It's okay, let's start for our last lecture. So this is the last recap. There won't be a recap of this lecture. Maybe, yes, maybe a recap. OK, so the last time we talked about the Libra, it's unbound in spin lattice systems. And we have seen that if we consider two observables, OA and OB, defining different subsystems, OK, that actually don't really only different subsystems, then you consider the commutator between the time of all the operator, ORT, and the operator OB. Then you find that this commutator is exponentially small. If the distance between the two subsystems is large, it's very large. So it is exponentially small in the distance and exponentially large in the time. So if you wait sufficiently long times, then this commutator can become significant, can become large. And this upper bound becomes completely meaningless, because it becomes larger than the maximal value that this quantity can assume. There is a nice corollary of this theorem. And it is the following that if you consider the time evolution of an observable defined in the subsystem A, and you wait given time t, then you can approximate this time evolving operator using just an operator defined in another subsystem S. And provide that the subsystem S is larger than the original subsystem A. And the subsystem S should be sufficiently large and proportional to the time. So if you choose your accuracy of the approximation, then you can imagine that to consider this kind of subsystem that grow. And you can always truncate the observable OOT in observables that act non-trivial only on S, and in such a way that the approximation is good. So we have seen this. And I want to say this is a very powerful result, because it's completely general. So this is independent of the details of the Hamiltonian. We only assume that the interaction should be local or quasi-local. It means that you shouldn't have maybe, I'm not sure whether this includes also Coulomb interactions. But anyway, issue decay is officially fast to zero interaction between the various sides. And we assume that the, or we assume here this, and the fact that the system has a finite dimension locally. So we are considered spin systems. We can put a spin in a given site. So you have a finite number of possible states. This is important, at least for the original proof of the theorem. But you see, it's very, very general. So we can choose our favorite Hamiltonians. Then, OK, we can see that we moved to quench dynamics. And we consider this kind of a situation where we imagine to prepare the system in the ground state of some nice Hamiltonian, like the Hamiltonian of the AC model. And then we consider the time evolution under a different Hamiltonian. So we switch on, switch off some interaction, or we tilt a magnetic field, or we change the field, or the coupling constant, whatever. And OK, in the case of the AC model, a simple situation is when we imagine to change the magnetic field of the transverse magnetic field. And in this particular case, for the AC model, what we find is that the state at the time t can be written in this form. And I guess you have seen this in the tutorial. No, my proof of this. OK, anyway, this is what is not difficult to prove that the state can be written in this form. So here, you have the Bogolubov fermions, all the final Hamiltonian. These are the fermions, the diagonalized h. And 0 is the, OK, here I forgot something. I forgot the sign. This is the sign. This is the vacuum of this Bogolubov fermions, of the b's. We find this way. And OK, we were trying to compute in the end yesterday the time evolution of the entanglement entropy only using the structure of the state. We were interpreting the state as the superposition of pairs of particles that are produced after the quench. And you see these pair of particles have the properties that, therefore, if you have a particle with momentum k, there is a particle with momentum minus k. And moreover, we were assuming, because, OK, this is what happens, for example, in AC model, that the ground state of the model has sufficiently fast decaying interactions. So we can assume that when a particle is produced at a given point, then only the particles that are produced very close by this point are correlated with the previous particle. And then, OK, in our approximation, we were just assuming that the only particle produced from the same, originally from the same point, are correlated. So you can imagine that you have all these pairs of particles. So we know that only the particle with the momentum minus k is correlated to the particle with momentum k. This we know, because we have this structure per structure. But now, OK, actually these two particles could be originated from different points because there is some correlation in the initial state in the vacuum. This correlation, as a matter of fact, is exponentially small. So what we are doing is saying, OK, no, well, we can just assume that only if this particle are originated from the same point, they are correlated. So we are seeing that different pairs are completely uncorrelated, no entanglement between this pair and the other pair. And the only quantum correlations are between a particle with a given momentum k and a particle with a momentum minus k if they are originated from the same point. This is the physical pitch. Is it OK? So can you see why I'm doing this? Just looking at the state. From this, I infer immediately that the pairs are uncorrelated because the density matrix of the state is written, is factorized in the pair, k minus k. It's completely factorized. So if I, for example, compute the reduced density matrix of a pair, k minus k, I find that this describes a pure state. It's completely uncorrelated from the rest. And this accounts for the fact that they are completely independent. Then, OK, again, the fact that they originate from the same point depends on the initials, on the vacuum. We should have exponentially decay correlation. So this is more or less where we stopped. And now we would like to compute the entanglement entropy between a subsystem and the rest in the easy mode. So I think the psi of t of k of cosine i. So the idea was this one. So this is our chain. It should represent the line. So this is the position. It's called x. Then we consider a subsystem, a. And this is the time, this direction. So we have all this. You should imagine all the pair originating from the initial time. And the entanglement between a and the rest is supposed to be somehow related to the, is given by the, is carried by the pairs, such that one particle of the pairs is inside a and the other is outside. This is inside, this is outside. So in particular, this means that this particular particle contribute to the entanglement only from this time to this time. Like I said, this particle contribute to the entanglement from this time to this time. Because when you go, if you consider time larger than this one, then you have both particles outside the subsystem. And so the idea would be that the entropy is just given by the sum of all these contributions of all these pairs. But OK, let's start from the contribution of a pair. What is the contribution of a pair of particles to the entanglement? What can be? OK, we are using here some classic arguments. So there is no mathematical answer to my question, I would say. So there is a mathematical answer, but OK, there is some physical argument behind that. And so the idea is that because we are considered in time, the entanglement between one part and the other is essentially given by the entanglement between one particle and the other, because one is inside. So when I say that the entanglement between the subsystem and the rest is carried by these particles, when one is inside, one is outside. So the idea is that maybe this is just given by the sum of all the entropies of the entanglement between one particle inside and one particle outside. So what is the entanglement of one particle in a pair? So given a pair, we just focus on a single pair of particles, which is described by the state psi, let's focus on the particular k, psi k of t, if you want, is equal to cosine of delta k over 2 plus i e to the minus 2i epsilon k t sine delta k over 2. So this is our, let's focus on this particular state. And this is the vacuum for the pair. So the idea is to compute the entanglement between a particle in the state k and a particle in the state minus k, because this is what essentially what our physical pitch will suggest. So this is the state. The idea will be to compute the reduced density matrix, corresponding to, it is just the vacuum of the two fermions with k minus k. So it's defined as b k 0 k minus k is equal to 0 b minus k vacuum k minus k is equal to 0. This is the vacuum. That is the vacuum of all the particles. This is just the vacuum of the two of these two particles. Not a pair. Yeah. So this is the vacuum of the pair. I could call pair or any type of pair. It's the vacuum of the pair. So we are just considering this simple problem. Instead of considering all the pairs now, because everything is completely factorized, so we can just focus on a single pair and say, what is the entanglement between one particle and the other? And then we know that we have to consider this contribution only when one particle is inside the subsystem and the other is outside. But that is a different problem. Now we can just compute the entanglement between one and the other particle, OK? First step. So I have this state. And so in order to compute the entropy, I have to find the density matrix corresponding to the state k of the fermion. So I have to compute the rho k. Then given rho k, I have to compute the entropy. The entropy of rho k is equal to minus trace of logarithm of rho k, logarithm of rho k. This is what we have to do. How is this defined? Rho k, we define it as the trace over the rest of the system of the total density matrix. Our rest of the system is the state minus k. We have just two particles, k minus k. So this is, by definition, the trace over the state minus k of psi t of k, psi t of k. The definition of the reduced density matrix. Let's see, our system is given by the particle with momentum k. So the rest is the momentum minus k corresponding. You can imagine this as the site of a chain. So this is one side, and this is the other side. These are the two states of our system, two properties. So we are tracing over the second property, the momentum minus k. If you want, yes, you should do it. But this is what I was writing for Britain, yeah? The density matrix is completely factorized when you write it. So as a matter of fact, when you focus on a single k, a single pair, this is described by the pure state. And this is the state, because everything is factorized. So you can write this psi as the product of all these states. It's the tensor product of all these states. Sorry, that's right here. So this is equal to, you can prove easily that this is the tensor product of all this psi t of k. And so this means that we can focus on just one state and compute the entanglement between the particle in the state k and the particle in state minus k. How can we compute this? Well, either we trace, we wish to write the density matrix of the entire system, then we trace out the degrees of freedom. Or otherwise, we compute all the expectation values of the observable. It's the dimension of the space is 2. So there are no many observables. We can actually compute the expectation values. Maybe it's easier. So we know that we have to compute the expectation value of which operator are in the state k. And which operator can you form with the BK? Clearly, you have BK. You can consider this. You can consider B dot k, that operator. And you can consider the operator number, B dot k, Bk. And then, clearly, there is the identity. These are the four independent operators in our subsist. Identity, Bk, B dot k, B dot k, Bk. Yes, because we want to compute the density matrix, the reduced density matrix. You can choose your favorite way. The one way is just to compute the expectation value of all these observables, then we reconstruct it because we have the density matrix from that. So let's see. The expectation value of the identity is clearly equal to 1 because it's a density matrix. So it's normalized. This is equal to 1. Now we have the expectation value of Bk. This is for the calculation. So let's try to do it in an inefficient way. So we have to compute psi t of k Bk psi t of k. This is what we want to compute. Let's call this A and this B. So we have that this is equal to A vacuum plus B minus A. This is equal to A. A is real and B is complex.  