 Let me let me start reviewing some of the things we have discussed yesterday. We have a state at least the problem We are interested in We we want to to calculate the entanglement entropy associated to a region, okay a special region suppose you have A state which is a quantum state which is described by a global density matrix Louder Like this. No. Okay. Now it's better No, I should do like this No, wait, I can Want me to try no, okay now Yes So the idea is that if you have an estate which is described by this global density matrix What you have to do is Once you choose a region this induce a partition in the global Hilbert space, so The complementary region. Okay, what you have to do is to trace Partially trace over the degrees of freedom that lives outside the region You get a reduced density matrix or local density matrix and from here you calculate The standard for new man entropy, okay It's just and this gives you the entanglement entropy associated to the region. Okay, and There are two points important points here first you need this kind of partition of the Hilbert space Otherwise you What you get? I mean you it's possible to define an entropy But it's not going to be a Quantum Entropy it I mean that the more general expression for this has Another piece a classical piece that the Shannon entropy, okay But in this case if you are able to do to do this kind of partition Then what you get is a pure quantum contribution to the entropy, okay? This is one point Another point is that if you want to give to this entropy and interpretation as a measure of Entanglement you need to start with a global pure state, okay? Only in this case What you get is a good measure of entanglement so I show you yesterday that in general, okay any Density matrix can be written as the exponential of an air mission operator called the modular Hamiltonian and in this way we can We can say that this entropy belongs to the same family as the standard thermo dynamical entropy, okay so today the idea for today is to see how How we can calculate this entanglement entropy, okay? Even if the idea is quite simple the method of calculation are very hard Usually we have in the audience experts Pacquale and Eric Tony. I don't know where it is so they They they have done many contributions in this field, so let me start with The simplest Calculation we can do in the lattice, okay? Where this entanglement entropy is well defined it has no divergences, so It's going to be Perhaps more clear okay the Result we can interpret the result more clearly, so yesterday Are you ready for this? We are going to consider Gaussian systems What I mean with Gaussian systems is that we are going to to think that The weak theorem I mean all all the information about the theory is contained in the two-point functions, okay, so In this case the density matrix takes a particular form, okay? The exponential of the modular Hamiltonian is is a special Otherwise, we don't know how how to do the calculation. Okay, so let's take set of variables canonical commutation relations and Let's put a name to the two-point function Pi and the two-point function of Pi and now This is our answer What we are going to say is that the modular Hamiltonian in this case Going to be quadratic Variables, okay, it's you can show an exercise that this form of the density matrix All the information of the theory is in the two-point functions, okay So you have this weak decomposition of the end point correlations in terms of two-point correlation functions so now Once Once you have this expression what what you can do is try to Diagonalize these modular Hamiltonian. I'm not going to to give you the details of the calculation but the idea is That you can rewrite this density matrix way are the eigenvalues of the Modular Hamiltonian reduce operators and Here you need Something in order to have the trace of rho equal to one. Okay, so just for normalization and Okay, the idea how to go from here to here is to propose, okay, Bogolubo transformation of the variables and In order to fix the the coefficients the free coefficients you use These I mean you use the density matrix to calculate these two-point functions Okay, so you can you can fix the the coefficients the free coefficients to arrive here and What is interesting is that? What you get is a relation You can ask me the details want to see the how to go from here to here but what is interesting is that You get a relation between these eigenvalues of the modular Hamiltonian and And the eigenvalues are eigenvalues of this quantity Where root times and why this is interesting because as the the entropy the entanglement entropy of course is given by the Can be calculated in terms of the modular Hamiltonian then if you have a relation between the modular Hamiltonian and this Quantity here. Let's call it you can rewrite Tanglement entropy in terms of correlators, okay, this would be the trace of C That's one half. So this gives you Very strong tool to calculate the entanglement entropy. What you need is just to calculate correlators Okay, two-point correlators These are going to be n times n matrices so it's It's much easier to to do the calculation in terms of correlators Than to consider directly the calculation of the density matrix for example, if you have a System of n spins your Hilbert space will have Dimensions two to the n so you will have matrices With dimension two to the n times two to the n so huge matrices But in this way you reduce your problem to n times n matrices, okay So let me the of course I this is for scalars But you can extend this result to fermions or also to a Maxwell field to any dimension Okay, so it's We have done many many calculations using