Jorge Pullin: Okay. So our speaker today is Chen Chen. We'll speak about entangled mountains and network states.
Qian Chen: Okay.
Qian Chen: sense. For in for introduction. And
Qian Chen: today. I I I would like you today. I would like to talk about entanglement in spin network basis Day
Qian Chen: before I move in detail. I would like to get. So I would like to get you. I would like to get you a briefly introduction to the to loop quantity, and in particular, I will focus on kind of a
Qian Chen: kinematical lubricant gravity which conties general relativity in the style of topological quantum field theory or latest condom field theory.
Qian Chen: The starting point is connection. Try, feel valuable
Qian Chen: in this fundamental in this fundamental bosom work
Qian Chen: in terms of connection, field and the Try field and the by this, by these 2 valuable
Qian Chen: which formalize the general activity in into a constraint system.
Qian Chen: We all the kinematics
Qian Chen: consists of coaching, constraint and spatial divorce and constraint, and, on the other hand, the a very complicated Hamiltonian constraint
Qian Chen: make out the dynamics of the of the loop converty.
Qian Chen: So the wave function of loop on gravity stay is based on some Oriented or in. Take off
Qian Chen: with, as you to Holonomy, with as few hormones as as value as as argument.
Qian Chen: For For instance, in this graph
Qian Chen: each holonomy, which, as you 2 phonomy, tell us how to parallel transport from 1 point to another point.
Qian Chen: and the loop content quantization is realized by heronomy flux algebra.
Qian Chen: which which represent which which which conties the
Qian Chen: fundamental bosom back into a
Qian Chen: quantum commutator.
Qian Chen: and for each for each, as you to for each component of su 2 function.
Qian Chen: we can implement Fourier decomposition. So in this way.
Qian Chen: so in this sense we can identify the
Qian Chen: as the position valuable and the half integer spin
Qian Chen: is identified as the conjugate momentum.
Qian Chen: but in other 2 we still need to made the wave function gauge environment on the local as you to transformation on the local, as you do transformation.
Qian Chen: To do this, we assign in the toilet tensor to each. Vertices. Tell us how to map in going the ingoing tensor state to the outgoing tensor state.
Qian Chen: In this sense, in the toilet among you, recoupling the spin into 0 angle, momentum State entangle this entangle the edge
Qian Chen: in tango at your state, naturally
Qian Chen: and no. So so
Qian Chen: we have a we we use, we we we use our traditional data to label us.
Qian Chen: spin that will basis. They They are graph spin, and in the twin
Qian Chen: so
Qian Chen: wave function of a spin-level basis, state is obtained by growing chronomy, and in the toilet.
Qian Chen: In this way
Qian Chen: a very important, a very crucial property is the females and invariance. Suppose we have supposed we have a spin our state.
Qian Chen: and in this graph, and we can move the graph a little bit on and move the for example, move this, move the edge and the mood of vertices a little bit, and due to the TV more and constraint the 2 spin now State are considering our stage equivalence.
Qian Chen: So
Qian Chen: and now. So. Now this is. This is a very brief introduction for loop kind of gravity, and I should say that loop-controllerity tail a piece
Qian Chen: to a tail piece that forwards, because this is a non perturbative approach which avoid a which avoid diverging.
Qian Chen: and we don't need you before any
Qian Chen: fixed backwards by time.
Qian Chen: So this is a background, independent proposal deal for loop congratulations to a position and quantum fluctuation of spin our business state, which go beyond the semi classical, which can be on semi classical quantity.
Qian Chen: However, we also we also there. There are also some challenges.
Qian Chen: the first. The first challenge is the popping of
Qian Chen: locality.
Qian Chen: Now we, since we can move that we can. We can move the we can move the graph, we can move. We can move the point around. How do we localize the subsystem in the sense of if you more is an imbalance.
Qian Chen: the prescription is to ask the relation of a study which emphasize the relation between subsystem
Qian Chen: and the second challenge is semi-classical limit. Now we are given a content.
Qian Chen: We are given a quantum theory which allow us to
Qian Chen: to have content superposition, superposition, and quantum fluctuation.
Qian Chen: How do we recover classical general relativity from this content theory?
Qian Chen: The prescription is the prescription is to ask the coast screening which kill many quantum fluctuation and quantum superstition.
Qian Chen: The search energy is a long-standing. It's a long-standing pop
Qian Chen: because the hamiltonian constraint is very complicated.
Qian Chen: so the dynamics of lubricon and gravity is super
Qian Chen: is a super complicated
Qian Chen: to outcome. To overcome this complexity.
Qian Chen: a possible way, way out is to to look at holography because the Council way charge of a deep, dimensional gravitational system can be defined on a D minus one dimensional boundary.
Qian Chen: And now I've I follow, I I I follow. I follow this challenge and prescriptions.
Qian Chen: I I should say that the quantum in quantum information theory become relevant for the Loop Chronic gravity.
Qian Chen: because we don't have any. We don't have any background, geometry, so the quantum geometry of quantum gravity should be reconstructed from quantum information.
Qian Chen: Indeed, in
Qian Chen: as the least as this list as this. This, this this
Qian Chen: we have a we study, we have, we have, we have, we have, we have studied the to model for Black Hole and the information parallel.
Qian Chen: and also we. We are going to have area law and war in law in some cases, and
Qian Chen: and and and and when we, when we come to, when we comes to who are open content system we have with the coherence, and because they see, because in our state it's the most, many body system we also have for.
Qian Chen: tens of tens of network prosperity.
Qian Chen: and also to be called typicality. And
Qian Chen: some more time.
Qian Chen: And nevertheless we will move. We will, we will go to a conference to consider the relation between the suburbs subsystem and and very interesting. We we have some content simulation for content
Qian Chen: gravity, and which can be implemented in a laboratory.
Qian Chen: And now I I I want to. I want to pay attention to the area law because this has something to do with holography.
Qian Chen: and it's very likely to have to have quantities dynamics holographically, because, as we know, that Black Black Hole entropy
Qian Chen: is proportional to the area of her horizon, and the second.
Qian Chen: and secondly.
Qian Chen: the costly local councilway charge, for example, across the local energy is defined. It can be confused on a twod surface.
Qian Chen: so I I prefer we, for I I prefer to refer holography principle as in this way the gravitation of the we've done in the 2D manifold can be in code in the
Qian Chen: D minus one dimensional boundary.
Qian Chen: But in this talk I don't I don't have enough
Qian Chen: to to to to talk about the dynamics, but I will pull the boundary bulk bungy relation into emphasize.
Qian Chen: and
Qian Chen: and also I will pay attention to the to the entanglement. This is the unffactorized ability of Condon State, which tell us there is that to a correlation
Qian Chen: that can be interpreted classically, so a composite content state will carry much more information than a local state.
Qian Chen: And now we are going to. We are going to have some additionally between content geometry and correlation, since we want to construct a
Qian Chen: quantum geometry from content information. So we have a some. We have now the distance in some time model, and we have. We also now the miti, which is, is, is express in terms of the correlation function of flux operator. And what part of the curvature and this is this is the
Qian Chen: go. This is one of the go of this talk.
Qian Chen: so I will also outline all right the the goal of this talk to study the entanglement
Qian Chen: in spin that was state
Qian Chen: in particular, the many.
Qian Chen: many partly entangled, measure the
Qian Chen: to quantify how much entangle, how how much entanglement
Qian Chen: the spin that will stay carried.
Qian Chen: and also we try. I I also. I try to reconstruct it the curvature from the notion of quantum entanglement and correlation.
Qian Chen: and I will, I will.
Qian Chen: and I will also
Qian Chen: to see how the bulk boundary relation may sense in the contest of entanglement.
Qian Chen: So this is the this is the part of introduction. Oh, do we? if if you have any question, please. please, don't hesitate to interrupt me.
Abhay Vasant Ashtekar: Okay, so can I drop this? I I I was very sympathetic with everything you said except for this holography.
