 Okay, thanks. First of all, I thank the organizers giving me an opportunity to present our recent studies here. Today, I talk about Gibbs paradox in stochastic thermodynamics. And this work was done in collaboration with Hewler, Nakagawa, and Yoshimura. Let's start with the main result. We study n-interacting identical Brownian particles in contact with the heat bus of temperature T. Such a system can be described by Langemann equations for the coordinate gamma, and gamma is a correction of the positions and momentum of the particles. If you like over time descriptions, you can ignore the momentum degrees of freedom. The interaction between the particles is described by the Hamiltonian. And the identical nature of the particles is represented by the symmetry property of the Hamiltonians. Sigma gamma is a transformation of the coordinate induced by the permutations of particle levels. For equilibrium systems, the stationary distribution of the system is given by the canonical distributions. Beta is the impulse temperatures, and z is the normalization constant, which is called partial function. For these systems, we can consider thermodynamics by introducing a control parameter alpha. The time dependence of alpha is given externally without mentioning time evaluations of gamma d. Such a framework is called stochastic thermodynamics. The first problem of stochastic thermodynamics is to define the free energy consistent with the second law of thermodynamics. Maybe I think some of you like shallow entropy so much, so you may assume that some dynamic entropy is given by the shallow entropy. And free energy is defined as the expectations of Hamiltonian minus ts, s is the shallow entropy. These expressions gives a formula connecting the free energy and the partial function, just given the normalization integrations of the Gibbs factor. This expression seems quite reasonable. The main message on my talk today is that there exists a quasi-static process such that the difference of free energy is not equal to the quasi-static work. This quasi-static process is illustrated in this figure, quasi-static decomposition, which I will explain later. This result implies that the definition of the thermodynamic entropy using the shallow entropy is not consistent with thermodynamics. So what is the correct definition of the free energy? We assume following two conditions. The first condition is that the difference of free energy should be equal to the quasi-static work or any quasi-static process. And temperature dependence of the free energy is given by the Gibbs-Helmholtz relations. Indeed, we can show that these two relations uniquely determine the free energy up to an additive constant, applying this result to simple cases, particles in a box. Then we obtain the formula connecting free energy that's defined with partial function. The most important point here is that independence of the free energy. You can see n factorial here. This means that n factorial naturally appears from the definitions of free energies. You may be familiar with this type of formula when you run statistical mechanics. For example, if you start with quantum statistical mechanics, then you first assume that the thermodynamic entropy is given by a shallow entropy or a hormone entropy for the density matrix corresponding to the chemical ensemble. Then you define free energy is given by expectation of Hamiltonian minus T s. Then we take the classical limit for these expressions. Then in that case, you obtain the formula connecting the free energy and classical integrations of phase space of the Gibbs factor. This argument is completely correct. Then from this fact, you may think that n factorial comes from quantum mechanics. However, this is not correct. I mean, this argument is correct. This formula is obtained from the classical limit of quantum systems. But to understand n factorial, quantum mechanics is not necessary. As I said before, n factorial comes from the definitions of free energy. Such that free energy difference is equal to quasi-static work. Now I give you the proof. The most important and most non-trivial point is the construction of quasi-static decompositions. This is the illustrations of quasi-static decompositions. You first prepare the systems and we want to decompose the systems with a specified number of n1 and n2. You may think that such a process can be easily constructed, for example, by inserting a separating wall. However, such a naive decomposition cannot fix the number of specified regions. For example, number of particles in the left regions depend on the confurations before inserting the wall. So we have to prepare a special type of confining potential to fix the number of particles in a specified region. I give some examples. H is the original Hamiltonians and v cone is the confining potential. This is given by these expressions. If the number of particles in the region d1 is equal to n1, then this potential takes a value 0. Otherwise, this potential takes a positive value. Then when we consider the case beta copper is sufficient to be large, then n1 particles is confined in the region d1. Lambda is a control parameter. When lambda equals 0, the system is at original systems and lambda, we assume the lambda is a functional time and when