With this, I mean B dot k B minus k 0. This is the meaning of the notation, k minus k is this. So k minus k, and then I have Bk Bk, and then I have the same here, A0 plus Bk minus k. Now let's consider this state. You start from the bottom, B, and I erase the bottom. So no contribution from here. Then when you apply B to the state, here you add k minus k. B removes the fermion with momentum k. So you remain with the fermion with momentum minus k. But now here on the left-hand side, you have either the vacuum or two fermions. So this is equal to 0 as well. So this means that it's equal to 0. OK. B annihilates the vacuum. So no contribution from this. Now you consider the other term. When you apply B, B destroys a fermion with momentum k. In this state, you have a fermion with momentum k and one with momentum minus k. When you destroy k, you remain with minus k. But now here on the left, you have either 0 fermions or two fermions. So when you consider this color product of the two, they are orthogonal. So we find 0. The same for B-dug. No, the same for B-dug because it can be obtained by taking the adjoint of this, the conjugate of this. So you also have the expectation value of B-dug k is equal to 0. We are left with this. Probably won't be from 0. Can I erase this? Because we have already this part. OK, let's do it without further ado. We have now psi t of k B-dug k B-k psi t of k. You have to compute this. This is equal to, again, A star 0 plus B star k minus k. Here you have B-dug k B-k. And here you have A 0 plus B k minus k. Again, when you apply this to the 0, 0, because this destroys the vacuum. When you apply this to this, what happens? This operator counts the number of fermions with momentum k. There is one fermion with momentum k. So this is equal to 1 when you apply to this state. How do you see this? Well, if you apply B-dug k B-k to this state, you see the antico-mutation between k B-k and B-dug k. Then you find that this is exactly equal to itself. So what I'm saying is that this is equal to A star 0 plus B star k minus k. And here you have B. And you have k minus k. Now 0 is orthogonal to k minus k. And so finally, we find that this special value is nothing but the absolute value of B squared. So absolute value of B squared completed. So we have this is equal to 0. This is equal to 0. And this is equal to the absolute value of B squared, which is sine squared of delta k over 2, which can be written as 1 minus cosine of delta k over 2. So now that we know the expectation values, we have to write, the idea would be to write a generic density matrix of the form some coefficient times the identity plus another coefficient, let's call alpha B-dug k plus alpha star Bk, because it should be our mission, plus gamma B-dug k Bk. These are the independent operators. So we are just expanding the density matrix in this space. And then we impose that the expectation values should be this one. And we see what we find. And we determine all the coefficient, lambda alpha and gamma. What do we find? Well, we find that alpha is equal to 0 because of this. Alpha is equal to 0. Then when we impose that the trace should be equal to 1, we have the trace of the identity, which is equal to 2, because you have two states. Either you have the spin or you don't have the spin. So you have the trace of rho is equal to lambda times 2 or equal to lambda. Then you have the trace of this operator. This operator counts the fermion. So either there are 0 fermions or there is a 1 fermion. So the trace is equal to 1. So you have plus, and this should be equal to 1. Then you have to compute the expectation values of B-dug Bk, which is equal to 1 minus cosine delta k over 2. This should be equal to the trace of rho k B-dug k Bk. What do we find when we compute this trace? So when we multiply by this operator the identity, we find the operator itself. And you set the trace of this operator is equal to 1. So that is equal to lambda. Then you must consider this term, multiply by these terms. And this term by these terms is an evolution. So B-dug is a projector, B-dug k Bk. So when you multiply it by itself, you find itself. So again, we have the trace is equal to 1. So if you have this 1 minus cosine delta over 2 should be equal to lambda plus gamma. We have to solve this very complicated system. And what do we find? We find that rho k is equal to 1 plus cosine delta k over 2 plus, minus, cosine delta k Bk. Is that correct? This minus this gives lambda 1 plus k. Then you write twice this minus this, which is minus cosine. I think it's correct. So this is the density matrix. Questions? Please ask. Yeah, all the states. OK, you have just one fermion. Which are the states of one fermion?  You cannot have two fermions because we are given states. They give a moment. So for the Pauli's question principle, so you have either zero fermion or one fermion. So the dimension is two of the states. And now you have to think of all the operators that you can construct with the Bogolubo fermions with these creation operators. Now, clearly, you cannot create more than one fermion. So you can have just one B-dug. The same, you cannot destroy more than one fermion. So you can have just Bk. But then you can also construct something like B-dug B, which is the number of fermions, which counts the number of fermions. Yes? It is, but it's right to this one. Because you know that B-dug is equal to 1 minus B-dug B. So these are the independent operators. Yes? OK, so he's suggesting, if I understand that, instead of doing this, so here we are using the second way to construct the density matrix that I told you some days ago. But OK, we can also simply compute the trace over minus k. So he's suggesting, because we know the states, the complete basis of states, which are either you have zero fermion. Now I can write here. So he's saying, the state is kind of simple, no? And then he's suggesting this rho k can be written as the trace over minus k. So it's equal to the state with zero fermions in minus k, psi t k, psi t of k, zero fermions in minus k, plus you have a fermions in minus k. So you have minus k, psi t k, psi t k, minus k, minus k. So this is what he's saying. And then so you have to compute just the scar product of this psi t with a state with zero fermions in minus k. And so what you find here, that this contributes, because you have the reason of fermions in minus k. In the other case, he said you have a fermion, so it doesn't contribute. And this is for the first term. The second term is you have a fermion in minus k. So this one contributes. And then you can compute the state. It's an alternative way. Maybe it's even faster. It depends whether you are more familiar with the partial traces or expectation values. I didn't consider this, because you know here you have to consider these kind of scar products where this is defined in a space smaller than the other. Maybe some of you are not familiar with this kind of operation. So this is why I prefer to use expectation to compute the expectation values. But that is fast. I agree. So if there are no mistakes, this is the density matrix of the density matrix corresponding to the fermions with momentum k. Now the entropy is defined this way. So how can we compute the entropy now? So what are you using when you say this? Because the entropy is written as a trace. So every time that you have a trace of a function of a matrix, then you can rewrite it as the sum of all the eigenvalues of the function of the eigenvalues. So we have that the s of rock k, the entropy, is equals to the sum over all the eigenvalues, one from the dimension of the space, which is equal to 2, of minus logarithm of lambda i times lambda i. Well, these are the eigenvalues. This is the density matrix. So we have the density matrix. We have to compute the eigenvalues. In this case, the eigenvalues are very simple to compute. Why? Because this is already diagonal, this matrix, in our basis. Because the vacuum of k corresponds to this operator equal to 0, instead the presence of a fermion corresponds to this operator equal to 1. So this is already diagonal density matrix, in the basis that we chose. And the eigenvalues are therefore given by, there is an eigenvalue corresponding to the vacuum. Let's write just to indicate it. There is a lambda vacuum, which is equal to 1 plus cosine delta k over 2. And then you have the other eigenvalue, which is lambda corresponding to the presence of a fermion, a momentum k, which is equal to 1 minus cosine delta k over 2. OK, if you have problems in computing this kind of traces and to see that this is diagonal or whatever, and you are more familiar with spins, you could use a Jordan-Biener transformation to rewrite this Hamiltonian in terms of spins, this Hamiltonian, this density matrix. So you can rewrite everything in terms of spins. And if you prefer so that you can understand whether it's diagonal or not, and compute the eigenvalues, it's completely equivalent. What I'm seeing is that we have that in this space, we know a complete basis for this space, and the base is given by the vacuum of the fermion k, so no fermion, or a fermion, a momentum k. This was our original basis. And my claim is that this operator here is diagonal in this space. Because this operator has a definite value on the state with a vacuum or a single fermion. So it's either 0 or 1. If it's equal to 0, then you have the eigenvalues 1 plus cosine. No, I repeat the words forever. If, in that case, 1 minus cosine. So we have the two eigenvalues of the reduced density matrix, or the entropy. It's simply given by S of rho k. It's equal to minus 1 plus cosine delta k over 2, logarithm of 1 plus cosine delta k over 2 minus 1 minus cosine delta k over 2, logarithm of 1 minus cosine of delta k over 2, which we can call, for example, give me a letter. OK, let's call this G of k. Yes. OK, it's the momentum of the fermion. And we are just considering the state of k pair, a single pair, k minus k. And this is the entropy of a fermion with momentum k with respect to the other, the entanglement entry, when it comes to subsistence. So you see, it's a different interpretation of subsystem. Because when we discussed subsystem originally, we said, OK, you have some special subsystem, some part of your system, which was a physical part, and the rest. Here, instead, we have a subsystem in the momentum space. So we fixed the momentum. We say that the subsystem is just the particle with a given moment. And the rest of the system are the particle with the other moment. So it's just a different point of view, but it is essentially the same. OK, fine. So this is the entropy of one particle with respect to the other. And now the idea is to just count all the pairs that contribute to our entanglement between the subsystem A and the rest, so all the pairs that are such that one particle is inside and the other is outside. And we weight this kind of cross section with exactly the entropy of one particle with respect to the other. So what I'm saying is that the entropy of A, the original subsystem A with respect to this one, is given by the sum of all the entropies of the pairs such that one particle is inside the other is outside. And what is the entropy of one particle with respect to the other for a single pair, this one? So we have to sum all these contributions. Is it clear? If there are questions, ask. So the calculations are simple, but it's important that the general idea is clear. The entropy of rho k is also equal to the entropy of rho minus k, right? This is exactly equal to s of rho minus k, because the state is pure of the pair. The state of the pair is pure. There is no entanglement between different pairs, so when you consider the reduced density matrix of a pair, it's a pure state. So you can treat it independently of