this order phase is basically the condensation of the domain walls. So domain wall doesn't cause too much energy or at most it comes from some corrections coming from near the top. So it's independent of the choice of subsystem sizes because it'll terminate within some length scale size. So in this case, free energy cost will be order one and therefore we'll have area law or the boundary raw of scaling. Yes. Oh yeah, so this screen is not working. I don't know what's going on. Actually, I see it from here. So I don't know how we can turn on this screen. Yeah, can you turn on this screen? So in this monitor actually, I see the camera is following me. But I think this screen is just not on. In the meantime, like any questions, yes. And sorry, I'll answer that later. Yeah, that I also answer because I'm going to go into the details of how we derived this. I think there will be answers. So any other questions about? Okay, sure. So after mapping to speed model, if you have measurement rate that's sufficiently large, you'll see it'll translate to the temperature that's high for the classical speed model. Therefore, your spins are just disorder. It'll be like plus minus random. So if you compute the free energy for two different cases and take the difference, it'll be very small because different boundary condition doesn't imply too much. The box pin will be plus minus random. That's in contrast to the volume of the case where due to the ordering, pinned boundary will just dictate the whole configurations. Here it's frustrated because you have a, on the boundary, you don't know whether it's plus or minus one. You have to have a domain wall that propagates through the box. That's what's happening, yeah. That's a very interesting question. That will start to happen when n is larger than two. n equals three. You will see that it's not just ising model, but we have a more internal state and you could have a different symmetry break. Here we only have ising symmetry so that's the only way you can break the symmetry. But n equal larger than two, there'll be different ones. As far as I know, it's not completely understood what are the tuning parameters for the different types of terms. Here we only have one parameter, q, but you could consider different symmetry break in principle. You're right, yeah. No, to answer, if it's three, it's a three factorial, so it's s3. So we have a permutation group of three. So we have a six element because it's three factorials. Let me tell you just one thing before I move on. So I told you that this height is basically depth, t, and width is the number of qubits n. We can ask what's going to happen in the ordered side, so volume of side, when the depth is not sufficiently deep. In that case, this domain wall could actually not propagate to the side, but actually it's better to propagate straight down. Because the length of the domain wall will be actually shorter for that configuration than this configuration if the depth is sufficiently low. This free energy difference will be dictated by the dominant term, the dominant, whatever that minimizes the free energy difference. So the typical configuration will be indeed what has a domain wall in the through the line. And if you increase the depth, at some point the height will become similar to the width and then you may want to actually have a domain wall to the side. But when you want to go to the side, you can either go to the left and to the right. If a is smaller than a bar, it's better go to the left because that's shorter. A is larger than a bar, like more than half, it will go to the right because that's shorter. So basically we want to draw the domain wall that's the shortest and satisfy the boundary conditions. The shortest cut. And that coincides with our expected behavior of the entanglement. So remember in the voluminal phase, we see the linear increase and then saturate. And that's reproduced. If the time is sufficiently short, shallow, you develop domain wall top and bottom and that domain wall length, so the free energy cost will linearly increase with time. And that is this linear increase. But once it reaches a certain point, it's better to go either to the left and to the right and then saturate the value. If the subsystem size a is less than half, you're dictated by the size of a half and free energy cost per domain wall length will define basically this alpha till that value. The alpha value. So this picture of time and entanglement plot can be reproduced for the voluminal phase. Also for the aerial phase, it's kind of trivial but this is also reproduced. So we can study how the entanglement grows. In fact, this picture is true already when q is equal to zero. That means it's a perfectly ordered phase. And this is true. And in fact, historically that was done before this measurement phase transition was discovered and there was the way to understand how the entanglement entropy in many body interacting system grows linearly on average in the generic quantum dynamics. As I'm kind of doing in reverse. Yeah, if you have a spatially periodic boundary condition, you just wrap it around and you get the same classical spin model. And the domain wall now will be of this form. I don't think that's possible but that's interesting direction. I mean, I would encourage you to think about it but my immediate reaction is like it may be difficult. Tell me if you could because I mean it might be possible. I just don't know. Any other questions? Okay, then let's get into the details. Let me actually tell you how to derive all those things using tensor network diagrams. And to do that, first I'll review the tensor network diagram very briefly. Okay, don't be scared. The tensor network diagram is