Abhay Vasant Ashtekar: I don't see any evidence for this holography. I mean the Black Hole entropy calculation. Yes, I mean there's a the degrees of freedom
Abhay Vasant Ashtekar: all right.
Abhay Vasant Ashtekar: We can code it in the
Abhay Vasant Ashtekar: quantum geometry of the horizon.
Abhay Vasant Ashtekar: but if I look at the Hamiltonian constraint
Abhay Vasant Ashtekar: that was.
Abhay Vasant Ashtekar: or even
Abhay Vasant Ashtekar: the
Abhay Vasant Ashtekar: the the
Abhay Vasant Ashtekar: there is no
Abhay Vasant Ashtekar: reduction right in the sense of it that I got a graph, and then I just look at the boundary. I I had to look at the entire graph
Abhay Vasant Ashtekar: which is so. I don't see any sense in which holography is realized in new point of gravity
Abhay Vasant Ashtekar: in the
Abhay Vasant Ashtekar: the dynamical sector. So perhaps you can explain what you meant.
Qian Chen: I mean.
Qian Chen: Yeah, I actually I don't. I I I don't have a I don't have a very confident
Qian Chen: I don't have a very confident answer to to to your question. Basically I mean, if we because if we we back to the 3D gravity, pure gravity, and the all the degree of freedom are carried by the bungee. So in this sense we say that the series holiday. And of course.
Abhay Vasant Ashtekar: if we move to the to a 40, that's not true. You look kind of gravity right? I mean, either in classical theory, or in Conduit theory
Abhay Vasant Ashtekar: that the degrees of freedom are all. All of them are in the boundary.
Abhay Vasant Ashtekar: It maybe, or you need something that I Don't understand
Abhay Vasant Ashtekar: we're not. As I was just saying, the Hamiltonian framework who don't see this that the degrees of freedom out of the boundary the Hamiltonian constrained by teama and mud or other I and so on.
Abhay Vasant Ashtekar: and it's in forms. Also, I got the whole graph. I don't look at, so to say, it just the boundary of that graph and that evolution, on the whole, graph, not just the boundary of that.
Qian Chen: Yeah, I actually I don't see. I I. I don't see any holography in the in the team and Hamiltonian, but I am. Why, I want to say, is that
Qian Chen: is it possible to to phone lines to formalize the general relativity into a holography way, and which me that you can find all the dynamical console way, charge.
Qian Chen: defy on the boundary so, and and then I try to. I I try to represent those cons of a charge.
Qian Chen: Those boundary comes away. Charge as the starting point for the quantization.
Abhay Vasant Ashtekar: all right. So this is a program. I mean, it's not that this is what it's realized it. Look on them gradually. You're saying that something like this will happen. Okay, Thank you.
Qian Chen: Okay, okay, Thank you. Thank you for the question.
Qian Chen: And now I so I I try to bet you the I try to buy to the entanglement and try to buy to the entanglement.
Qian Chen: But so the second part is about the coverage for the relation between curvature and the multipartite entanglement.
Qian Chen: But before but before moving into
Qian Chen: into detail, let's look at, let's look at a very simple, a very simple cases.
Qian Chen: So I would like to in this 2 slide I I would like to distinguish the 2 notion of entanglement in the twin entanglement and the bungee entanglement in the toilet entanglement is the entanglement between the 2 vertices
Qian Chen: between 2,
Qian Chen: and the bounty entanglement is the entanglement between the boundary state.
Qian Chen: so you can see that they refactorize the hib the spay in this way, and then we study the entanglement and the bounty entanglement.
Qian Chen: But they are different. I need to. I I need to emphasize. They are different
Qian Chen: and
Qian Chen: based on this single, this based on this single link graph, we can see that a spin that would basis stay the in the toyness, the in the entanglement of the spinning. Our basis state will vanish because we don't have any sort of sensible position, or in the toilet collaboration.
Qian Chen: and
Qian Chen: but for the boundary entanglement, the spin network basis state have for non-violent non-vanishing
Qian Chen: spin login contribution.
Qian Chen: and the in the twin and the and the in in Tankerman Doesn't depend on the of polynomial between between. along this edge, because you can see you can do some. You can do some local unitary
Qian Chen: local local as you to get you transformation on this on this vertices, and then you can. You can move the All on to Bungee and this boundary, actually a local, a local unitary transformation for the for this subsystem. So it doesn't it doesn't depend on the
Qian Chen: honomi of folk.
Qian Chen: But this is this is this this conclud conclusion? Only only work for single, this single graph. If we consider we, if we consider some loop structure, this this this holonomic this conclusion for phonomy doesn't help.
Qian Chen: So this is this is
Qian Chen: This is why why I need to do. But I
Qian Chen: But in this part in this talk I will focus on the in the final entanglement
Qian Chen: which, defined by the unfectualizability between the spin network state
Qian Chen: from the perspective of what? Of tensor product of
Qian Chen: what what is state?
Qian Chen: And let's let's get you.
Qian Chen: Let let me let me get you. Let me get you a very simple, very simple cases. So consider a candy graph in this way
Qian Chen: and this can disk, and we we start with the initial. We start with the initial state.
Qian Chen: which is a spin level basis. State
Qian Chen: and this State doesn't carry any in in the total entanglement, because all the spin and in China are fixed.
Qian Chen: And now we can implement a Loop polynomial operator.
Qian Chen: which will introduce some spinsual position
Qian Chen: along the bulk edge.
Qian Chen: So the 2 word is gay entangle.
Qian Chen: And now in this
Qian Chen: and now in this part I will I will focus on. I will focus on the generalization of this picture. I will consider the tree well and spin. I will
Qian Chen: with one topological loop, and in this sense
Qian Chen: the code, the the curvature, your Gary fact in the loop, her nominee operator on this bulk edge.
Qian Chen: And now it's a time to look at the action of Loop Holomi operator.
Qian Chen: The Loop polynomial operator is defined as the classical as a classical as a class function of su 2 representation, and
Qian Chen: along along the edge they are. They are the
Qian Chen: they are the product of we can our dmitry
Qian Chen: and the animal is the spin of the Su. 2 representation
Qian Chen: this operator add on spin that will
Qian Chen: state by by multiplication.
Qian Chen: But I should mention this is a
Qian Chen: for each piece wine
Qian Chen: the we can, our Dimitri will introduce a spin sewer position
Qian Chen: due to the spin, a spin recoupling.
Qian Chen: and this the upb and the lower bound is the is determined by the triangle condition of recover of su 2 recoupling.
Qian Chen: So
Qian Chen: we can, we can, we can. We can represent. We can represent a loopholeonomy operator
Qian Chen: in pencil for a graph.
Qian Chen: and
Qian Chen: it's turned out that the that the representation will will be a product of 60. Simple.
Qian Chen: Because of the trace we will. We will connect it to the open the open edge together, so we will go, and we are going to have some corner.
Qian Chen: and this corner will get us some 6 simple.
Qian Chen: So the transition Mp. 2. So, finally, the transition Mp. 2 is the product is to put that off for 60 symbol and some spin Mp. 2 and a face. In this way.
Qian Chen: moreover, we can we can use. We can use the asymptotical formula for 60 simple.
Qian Chen: and this and
Qian Chen: and when the spin. When spin. When the loop is spin, it's very small compared with other spin.
Qian Chen: In these cases we can write the 60 symbol into this form, and we the d is a we're going to
Qian Chen: It's a small, weak wigan at Dimm tree, determined by the edge.
Qian Chen: By the end of that, by the end of the
Qian Chen: so asymptotically
Qian Chen: the transition Mp. 2 is, it is passed in terms of angle Theta and the spin shifting.
Qian Chen: and the
Qian Chen: which is the product of we can not of small wig, not Dimitri.
Qian Chen: and now it's a 10 back to the
Qian Chen: multiparta entanglement in other 2, in order to
Qian Chen: quantify the entanglement of a spin. Our state. We need to choose the many.