the rest. So indeed, this is important. Otherwise, how could we quantify the entanglement between one particle and the other if you find different results here? Yeah, let's say. OK. So we have this, and let's try to carry out the calculation of this to write the question that we have to solve. So what is the question of motion of a particle? How does it evolve? So let's assume that I have a particle here. It is pronounced here, and this is x, and this is time. And I want to follow the termolution of the particle. What is the question of motion of this particle? This is a fermion. It doesn't interact with anything else. It doesn't interact in fermions, which is working. And at a given velocity. So we have just to compute the velocity of the particle, and then this is just a linear trajectory. OK, we'll follow. So what is the velocity of a particle? We know the dispersion relation of the Bogolubov fermions. So we compute the dispersion relation. The velocity of the particle is the derivative of the dispersion relation with respect to the momentum. Is this clear? This should be clear. If you want in classical physics, if you write the Lagrangian equation where you compute q x dot is the derivative of the Hamiltonian with respect to the momentum. This should be a linear equation. So this is the velocity of the particle with momentum k. And then we have to follow the time evolution. So again, we have to count the particle inside and outside. Well, we have to do in practice. So we have to. First of all, the pair can be everywhere in our chain. So we have to integrate over all the position the x from minus infinity to infinity of where the pair is originated. The time t, the particle is at x plus v of k t. This is the x of t, the position of the particle. The time t is the position at the initial time, which I call x, plus vt. What is the position of the other particle of the pair? This is xk. Now we have x minus k of t, which is equal x plus v minus k. But in the particular case of the easy model, we have that the velocity is not the function of k, because epsilon is even. So you have that this is equal to x plus minus v of k t. This is the question of motion of the other particle. Then we want the one particle should be inside the subsystem. No. So we have to put some delta function here. Yes, delta function. Let's write a theta function, a characteristic function, where one particle x plus v of k t should belong to our subsystem A. I'm just using this with notation. Now it's not the mathematical notation. I'm writing what I'm telling you. So the particle at the time t is in our subsystem A. And the other particle should be outside. So here, I have to multiply by the other particle, x minus v of k of t times t. That shouldn't belong to A. The theta is the characteristic function. No, this is no problem. Then I have to integrate over all the contribution from the pair, the contribution from the pair. So yeah, I have to integrate over all the momenta. And I have to put the contribution of a single pair, which is this one. So yeah, this expression times g of k in the x. Because I have to consider that the pair can be produced everywhere. But then I want the only condition that I have to satisfy is that they can be everywhere. But when I consider the time t, there should be one particle inside the subsystem, which is this one. But the mate should be outside of the particle. My initial state is just this superposition of the pairs everywhere. You have pairs originating everywhere in your system. Yes, any time moment. Because it's written here. You have every time, every moment here. You can have every k in each point. Because indeed, this is a translation invariant state. So there is no privileged point. So each point is equivalent to the other. So physically, the idea is that these pairs are produced everywhere in this pitch. Yes? Yes? Exactly. Because this is the structure of the state. So we can always, this is exactly the initial state corresponding to a quantum point g that is in model. When you change the magnetic field, the initial state is written in this way. So this is actually our problem, which is very complicated. I want to stress that if you try to compute this kind of quantity analytically, it's very complicated. And I did it 10 years ago, but it's not simple. But anyway, you find exactly the same. You obtain the exact solution if you use just this semi-classical pitch. Ah, this one? OK. OK, OK, sorry. Theta of x in A is equal to 1. If x is in A, it's equal to 0. If x is not in A. Yes, it's there. And now we have indeed, OK. This was just an informal way to write this. Now we have to write explicitly. Yes, this is the entropy of A. This is the, OK. This is the entropy of the subsystem A at the time t. Now we can define what is the subsystem A. OK, now the subsystem A can be, for example, the spins from 0 to L. Everything is inherent under translation. So this is arbitrary for AMBA. So we can just say 0 to L. And so what does it mean, this theta function? This theta function means that the x plus v of kt should be between 0 and L. But we have also this other condition. The other condition is that x minus v of kt should be between 0 and L. So what does it mean? This means that x minus, sorry, sorry, whatever, should be larger, OK, no, this other condition means that it should be either larger than L or smaller than 0. And so we have to solve this problem. I'll write this integral. I give you four minutes, and then we'll do it. I'm going to sequence this calculation. So we have to compute this integral when we have this condition. So the deposition, x plus vt should be inside A, which means that they should be between 0 and L. And instead, the x minus vt should be larger than L or smaller than 0, OK? We have to fulfill all these conditions in our integration. So we have to find the correct domain of integration. Domain of integration, sorry. Hey, if you're not interested, you can just go outside. And it's your choice. So from this relation, we have that x plus vt is more than L. So this means that the x should be smaller than L minus vt, right? So x should be smaller than L minus v of kt, this. From this relation, we also have that x should be larger than minus vt from this. But now we have also to impose these other conditions. This condition means that x should be smaller than vt. It should be also smaller than vt. So we must remember here. And this means that it should be smaller than the minimum between the two, v of kt. From this transmission, we have that x should be larger that L should be larger than L plus v of kt. And so this means that it should be larger than the maximum between this and this. You are left and right, OK? You're right. So we first have to write this, OK? Yes. You're right, OK. This is an end. And then we have an end with one of the two. OK, sorry, I need to postpone it. So we have this, and we have either we have that x should be smaller than v of kt. So it should be this. But then, OK, what do you see? You have this. What you have? The other condition is that it should be larger than L plus vt. So we have the same maximum minus v of kt and L plus v of kt. And here, you have L minus v of kt. But clearly, this should be larger than this, because it was an end condition. So this means that this is possible only if v of kt is larger than minus v of kt, OK? So indeed, this is possible if theta, if there is a theta of v of k, it should be much larger than the minus itself. So it should be positive v of k. Tell me if there is something wrong. And here, instead, you have the L minus v of kt should be larger than L plus v of kt. So again, it means that now minus v should be larger than v. So it should be negative, plus theta minus v. And here, you have the whole integral in dk, and so on. And the function g of k here, I think we can simplify this expression. Now, how do we simplify this expression? Let's start putting some absolute values, just because I know the result. So because here, you have theta of v, then v is equal to its absolute value. I can write this. Now, v here is negative. So v is equal to minus its absolute value. I have to change the sign. So here, there is an integral in dk of 2 pi, or minus pi pi. OK, of g of k. Now, the integral over x of this domain is equal to minus this. There is no dependence on x in the integral. So we can carry out the integral. So this is equal to integral, minus pi to pi. In dk over 2 pi, theta of v of k.  Then we have the other part, which is theta under the g of k, plus theta of minus v of k.  Now, the minimum of this plus this is the minimum of this plus this, and this plus this. Minimum, the minimum between a and b plus c is equal to the minimum between a plus c and b plus c. So I can plug this inside the minimum. So yeah, I can simplify this. And here, I get a factor of 2 here. Then minus the maximum is equal to the minimum between minus the maximum between a and b. It's equal to the minimum between minus a and minus b. So here, I can plug the minus sign inside and transform the maximum into a minimum. So here, it becomes plus, minimum, minus, minus, plus.  So if I write everything, now it becomes equal to the integral dk over 2 pi theta of v of k. Then here, we have the minimum between l v of k t. There is a g of k. Plus minus v of k.  G of k. Oh, look. This is exactly equal to this one. So we find the integral dk over 2 pi theta v plus theta minus v is equal to 1 because v is either positive or negative. So minimum between l and 2 v of k t. And we have this function, g of k, that we computed before. This is amazing because we computed something extremely complicated. You don't realize it, but we're going to. So we found this very simple result for the entanglement between a subsystem of l and l and the rest in the easy model. And what is the behavior? If I plot this function, what do I find? No, what should be here? Yes, minimum, yes, it's right. What do I find? Now let's plot this. This is the entropy of the subsystem a at the time t. This is the time. So at time equal to 0, the minimum between these two numbers is 0. So it means that at the end of this, we find 0. Then what happens when you increase the time? So you have to consider the minimum between the l, the subsystem length, and something which is multiplied by a small number. Now the velocity, you remember, is bounded in the easy model or in the spin chains. You know that there is a liberal bits on velocity, which is the maximum velocity that you can have in the spin chain. So this means that as long as the time is not enough for the maximum velocity to be such that l equal to vt, then the behavior is just linear in the time. You have the minimum of this. It's always equal to 2 vt. So what you find is that there is a linear increase, exactly a linear increase, up to which time? Up to the time equal to tf is equal to l divided by 2is the maximum velocity. So here our prediction is an exact linear slope. But then what happens then when you increase the time, then there will be momentum for which this is still smaller than l, and there will be momentum which instead you have the opposite relation. So what happens is that this will start curving like this. And what happens when the time approaches infinity? When the time approaches infinity means that you have to compute the minimum between l and the velocity multiplied by an extremely large number. So this number can be smaller than this only if the velocity is close to 0.  