Method and in a way Okay, we are doing the calculation in the lattice, but it's the way we have to measure Okay, the entanglement entropy and to test some results that as I will show you analytically are very hard to to calculate then This tool is is very strong to to test if our results or super or or our Answers are true or not. Okay, so let me show you Some results. I'm sorry. I don't I don't hear you Because these correlators are restricted to your region. Okay. Yes. Thank you Yes, otherwise the result you are going to get is zero As you are as your global state is a pure state if you take The the the whole correlator you get a zero. Okay, so you have to restrict your correlator to the region Thank you, so Here, oh, perhaps I can see it here Okay, here we have an example for a scalar field, okay a Musless scalar field and What you have written there is the discrete version of the Hamiltonian. This is the real Hamiltonian Okay, this is not the modular Hamiltonian and once you have written your Hamiltonian in this discrete version, this is a You are thinking in a two-dimensional lattice, okay And we are interested in calculating the entanglement entropy of a square The idea Is that for a for a for a scalar field you can prove that if your Hamiltonian the real one is Has this form let's say the correlators for the vacuum state Are given by this matrix here and the restriction to the region is that this indices have to live Within your region, okay So once you have the expression for the Hamiltonian you have to identify This k matrix here and you calculate the correlators X and P once you have X and P you have This this matrix see and then You can calculate the entanglement entropy. This is the the steps you have to follow and what you get Answer for this for this case for a square You have an area term, okay, and a logarithmic correction Both divergent when the lattice space goes to zero, okay, so we expect that in the continuum We are going to have a divergent entanglement entropy But pay attention to the numbers The coefficients that we have for the area term and for the logarithmic and Let's see what happens We Change a little bit our square in this way and we see that The coefficient of the area term is the same Okay, because the perimeter of the of the square is the same But the coefficient of the logarithmic term has changed. Okay, so you Perhaps now is difficult to imagine but in the beginning we didn't know exactly How how the expansion of the entanglement entropy Goes so this was the only way to test our guessings, okay, so now we know how the entanglement entropy can be expanded in in powers of The ultraviolet cutoff but at that time We just guess so this kind of test were very very Important just to test what happens for example if your region have vertices or is a smooth boundary or I Mean you you you can you can play with different geometries, okay At no cost because it's it's just to to follow these these steps. Okay, and Okay, our conclusion with this experiment if you want is that The general expression for the entanglement entropy is a coefficient Times the perimeter so an area term the coefficient of the area term is not universal Why because if you for example move your square You rotate your square in this way the the coefficient of the area term Change so the the area term depends on the regularization You are using depends on the lattice you are using. Okay, so this gives us in this this this gives us We were sure that the area term is not universal, but the logarithmic term Is is contains relevant information about our theory. Okay, and In this plane with this With these geometries different geometries what you can you can guess is that The coefficient the logarithmic coefficient has a special relation with the angle in the vertex Some years later We we were able to find Exactly, which is the function this c function of the of the of the angle. Okay, but it took us at least some years to arrive to the to the right Answer so Let's see Let's see now what happens in the in the continuum. It's Clear that that the the entanglement entropy Even if it's a very interesting quantity it has problems in the in the continuum What we know now is that entanglement entropy admits an expansion in powers of of the let's go now Continuum don't see anything Call it Epsilon the coefficients depends on the boundary of the region Okay. Also, we have the contribution final part Okay, what we know now is that only The logarithmic term has a universal meaning and contains information about our theory In general, we know that this this expansion is true for a conformal field theory But also if you consider for any quantum field theory The diversion part will look Like this, but of course you can have corrections coming from the other dimensional parameters of the of the theory Another thing which is interesting is that this diversion part of okay I should mention that this is known very a law Okay, and this makes people to think that the entanglement entropy was a good candidate to explain black hole entropy These coefficients Depends on the boundary I mean the Contribution The biggest contribution to the entanglement entropy comes from the UV. So the character of this of these coefficients That that's why these coefficients depends only on the boundary Okay, you are local and extensive on the boundary are local and extensive functions of the boundary and We we know now that in this In this coefficient the logarithmic coefficient has information for example for depending on on the on the region you consider for example if you take Circles or if you take In general for a spherical regions what you get here is The anomaly of an informal