Qian Chen: many, many, many part in tech, I would measure
Qian Chen: and but to remember the oh yeah.
Qian Chen: but remember they are. They are they they they have they
Qian Chen: many, Many better entanglement has to reach a structure than Bipart entanglement. For
Qian Chen: for instance.
Qian Chen: we can't have a Zoom Meeting to decomposition even for the tree part in a trip part entangled state.
Qian Chen: So we don't have this equation anymore.
Qian Chen: and we have for some larger entanglement and no further entanglement.
Qian Chen: For instance, we can. We can write a Ghd state and double state.
Qian Chen: and if we
Qian Chen: we can we can. We can forget the the the, the the circuit will be of G at the State, and the
Qian Chen: and the and the and the rest of to the rest of of bipart in tank. A pipel stay is on entangled.
Qian Chen: but the 4 W's. Day, if we we trace our the silk will be the second, the the first and the second to be, are still entangled. So this is a long to see the meeting of non- factual entanglement.
Qian Chen: And I should say that we have many, there are many. There are many equivalent entanglement measure for many part of entanglement.
Qian Chen: In in this past I allow the geometrical entanglement as the entanglement measure.
Qian Chen: Suppose we have a few steps.
Qian Chen: and the geometrical entanglement is defined by this equation.
Qian Chen: Where the P site is table. The precise taste from all unentangled state. So the geometrical entanglement is defined
Qian Chen: by the mess
Qian Chen: that it might seem more projection
Qian Chen: of
Qian Chen: maximum progression on to the on that
Qian Chen: onto the contents on the site of Content
Qian Chen: on entangled quantum state.
Qian Chen: and the goal of this part is to study the evolution of a geometrical entanglement. Now we we are we initial? We? So we need initial. We so product, state. So we don't have any. We don't have any
Qian Chen: geometrical entanglement. But now we implement some exponential evolution of heronomy operator. So we are going to have a entangled state, and now we
Qian Chen: we are. We are going to look at the evolution of geometrical entanglement we.
Qian Chen: And so
Qian Chen: The My result of of this part is
Qian Chen: is about 3 valent, one-loop spin network.
Qian Chen: and we can compute it. It's my government institution.
Qian Chen: The first.
Qian Chen: the first, derivative with respect to time.
Qian Chen: must be 0; but the second order derivative. Of of geometrical entanglement are equal to the
Qian Chen: dispersion of Holomi operator.
Qian Chen: and we can
Deepak Vaid: question, if I may.
Eugenio Bianchi: I have a question.
Qian Chen: Oh, yeah.
Deepak Vaid: where is the what you're taking a derivative with respect to time, right. So where is the time coming into this picture?
Qian Chen: because the time is the the time coming into the is financial evolution.
Qian Chen: Okay, okay? And and what Hamiltonian are you using? I mean, I I I I I choose the Hamiltonian as the Holonomy operator.
Deepak Vaid: Okay.
Deepak Vaid: All right, thanks.
Abhay Vasant Ashtekar: Okay. Since you had this slide, the the last slide, please.
Abhay Vasant Ashtekar: So you just have this on this slide. Yeah, yeah, this lot. So I just I don't understand symbols. So G stands for? What for the graph or
Qian Chen: what is it?
Qian Chen: Osmotry is me the geological
Abhay Vasant Ashtekar: or geometric Isis, and that so? What is the Major here? The May, I mean you are given a fixed number of vertices. What what is the system? What is the size supposed to be a state
Abhay Vasant Ashtekar: of what I mean on a given graph or on what
Abhay Vasant Ashtekar: fixed. Well.
Qian Chen: it's. It's supposed to be a spin. That will.
Qian Chen: It's it's supposed to be a spin that will stay
up.
Abhay Vasant Ashtekar: Okay.
Abhay Vasant Ashtekar: it's a spin network state. And and can you explain again what 5 was? I just was just too quick. This is a very general. This is a very general
Qian Chen: definition.
Qian Chen: So we want to look at the entanglement. We want to quantify or entanglement of side. And now we choose.
Qian Chen: and and you can, you can separate the hip, the sp of of of of content state into a 2 part. One part is the entangle state, and another part is the on entangled state. And now now the geometrical entanglement is to quantify the distance between the between in t or
Qian Chen: entangled state, to the to the site of an entangled state, so precise is to choose as an arbitrary spin that will stay.
Qian Chen: and the psi is the on entangle spin that will State.
Qian Chen: so the distance will be will be will be measured by the by, the by, the position of a entangled state to a an entangled state.
Abhay Vasant Ashtekar: I I.
Abhay Vasant Ashtekar: Okay. So I
Abhay Vasant Ashtekar: These are not the subspaces of the
Abhay Vasant Ashtekar: We. We just look at. C: yeah, these are not subspaces. They are just
Abhay Vasant Ashtekar: every and and and
Abhay Vasant Ashtekar: you you're wearing on all our unentangled states for a given entangled state. Is that what you're doing here?
Qian Chen: Yes, yes, yes.
Qian Chen: yeah, it's it's no, it's not, it's it's it's No. Because
Qian Chen: if you super, we compose a to 2 or 2 for that day.
Qian Chen: and you will you? You are going to have what you are Mostly you are.
Qian Chen: Most probably you are going to have a entangle.
Abhay Vasant Ashtekar: and and conversely, I can stick. Read your accommodation and titles when I, and tackle State to make them unintended.
Abhay Vasant Ashtekar: So here size is a given
Abhay Vasant Ashtekar: internal state, and 5 are all possible
Abhay Vasant Ashtekar: unentangled states, and you are just looking at this in the product and then taking it. Yeah, yeah, yeah, yeah.
Abhay Vasant Ashtekar: Thank you.
Western: Sorry I have a related question. maybe you said it before, but I may have missed it. So is there any dependence on the graph that you're using to define your state? Or are you considering a a superposition of all possible graph.
Qian Chen: that's pretty I
Qian Chen: in this set. In. In this part I only consider I only consider this kind of
Qian Chen: this kind of graph.
Qian Chen: because for the for the simplicity of computation, but in the last in the in, in the, in the in the next part I will try to generalize to the actual graph.
Qian Chen: So so the conclusion for the for this part I only considered
Qian Chen: this kind of graph.
Qian Chen: but actually this, this definition for entanglement Doesn't: Rely on which graph you choose.
Western: You you mean, just to the first graph or all the graph of the kind or like the okay. Okay, Thank you so much for that.
Qian Chen: Yeah, yeah, I will. I will try to generalize the I will try to generalize the graph in the in the in the nest in the nest part.
Eugenio Bianchi: Can you Can you say also what's the set of subsystems when you say multi-partite?
Qian Chen: Oh.
Qian Chen: it's the
Qian Chen: to stop the system. The subsystem is is decomposed by the by, the vertical state. For example, in this candy graph the subsystem is the what has the in the twin of State. I tends to put that in the twin
Qian Chen: in the twentieth day fee.
Qian Chen: And in this graph in this triangle was the the sub subsystem is the is, the is. The tensor product is that it's a 10 to put out here. Display of a, B and C, and is that for arbitrary
Qian Chen: number of votes.
Qian Chen: So I separated. I separate the subsystem by the with respect to the
Qian Chen: but what is?
Eugenio Bianchi: And the and the question was asked already before I if I can ask again time evolution, you said that you're considered a as a Mytonian, not the with so loop on this single loop that you have here. Is that right?
Qian Chen: yeah.
Eugenio Bianchi: Okay.
Abhay Vasant Ashtekar: this is a mathematical.
Abhay Vasant Ashtekar: and it's not a physical admin.
Abhay Vasant Ashtekar: You: just
Abhay Vasant Ashtekar: yeah, yeah, it's just I mean, you're just saying that
Abhay Vasant Ashtekar: any function or any operator self as an operator can be taken to be
Abhay Vasant Ashtekar: I me, too, again.
Abhay Vasant Ashtekar: and it just
Abhay Vasant Ashtekar: using that it's not the
Abhay Vasant Ashtekar: it is. It is not a physical time.