So this means that in the end you don't see anymore the contribution from these low energy modes. And you only find this term. So in the end at infinity time, what you find is that this is the entropy. What you find is that it approaches l integral in dk over 2 pi of g of k. Yeah, let's assume that the time is equal to 1 million. And the maximum velocity is equal to 1. And then you have that the end. So the velocity is something like this. This is something like this. This is the velocity as a function of k. Now the momentum for which for large time this becomes larger than this are confined around here and around here. So you are a strict in this region, increasing the time. This region becomes smaller and smaller. And so the integral over this region becomes smaller and smaller. And in the end they don't contribute anymore. And so you have always the opposite situation where this term is larger than this one. And so because there is a minimum here, you always pick l. So this is the quality of the behavior of the curve, not the entropy. And what you find that if you do the calculation exactly, you find that this is exactly what you get when you start the limit l goes to infinity of the entropy of a divided by l. Let's try this. So what I mean, yeah. So if you start this kind of this limit, so you divide this expression by l, if you want. So you see that this becomes a function of t over l. Now if you take the limit of large subsystem at even a ratio t over l, you fix t over l, and you take the limit of large subsystem, which also mean large times, then you obtain analytically this expression. So this is the exact result. So we obtain this result in a semi-classical way and we obtain the exact result in the asymptotic behavior. Why this is semi-classical, this approach? Essentially, it's because our particles have a given position and momentum. So we are following the trajectory of this particle. I say, OK, at the given time, the particle is here. Quantum mechanical, this is impossible because you have the Eisenberg determination, for example. OK, so this is a semi-classical calculation. Nevertheless, we obtain the correct result. And then, what is interesting here is that the semi-classical, because in order to obtain these results, we follow the time evolution, the trajectory of the particles. But the trajectory is defined only for classical particles. And it's not classical, but semi-classical because we started from a quantum model and we obtained the correct dispersion relation. We checked that the state is factorized, so we didn't start from. OK, and do we see from here? One interesting part of this result is that we see that it's sufficient last time. The entry becomes independent of the time. So you see that the approach is a stationary value. And this is an example of what I was telling you yesterday that when you consider the expectation values of observables after a quantum quench, then in the thermodynamic limit, the limit of infinite time exists. So you don't have this kind of quasi-periodicity. The limit exists. And all these observables behave in some stationary way. Moreover, as you can see from here, the entropy of the subsystems in this limit, in the limit of last time, becomes extensive, proportional to the subsystem length. We should remind you what happens in statistical physics when you have a finite temperature or whatever. So just from this picture, you can measure. There is some hope that we can describe the system at large times using some kind of statistical way. For example, introducing some statistical ensemble. This is what we have done in the past. OK, but it's very late for any calculation of this kind. And so let's see. I just give you the general picture. And maybe if you're interested, I tell you very briefly, just a matter of two minutes, how can you do this kind? How can you see in the easy model that indeed the limit of infinite time exists? It's just two minutes, what you have to do. So I told you that in order to compute the expectation values of observables, you can rely on the big theorem. Big theorem means that you just need to compute the expectation values of the Majorana fermions. Now you are interested in the time evolution of the Majorana fermions. So you actually compute the expectation value of the Majorana fermions at the time t. I told you, I wrote the relation between the Majorana fermions and the Bogorimov fermions again. So what you have to compute is this kind of correlations. In the end, you find the relations like this, one, for example, 2l minus 1 of t to n minus 1 of t is equal to delta ln plus 1 over n sum over k over e to the minus i l minus nk sine delta k sine of epsilon kt.  Then when you take the thermodynamic limit, again, this sum becomes an integral. So you have delta ln plus the integral over dk over 2 pi of the same expression. Now you see that this is an oscillatory function. Use the Riemann-Lebesgue lemma. And so in order to prove that this integral and the limit of large times become approaches 0. And you find the same also for the other correlations. So you always find terms that survive the infinite time limit and you find terms that oscillate. So you have to cancel that. And you prove, indeed, that the limit exists. This is the Riemann-Lebesgue lemma. The fact that this limit exists. Limit of infinite time. This is, OK, first, I can see the limit on Gaussian feet and then the limit equals to infinity in order to simplify that term. OK, this is just to show that there is relaxation in this model. OK, what is the general picture? Indeed, OK, just a moment. Here, it simply means that we are relaxing. All the expectation values are relaxed to some values that we don't know. And then the question becomes, how can we characterize these expectation values? And the picture is the following. So in order to characterize the behavior of the state at very large times, you don't need to know everything. All the expectation values, all the time evolution. You just need to know the expectation values of the conservation laws. So the quantities that are conserved. So if in your system there is a quantity that is conserved, it means that its expectation values is the same at any time. So if you will consider, let's assume that you find a quantity, q, that commutes with the Hamiltonian. Now, let's consider the time evolution of the expectation values of this quantity, q e to the minus i h t side 0. Because this commutes with q, then this is equal to side 0, q side 0. So without doubt, you know the expectation values of this quantity at any time. It is just given by the expectation value at the initial time.  Because you know that whatever the final description will find at infinite time, it should be compatible with all these constraints. Then, OK, let's just imagine that we are in a generic situation. Let's imagine that the Hamiltonian is completely generic, no symmetries. There are no symmetries in the Hamiltonian. So if there are no symmetries, you don't expect the existence of this conserved quantities. So you can just say, OK, the energy is conserved. That is true, because we consider an isolated system. So h is conserved. And this is a derivative respect to the time of the expectation value of h is equal to 0. But OK, we can't find all these other quantities. So the idea is that when we are considered the limit of infinite time, we are somehow losing some information about our system. For example, we are losing the information that could allow us to go back in time. Because we are replacing, for example, all this curve, which is a kind of complicated curve, with just a line, just a value, the asymptotic value. So we are not able to go back anymore. And so the picture is that, well, as a matter of fact, the time evolution is making us lose all the information about the system. And the only thing that we know is this one. It is given by the constraints to the dynamics, given by the conservation laws. So in this case of a generic system, the constraint is the energy. And which is the quantity that measures the information about the system is the entropy. So the conjecture, it was a conjecture, is that you can describe the stationary values at infinite time after this kind of quantum quenches by replacing the state. So here we are considering the time of our mistake, psi t. And now I'm saying, because I'm interested in the limit of large times, so I can replace the state by density matrix rho. And how do I define the density matrix? The density matrix is that operator that maximizes the entropy, sorry, not the entropy. So this is maximum. But you have to follow all the constraints. You have to impose all the constraints. So in particular, you have that the energy should be conserved. So the trace such that the trace of rho h should be equal to the expectation value of the Hamiltonian in the generic case. But this is nothing but the thermostat. They keep the distribution, Boltzmann distribution. Because this is how you obtain, for example, the canonical distribution. You maximize the entropy, and you fix the energy. When you do this, you can, for example, use a variational approach. Anyway, what you find when you impose this is that rho is implies that rho is equal to some constant times e to the minus beta h, where beta is a parameter. Generally, beta is the inverse temperature. Now beta is a parameter which is fixed by this condition. And this is what we call thermalization in closed systems. So we are in an isolated system. We focus on a part of the system, because we are in the thermodynamic limit. And we are saying that I can replace my extremely complicated time-evolving state by a simpler density matrix, which is essentially the density matrix of the canonical ensemble in the generic case. This holds when there are no symmetries in the Hamiltonian. Let's assume now that you have a symmetry like the irrotational symmetry for a spin chain. For example, let's assume that what happens if sum over L over sigma Lz is conserved? Or the number of operators. If you have the number of particles conserved, then you have to do the same as you did in your course of statistical physics. You have to impose all the other constraints here. And for example, here you find something which is the canonical, the grand canonical ensemble. So if you have particle numbers conserved, so you can expect an expression like this, grand canonical. Now, OK, and then you can compute this using this partition. You know how to obtain this or not. OK, generally, when you want to maximize a function given some constraint, you introduce Lagrange multipliers. I don't want to enter into details, but this is what you find. What happens if you now have more constraints? Yeah, you have to put. You see immediately that you have to include here all the conserved quantities of the Hamiltonian. And other cases when you have more conserved quantities than, for example, the Hamiltonian, the particle number, the angular momentum, or the momentum are there. Well, if you read the Landau leaf sheets, then in the first page they said, OK, the integral of the motion are just seven. So the three angular momentum, the energy, then the momentum. But this is for generic systems. If you consider a very special system like easing, you can have many more conservation laws. As a matter of fact, in the so-called integrable systems, like easing, you have infinitely many conservation laws. So it means that here you have to place infinitely many operators. And instead of considering just a thermal ensemble, you have to work with something which has this form, the exponential of minus sum over j up to infinity of lambda j qj. So when you have infinitely many operator here, this is called generalized Gibbs ensemble. So it's a Gibbs ensemble. As far as we have just a finite number of conservation laws, it's called Gibbs. Instead, you have to deal with infinitely many operators that this is called generalized Gibbs ensemble. I think that you have done. And I'm happy that the last words that I said are generalized Gibbs ensembles. So what are the questions? This was a conjecture. And the idea is that when you consider the expectation values of observables, that you have all the, when you expand in the basis of the Hamiltonian, you have all the phases. And you expect that there is this kind of defacing mechanism where all these phases in the end simplify like in this case. And then this means that actually you are losing all information about the system. But the part which is clear, you should be clear and satisfied from the time equals 0. If you want something else, I can tell you that because you are interested in the limit of infinite time, instead of considering the infinity, you could consider just the time average. Although here there are problems of limits. But anyway, if you consider the time average, then you will find that you can describe everything you see with the diagram ensemble. And then you have the eti, if you apply eti, etch, and so on.