field theory in the informal field theory you get the anomalies That gives you That tells you what is the theory you are you are taking into account. Okay, so let's see now Another method now not in the lattice in the continuum using the The replica method that was introduced. I don't know many many years ago for different different to solve different problems It is very useful to Calculate the entanglement entropy so let's see now the replica method Perhaps most of you already know what is it about but The idea is that the entanglement entropy can be can be Calculated from what is known as the rainy entropies These are associated to powers of the of the density matrix and Okay, let me define Are the rainy entropies? Okay, so if you take the limit And go into one you get the entanglement entropy, okay? You can ask yourself, okay, why It's easier to calculate rainy entropies instead of to calculate Directly the entanglement entropy the the advantage of this is that This quantity here Has an integral functional representation so we know how to do the calc well We know at least formally we can write an expression for the traces of powers of raw. Okay, and Functional representation of course I mean it's not even if we have this functional representation It can be very hard to calculate the rainy entropies But even if you succeed here you have a second step That is to take this limit can be very hard also, so In general this this calculation is very difficult. Okay only in some cases and for some geometries you are able to To arrive to the end, okay So let's let's see How does it work? Let's start with the Vacuum For this we have no you can you can write the way functional of the vacuum with the with the functional Integral over the lower half plane This is Normalization this is just Write it like this and here the Euclidean action Okay, this this this expression comes from in quantum mechanics The the reason why we are integrating and integrating over the lower half plane This is a way to select only the ground state. Okay So Once you have this expression for the for the vacuum that that's why We are choosing the vacuum because we know how to write this expression now We can write What we have done here is to integrate well zero here They are thinking in one plus one dimensions But of course can be done in any dimensions and You need the boundary condition and Now we can write the density matrix Just taking twice this expression. Okay, what I'm doing here is just thinking By two and five one are the boundary conditions at Well, zero how to arrive here. Yes. Yes Let's write it here suppose you you Who who asked me? Okay. Suppose you start quantum mechanics. Okay. Suppose you start with this Transition probability Final and then if you use them Both you can rewrite this. I mean using agent states of the Hamiltonian. You can rewrite this 13 this is the and this Yes, and now let's Agen states of the Hamiltonian it is They and they're also Okay, and you have to sum over n Go to the Euclidean and and now it's clear that when How goes to infinity? Yes, the only the only state that survived is the one with energy zero so the same thing can be done in With quantum fields and That's why you have to integrate over the lower half plane. Okay this is the way to select the ground state and Okay, once you have written your your density matrix the powers of No, there's one step first You're interested in a region. Okay, so you have to trace over the degrees of freedom Which are outside your region? So you have to add One integral and what you get at the end the answer is What you have is an integral home plane With a cut the boundary conditions zero plus call it By two or or from below your boundary condition is by two Okay so what what you have is two copies of The half plane. Okay, which are glue Everywhere except along the region. Okay, so you have a plane the complete Euclidean plane with a cut along the region and the boundary conditions with boundary conditions, okay along the Along your region and and now you can think What happens if you are interested in raw square you have simply Game this will depend have to take one Prime, okay So you have two copies here or the lower part part of the cat Inside with the upper Part of the cat in the in the second copy Like this. Okay. Now if you want the trace raw square Integral with traces Get occasion Okay, so in general the the representation of of the traces of powers of wrong Okay, we'll we'll be given by by this function functional integrals In a very complicated manifold, okay It's it's you have n copies in general n copies of the plane with conical singularities In the boundary, okay Why why I'm saying conical singularities in the in the in the boundary because Perhaps one way easy way to see conical singularities if you think for example You take one path you fix a point P and move from P let's say Have two copies. Let's take raw square and Move from here and you appear Immediately here in the second copy you move here and you appear again here. Okay, so let's So if if your path Doesn't contain the boundary you don't see the cat. Okay in this manifold, but your way from P Contains the the let's say you go again This You appear here But now I decide Then I will appear here Go back to P. I have to round Foundry twice a four pi angle To go back this means have a singularity a two pi and Singularity in the boundary So what you have to do is to calculate the partition function at the end in a very complicated Manifold, okay, and and you don't know how to do it. I mean, there's no a systematic way to do this calculation So let me let me show you What happens if you consider free fields in the case of free