literally drawing diagrams to represent array of numbers. For example, suppose I want to represent a quantum say psi and you can think of it as C i some coefficient in some basis, basis state i. Or in my mind, I always think of like this as an array of vectors. Maybe the maximum number is like the D. So I'm going to represent this array as some diagram that has one box or circle with the lag attached to it. The fact that it has one lag means this is a one-dimensional array. This array index i will run through one from one to D. Let's consider the quantum state, you know, describing bipartite system part A and part B. The wave function can be written as i j C i j and i for the basis state over A and j for the basis state of part B. Now this is a two-dimensional array. You can think of C i j as a matrix, right? And this matrix will be represented by some box or circle, it doesn't matter, where it's running over i one through D and j one through D and we have a two index. And in some sense, there's no distinction because we can always define a composite index, i comma j and treat them as this one index and then write it in this way. So we should understand that whenever there's a single lag running from one through D, we can split it into two lags running from one through D one and one through D two when D one times D two equals to D. So we have some kind of flexibility there. So we talked about stable, we can also represent the operator, say density matrix rho. It's explicitly operator, this is like sum over i j, rho i j, i and j. This is very similar to this guy and we can draw the diagram that looks like rho, we have i and j. So in some sense, the same diagram, just wrote rho, it's not so important. One downside of this approach is this diagram and this diagram is basically the same, some object with the two lags. So we do not distinguish whether it is a bra vector or the ket vector. So we need to keep track of that in our mind. However, in some sense, that's the beauty of this. We have some flexibility of describing different observables without having to worry about these two too much. But however, we need to make sure things are correct. Okay, a couple of more examples. If you consider the identity matrix, that's basically delta i j, i and delta j. And these kind of operators I can represent as a line, j and i. So in this tensor network diagram, the convention is whenever of a line, basically the top and bottom index has to match, otherwise you vanish. So that's the delta function. We can multiply matrices A times B. So basically we represent the matrix A with the tensor network and represent the matrix B with the tensor network. And this matrix multiplication is a contraction of the lag. So we ends up getting the new matrix. So this is equal to the box of matrix A times B. Yes, that's the multiplication. Let's do a little more things. How to represent trace of rho. So rho is going to be a sum box with two lags. And trace means we are summing over these two lags when they are equal to one another. So we are just contracting them and this is a diagram. Okay, so in other words, tracing operator can be represented as these components. And the rho is rho, and tracing rho is just contracting those tensor network. So let's consider reduced density matrix. Suppose I want to get the reduced density matrix by tracing out a bar from some quantum state psi psi. How to represent this? We represent psi by this. And it's a bipartite system. Maybe we have our A and A bar, so with two lags. We have our bra vector, the cap vector, and the bra vector, which is psi and with two lags, that corresponds to A and A bar. The trace cell, but only partial trace cell, we contract A bar part. So we just contract this guy with this uncontracted guy, that's rho A. So this is basically the same as rho A. Is everything clear so far? Okay, and the final example. Let's compute properties such as rho square. This is called a purity. Trace of rho is normalized to be one. Trace of rho square could be any value from zero to one. And this is small when rho is mixed. And this is large when rho is near pure. And diagrammatic representation is easy. Basically, you can think of it as trace of rho times rho. Therefore, this is rho and rho, multiply them, and then trace it out. So this is a diagrammatic representation of trace of rho square. So is everything clear so far? Okay, now that we are the masters of tensor network diagrams, and that's all we need to do. So let's go through actual step of deriving the spin model. Step number one. So I'm going to write down the expression. Don't be too scared, it's a little long, but I'm going to write down the expression for Sn. So I'm going to define Sn bar to be minus one over m minus one. Okay, so this is my Sn bar. So let me unpack what I mean. First of all, this n is an index n. There's a funky things going on when n goes to one. That's why we need to take the limit eventually, rather than just directly defining it. And there's a log, and there's averaging over a unitary gauge, okay? And inside this averaging, we have this object. So we have a quantity rho tilde a of m raised to the power n. We take the trace of that, okay? And then sum over m. The m is going to the measurement trajectory. So we need to know what this rho tilde means. So rho tilde m is defined as our previous all-normalized wave function, psi of m. So this also depends on the unitary, but I'm just dropping that dependence. This all-normalized wave function for the particular trajectory, and we can correspondingly define the density matrix. Now rho a of m tilde is defined as partial trace of a of this rho. So we literally get this wave function, it trace out, so it leads to the partial density