Alrighty, so we have our new construction update for both Carowinds and Canada's Wonderland. The construction update pictures provided today are from Carowinds fans only. Go check them out on YouTube and Facebook. Thanks for letting me use your pictures. They're super high quality, so thanks so much and absolutely love them. So if anyone doesn't know, I'm combining the construction updates now just so we're not making two separate videos for two separate projects. And it's a little fun to kind of have a mini little race, I guess, on the construction project. They're very similar in timeline. So as you can see, it does look like a few more track pieces have arrived at Wonderland. I also was able to film Sledgehammer doing its little, I guess, like boot up test cycle. So they spin each compartment separately, one at a time, I guess, to check to make sure they're working. Then they'll raise and lower the arms to make sure they can do that. Then they bring it back down and they start spinning them all, the whole frame, all on the base, sorry, it spins. And then they'll put it up and do the same thing up there. And then I guess that's how they tested and get it ready for operation. So that was cool to watch. You'll see that throughout the video, kind of them, getting that set up and ready for Halloween haunt tonight. In terms of construction though, the exciting news is half of the drop is up. So as you can see here, the drop is on its way up. And I'm hearing that maybe they'll even be done, the lift, drop and top portion by next week. That's what I'm guessing by next Friday, they may have it all topped off, which would be super awesome. I mean, Carolyn's is off to an amazing start too. From these pictures, you can see, they've got the brake run in, they got the station done, the station track, obviously. They got the JoJo roll, they have the turnout of the JoJo roll into the launch or that little pre-show building. And then they have the launch done all the way up to the vertical loop. And the supports are half done for the vertical loop, as you can see right here. So both projects are off to a very quick start. From what I'm hearing though, Carowinds may also not be in a delay because nothing's gone wrong, but we may not see too much action over the next couple of days or anything because the parts for the vertical loop, the actual track pieces for it are not on site yet. That's what I'm hearing. Again, I have no idea. I'm not from Carowinds or South Carolina or North Carolina. So I have no idea if that's actually true. I can't go check the store shop but that's what I'm hearing parts for the vertical loop are not on site quite yet. Whereas Yukon Stryker is now out of its delay and in full force. And again, they help from what I'm hearing and seeing to have this topped off by next Friday. So that would be awesome too. As you can see that little V frame, looks like an A frame, but it's a V frame support on the drop and they'll get the rest of it put up next week. They got the pieces already ready and assembled again catwalks and all right there. They'll probably bring in the turn pieces for the top of the lift hill on Monday and they'll get that all set up and ready to go. So both projects are off to a very quick, very quick start. Again, Yukon Stryker had its tunnel and track pieces installed very early in the season and Copperhead strike is just taking off as of this week with an impressive start. So it is very tough to call which coaster will be done before the other because both coasters are still pouring footings for the actual coaster. So both coasters are still pouring footings, Wonderland is digging theirs, Carolyn's is digging theirs, they're pouring them. In fact, if you look closely in this video, when I pan over to the left, you'll be able to see the amount of construction going on over there and I'll be in the park tomorrow with my team, grabbing some footage and zooming in on some things, getting you some really awesome shots and all that. And hopefully I'll actually have some exciting news to announce to you guys shortly. So stay tuned for that. Hopefully some exciting news to announce to you guys shortly. I'll know very shortly what's going on with that and then I'll provide an update on this exciting news. But yes, so not much else to discuss. Most of you already know what's going on at Carolyn's because the live cam, Wonderland doesn't have a live cam. So most of the updates actually come from me. But yeah, as you can see, Yukon Striker is looking really good. They're in the 90 degree portion of the drop and it looks phenomenal. You can actually see that 90 degree drop, the little bit of it from the V frame up to the tip there from the parking lot and from Jane Street. You can see it from all the way over there. So it is visible now in the skyline and as of next week, it'll be even more visible. I also want to take some time to apologize for my nasally voice for those of you that know me. I am sick. So I'm so sorry if the sound sounds really because of my nasally voice, but I just love the fall colors on Wonderland in the left there. I absolutely love it. It's just gorgeous. And the crane actually blends in with the fall colors, which is funny. And so does the colors of Yukon Striker. So it's just, it's a really beautiful shot. But as you can see, they're digging footings over there in the Timberwolf area. So awesome shots. I believe in a second, I'm actually gonna start. I included some of the cinematic footage I'm gonna be putting in the documentary. Yeah, I think this is my cinematic shots that I'm getting right now. I can't tell. There is some cinematic shots in here where I just slowly pan and grab some really awesome shots. This is it. I'm gonna end up talking here because there's nothing else to talk about, but I'll be in the park tomorrow to see if I missed anything up close and personal. And you'll have an update tomorrow. Anyways, thanks for watching guys. Hit that like button, subscribe if you haven't and share this video for others to watch. Hope you guys enjoy this. Carolin's versus Canada's Wonderland construction updates and stay tuned for more. Have a good one guys. Bye.


This video will show you a quiz. It will also walk through different types of questions that you will see on practice quizzes in Canvas. I'm going to begin by clicking on the blue button that says Take the Quiz. The quiz directions are up here at the top. This quiz says to follow the instructions to practice answering different types of quiz questions. This is a practice quiz. First, we have true-false questions. For these types of questions, we read the sentence. If the sentence is correct, select true. If the sentence is not correct, you select false. So question one. Practice quizzes will help you learn English. That's correct, so we're going to mark true. Question two. You can take a practice quiz only one time. That is not correct, so let's mark false. Next, you have multiple choice questions. For these, you read the question, and then you select the correct answer from a list of answers. So the question is, what is a common type of quiz question? Our options are multiple choice, hide and seek, and discussions. Multiple choice is the correct answer, so I'm going to click on that. Another type of quiz question is fill in the blank. The sentence below is missing a word. Read the sentence and type the missing word into the box. So the sentence says, practice quizzes will help you learn blank. I think the answer is English, so I'm going to type English. Another type of quiz question is a short answer question. You answer the question and you type your response in the box. So the question is, do you like practice quizzes? Why or why not? I need to click on the box and then I can type, yes, I like practice quizzes. They help me to learn. Another type of quiz question is matching. For this, you match words together. You read the word on the left and then you click on the box to the right to select the correct description. For this question, you're asked to match the type of quiz question to the description. So the first type of quiz question is true or false. If I click on this box, I'm given a variety of answers. One is, select the correct answer from a list of options. Another answer is type your answer to a question. The next is, select if a statement is correct or incorrect. And the last one is type a missing word. For true false questions, we select if a statement is correct or incorrect. So I'm going to click on that one. And you do the same thing for the other types of quiz questions. So multiple choice, we select the correct answer from a list of options. For fill in the blank, we type a missing word. And for short answer, you type your answer to a question. Once you're done with the quiz, hit submit quiz. Once you've taken the quiz, you can scroll down and see your answers. And then you'll see feedback. If you got the answer correct, it will say correct. If you didn't answer a question, it will tell you the correct answer.