Qian Chen: Yeah, it's no physical time. It's just a parameter time.
Bye.
Qian Chen: Yeah. Yeah.
Qian Chen: And now.
Qian Chen: in particular. I can bet you the I can bet you the bipart entanglement.
Qian Chen: and then and now we can try to. You can try to compute the linear entropy.
Qian Chen: and we can. We can see that this is the just so twice as the entanglement isitation.
Qian Chen: and in large
Qian Chen: and in particular in large spin limit. We can use the asymptotic formula to be to to rewrite the the geometrical.
Qian Chen: the geometrical entanglement are to the second order
Qian Chen: of time.
Qian Chen: and the pl are the legend Polynomial.
Qian Chen: and this
Qian Chen: firmly. It's very simple in this way, but I only are to the second order probability of time, second order of time.
Qian Chen: And now I try to. I remember all this. We resolved a work from work for
Qian Chen: a tree or tree valve, one loop graph, so I try to
Qian Chen: generalize the picture.
Abhay Vasant Ashtekar: Are you confused about dim dimensions here?
Abhay Vasant Ashtekar: H. Times t has a dimension is dimensionless right? Because you are exponentiating it
Qian Chen: edge
Abhay Vasant Ashtekar: times. T: I mean you are exponential in it. So h times t is dimensionless.
Abhay Vasant Ashtekar: maybe you can. You can. You can. You can divide. You can pay case on h bar over. no, that's not right, whatever it is, but h times t Your framework is dimensionless, but in the next line
Abhay Vasant Ashtekar: you got a superposition
Abhay Vasant Ashtekar: in which
Abhay Vasant Ashtekar: sorry, maybe
Abhay Vasant Ashtekar: maybe next next one or
Abhay Vasant Ashtekar: somewhere. You are a superposition just just just now. Here.
Abhay Vasant Ashtekar: Okay, this is this all fine. But can you? With very showing the next one next transparently just now?
Abhay Vasant Ashtekar: Yeah, in this, in this transparency
Abhay Vasant Ashtekar: you got h squared times t squared, so that is dimensionless.
Abhay Vasant Ashtekar: but the first term has only h square, and so the dimensions don't match. So what's going on here?
Abhay Vasant Ashtekar: previous transformation trans transparency. Everything was matching. But now suddenly you did something that i'm confused about.
Qian Chen: actually, I don't see any
Qian Chen: any. I don't see any, and I don't see any problem by the dimensions
Abhay Vasant Ashtekar: because I Everything has dimensions of it's square. Is that right? Everything is dimension. It's square.
Qian Chen: Yeah. So linear and entropy has dimension of it square. I
Qian Chen: Hmm. I think that's what I I I I I to to to take the
Qian Chen: another distance by the dimension, because you can pay the t as the dimension of a as a dimension of inverse of.
Qian Chen: So in this sense the
Qian Chen: exponential will will be dimensionally.
Abhay Vasant Ashtekar: but it is, it's not Hamiltonian dimension, right? It just a whole lot of money.
Abhay Vasant Ashtekar: so I don't think plan constant helps here at, and t have to be opposite dimensions.
Abhay Vasant Ashtekar: and i'm just confused as to why
Abhay Vasant Ashtekar: and tag of me, and properly as dimensions of it. Square.
Abhay Vasant Ashtekar: I thought it was going to be dimensionless, and I I got confused account. But okay, go ahead.
Deepak Vaid: Okay, okay, I'll see. I'll see is dimensionless. Then it's fine, because in that expression for the entropy.
Abhay Vasant Ashtekar: No, no. I agree that that every term has dimension of it square.
Deepak Vaid: I can see that I I saw a factor of X T. Over Square.
Deepak Vaid: maybe. Yes, I
Deepak Vaid: can. You so that
Abhay Vasant Ashtekar: yeah, okay. So if you say it is, it's square t squared. That is dimensionless.
Abhay Vasant Ashtekar: But the first term has only 8 square.
Deepak Vaid: No, the whole thing is multiplied by t square.
Jorge Pullin: There's a parentheses.
Abhay Vasant Ashtekar: Oh, this oh, I'm: so sorry. I just missed everything. Okay, I I take it all back.
Abhay Vasant Ashtekar: Okay, Thank you.
Qian Chen: Okay, you're welcome.
Eugenio Bianchi: can I? How that formula relates to the one on multi-partite entitlement that you have before in particular what is, say? Not
Eugenio Bianchi: before you had fi inside. And now there's only to say not to. Can you? Can you say how they are related?
Qian Chen: you meet the site.
Qian Chen: you me, the Psi. Here.
Eugenio Bianchi: I understand this is the initial one
Eugenio Bianchi: what I'm asking is, where is the subsystem that's? That's what i'm asking
Qian Chen: the subsystem. The subsystem is the is, the entangled is the in the pointer state at each vertex.
Eugenio Bianchi: Yeah, yeah, one page where you had this way formula.
Eugenio Bianchi: And then you had the linear entropy. I'm asking, Are you using the same definition? what? What?
Eugenio Bianchi: What is the relation between the 2?
Qian Chen: You mean a linear entropy.
Eugenio Bianchi: Yes, for instance, the entropy in this page you have the definition on top.
Eugenio Bianchi: and
Qian Chen: let's pat you the this graph.
Qian Chen: The linear linear entropy is is the is the reduced density matrix of a or B.
Eugenio Bianchi: Okay.
Eugenio Bianchi: So we always reduce into a single node. That's what i'm asking you, always reducing for a single node.
Eugenio Bianchi: or to some of the notes. I just cannot read it in the formula, because that format doesn't
Eugenio Bianchi: say which subsystem it is.
Qian Chen: yet the subsystem is depend on its detail.
Qian Chen: which is determined by the by, the by, the product of who has stay. And and I I I say, I I say the candy graph or or the linear and it's just for this. It's just for this for this, this, this particular cases. But if we move to the triangle graph
Qian Chen: or script you you you will not have, for you will now have the definition of linear of linear entropy, because the because.
Qian Chen: because
Qian Chen: because you you don't have a Schumeter decomposition for the many, many partite entanglement.
Qian Chen: So I I I just say that
Qian Chen: if you back to the better bipartite cases, and the linear entropy is proportional to the, to the
Qian Chen: geometrical entanglement dissertation.
I see.
Okay.
Qian Chen: Okay.
Qian Chen: And
Qian Chen: And now I try to. I try to. I try to generalize the feature, and this this lead us, lead us to the to the entanglement. Co. Co-screen. And in this talk I I refer Co-screen in. So we do that in of.
Qian Chen: of course, of graphic structure.
Qian Chen: For for this illustration you can see that the
Qian Chen: we can we can choose which we can choose to choose a separate, we can choose some subsequent and and we can trace our. We can trace out what the bulk of the normally which defy which define some boundary density matrix.
Qian Chen: Yeah. And
Deepak Vaid: yeah, can I ask a question?
Qian Chen: Yeah, yeah, yeah.
Deepak Vaid: so can you go back to that expression for the geometric entropy.
Qian Chen: you me this equation.
Deepak Vaid: There is the other one where you have a
Qian Chen: maximization over as well.
Deepak Vaid: So so this is this is the so you are maximizing over all of the product States, right? And then
Deepak Vaid: in the next in the next step
Deepak Vaid: you take Si of t. I guess I of t is your State with whose entropy you want to measure right?
Deepak Vaid: and and then what what are the unentangled States with respect to which you are taking that
Deepak Vaid: that projection.
Qian Chen: Yeah, because I choose the initial State as an entangled State.
Deepak Vaid: and that is, that is what is that? That is just the trivalent node
Qian Chen: if you you me, why, the first order? Your question is, that. Does your question me that why the first, all the derivative? Should be 0?