matrix. It's non-normalized. So if you compute the trace of rho tilde a of m, we obtain the probability of getting that particular measurement trajectory history, and that's what I'm going to write as a p of m. And also that p of m is this p of m. And we are summing over all different measurement trajectories, and then we have averaging and the logarithms and the normalizes quantity. Sorry, absolutely, that's right. So if I use the normalized wave function, we'll get the normalized density matrix. I'm taking unnormalized wave function, so your unnormalized density matrix, but trace of rho tilde is just the norm of this state. Therefore, it's a probability, okay? This is not equal to one. This is the probability of getting a particular measurement history m. It can be anything between zero. It's going to be actually exponentially small because there are exponentially many different measurement outcomes over the multiple layers of circuit. So individual p value will be exponentially small. Yes? We'll do that now. Yeah, we'll do that. So let's check that this actually reproduces the relevant quantity, yes. That's right? This en, that's typo, that's u. Thank you. Roles of a, that's right because that's the same. This is also the same as trace of rho tilde. So I mean the rho of a just partial trace and then trace the rest component. So this is tracing over a part. Okay, cool. So why is it reproducing the desired quantity? Let's quickly derive it. So I'll do the calculation pretty quick because it's a bit technical, but it's worth following it. And it's a popular technique in many fields, not only in these measurement and use situations, but also in many of the spin glass problems where we cannot compute logarithm efficiently. So one beauty of this is this. So we perform averaging over a unit right here inside a log, right? So once you compute everything inside a log, you don't need to take any logarithms. And in fact, this is just rho to the power n. So it's kind of a monomial of the rho. That's something we can calculate easily and perform averaging. And then apply nonlinear function which is log. However, in a way that if you take the limit n goes to one, it reproduces the quantity where averaging is done outside of log. And I want to show you that. But that's kind of like very useful and relevant. So just for the record, these whole thing is what I'm going to define as a Z a for n. And these quantities what I define as Z of n. And that's the partition function that I described for the spin model. The two different boundary condition. And boundary condition will come from difference between rho a tilde versus Pm. In fact, if you choose a to be nothing, partial trace is basically tracing out everything, right? Then you get the Pm. So actually these two terms can be understood as different choice of a. That's why we have a different choice of the top boundary condition where in one case, a is just nothing. Everywhere is positive. Okay, so let's derive it quickly. Let's see what happens when n goes to one. First thing to notice is that denominator goes to zero. So this diverges. However, actually that's okay because when n is to one, rho tilde to the n is one. And the rho to the n, so it's a trace of rho. So these quantities are P of m, right? And this is also P of m because it's n equal one. And averaging over unitary, averaging unitary, they are exactly the same quantity. So these quantity actually cancel each other. So the numerator is also zero, right? So denominator and the numerator both vanishes. Therefore, we should evaluate this by taking the derivative of both numerator and denominator. A derivative of denominator in n is just one. So I'll skip that calculation. And we only need to evaluate the derivative of the top component. Limit n goes to one. S a of n bar is equal to minus the derivative we respect to n. Let's only consider this component because the second component will look very similar. We'll make a derivative, we have a log function. So what we are going to get is expectation value of u. Actually, we have a derivative where we have our expectation value of u and then summation we can exchange because that's a linear operation. Do you agree with this expression? Please correct me if I'm wrong because sometimes if you are standing in front of the blackboard, your IQ drops by one auto magnitude. So I easily make mistakes, so tell me if I'm wrong. I believe this is true and then we need to handle making a derivative here. But here, let's do this calculation. Basically, we want the derivative of something x to the n, right? You can understand this derivative of e to the n log x and evaluate n equal one. Evaluate n equal one. And that's equal to log x times e to the n log x. Evaluate n equal one, which is x log. So we can do that. And that's why this rule give you the factor average over unitary sum over m. We have rho a of m tilde log rho a of m. Log rho a of m tilde. And in the denominator, so we already evaluated n equal one. Denominator n equal one, this is basically rho tilde and then trace that's a p of m. Your sum over all m and probably adds up to one, so it's one. So we don't even have to calculate average over unitary, just one. Right, the constant is average over, so there's no denominator here. And then there's a minus sign coming from here. And here we have a term with the negative sign, basically the same term when rho a is replaced by p m. So that quantity is going to be average over u or I just pull out this averaging and put it here. Minus p of m log p of m. And final things to notice is that rho a of m tilde can be normalized. We know the trace is equal to p of m, so what we are going to do is pull out p of m and right introduce a normalized