Picasso is stretching and getting ready for his debut. Are you getting ready because we're about to tell a story about you and what you have been up to lately? Oh, yeah. Yeah. Hello, my fellow parrot lovers and sniffers. Do you guys know what sniffer is? Sniffer is someone who loves to sniff birds. Which most of you probably are. Or maybe you don't know anything about birds. Maybe you're not a sniffer. If not, welcome to the channel.  Although he's kind of shocked, he's not so sure what it's going to be about. But I bet he has an inkling when I remind him he's going to be very embarrassed. So we are about to out Picasso. Now, before I go on, I kind of feel like it's been a while since I sat here and did a story time Sunday with you. Last Sunday, I didn't get to put a story time Sunday out because so much was happening that it was just so difficult to live life properly and be healthy and put out a video at the same time. Picasso is like, get to me, get to me. But I have been doing some videos of the day so you guys can see my birds and their antics. And if you guys are watching consistently, you know that I went to Texas this past week. So when I came back, I made a video of the birdies being really excited and seeing me afterwards. Which I think is really cute. Jersey, Cody and Vinnie are dancing. Somebody left me a comment and said, can I please see Picasso scream? I want to know what he sounds like because he's always so quiet. Well, here he is screaming. He's probably not going to be quiet throughout this video because it's a little bit past his bedtime. And Picasso is very, very specific about his bedtime. Picasso does not like to be out after his bedtime unless he's watching a movie with me and he is getting sniffed. I'm not kidding you. So if I sit here and maybe sniff him, he will be okay. No, see, he wants lip gloss. You cannot. Why do you always want lip gloss? Oh, he smells so good. So I'm not going to sniff him because you guys know what happens when I sniff him.  Okay. Okay. I'll try to stop. Okay. Before we go on, let's do the Engaged Not Caged Featured Bird of the Day and also of the Holidays.  If you would like to see your bird featured on the Instagram page or maybe on this channel, remember to hashtag Engaged Not Caged. Meaning that you will even engaging with your birdie, letting him be free, letting him hang out, and not caging birds. So today I have chosen... I really had to go for a festive one this week because it's the holidays. So out of all of your holiday Engaged Not Caged birds, the Engaged Not Caged bird of the day is lolly of horror. Seriously guys, look at this macaw. How can we not choose this extremely festive family? Everything about this picture is amazing. Down to the sweater, the bird, the Christmas tree, the whole picture spells out holidays and I love it. So I think that this bird right here is a winner. The bird is also in festive colors. Red, green. Festive colored feathers. Perfect. I just love it. So go check this bird out. He's got a really, really cute life with his family and I think you guys will really enjoy the profile because it's quirky and funny and I've been really enjoying it too. So that is our Storytime Sunday Engaged Not Caged bird of the day. Let's get on to what else is happening. First of all, Harry is in here. Second of all, you see this toy here? Okay, this toy. Yeah. Well, Harry has stolen that toy from Bella. Bella is a human baby that belongs to my best friend Tracy. You guys probably all know Bella anyway. Beautiful baby. And my mom wanted to get Bella's attention. So she wanted a squeaky toy for Bella.  It doesn't matter. It's a stuffed pet. It doesn't matter what section it comes from in the store. So we get that for Bella and the moment Bella dropped it, it became my dog. And obviously it's not going back to Bella, but Harry has been following it all over the place. Now, what about you, mister? What do you think you have done that is worse than Harry stealing Bella's toy? You look so innocent. You also look like you forgot, but I do not believe that you did. Guys, I don't even know if I should tell you the story because it's actually so traumatic to me that I don't think I can tell you the story. I think I just need to show you what happened. I'm sure you know, this is not the first time this has happened. I'm very reluctant to show you guys this. Well, no, you guys should see, but let's see if the guilty party feels guilty. Does he look guilty? What do you mean no? It should be very guilty. This is what has happened. Can you see it? Do you see any keys on this keyboard?  This is what my laptop looks like now. I'm very upset about it. Cute little laptop case cover. I love books. Yes, I'm kind of book crazy. So Picasso did that and I'm very, very upset about it. It's made everything hard. It's made everything hard. It's made it hard for me to edit any videos for you because obviously I can't use my laptop. Not only that, but at the same time that happened, my tripod broke because I took it to Bird Mart, which I'm sure you guys watched that video.  Actually, the reason I'm really reluctant to tell the story is not because Picasso did this. It's not because I'm depressed about it, although you guys should never get depressed over material things. These things are not a big deal. It's fixable. Obviously you should never get upset over material things, but I will admit, I feel kind of handicapped right now without my computer.  What if it's the same for yourself, anything? So this is what happened. This is why I'm so upset at this. It's not really Picasso I'm upset at because he's just like the shredder as you guys know. So I'm not really mad at shredder for doing this. Okay, yeah, I am. Look at him. He's all offended. He's mumbling in my ear. Mumble. I'm mad at myself because I know better. I'm the type of person that like I live in Los Angeles. There's earthquakes. This is me. Like there hasn't been an earthquake. Okay, lies. We had an earthquake a few months ago, but before that, I don't think I felt anything in a few years, but I'm still the person that like moves the glasses off the table when they're full and I won't leave the house or the glass anywhere or anything in case like of an earthquake. I put all my equipment back all the time in case of an earthquake. I'm that person. I know never to leave a laptop open around a bird. If you guys haven't learned that yet, you will. I know all of these things. That's why I'm so upset, but I just don't know where I lost my conscious thought. So here's what happened. I was downstairs having lunch with the birds. Lunch with the birds is very important because I want to make sure that they're all full and they're all healthy and that they all have eaten something and also that they've eaten something really good for them and not just the pellets or seeds that are in their bowls. It's really important for me to be conscious that they are eating. So I always have my lunch with the birds and it's something that we can all share. Yes, Harry. Why don't you just step down on my laptop too? And what are you looking for? Oh my God, guys, you don't even know what Harry is doing right now. This dog is so smart. Along with this toy that my mom got for Bella. I know I'm going back here. My dad got a lamb from Ireland that he loves to get babies. It's like this lamb. And like if you ever shop in a tourist store in Ireland, these lambs are everywhere and they're so cute. So he got it for Bella and he's really excited except my dog has fallen in love with that too. And I told Tracy, you can't just leave Bella's stuff all over because dog toys, kid toys. There's a lot of confusion here for my dog. But why he left is because he sees the lamb on top of the dresser. Like he sees it high and it came down to see if he could like jump up there and get it. I don't even know. What do you have to say for yourself? Get back to you. Okay, we'll get back to you. You are naughty.  So here's what happened. I ate with the birds. Everything was fine. And then they were on their stands and Jersey gets all like food coma-ish which I love because it means that she's like content and not going to pluck and then Picasso generally sits there with her. And I guess I brought down my computer to answer comments from you guys and left it open. And from there I have no idea how I lost conscious thought, ended up upstairs. And I actually remember kind of zoning out upstairs and looking at my phone which is not normal for me. And then I came back downstairs and the laptop was open. And she was where she shouldn't have been. The keys were all over the place. And I think I went into severe shock because they had damaged other things too. And I just couldn't. And I just lost my emotions. I didn't yell at them. I just started crying. I didn't even know. I was like, why a laptop? Now, here's the funny thing. We have an IT company. So this fixing this kind of thing is like extremely easy. When I say we again, I use the term loosely, George is going to fix it. But if any of you know anything about Macs and opening them up to replace this keyboard, there are like 80 tiny screws like itsy bitsy screws. Like it takes forever, extreme precision, extreme carefulness. And so I know that by the time anyone has the time to fix my laptop, it's going to be a little while. But what frightened me most that I got so upset about was not that the bird did that because I should know better. It's because I should know better. I am extremely cautious. So that wasn't what upset me. What upset me was the loss of consciousness. Because when I got downstairs, I hadn't even remembered bringing my laptop down. And if I ever brought my laptop down, which I did because I was the only one in the house, I don't know when or where I lost any kind of consciousness to think that I shouldn't close it. Or to just leave it open. Like, I know not to trust the birds. No one knows better than me not to trust these innocent little feather heads that look so beautiful and sweet and steal my heart and make me forget everything that they've done. I know not to trust them, but I just don't know where or what kind of stress I was under that I can't even put the piece together of me bringing that laptop downstairs at all and opening it. I definitely would never have opened that laptop. And I think the bird did it. I think Jersey was the accomplice. I think she lifted it up and opened it. I know she could do that. And I think he went in there and just went deep in it. So I was really upset mostly because I was worried about myself and why I wasn't like mentally present. It's just so weird. It was just the weirdest thing.  And guys, I kid you not, before I made this video, I'm like, it'll be fun. Like Harry will be in here. And what reminded me was Sandy was in here. And so I brought Harry in and I'm like, what could he possibly do in the video? Especially since Sandy's not here. He can't find anything to do disturbing. There's no way.  Like how do animals always find something to do? Now, if you want to know really what happened after they ate my laptop, I went into some sort of shock. Now, you guys got to know something about me. I'm an extremely motivated doer type person. I don't zone out very much and I don't like sit around on my phone at all. In fact, a lot of people think I'm not very communicative on like responding to texts and stuff because I just have like better things to do like make videos for you guys. Honestly, I can't accomplish everything in the world at once. So, see, I forgot what I was going to say. So maybe I do zone out. So I remember what I was going to say. I kind of like go, go, go during the day. That doesn't mean I get out of this house. It just means like I don't stop doing stuff because I got to make videos for you guys. I got to take care of the birds. I got to keep the house clean. I got to do whatever else I'm trying to accomplish in the land of me making my dreams come true like movies and music or whatever. So I got to figure out how to do all that per day. So not much affects me, but if I get any kind of shock or fear, it's like my body freezes up and I freak out. And the only thing I can do is go to sleep. Like I get like extremely stressed, upset and then narcoleptic.  