Deepak Vaid: No, I'm. Just asking what are all the and unentangled States with respect to which you are
Deepak Vaid: taking that projection
Qian Chen: part is itself, because
Qian Chen: because when you look at a first-order, derivative
Qian Chen: you just you, you just lead the when you consider the polynomial operator. You just lead a
Qian Chen: lead a spin that will stay a little bit, and then you purchase onto the itself
Qian Chen: onto the onto the initial state.
Deepak Vaid: I see, I see. So the Holonomy operator is an entangling operator.
Qian Chen: Yeah.
Deepak Vaid: you are evolving. You are starting with an unentangled state to begin with.
Qian Chen: Yeah, Good morning with an on entangling operator.
Qian Chen: Yes.
Deepak Vaid: And then you're measuring the
Deepak Vaid: entropy of that evolved state with respect to the initial and
Qian Chen: yes, yes.
Deepak Vaid: okay.
Deepak Vaid: Okay, I I guess that's somewhat the other. Yeah, thanks.
Qian Chen: Okay.
Qian Chen: and no.
Qian Chen: and no. But but but this talk I don't. I I I don't. I don't want to talk about the trade, but by the first co-screen I I I focus on the gauge facing
Qian Chen: which means that we can get this: the polynomial and and have some, and then have some loop is when I walk.
Qian Chen: and this.
Qian Chen: and and then I will. I will look at the evolution of co-
Qian Chen: and study the entanglement between the spin supper network.
Qian Chen: This is the generalization of in the twin entanglement.
Qian Chen: I know.
Qian Chen: and as the as illustration, you can see when we choose a some spin some, some some of the region, or spin up on that, we will nevertheless, to introduce some boundary. So we will. We will study this, we will start the the formulation
Qian Chen: for the for the spin network with boundary, which means that we will study some open spin level, and this picture is is actually motivated by the study of playhole entropy.
Qian Chen: and you can see that in this picture the in this picture the open spinner will have some puncture of a non trivial puncture on the
Qian Chen: on, on the on the twod surface.
Qian Chen: and so
Qian Chen: it's very natural to define it. It's very natural to define a boundary cube display by tensor in those open edge.
Qian Chen: and we can see, and we can consider it the spin network wave function as some bulk boundary map
Qian Chen: in tons of
Qian Chen: holy, allegedly as functional. And then we we are going to have a boundary state.
Qian Chen: and this boundary state is transformed under under the under, under the local as Youtube page transformation.
Qian Chen: So
Qian Chen: when we do where we do the date, a local transformation on the bulk, we will have some boundary hormone.
Qian Chen: Of course we can. We can define a dual boundary here, because this is the this is the this is the boundary state, and this book, the the scalar put out of this dual boundary. Here. The space defined. It's just defined by the scalar part of the spin that will hit the Sp.
Qian Chen: And now remember that this is the boundary state, so we can we, can we? We are going to be a me.
Qian Chen: I mean some decomposition which me that we can decompose. You can decompose a boundary state into some into a total spin, and
Qian Chen: ties some in the twentieth state, and this total spin must be integer because because of the page invariance of spin. Now, State
Qian Chen: and in the twin tell us how to recouple how to recouple countries, boundary spin into this
Qian Chen: total spin.
Qian Chen: and now it's a time to to to to to return to the top of of Co-screen. And this coast screening is done by gauge, facing, which means that we choose for any. For any bulk we choose a. We choose a row, word, test, and choose some mass mode tree.
Qian Chen: But remember, we don't want you. We don't want you from any loop.
Qian Chen: and then we can get. We can implement data transformation, one by one.
Qian Chen: and to to, to, to have to have to have a trivial
Qian Chen: trivial horonom is along along the mass mo tree, and then the
Qian Chen: we can go the boundary. We can go the boundary edge together
Qian Chen: along this maximal tree, because the heronomy now are trivial.
Qian Chen: So we are going to have some loopy spin. I will and the loopy, and the and the loop edge is the edge out of
Qian Chen: at the at the bulk edge out of the maximum tree.
Qian Chen: So we are going to. So we are. We are going to have a have a co-screen in coast green graph.
Qian Chen: So the gauge facing so under the gauge facing under the gauge fee under the gauge facing. We will have. We will have. We can write all
Qian Chen: pong. You stay in this way.
Qian Chen: and we we and and in phone off in form of the
Qian Chen: country state.
Qian Chen: With respect to the co-screen graph, we will have. We will have some boundary chronomy.
Qian Chen: and we can see that
Qian Chen: from the by, the this, this go to physician to serve the scale of product because we can, we can. We can remember we can remember the scanner product of you up on your Hube display.
Qian Chen: So in this sense we are,
Qian Chen: we define. We define a local unitary transformation between between a final graph, and the coastgu in the graph.
Qian Chen: So, because
Qian Chen: due to the due to the definition of entanglement, this is the local unit, a local unitary transformation with respect to the subsequent, so
Qian Chen: the spin and entanglement will no change
Qian Chen: will not change under the under under the cold screening.
Qian Chen: For instance.
Qian Chen: if we want to compute the entanglement between between between the 3 supper region. We only need to look at the code screen, the graph and this
Qian Chen: which me, that the
Qian Chen: entanglement between spin up, and they will equal to this
Qian Chen: equal to the entanglement between those 2 those 3 in the twin
Qian Chen: loopy in the twin.
Qian Chen: And no, we can we? Can. We can.
Qian Chen: Oh, okay.
Deepak Vaid: So there might be more than one maximal tree
Deepak Vaid: for a given grow.
Deepak Vaid: And so you you you you could end up with with inequivalent
Deepak Vaid: for screening right?
Deepak Vaid: Is that is that not a possibility?
Qian Chen: Oh.
Qian Chen: you me that
Qian Chen: this coastguarding
Qian Chen: it's equivalent to other co- screening.
Deepak Vaid: Yeah, I mean that that could be more than one.
Qian Chen: You have an equivalent cost
Qian Chen: it's it's very likely to have an equivalent of squinting
Qian Chen: because it doesn't doesn't, we move any information. But if we you consider to trace our the polynomial email, you will we move some. You will we move some information
Qian Chen: So there
Qian Chen: Basically, they are. Again, it is not necessary to have an equivalent feature.
Deepak Vaid: No; but I mean with with the cancellation of entertainment entropy remain the same for in equivalent course training.
Qian Chen: I know. Sure, I know. Sure, basically I don't see. I I don't see any. I don't see any hints to say that they are equivalent
Deepak Vaid: okay, maybe i'm not. Anyway.
Deepak Vaid: we I can last later.
Qian Chen: yeah, because basically the a another approach. We we are
Qian Chen: another port. You will. We will. We will get us some
Qian Chen: mix or boundary state.
Qian Chen: but the coast screening
Qian Chen: but that co-screening by page fishing doesn't doesn't.
Qian Chen: but but that doesn't get us
Deepak Vaid: what I was saying was
Deepak Vaid: that's
Deepak Vaid: You are choosing a maximum 3 right
Deepak Vaid: and you could have more than one maximum 3 for a given graph.
Qian Chen: Yeah, yeah, of course, of course.
Deepak Vaid: And so if you have more than one maximum 3,
Deepak Vaid: the resulting loopy graphs, the resulting 4 screens with the graphs.
Deepak Vaid: Yeah, yeah, that could be more than one for the same. Starting from the same original crowd.
Qian Chen: Yeah, I will. I will talk about this question. I will ever talk about this question. I will show you that that they are equivalent in the sense of entanglement.
Deepak Vaid: Okay, Great thanks.
Qian Chen: Okay, okay, I will talk about this. This is a good question.
Yeah.
Qian Chen: And now we can. We can. We can. We can make some we can. We can make some gauge we can make some page facing for Holonomy operator, and this one will tell us that
Qian Chen: the the the co-screening of Holonomy operator.
Qian Chen: So let me give you an example so supposed to be so Star is supposed to be start with a squares graph. And now we try to compute the entanglement between the side and the right hand side. And now I I I need to. I need to call screen. I need to call screen
Qian Chen: Co-screen in the the right hand side and the left hand side. And so we are going to have a candy, graph.