version of rho tilde and then replace this rho tilde with this p times rho and p times rho. And what goes into the inside the log will factorize with the component p and the log p and the log rho. The log p component will cancel with this guy because, oh sorry, I'm missing trace here. Because we are tracing over rho and then we have another p and that will precisely cancel. So after cancellation we will get the desired quantity of averaging over u minus sum over m trace of p of m times rho of m log rho of m. And this is precisely why we want it to have. So the lesson is, it's not so complicated calculation. The lesson is the log factor is now inside the averaging over unitary. How did it happen? Because we are making derivative with respect to n, but the derivative is bringing log factor from this guy and that's already inside the averaging. So that's why the log factor can be pulled out. And this is, if you want to learn more about it, this is a replica trick. Okay, any questions? Rho 2. That's right, just standard log tau rho. So why? So this whole thing is the numerator and goes to zero when n equals to one. The denominator is one of m minus one here. So it goes to zero when n equals to one. Yeah, you factorize this guy into two of these and then you have a trace which acting on this rho copy and then it gets p m. That cancel this guy, right? That's why you have p m here. So this is literally averaged quantity. Okay, good so far. So we learned about replica trick and then let's go to step two. So we want to write down the 10-centimeter diagram. So let's remind ourselves how the purestase i m look like if you draw in the circuit diagram. Circuit diagram is a version of 10-centimeter diagram. So it's good. So let's have for now like six qubit zero, zero everywhere. And then we apply some unitary u one, unitary u two, unitary u three, unitary u four, unitary u five, dot, dot, dot. There's some other unitary here to the left. Some other unitary to the right. And this is a 10-centimeter diagram. Yeah, dot, dot, dot, dot. This is a 10-centimeter diagram for our pure state psi of m in the absence of any measurements. Let's introduce some measurement, okay? I perform the measurement here and therefore I obtain some measurement outcomes at m one and that's equal to inserting a projector m one, m one. You should project to the state m one and then state is in that state p one. You don't measure this guy but maybe you measure another guy here, m two, m two. So it's a projector inserted there. So we probabilistically choose certain sites and insert the measurement outcome, the projection to the measurement. That's psi of m tilde, all normalized. This is two complicated diagrams. I'm going to just simplify by saying r u c plus measurement. So whenever you see this diagram, you just imagine this whole thing just shun in this way and then we have many lags, many lags. Inside there's like here and there are some projectors associated with the measurement outcome. So this depends on the choice of u, this object depends on the measurement outcome m, okay? So how do we write down a density matrix row of m? Very simple. You have this the same thing, copy it down but it's a broad vector so you need to take the complex conjugation and that's this. So this diagram is basically r u c plus measurement with the lags and lags and here we are just contracted to all zero. And we have another box, r u c, let me put star here, that means complex conjugation, otherwise everything the same. If u one appears here, exactly the same u one appears here with the star with the complex conjugation. So they are correlated now. So individual u here are uncorrelated random unitary but same u appears in two different boxes that are perfectly correlated unitary because we are considering identical copy for a given set of u's and measurement outcome. R u c star for the measurement and here we have an impulse state that's a broad vector of zeros rather than count vector of zeros and then output lags. And let's now try to, it's a tilde and on normalize. Suppose we want to do the partial trace of this row tilde so that we get the reduced density matrix for the on normalize reduced density with an A part. That means we need to trace out a complement. So let's divide the system into half, say this is A and this is A bar and half this is A and A bar. We are going to contract the indices on the A bar one at a time for every qubit. So that's going to be a contraction of this line. And you should be careful about the ordering though the same qubit are contracted because we are just tracing out the degrees of freedom. And this diagram is already complicated so I'm going to compress them by having double rectangle. So we have these two things lying on top of each other and then now every side we have two lags where one from cat vector and one from the bra vector. So the initial state is zero, zero here and zero, zero here, dot, dot, dot. And this is RUC measurement tensor RUC star measurement. Just two copies of this box on top of each other. And then we have two lines per site, one from bra vector and then one for the cat vector. And then we are tracing it out only for the part A bar, complement of A. And this A part is left uncontracted and this represents unnormalized reduced density matrix of row. And finally, we are talking about N equal to case, the special case of N equal to. We want to evaluate row A tilde square. Yeah, measurement record M is encoded by different choice of locations of inserting the projectors and zero and one. So that's all inside this box. Okay, how do we compute row A tilde square and then we want to take the trace of this quantity? That's what we want, right? Remember, so far here, this is