Like just like, like some sort of weird narcolepsy. Okay. Like, like, oh my God, the day is over. I can't deal with it and it doesn't happen to me often. But if I get any kind of extreme emotion or if I get hurt, like let's say I, I don't know, went to the dentist and nothing big happens. But like, you know, it was just too much and after that I fall asleep. Okay. Just like even if I get frightened, like my body stiffens up, something happens and I just like want to fall asleep. Even my muscles permanently for the rest of the day tense up so bad that I can't move. That's happened to me when I was on a bicycle.   You know, like those skateboarder type people, whatever he does, all that kind of stuff. And he jumps his bike right in front of me.  And so like that was it. I had to get off the bike and walk it and I couldn't, my muscles couldn't even move. It's like I can actually feel like some sort of adrenaline get released in my body and paralyze me. It's the weirdest thing. So something like that happens when he ate my keyboard because I was in shock of myself. I was like, where was I? Like where was I? Like how did I just zone out? Like I don't even know. I think I just got shocked because I am the bird expert. Picasso is like, yeah, you weren't ready for me. No, because like I'm just like an expert at hiding stuff and keeping everything perfect. And if you guys remember the last time Picasso ate my keyboard, which was my desktop and also ate my memory cards. Somehow it was because my brother and mother and father and whoever, they all blamed each other. They were negligent. And apparently Picasso figured out how to get upstairs, get into my bedroom, get into my editing room and get all that stuff. That right there is negligence. What happened to me? I don't know. I think I just zoned out. So I was basically long story short or not short frightens my by itself. And then I just panicked and then I was afraid to tell George what happened because he was going to be the one to fix it. And it was before I did the booth and he said, just focus on what you have to do and ignore this or make a video about it. And my video would have looked like the crying cat lady. I almost turned on the camera and then I'm like, we're gonna get a grip on yourself. It's just material stuff. You know that it can be fixed. It's like just it's just also going to cause me so much difficulty in editing and everything now because I can't use my laptop. So it just makes things a lot harder for me than they need to be. Yeah, you. Yeah, I see you. But he smells so good. So hard. So that's my story. My birds are naughty. You guys know that Picasso is a shredder. He looks really innocent here on my shoulder. I know you guys love him. I love him. I'm crazy about him. But owning a bird is very, very interesting. Let's just say a lot of you write into me and say, how do I prevent my bird from eating my laptop or my wires or anything? And I'm like, you got to hide them. You got to move them. That's kind of what you got to do. But if it's a permanent thing, then I have other tricks and other videos that I have given you guys. But that is the story. Let me know in the comments if you think Picasso is naughty or not because, you know, Santa doesn't come to naughty birds. What do you think? You don't care? Yeah. So that's my story, guys. I hope you guys enjoyed my traumatic story. Life goes on. Don't get wrapped up in material things. It's all fixable. Right? This is life. Don't sweat the small stuff. Harry, don't sweat the idea that you cannot eat the lamb that belongs to Bella. He's pretty upset about that. Now, before I go, there's some things that you should know. One, if you are a new bird owner or you are a longtime bird owner but want to see the products that I love for birds, check the link below. It's called Products I Love. I have so many exciting things in there for birds. Some of the things I have in there you've seen in my aviary for the birds. Things that Picasso loves. So check those things out. Bookmark the link so every time you want to remember what they are. Everything from food to cleaning supplies to fun things to toys to everything you can think of for your birds is in there. Even vacuums because I'm a psycho and I like to make sure everything is clean and I can't live without my Roomba. So check it out. See what's in there. And Parrot Station is approaching 40,000 members. 40,000 members on Parrot Station, guys. That's the Facebook group for parrot lovers like yourself. If you're not on it, come join it. If you're not on Facebook, get on Facebook just for these birds. And if you often get depressed by everyone else's Facebook posts, you will love joining Parrot Station because your whole news feed will become just about birds, just like we like it. Right Picasso? Yes. So that is another exciting thing and even more exciting. My channel is approaching 50,000 subscribers. That's really exciting to me and birds and Harry right now. So please, please, please, if you're watching this, subscribe, like, comment, share, educate people on birds. I love it. Remember to spread the message about birds being engaged and not caged. And on that note, if you want to follow me on social media, it's at Marlene McCohen on Instagram and Twitter. And we also have at Engage Not Caged on Instagram. I look at Picasso like he's going to speak.  You want to leave your socials? No? Okay. And if you have not gotten your sniffer shirt yet, there's still time to order it. Then hopefully it will come to you before Christmas. So check out the link for sniffer shirts below. We have sniffer, we have sniffer's anonymous, which kind of goes with our sniffer's movies and we have parrot station shirts. So check them out. I still have some left. I don't know who's going to get them. I can't, I can't even focus with this word. That is it guys. I love you so much. Thank you all for watching. Thank you all for subscribing. Thank you all for being so much fun and so engaging. I love you all. Bye.


Welcome, Dr. Mark Levine, commissioner of health in the state of Vermont. We wanted at CCTV, which is a community media center based in Burlington. We wanted to do some opening of our facility and in the process of deciding how much to make our facility available to community members to produce video, we thought we better ask Dr. Levine really what, how to turn the spigot. So I wondered if you could start with where we are today on September 17th in terms of the well-being of Vermonters in relation to COVID-19. Does that all? Sure.  Yep. So you know, I will piggyback onto the esteemed Dr. Anthony Fauci who of course was here on Tuesday and basically said he would like to bottle up whatever we're doing right and take it with him to other states and that we are a model for the country. So that's his language, not mine, but to support what he said, the fact of the matter is, we have succeeded in achieving a level of virus suppression that would allow us to turn the spigot in the ways that the governor and our whole leadership team have chosen to turn it. So opening up aspects of the economy, increasing the size of groups of people who can gather together, opening up colleges, opening up schools now, all of these are being done very deliberately and very cautiously, but I shudder to think how we would do them in a state that had a much higher level of virus prevalent in the population. So when we are touting ourselves as the case, the state with the lowest new cases per day and the lowest percent positivity level of our tests being well below one percent with abundant testing occurring, we can feel pretty confident in making some of the moves that we're making, but I will again reiterate Dr. Fauci's counsel, which is, Vermonters, don't let up, don't let your guard down, continue to do all the right things and we can continue to enjoy the benefits, if you will. That must have been so encouraging for you to have Dr. Fauci give us this vote of confidence. You've worked so hard, you and the governor in the state, so it must have been kind of thrilling to hear from him. It was actually, yeah, we didn't like put the words in his mouth that he said what he wanted to say, but the bottom line is again, it is hard work to get to where we've gotten to and it's not my hard work alone, it's the work of Vermonters and we really need to emphasize that. If people did not adhere to all of the guidance that we've been providing as a team, I don't think we'd be in the place we're at. So I think it's hard psychologically for people to keep vigilant when they see that the numbers are low. What do you think about the fall and winter, the prognosis for the surge? What's Vermont, how are you seeing that, what kind of impact do you think it will have in our state?   One thing Dr. Fauci was very strong about saying was the virus does win and if you do let it enter into the picture again, it will do what it normally does. So again, we have to be as vigilant as we can. We know that one aspect of summer ending fall beginning is people will congregate more indoors than they were outdoors. So already that increases the chance of more cases occurring. Today though, there's still occurring in a low prevalent state where very few of the people who are congregating indoors will have the virus. But of course, we're a tourism state. We will have lots of people coming in to do leaf peeping and eventually skiing and we want those activities to continue. So again, when we do have people congregating indoors, we need to be very vigilant still because it's not just remoders who'll be doing that. It'll be people from all around who aren't necessarily ill-willed and have poor intentions, but this is how things happen. So, you know, opening something as big as college or all of our K-12 schools, all of a sudden takes populations of younger adults as well as those who are in their supervisory and teaching positions from being in a stay-home setting or a stay-pretty isolated setting to all of a sudden every day being in a more congregate setting. So we've always forecast the fact that we do think numbers may go up slightly when we do this, but slightly is different than we'll have a real problem on our hands. And I think, you know, we've already seen the number of college students that tested positive was very low. You know, we're talking less than 40 students in the arrival testing out of literally 40,000 tests. And in our K-12, you know, we've gone through a period of time now and we've had only two schools that had very small numbers of cases. And frankly, those cases were students who were just coming off their summer break. So it's not like we put them all into a school and we had a whole bunch of people contract the virus. These were really people just who were ill as they were coming out of the summer and had a day in school when they were feeling fine and then that night they weren't. So we need to watch closely on the schools, but so far so good. Everyone's really upbeat about the way the reopening of schools has occurred. You mentioned the rapid testing or the dissemination of testing.  