fields So let's let's write the the Result here races. This is the general result. So this is the normalization So It is manifold Something you don't know how to calculate except for some particular cases. Okay, and then Iranian dropies This is the final result What is difficult also is to find the continuation Partition function is only defined for integer n Okay, and and you need to take the limit and go into one Find the analytic continuation of this. So this is also difficult to arrive to the entanglement entropy but Good news Free fields are easier. You can map this difficult problem in a different one that you can solve exactly at least for fermions and scalars Massive fermions massive scalars on one segment and if you have multi-component systems only for fermions So this gives you an idea of how difficult it is In fact to get exact results for the entanglement entropy I have ten ten minutes more or or Yes no Then at the end You see that the the important thing is that what you have is This discontinuities in the boundary Okay, what happens along the cat is not going to be important at the end of the calculation But it's important. It what happens in the boundary You will you will see in the in in this Example I will give you at the end what you need is to know what happens here Okay, these singularities are the problems. Okay The the boundary conditions you put here It's I mean it's your problem will not depend on the boundary conditions you put here, okay So in the case of free fields you can map this difficult problem to another problem which is At least we know how to how to solve The idea is the following Let me find up too many papers here This one So for free fields Yeah, we are we are changing the problem, but always difficult problems, okay We start asking how to calculate entanglement entropy. So we say, okay We have a method this replica method and instead of calculating the entanglement entropy I will calculate a partition function in a very complicated manifold. So I don't know how to do it. So let's try Now instead of calculate the partition function We are going to map this problem to a different one That can this can be done for free fields only. Okay. The idea is that Of having n copies of the plane what what I'm going to do I'm not going to give you the details, but the at least the The idea is that we are going to introduce a vector field, okay with n components So let's say Like this this field lives in in one plane, okay with with a cat but what we know is that In this in this new picture The conical singularity that appears in the original in the original poor problem will appear appears let's say Boundary conditions of The vector field I mean to be to be more precise the the problem with this vector field is that Not going to be single value. We can introduce a matrix This matrix. I don't know if it has sense to to write has to give you that has once So this matrix when when approaching, okay? The the boundary the this vector field is multiplied by this by this matrix You see that the the work that that done That do that as these these these matrix is to go when from one copy to the next one Okay, that this is the the the work that these ones okay, and You can diagonalize this problem and what you get is a set of fields that Satified equation Approaching the cat from below or from above Sorry No In in in this picture. There's no end copies You have only one plane and This vector field lives in one plane a single plane with a cat But this field is not any more single value. This is the price. I mean you pay and Satisfy this condition here Yes, and there's also There's also a one here depends on if you are if you are dealing with fermions or scalars you have also here like this and and The values of K For scalars Calculate the values of K scalars and for fermions, okay? So this is the this is the answer. So what what you obtain is and Decouple fields would satisfy this funny condition, okay, and In the case of fermions, you can solve this problem And what you get at the end is that the only thing you have to calculate using Bostonization and and and another tools is that you have to calculate two point correlators of vertex operators With this Given by by by these values of K So you can solve exactly the problem even if you put for If you consider you can solve the problem for one segment and also for multi-component regions If you put a mass you can solve the problem exactly only for one segment for a scalars It's more difficult. You can solve only for one segment, but you can put a mass also, okay so Questions. Yes. Oh here What did you use that? The the fields has to be free You said that this technique is just for free fields. Yes, the point is that up to here you are not using the The field is free, but when you want to solve this problem, I mean this is just formally Written then you can you have to solve you have to find the density matrix for this For this case, I didn't show you how to solve exactly. I mean For scalars and for fermions, then you need the density matrix We didn't do it Okay, we just state the Problem and then we are going that's why for free fermions. For example, it's easier than for scalars. I mean this Just state the problem. Then you have to solve it. Okay? In the replica method You mentioned that this method all always it just works for the vacuum states Did I get you correctly? My question is exactly about the representation you are constructing the wave functional of the vacuum state here But I think in principle you can do it for any state So in principle the machinery works, but maybe difficulty makes you stop at some point Okay, any other question, thank you we can continue