Qian Chen: and we can see that the transformation of Holonomy operator, I mean the transition matrix of hlynomial operator is can be written in this way.
Qian Chen: and this W. Present the
Qian Chen: of square Graph and the L presenter
Qian Chen: on this kind of
Qian Chen: So we are going to have for this
Qian Chen: transformation room.
Qian Chen: And so in this way.
Qian Chen: in this in this transformation tell us
Qian Chen: that the evolution will no change.
Qian Chen: so we can study. The spinner will intend to spin level entanglement
Qian Chen: on this
Qian Chen: on this, on this coast screen the graph
Qian Chen: A. As long as as as long as the evolution operator is choose is choosing us. it is choos by the by heronomy operator.
Qian Chen: So
Qian Chen: I will get you. I will get you some. I will get you some conclusion from from from this, from from this, from this from this.
Qian Chen: Let's bet you let's be a let's bet you let's let's bet you the let's bet you the entanglement institution in previous in previous part.
Qian Chen: So in this graph we can gauge fees. We can gauge fees the spin level in this way.
Qian Chen: and now the structure is.
Qian Chen: it's a single. It's a single vertices plus a single loop.
Qian Chen: and we, in order to have for entanglement institution, we only need to compute the expectation of Holonomy and the test, pay
Qian Chen: dispatch the dispassion of her. No.
Qian Chen: So
Qian Chen: in this way we can write down the boundary state.
Qian Chen: We can write down the boundary state, and this is the distribution of total spin.
Qian Chen: and we can. It's very easy to to to. It's very easy to see that the second order derivative
Qian Chen: is,
Qian Chen: is first in this way.
Qian Chen: And now
Qian Chen: a simple example is is to look at a candy graph.
Qian Chen: and this
Qian Chen: this will back to your question.
Qian Chen: and we can we can. We can choose, we can choose 2 different way to gauge fees, the the candy graph we can. We.
Qian Chen: for example, in the first, in the first gauge facing. We choose the we choose K. 2. We choose the edge associated with K. 2 to gauge fees, and now we have some 100
Qian Chen: safe loop with in K one, and we can also gauge the edge associated with K. One. And now we have some
Qian Chen: we have. We have some loop. We have a safe loop with in K 2,
Qian Chen: and now we can see that the 2
Qian Chen: probability distribution are no equal to each other.
Qian Chen: But it's turned out that the entanglement institutions are equal to each other.
Qian Chen: so you can see that
Qian Chen: it always has this has this phone.
Qian Chen: So
Qian Chen: this I I I hope this answer your question.
Qian Chen: So the gauge facing Doesn't so the entanglement
Qian Chen: doesn't we like that doesn't depend on the
Qian Chen: on the on the trice of cage facing.
Qian Chen: And now I I move you, my my, to my my last, my last example.
Qian Chen: So suppose we we we we want to know the entanglement dissertation, or for
Qian Chen: of a and B Z,
Qian Chen: and we have we. We have now. We have a. We have a 2 option to add to atom at the Holonomy operator. One is along this path and another is along this path, and now we can, since we can, since we can get, since we can gauge the we can co-screen them.
Qian Chen: We we hand co-screen in the
Qian Chen: we are going to. We are going to compute the entanglement between the 2 vertices.
Qian Chen: and that the difference the the only difference between the 2 graph is the is the loopy spin.
Qian Chen: and we can. We can see
Qian Chen: that the the entanglement institution
Qian Chen: depend on the past
Qian Chen: is now can convey, is now conveyed to the dependence of spin along the
Qian Chen: or along the safe loop.
Qian Chen: And now, yeah, this is the end of
Abhay Vasant Ashtekar: this is the so in this example, then they did the
Abhay Vasant Ashtekar: yeah. So
Abhay Vasant Ashtekar: the the second derivative of this is, as
Abhay Vasant Ashtekar: will be 7, 9 and 26 about 27, right?
Abhay Vasant Ashtekar: So they are different in the previous example candy graph, they were the same. But here is a different is that the point that you're making on.
Qian Chen: Yeah, yeah, they are different, because you, because you add in on a different path.
Abhay Vasant Ashtekar: Okay.
Abhay Vasant Ashtekar: So it is not.
Abhay Vasant Ashtekar: So. You based on how
Abhay Vasant Ashtekar: it depends how you course. Great is it?
Abhay Vasant Ashtekar: It depends on what it is.
Abhay Vasant Ashtekar: no, I'm still understanding that you front. But yeah, just explain to us the difference between the candy graph and this graph
Abhay Vasant Ashtekar: and the results on what is the same and what is different.
Qian Chen: You mean the candy graph and this one yeah. So this one you explain. You only have one. You only have one loop, so you don't have any try choice to to to choose a a different way to add in on your loop Loop Holonomy operator.
Qian Chen: But you can. What what you can do what, what what you can do on this graph is to choose a different way to gauge fees.
Abhay Vasant Ashtekar: So you have.
Abhay Vasant Ashtekar: So are you saying that this gauge, fixing this can be graph result that you showed us that
Qian Chen: that second day with you doesn't depend on how you gauge fixed right. That's what you just showed us. No, no, no, no, I mean it doesn't it Doesn't. Depend on how? How you gauge fees.
Abhay Vasant Ashtekar: But is that is that result?
Abhay Vasant Ashtekar: Completely, general, or is it only for candy? Gr.
Qian Chen: Oh.
Qian Chen: Yeah, it's it's a very general. It's a very general conclusion
Qian Chen: for for for for for one, for one loop cost.
Qian Chen: But I mean in this graph it's a 2 loop graph because you can do the gauge fees, and you can see that this graph has a 2 loop, one
Qian Chen: and and this will you, I mean. Oh, okay.
Abhay Vasant Ashtekar: So the key differences between one over 2 group, 2 loop clubs
Abhay Vasant Ashtekar: in the one loop graph.
Abhay Vasant Ashtekar: The secondary video did not depend
Abhay Vasant Ashtekar: on how you gauge fixed. And here
Abhay Vasant Ashtekar: the second derivative in the 2 methods that you showed us up here
Abhay Vasant Ashtekar: has a does change.
Qian Chen: No, no, no, I I mean it Doesn't. Depend on how you go to fees to the the spin. That will it's just depend on the on the spin along along the loop, for example, in this, in this graph you choose a. If you choose a pass, you will you.
Qian Chen: you polynomial operator. We are adding on some spin with initial spin, one
Qian Chen: and another. Pass
Qian Chen: your your her. Normally we are adding on initial spin. We've been half so they're different because because because of the the the spin are different.
Qian Chen: but it doesn't matter to the co-screen it doesn't matter to the gauge facing.
Qian Chen: I mean, I mean
Qian Chen: DC. because you
Qian Chen: if you, if we, if we you are, if you are
Qian Chen: in boundary, you don't know how
Qian Chen: you you don't know which past you you're adding on, but you can
Qian Chen: pass in the past dependency to the spin dependency.
Qian Chen: but actually the the in the to the the entanglement will never, will never depend on the it will never depend on the page facing.
Abhay Vasant Ashtekar: Yes, okay, okay, that's okay. Thank you.
Qian Chen: Okay, okay. Thank you for the question.
Qian Chen: And and and I, I will try to make a conclusion.
Qian Chen: and along this but along this talk I will try to. I will try to present the spin now entanglement
Qian Chen: for the first thing I I do is to define a
Qian Chen: geometrical entanglement are so many, many apartheid entanglement measures for lupoon gravity.
Qian Chen: and the second, I I reconstructed, or find a relation between the second order, entanglement, dissertation to the
Qian Chen: dispersion of loop hormone operator. So in this sense, as we reconstruct it, I reconstructed curvature.
Qian Chen: in terms of entanglement institution.
Qian Chen: This and the so I
Qian Chen: I can. I I show you the spin network entanglement, actually get getting fat in loopy or in loopy in the china.