Do you see that coming to Vermont? I do because the federal government is going to be providing a fair amount of rapid testing, access to rapid testing. They're targeting places like long-term care facilities right now. They envision, I think, them going to schools and other settings. So it won't be like every person has their own set of test materials and can just at will do a test before they go out and do something. But the reality is we need to learn a little bit more about the performance characteristics of these tests and we want to make sure that people aren't falsely reassured by doing a test that might miss 10 to 15 percent of the positives. So we have to be very careful as we walk this new into this new territory. And indeed, we're learning more about this all the time, but most of these new tests are so new they don't have a huge track record and a lot of experience. So we'll have to be very careful, but I do see them entering use in Vermont for sure. So I think it's starting to dawn on all of us that this will be in this. I'm not sure what the word is, but this period of time, probably through 2021 or maybe a year from now that we really have to keep up this vigilance for another year. Do you think that's a correct assessment and can you talk about the impact the vaccines might have on our way of life right now? Yeah, so you're referring to what many of us term the new normal. For me, the new normal is we will identify cases throughout the state, but not hopefully large numbers of cases. For you, the new normal is this life that I've had to adapt this lifestyle isn't over yet and I can't act like I used to act before the pandemic. So though the most optimistic of political figures will talk about, you know, November and December, the more reasoned and cautious ones will talk about the fact that even if we have a couple of great vaccines that are ready to go, they're not really ready to go. In November, December, they're just going to be available to maybe 10% of the population. And it's going to take a while to rev up manufacturing of those to get that to the point where everyone who wants one will be able to have one. And that's where you're talking maybe a year from now. I would like to be more optimistic than a year from now, but probably not just a few months from now either. So somewhere in between. And so it's really challenging because we want to have the population be optimistic and see the light at the end of the tunnel while at the same time not letting their guard up and not assuming that the quick fix is here and it's just a little more time and I'll be fine. So I think this change in lifestyle is going to last well into 2021, probably towards the second half of it. And can you just say a little bit more about rolling out vaccinations and sort of how that works? Because again, as you said, it's not as if a vaccination comes out and all of a sudden everyone is vaccinated. What do you see as a rollout in the adoption of the vaccines?  First of all, the trials have to come to a conclusion and the trials have to conclude in some in such a way that the public will have confidence in the scientific integrity of the trials. The effectiveness of the vaccine will be visible to everyone in terms of how effective it really is. And if there's any safety issues, they will have come to light. Knowing that unfortunately some safety issues only come to light when you start disseminating it to large numbers of people because none of the trials can be so large that they might pick up a very rare side effect. But the confidence level of the public has to be high. We also know that people are cautious in their own lifestyles and many want to really wait on something. They don't want it right away even if it means they have to prolong this period of change in lifestyle. So we have to respect that. And we have to understand that while a lot of money is being sunk into these vaccine candidates and some of them, even before the trials are over, are having a lot of manufacturing going on in the hopes that this will be the one and they'll have some ready to go, the reality is think about the number of people in the United States. You know, it's in the 300 plus millions of people. And that's one country in the world. And then you think of all the developed nations, European Union, Canada, etc., some of the Asian countries and millions and millions and millions more people added to that, all of them wanting the same vaccine. And how much can you produce in a period of time and do it safely, of course? And then even if we had 300 million doses available tomorrow, how would we actually do the logistics? Logistics meaning have all the right materials to inject with and store with and any temperature considerations. Have a communications program so that people understand the vaccine better and know it inside out as to what it can and can't do. And if there are any potential adverse effects, how often do they occur? How serious are they or how mild are they? And then have an entire information technology infrastructure because especially if the vaccine is a two dose vaccine, which most of them are going to be, you need to actually be able to track the right dose for the right person, the right vaccine for the right person and not make mistakes about that because you want them to get the maximal benefit. So you have to really have that technology down. That's only the scratch on the surface perhaps of the complexity of this because nowhere on the planet have we done something to this degree before. Doing the annual flu shot is a piece of cake compared to this. Though the reality is we're learning a lot this year as we prepare to be more aggressive with the flu shot and try to make it more accessible to a larger portion of our population. And we're thinking and implementing lots of innovative ways to deliver that vaccine to Vermonters, whether it be traditional like in a pharmacy or a doctor's office, whether it be in a clinic that's set up outside of school, whether it be through a drive-through like some of the testing protocols are done, whether it be through some other innovative partnership where the vaccine can come to the person, meeting them where they are as opposed to some complicated structure that makes people show up somewhere that they normally would never go. So we're going to learn a lot just in the next month and a half about how to deliver a lot of vaccine innovatively and creatively to a lot of people. And why is it important to get a flu shot this year? We think it's really important because we know that the way these viruses behave and especially because they prey on this indoor assembly of people that they will exist at the same time in our cycle, whether it be the fall or the winter or both. And it would be much nicer to have only one of them active and the one we can do something about today is the flu. And if there is a 50 to 60% efficacy rate of this vaccine, which we'll find out but we would hope for at least, that would mean we'd have a lot of people who, if they presented with respiratory symptoms, probably don't have the flu. And we don't have to worry so much about that. And we can then not overwhelm the healthcare system with lots of people sick with both COVID-19 and or the flu at the same time. Plus all the other usual respiratory viruses we see in the wintertime. So it'd be very nice to have done as much as we can to prevent what we can prevent. So I just have a couple more questions. The nuance and complications of public health.  There's so much to think about. Yet, when you're the governor, you need to boil it down and make it as simple as possible. How has your relationship with Governor Scott, you know, bringing the detail that needs to be understood with the political need to communicate? How has that worked for the two of you? Oh, marvelously. I learned a lot from him. So that's an important take-home message. Not just in how to communicate, but really, he's very thoughtful about all the issues. And it's very good to get his point of view on things because it's so often reality-based and almost to the level of a vermonter, and we need to hear that all the time. But Governor Scott has done an incredible job of educating himself in science, in data, in COVID. Admittedly, we play a role in that, but the fact of the matter is he's taken this as job one. So he's become quite educated. We have a great working relationship. He's assigned a whole team to this opening of the spigot process, and we always have the public health input to that process, but not at the same time without the input of all the other sectors of state government. And people in Vermont need to know that their governor really puts health and safety as job one. And that makes it so easy for me and my entire team at the health department because that's what our job is. The mission of the health department is to protect and promote the health of all vermonters. Well, a governor that makes that their job makes my job so much easier, and vermonters should find a tremendous amount of comfort in that. Because just looking around the country, that's not always as obvious as one might like it to be from state to state. So I think in that sense, things have worked very, very well. And I think the communication in terms of communicating our messages out, we get a chance to do that twice a week. We used to do it three times a week. It's been somewhere between 60 and 70 times already this year. And if we don't make the message as clear as we can, we get asked a question that allows us to clarify even further. So our vermoners really get a chance to understand things. Plus, they get to understand things happening on the national front through the eyes of us in Vermont and how we interpret it and try to bring it home to them so that not only is it understandable, but it's applicable to experience here in Vermont. What's the benchmark that you're using to turn the spigot off? What danger signs are you looking for? So every Friday, we do a modeling presentation. And the modeling presentation that Commissioner Pichek does from finance and regulation covers numbers of new cases, the rate of new cases developing, the percent positivity rate, the what we call syndromic surveillance, which means how many people are presenting to ERs or to urgent care centers with COVID symptoms. We look at hospital capacity and how our health care system is doing in terms of numbers of beds available, ventilators available. ICU rooms, et cetera. A whole host of factors like that. And we look to see not only are they stable or are they changing, but did something we did with a turn of the spigot have any impact on them?  It's because we really do need to respect the virus. And if we do something that creates more opportunity for more people to be together, we need to give it at least that two-week period of virus incubation to make sure that we haven't changed anything in a dramatic fashion before we move on to the next turn of the spigot. And that's really been our philosophy all along and taken us really far along this continuum. So in closing, I would just thank you so very deeply for your work. I know that you have been working on this morning, noon, and even in your sleep, I imagine. But what makes you hopeful in this period of time? The thing that makes me the most hopeful is the performance of our state. If there weren't even a vaccine available, the state has done so well. And though people may not regard their life as back to normal, whatever normal was, who remembers anymore, they at least understand that there's more that they can do than they could do originally. And we've worked with the virus in the background to try to get them back to some things that they enjoy doing, whether it be doing recreational things in the summertime, whether it be sitting out on Church Street in Burlington at an outdoor restaurant, whether it be sending their kids back to school, et cetera. We've been able to do a lot in that context. And I feel pretty confident from the scientific point of view that a vaccine is on the way. And though we will all be very impatient and want things to move faster, the reality that they will at least get us to a point in time that we haven't been to yet and there may be even more of a bright rainbow on the horizon, I think is really, really important for everybody to take into consideration. Well, Dr. Mark Levine, Commissioner of Health in the state of Vermont, we really thank you for your time today and for all that you've been doing for our state. Thanks for having me.