a matrix. Therefore, we have uncontracted lags, open lags. This is a number, which is a one real number. So we will not have any uncontracted lag when everything will be contracted. How do we do that? Simple, we just copy this twice and then contract them exactly the same way we have done here. So use your imagination and I'm going to just draw the final results. That's going to be, we have four boxes. So RUC measurement, RUC star measurement, RUC measurement, RUC star measurement, like four copies of the same tensors. So inside each U appears four times U, U star, U, U star, like four times. All the measurement copies happens at four times M1, M1, M1, M1, M1. And every qubit is now associated with four lines, one, two, three, four. First two is from the first copy, bracket. Second two is a second copy, bracket. So if I write like zero, zero, tensor, zero, zero. And that's the initial state, some initial state. Four lines, one, two, three, four. One, two, three, four. One, two, three, four. So that's the initial tensors. And then on the top we have a one, two, three, four. One, two, three, four. One, two, three, four. One, two, three, four. Let me only draw four. Here we contract in this way. So what you see is within copies, say copy one, we are contracting by themselves. A copy two will be contracted by themselves. So contraction will look like this. However for the A part, our contraction will be crossed a copy. Therefore this contraction will look like this. And this is part A, and this is part A bar. So already start to see some structure here. So we see the bottom boundary where the tensor and the contraction is contracting to the zeros like product state. Eventually there will become an open boundary condition for the spin model. Here we are contracting a particular form of tensors and that will correspond to pinning the spins. Here are another tensors and that will correspond to pinning the spin with the different configurations. And we will see that in more detail soon. So that's the overall structure. And what it means to do is to let you zoom in and look at what happens in the bulk of these tensors. Any questions until then? So everything is clear? Okay, so let's do that. So if you zoom in this box, like four copies of the box, you basically have about two components, yeah? Yeah, that's right. I didn't draw many dots, but everything has to be contracted to product state zero. So it's not an open line. You are contracted a particular product state. I just delayed it, but the pro products, yeah. So if you zoom in this RUC box, we have about two components. So now it's like step three. We are going to perform averaging. So if you zoom in, every U, U, and their complex conjugate U star, and U, U star appears as a four copies always together. So to simplify this, I'm going to draw again, that's one box. U tensor U star, U tensor U star, about one, two, three, four, one, two, three, four, one, two, three, four. So remember a single U is acting on two qubits, left qubit on the right qubit, input, output. So here we have a left qubit, right qubit, input, output. Copy one, copy two, we didn't copy one, cat bra, cat bra. So I think it's everything clear. So we have many legs, it's how many, like 16 legs. What we want to do is we want to average over this quantity over all choice of unitary. So why do we rearrange in that diagram? Because I want the same U's to be on top of each other because they are perfectly correlated, so they have to be averaged together. However, each U appearing at the different part of the circuit are identical and independently distributed, so each of them can be evaluated separately. So in the bulk, we will have to evaluate this quantity and I'll do it later. Another component is a measurement component. So whenever we perform measurements, we basically project to the M, say M1, and do the same projection for the next line, M1, and the same projection and same projection. So these four lines are basically those four lines, I just expanded, and we need to sum over this M. So we do need to sum over M, we are not evaluating P over M. So just diagrammatically, we are just summing over the projectors. And that's possible because we are basically utilizing Tesla networks. We just average at the level of diagrams without having to evaluate all the quantities. And summing over M is basically an object that I'm going to draw like this. So this means I just define this object by this. There's a one index M and it runs from one through D. In our case, D equal two, so one or two is zero and one. And whenever this is zero, all the indexes is non, you know, the index has to be zero, otherwise it vanishes. So this is a simple tensor and a torque diagram. So it's just a copy tensor. The single classical variable at the core, you know, zero and one, we'll just copy. So let's say zero and one, okay. Zero and one. And that basically means that eight of the legs has to take the same value and that value could be either zero or one. Okay, that's this tensor. So measurement can be taken care of in this way. Remember, we perform measurement or not with a certain probability Q and they'll handle that later. Let's talk about this one. So this one has nothing to do with our specific problem. This is exactly just some mathematical object where you take four copies of unitary and consider their average over unitary. This quantity is called a second moment. Or depending on the community, sometimes it's fourth moment. Because second moment, because we always have a U and U that go as a pair. So the second moment means like two pairs of pairs. So the second moment, the fourth moment because we have a power four in unitary equation. So each box contains many unitaries