Qian Chen: and the spin level entanglement is preserved on the page fees on the coast green
Qian Chen: up
Qian Chen: as long as the evolution, as as long as the evolution is generated by the Holonomy operator
Qian Chen: and the and the side
Qian Chen: And the last point, I want to show you that the code screening implied the degree of freedom you have to the spin that will entanglement are in code in the boundary information and the bunk and the bulk, top logic, the top lot.
Qian Chen: the box top top log here is me at the same.
Qian Chen: It's me that a safe loop.
Qian Chen: because you can see that even for for given given folk even you, even you, have a many possible way to to define the co-screen
Qian Chen: you define the page facing you. You are the both. Both top of logic is the is the to my is due to mine by the number, by the relation.
Qian Chen: by the relation of number, or by the by, the relation of number of edge and number of vertices.
Qian Chen: So that's all that's what I want to show. That that's why I want to talk today.
Qian Chen: That's for your attention.
Questions.
Qian Chen: Okay. Hi, John. oh.
Deepak Vaid: no question.
Deepak Vaid: This entanglement operate. sorry. This
Deepak Vaid: phonomy operator. Right?
Deepak Vaid: it it
Deepak Vaid: it is an entangling operator. You said.
Qian Chen: This is a yeah, this is a entangle and operate because it's operator or so many, many many so many vertices, and we will create the spin to a position between the 2 vertices. So you'll create a
Deepak Vaid: and what what is the precise form of this operator. I think I can catch that.
Qian Chen: Here. This is the
Qian Chen: formally for this operator.
Qian Chen: There you are.
Qian Chen: Yet this operate operator is the trade is the trace of Wiganard Dimitri.
Qian Chen: Yeah. And so then to each other. So you will go. You are going to have a Dl. And a G one times 3, 2 times street in, and then you will close the maneuver
Qian Chen: by the sin.
Qian Chen: by the sin. And one so is the choice.
Qian Chen: Oh.
okay.
Deepak Vaid: Oh, okay, I see. So
Deepak Vaid: And okay.
Deepak Vaid: All right.
Deepak Vaid: Can you show me the the the the graph picture.
Qian Chen: the there's a graphical representation like this.
Deepak Vaid: Yeah.
Good.
Okay.
Deepak Vaid: I see. Okay, so
Deepak Vaid: Oh, all right, thanks. Thanks for your
Qian Chen: Oh, you're welcome.
Qian Chen: So basically if you tense
Qian Chen: we cannot demonstrate along this
Qian Chen: along this edge you will have some to open end, and then you are you are. You are summing the Mega negative number. So we connected them to each other, and finally you will connect it all the open end. So we are. You are going to have some corner, and this corner will get you a Secretary. Simple.
Deepak Vaid: right? I I see that now. Yeah, no, thanks. That's very.
Eugenio Bianchi: Can you? Can you see also out the parameter t your time enters? Are you exponentiating this operator? Or are you put in T. As a parameter in a new?
Qian Chen: You? You may need this, this this parameter.
Eugenio Bianchi: Yes.
Qian Chen: Oh, yeah, this is is just a just as up. I say, this is a parameter. This is a just parameter time.
Eugenio Bianchi: Yeah. And the question is, what is age? Is the
Qian Chen: H. Is the whole nomi operator.
Eugenio Bianchi: I see.
Abhay Vasant Ashtekar: So it's that. Use Hamiltonian, and time is really in there
Abhay Vasant Ashtekar: in the general sense, like I mean, given any excel for job Operator.
Abhay Vasant Ashtekar: Yeah. And and the parameter that flow is called
Abhay Vasant Ashtekar: I mean, it's not really the physical.
Qian Chen: Yeah, actually, you can.
Qian Chen: You can place any operator in this formulation.
Abhay Vasant Ashtekar: Yeah, any. It's just to the second chance. So so I mean the the the motivation was a those of us who are not exporting this thing was was a little bit
Abhay Vasant Ashtekar: and clear. So let me just make sure that
Abhay Vasant Ashtekar: that we understand your motivation
Abhay Vasant Ashtekar: of using. I mean, Why, why do you use a lot of me as opposed to anything else?
Abhay Vasant Ashtekar: And is it because you want to
Abhay Vasant Ashtekar: sort of see how curvature
Abhay Vasant Ashtekar: is generating entanglement.
Abhay Vasant Ashtekar: Is that the idea? Or
Qian Chen: yeah, yeah, yeah, basically, yeah, I want to, because I size as I say in the introduction, I want to. I want to see how the I want to see the relation between curvature and the entanglement. So I choose the curvature
Qian Chen: we choose. We choose curvature, or we choose polynomial operator
Qian Chen: to to study.
Qian Chen: And the
Go ahead.
Abhay Vasant Ashtekar: go ahead. And
Qian Chen: yeah, and actually
Qian Chen: actually this yeah, as I as I mentioned and the question, this operator. This operator is the entangling app, which means that you it will create some entanglement.
Abhay Vasant Ashtekar: What is the idea that supposing I give, I mean, I wouldn't understand the physical. I
Abhay Vasant Ashtekar: it's a very nice, automatically clean thing.
but
Abhay Vasant Ashtekar: physically is the point of view that's supposing I've given really some state of the gravitational field.
Abhay Vasant Ashtekar: and it is really a presented by some spin that to work to begin with.
Abhay Vasant Ashtekar: that somehow.
Abhay Vasant Ashtekar: if there is type, there is a physical time evolution in presence of curvature.
Abhay Vasant Ashtekar: I mean, you're not there. You're not telling us the physical time evolution. But I come from outside and say, Well.
Abhay Vasant Ashtekar: in us is going to this particular solution of quantum gravity solution is going to evolve in presence of curvature.
Abhay Vasant Ashtekar: and therefore you are saying that well, if that happened.
Abhay Vasant Ashtekar: then indirectly, you, you'll come and say that. Well.
Abhay Vasant Ashtekar: there's going to be entertainment generated
Abhay Vasant Ashtekar: between the subsystems that we defined.
Abhay Vasant Ashtekar: Is that the idea? I would
Abhay Vasant Ashtekar: Oh, or the physical idea.
Qian Chen: Yeah, the physical, the physical idea is, try to.
Qian Chen: It is try to, because we call loop converity is a background, independent proposal.
Qian Chen: and we don't have any. We don't have any background. Geometry, so we don't have any. We don't have any
Qian Chen: we don't have any. We don't have any notion of quantum geometry
Qian Chen: so
Qian Chen: but but but we but but as as fast I can consider
Qian Chen: that we have a relation of prosperity which emphasize
Qian Chen: the the relation between subsystem. So we can define, we can define the quantum geometry or
Qian Chen: geometrical notion from the from the relational perspective at the end of the most
Qian Chen: most direct, the a straightforward way to to have for to have a relation of prosthetics is to consider the entanglement between the between Hib. This here but here but the subway.
Qian Chen: So so I try to. I try to. I try to connect. I try to connect the relation
Qian Chen: of entanglement
Qian Chen: and
Qian Chen: quantum geometry.
Qian Chen: So that's why I choose to study the entanglement.
Abhay Vasant Ashtekar: That that's right. But is that the physical idea that
Abhay Vasant Ashtekar: I give you some state of quantum privacy. It says, like, I give you some spin network.
Abhay Vasant Ashtekar: Oh, yeah, actually make any evolution.
Abhay Vasant Ashtekar: And if that's been there to work has not to hold on on this
Abhay Vasant Ashtekar: we would say that well, physically, that corresponds to
Abhay Vasant Ashtekar: encoding some curvature.
Abhay Vasant Ashtekar: and then
Abhay Vasant Ashtekar: or aren't you saying that somehow.
Abhay Vasant Ashtekar: if there are these non-trivial holoma means.
Abhay Vasant Ashtekar: then
Abhay Vasant Ashtekar: the subsystems will get them tangled.
Abhay Vasant Ashtekar: So I am talking about futuristic point of view. I mean, this is kind of
Abhay Vasant Ashtekar: mathematical facts here. But
Abhay Vasant Ashtekar: point of view of physical application
Abhay Vasant Ashtekar: would be that. Well, if you give me the initial state, and that corresponding to such a flat space.
Abhay Vasant Ashtekar: then would you say that that means that as time evolved.
Abhay Vasant Ashtekar: this entitlement operator, if you like, will be trivial.
Abhay Vasant Ashtekar: It's just. There's no in there. There's no hologram is all trivial is identity.
Abhay Vasant Ashtekar: and then subsystems would not be evolved. It would not be entangled.
Abhay Vasant Ashtekar: But
Abhay Vasant Ashtekar: if there was curvature in the initial State.
Abhay Vasant Ashtekar: and somehow somebody gives you a lot of evolution of this initial state
Abhay Vasant Ashtekar: with respect to some internal clock, for example.
Abhay Vasant Ashtekar: then under that evolution.
Abhay Vasant Ashtekar: if there was
Abhay Vasant Ashtekar: subsystems to get an entangled. I mean, i'm trying to understand the physical message.
Qian Chen: if I want. If you want to look at a physical message, I I can
Qian Chen: let me. Let let me give you a very simple example. Oh, for example, I can consider a 2 particle interrupt to each other by nude and gravity.
Qian Chen: and in these cases
Qian Chen: you you are going to have with some correlation function or for a reality system, and you can. You can find that this relative distance
Qian Chen: encode the information of entanglement.
Qian Chen: So in these cases we don't have any
Qian Chen: a spin that spinner.
Qian Chen: but we have some we have, we we have, for we have we have a basic notion of for this and the entanglement?
Qian Chen: So in this sense I can try to. I can try to reconstruct it. The distance from the entanglement.
Qian Chen: So this system, this is the physicist. This is the physical motivation.
Abhay Vasant Ashtekar: So you're saying that you can
Abhay Vasant Ashtekar: reconstruct curvature from Nintendo.
Abhay Vasant Ashtekar: Yeah.
Qian Chen: Yeah, yeah.
Abhay Vasant Ashtekar: Is that what you're saying? Okay, yeah. Okay.
Qian Chen: yeah, yeah, yeah, this is the goal of this talk.
Abhay Vasant Ashtekar: But you can also do the other way around, right? And you can just say that in a in part example, you told me
Abhay Vasant Ashtekar: that I got
Abhay Vasant Ashtekar: this this interaction, which is the interaction Hamiltonian, which is the coulomb, one upon r, one minus r, 2
Abhay Vasant Ashtekar: potential, and that
Abhay Vasant Ashtekar: generates and time I mean the cause and effect
you can.
Abhay Vasant Ashtekar: We can switch right. We can say that entangle many generated by this particular
Abhay Vasant Ashtekar: the Newtonian potential.
Abhay Vasant Ashtekar: So you're also one could say that entanglement is generated.
Abhay Vasant Ashtekar: My
Abhay Vasant Ashtekar: the
Abhay Vasant Ashtekar: by the presence of curvature in the initial state. If the initial State had no curvature, then entanglement would not be generating.
Abhay Vasant Ashtekar: The initial State had curvature
Abhay Vasant Ashtekar: then
Abhay Vasant Ashtekar: by initial, I mean physical emission. Now.
actually
Qian Chen: actually I I I I I I didn't say that the the initial state have have Have I have no Covid. I I just want to say that the curvature is scary fat in the entanglement. Hesitation is is based on some initial state which have to know any entanglement.
Bye.
Abhay Vasant Ashtekar: I I think we're not communicating. Okay, Thank you.
Qian Chen: Yeah, yeah, you're welcome.
Western: Hi, I i'm gonna ask a question regarding the connection with the Black Hole line to be. Because so you started with a graph that it's a very similar to the kind of things that are considered when you you have a the colourizon with punctures. But I was wondering if in the end of your procedure, can you see that your calculation leads to an area low.
Western: and not just that, but in a deal with the location corrections as we have for black holes.
Qian Chen: Hmm. In this talk I didn't. I didn't show you. I didn't show you. But actually I seen I have some. I have some relation. I I have some intuition about about this question. yeah, since you mentioned the area law, and I can show you that,
Qian Chen: due to the coast grinning.
Qian Chen: you can see that
Qian Chen: that that that the entanglement will depend on some boundary spin, and the loop is spin and this loopy. This loopy spin
Qian Chen: actually has something to do with the you in film world, and in this you you in finger. I remember that in this, in this paper.
Qian Chen: and in this sense, in the in in this paper they study the you in framework, and which me that they you support your you. You have a you have a
Qian Chen: one word is a spin level, and you have some open.
Qian Chen: You have some open edge, and then you you look at the in the toilet space.
Qian Chen: You you look at it in the twentieth way of this vertices, and then you are going to have some area law
Qian Chen: if you make some assumption.
Qian Chen: And and now in in my
Qian Chen: in this study, I can show you
Qian Chen: that order, for all the information for entanglement
Qian Chen: is carried by the loopy by the loopy spin and the boundary spin.
Qian Chen: So you can imagine you can imagine a 2 vertices of a two-week voltage graph and they are many, many edge arbitrary arch link to each other. And then you study the entanglement
Qian Chen: between the 2 edge. And now you you.
Qian Chen: This study, will tell you that this study will tell you that in
Qian Chen: entanglement will depend on the will depend on the
Qian Chen: your depend depend on the on the model of you and feature.
Qian Chen: and the the issue is, Try to remove the loopy, try, try to remove the loopy, the
Qian Chen: the the bulk loop.
Qian Chen: So it's possible to have a aerial law.
Western: Okay.
Western: it seems so I I i'm struggling a bit, because when you were mentioning this graph with tunnels and
Western: and many meetings attach it to them. and connected with just oneing seems a bit of the opposite situation with respect to what you have with the black holes. So i'm confused the how you physically interpret the one situation with respect to the other.
Western: but but but but but but but but so you're saying that that there is a way from what you have to connect to to to the caller that results there is
Qian Chen: you made this graph.
Western: Yeah, I mean, this is what you were referring to. Right?
Qian Chen: Yeah, yeah, yeah.
Qian Chen: I I I I I I to emphasize this is just a very particular cases, and some conclusion can be generalized.
Western: Okay. So I mean, this will be probably a very interesting continuation of your project. This is something we need to see from what you're hand done. So thank you.
Qian Chen: Yeah, you're welcome.
You book
Jorge Pullin: and the others.
Deepak Vaid: yeah, Hi. So I have one question this the fact that this entire this of Colombia operator is entangling wasn't. It noticed by anybody before this.
Qian Chen: Oh, sorry.
Deepak Vaid: The fact that this Holonomy operator
Qian Chen: Yeah, he's an entangling operator.
Qian Chen: Yeah.
Deepak Vaid: this. This fact was not noticed by anybody prior to your work.
Qian Chen: I think this is a very DC. Did. This entangling
Qian Chen: property is very, very straightforward.
Deepak Vaid: No, I know. I know I know it's a beautiful property, but i'm just wondering if if if if anybody else had mentioned it in the in the literature in the past.
Qian Chen: I I I don't know. Actually, I don't know.
Qian Chen: I don't know. But I think this is a very. This is a very straightforward this. You can create some principles. Position.
Qian Chen: Yeah. Actually, if you find some operator? Can we a spin to a position? You will we? You will. The 2 Sounds
Qian Chen: entertainment institution
Qian Chen: and the special.
Qian Chen: the Loop polynomial operator is special because you prefer the gauge equivalence
Qian Chen: How to so the gauge Invariance.
Deepak Vaid: Right?
Deepak Vaid: Right? No, I mean, i'm. This is very simple and straightforward, but it's also very, very beautiful, and
Deepak Vaid: i'm just surprised that nobody noticed it in the past.
Deepak Vaid: or maybe they did. I don't know but yeah, it's very nice
Jorge Pullin: any other questions.
Oh, goodness, thank you, Speaker again.
Qian Chen: Oh, thank you.