within the box. And we will have a many averaging. Here we are averaging over U1, for example. U1 over there. We will do the averaging over U2, U3, U4, all separately and independently. So let's do this averaging. So this derivation is not so difficult but let me just write down the answer. U, U star, U, U star. This averaging is what makes a random unitary very useful. Turns out this can be written in the following way. Some over some classical label or variable sigma, which could be either plus or minus. This plus minus is the label that I came up with. Like you just label like zero, one, plus or minus. It doesn't matter. Red or blue, it's okay. And then some over tau, which is another classical variable like plus minus. And then there's a diagram. And I associate a label to a diagram because I'm going to talk about two different diagrams. Tau diagram, one, two, three, four. One, two, three, four. And sigma diagram, one, two, three, four. One, two, three, four. So this means we are summing over four different diagrams. Sigma plus minus one, tau plus minus one. But we weight them by some coefficient. And this coefficient has a name, sigma and tau. And also it depends on local Hilbert space dimensions of this unitary or let's just be explicit. The dimension of unitary. And this coefficient is called a Weingarten function. And for the, for Weingarten, and it's just a particular function that I can write down now. Sigma tau is equal to one over d squared, d, let's say d, square minus one. If sigma and tau are equal. And minus one, d times d, square minus one. If sigma is not equal to tau, okay? So we have four, four diagrams to be summed over. And coefficients depends on the relation between sigma and tau. And averaging over tensors leads to the summation over tensors where each term contributing to the summation can be enumerated by the variable sigma and tau. So when sigma is equal to plus, this diagram is a particular tensor and that particular tensor looks like this. What it means is these indices have to be the same. Otherwise it vanishes. And these indices have to be the same. Otherwise it vanishes. But these indices and these indices need not be quarter-wide. The same is true. These pair has to be the same. This pair has to be the same. But closing doesn't have to be the same. That's the contraction rule of this particular tensor. And it's the sigma plus tensor. We have a sigma minus tensor. One, two, three, four, one, two, three, four. That's this diagram. So that means the first and the fourth has to be paired. Second and third and same here. And tau tensors look the same. One, two, three, four, one, two, three, four. Plus case is equal to this. And tau minus case is equal to this. That's right. So the question was, what determines this function? This function is determined by the distribution over which we are averaging over a unitary. So we are assuming that we are averaging over high random uniform distribution in the unitary space. If you're averaging over a different space, these entire results on the right could be different. It's just not just modifying the tensors coefficient. It would be just different. So what we should do is we apply this rule, this averaging rule locally to every unitary that appears in this large box. And also we apply this rule for every measurement and then apply to the box. But before doing there, let's build up a little bit of an intuition. Suppose we have two copies of density matrix row and then contract with sigma plus tensor. For example, this portion of sigma plus tensor. And that looks like this. And that's equal to trace of row whole thing squared. But if you have a copy of tensors like row and row and then contract with, say, sigma minus case, that's like contracting in this way. That's equal to trace of row square. It's a purity. So this identity component, this plus component, is sometimes called identity. Just tracing out whatever the things that are coming in. And this minus component is called a swap component in the literature. Why? Because you can understand in this way. So we have a two lags, bra cat and a cat bra, cat bra. I'm going to exchange the cat component and then trace. And this is equal to one another. So you swap and then contract. And this way we can compute the purity. And this way we can compute the nonlinear entities that are nonlinear in row. And that allows us to compute the entanglement entropy eventually. So there was a question about what's the meaning of n. Here we have an n equal two. That means the quantity we are evaluating is nth monomial of the density matrix. So here we are only talking about the second moment. If you go to the higher moment, like nth moment, we will have two n unitaries, two n pairs of unitary. And then we can apply the same rule. And basically we get exactly the same formula. However, sigma and tau will be enumerated by permutation entries. How n copies are permuted, similar to here. So it's just either tracing versus swapping tracing. But if you have three copies, there are more combinations to permute them. Like if you swap the first two or the second and third or like cyclically permute, there are six different combinations. So sigma and tau will run over the entries of permutation group of entity n. N equal two case that happened to be just identity of swap only. And y garter function has to be suitably modified according to that. But then when you can just look up the textbook, like math textbook and then look up the table and then you get the matrix. So that's the meaning of a large n. So we are, by choosing n equal two, we are taking some minimum components that allows us to compute the nonlinear, the quantities in row. Good, equations. That's a very good point. So if you want to compute out of time order correlators, what happens is you have a forward evolution and backward evolution and forward evolution and backward evolution. You have four copies of U. Just draw the transnetic diagram according to this. This U star can be understood as a backward evolution. Draw the diagram and insert the operators in OTOC case, it will be one operator on the bottom and the one operator on the top, average of the unitary and you will get exactly the same spin model. In fact, historically, this technique was used to compute the development of entanglement entropy over time and exact calculation of OTOC in the random unitary circuit even before this measurement in this phase transition was discovered. The new component is in some sense, like new component relevant to the measurement in this phase transition is basically adding this measurement component and turns out introducing this component has a tuning knob, which corresponds to tuning the temperature of the spin model because without this, we have just one spin model without any parameters. But now we have parameters and that allows us to do the phase transition. Okay, cool, so let's stitch them together. Let's combine everything together. This is original circuit diagram, so I'm going to keep it as it is. This is basically four copies overlaid on top of each other. Let's zoom in a part of this diagram. Remembering the geometry looks like this. So the four copy version of this will look like this. After averaging over unitary, each four copies of U will split it into sigma and tau. Therefore, we'll have a diagram that looks like sigma, tau, sigma, tau, sigma, tau for these three unitories. And then sigma, tau, sigma, tau, sigma, tau, sigma, tau. I'm kind of matching them so that we can compare them each other, tau. And we have four lines, one, two, three, four. And here, the bottom boundary condition is basically they all have to be zero. It's very similar to measurement, but not summing over M, it's just like all zero. And here, one, two, three, four, all zero. And then if we have these connections, we have four lines and four lines and they newly contracted together because this is contracted. However, we have a measurement inside. So measurement tends to look like this after summing over M. So this contraction will be one, two, three, four, one, two, three, four, and there's a value M and we are going to sum over M. Here, we have a one, two, three, four, one, two, three, four. I guess you get the point, right? Maybe I don't have to draw the whole thing. So you just evaluate this whole diagram. And then what? We need to sum over set of sigmas instead of tau. And then this whole thing is weighted by a wine garden function, omega of G, sigma and tau for every pairs of unitories. So you need to multiply them and that's the weight associated with them. Right? So what I see is already something that looks like a partition function. We have a classical variable that's plus minus and assigned the configurations. Once assigned the configurations, this is a product of the weight and this is determined by local configuration of pairs of sigma and tau. Only remains to evaluate this diagram. However, this diagram is very simple because for a given assignment of sigma and tau variable, say we have a plus and minus. We have a certain assignment of sigma and tau and those diagrams will factorize because we have a left and a right that's factorizing this way. So this part and this part will factorize. Furthermore, the diagram will be very easy to compute for example, sigma equals plus and tau equals minus. And if you look at this particular component, that's equal to contracting a diagram that looks like this and this diagram is basically a loop and loop is actually summing over indexes and if the dimension is two, this is evaluated to dimension two because let's recap. So if you have a line, it's identity. It's identity matrix and if you do this, that's a tracing of identity which is a dimension of the identity. So this is a dimension. In this case of a one loop, however if sigma is for example minus and tau is also minus, we get a different diagram, namely we have a diagram that looks like this with two loops, so we have a d square which is four. So what happens is depending, in fact we can just fill out the computation and depending on they are equal or not, it's either four or two, it's d square or two. They have a higher weight when they are equal. That means every pair, you look at the weight and define this weight as some diagonal weight that depends on sigma and tau in the absence of, you know, with no measurement and then this weight is higher when this configuration is the same. That means there's a ferromagnetic interacting. The fact that it only depends on their relative relation equal or not means it has an ising symmetry. Plus plus and minus minus are not distinguishable. In the case we have a measurement, we can also evaluate the diagram and you will see that in this case it's four versus two but if it will all contract it's always two. That means regardless whether they are equal or not, they take the same value. You can see that because once you have diagram, it doesn't matter if you contract in this way or this way because they all have to the same value and this whole value evaluates to this d or in our case two. So in this picture, we have a ferromagnetic coupling for every pair on the diagonals. We have a wine garden function for every pair on the vertical. And whenever we perform measurements, we are weakening the ferromagnetic interaction. What we could do, we define averaging over whether you perform measurement or not and then evaluate the average of the weight between these two cases. So in the case of measurement, the weight is independent of their relation. No measurement, they have a ferromagnetic coupling. So if we tune the queue, average weight will interpolate between these two. Because of that, eventually this whole quantity which I define as z, this whole thing as z will be the following like z, a will be sum over variables for sigma and tiles and product over this diagonal component of the sigma and tile that depends on queue and product over all this wine garden function, sigma and tile. And then finally, we have a top boundary conditions where top boundary condition can be understood as basically plus configuration and minus configuration. Just look at the diagonals, it's identical. And different subsystems, choice A or nothing dictates top boundary condition to be how much would be the positive and negative. So there's one problem, I'm sorry I'm already over time but let me try to wrap up. There's one problem which is that this wine garden function has a negative value in their component. By the way, here I use a D instead of D because D should be the dimension of the unitary which is D square if the D is the local people's dimension, it's a two qubit gate. So you just put the D square here, so it's a D to the four minus one. This negative weight is problem because that means for certain configuration our Boltzmann weight contains a negative weight and that doesn't really have interpretation as a classical spin model. And that problem persists if that problem is really serious when n is larger than three, larger than equal to three. However, in the special case n equal to two we have one way to eliminate the sign problem. One way is here we have a classical variable that's a sigma and tau but we can explicitly integrate it out tau degrees of freedom. So if you integrate the tau degrees of freedom you basically induce three body interaction among them and get the triangle interaction. And turns out that triangle interaction still respect the ising symmetry because this integration will not change the ising symmetry. However, now there's no sign problem, we can eliminate the sign problem. And that's how we eventually obtain the classical spin models, different boundary conditions and thankfully we already talked about the interpretation of classical spin model to the measurement induced phase transition. All right, that's it, thank you so much. Yeah, question. A and A bar, that's quite different. So what I'm doing is I'm using this particular identity. This is an identity, this is an important identity rather than random unitary circuits. So we are performing averaging at the diagrammatic level for each unitary and that gives us the splitting into the tau part and the sigma part. These will eventually become the classical degrees of freedom. I don't know if they have a name, yeah. That's right, yeah, thank you for asking that. So the quantity we are evaluating, I told you, I should have mentioned it. This S, A, two bar involves averaging over measurements and also, it looks like similar to Renny entropy, but I told you that it's not. The reason being, this is A averaged inside a log so that you want the first. And furthermore, you are averaging over P of M square and trace of row A, M, square. Previously I wrote it as an unnormalized wave function, but if you rewrite it in the normalized wave function, normalized density matrix, you are computing something like purity, that's okay. However, you are weighting them by probably square. That means, relatively speaking, you are weighting more, you are giving more weight than you should for the more likely measurement outcome and they skew the statistical properties. In the limit angles to one, this will be fixed. However, that's the reason why we do not trust the universality of behavior of these guys will be the same for the universality behavior of the original model. So more technically, this does not capture the typical behavior because this will be dominated by a few instances that happen to have a larger prevalence. Yeah, I'm not familiar with that, maybe you're right. Yeah, I don't know about this. I see, okay, so I don't know the answer to your exact question, but this particular classical spin model, you can exactly compute a critical point analytically. So it turns out there's a little, this particular model has been studied. So what's going to happen is if a triangle interaction will be the combination of ferromagnetic and anti-ferromagnetic interaction, but overall the net anti-ferromagnetic interaction, even though naive coupling has different sign, and that particular model has been studied in the literature, so we can just quote a value. Yeah, absolutely, that's how the community is moving and has moved. For example, here the magnetization is an order parameter. So what we want to do is, for example, two point correlation function, two point long distance correlation function that we want to evaluate. And how does it translate to the random circuit model? Turns out it has an interpretation. Rather than just doing random circuit, you bring some extra insular degrees of freedom and couple to the system and study how the purity of this particular insular changes after a long time. So that kind of situation is meaningful in quantum information theory, but originally motivated by this type of calculation. But at the same time, you should recognize that eventually we are measuring free energy difference coming from the boundary domain walls, right? That's conceptually super clear, but obviously that's not local order parameter, right? That's not local quantity, because like you said, so that's why we could capture the non-local behavior. Okay, thank you. Okay, thanks.