lambda equals 1, we have the decomposed state. Then we can easily construct it, slowly burying functions of time lambda t. So in this way, we construct quasi-static decomposition using rather special confining potential. For this process, we can have quasi-static work. This is the standard formula and the important thing is that quasi-static work is expressed in terms of the functional functions at the final state and initial state, where the lambda is a partial function associated with Hamiltonian h lambda. The initial partial function is nothing but the original partial functions and partial function at the final state is estimated as this form when copper goes to infinity. This expression takes a slightly complicated form, but the essence is very simple. This is a partial function of this dictated ensemble such that n1 particles are in the region d1 and we don't know, we cannot specify which particles belong to d1. Now, starting from these expressions, we can calculate these partial functions using the symmetric properties of Hamiltonian. As I said before, Hamiltonian is symmetric for permutation of particle levels. Using this property, we can immediately obtain that the result is very simple. Partial function is given by the product of partial function of each box multiplied by the number of combinations of choosing n1 particle, rome n particles. Summarize the result. We obtain the quasi-static work as follows. Now, looking at these expressions, you can immediately confirm that. If you use the sum entropy for the sum entropy, you can find that free energy difference is not equal to this quasi-static work. On the other hand, if we use the definition of the free energy, such that free energy difference is equal to this quasi-static work, we can uniquely determine the independence of free energy. More careful and complete derivations of including the T.V. dependence of the free energy was discussed in the paper. I think this is a very basic issue in stochastic sum dynamics and also in equilibrating statistical mechanics. Now, this session is for information processing. Now, I will show you this result is closely related to information sum dynamics. As I said before, just inserting the separating wall cannot fix a particle number. I prepared special confining potential to fix the number of particles in the specified regions. But instead of preparing such confining potential, we can also use measurement and feedback process. I mean, we first measure the particle number in the region. This case is just one. And we insert the wall. Then in this way, we can fix the particle number in the specified region. This is measurement and feedback process. Now, we know that sum of dynamics can be extended involving such measurement and feedback process. I don't explain the formula itself, but according to the formula of the information sum dynamics, we can calculate free energy difference. And actually, we obtain this result. So, if the free energy in the information sum of dynamics is consistent with some dynamic relations, then actually, this relation gives the free energy. But to prove the statement, we have to construct first isotropic process associated with measurement feedback process. Indeed, this was discussed by the paper by Horowitz and Pound. And actually, they give such examples in sentences. But as far as I see, there was no mathematical expressions of the protocol. And recently, expressed presentation of the protocol by Horowitz Pound was presented in the paper and with discussion of the relevance within factorial programs. In this viewpoint, our results give another construction method of such precise static work corresponding to the measurement and feedback process. Finally, I give you another approach to n-factorial programs. The first, there were some references mentioning the partial function of the composite systems. This expression is quite similar to our result. But the problem is that what is the definitions of the partial function of the composite systems? In classical mechanics, a partial function should be given by the normalization constant of the counterbalance sum. So, we have to specify the confining potential fixing the particle numbers. As far as I checked, there was no argument. So, in this sense, our result gives a precise answer what is the definitions of the dispersion functions. As another approach, free energy is defined by the fluctuation property from the fluctuation properties. For equilibrium cases, large-dimension function characterizing macroscopic properties of thermodynamic variables is equal to thermodynamic functions. Therefore, one can define the free energies using the fluctuation properties. Indeed, in the case of macroscopic systems, we can identify any dependence of the free energy. However, this argument cannot be applied to finite systems, except for non-interacting particle cases. And quite recently, n-factorial program was discussed from the viewpoint of solute irreversibilities. As you know, the free energy is also calculated based on non-equivalent processes using the Zierczynski equalities. And if the reverse path is similar, Zierczynski equality is modified. And we can consider modified Zierczynski equality for free mixing processes. Then we compare the difference. We discuss the difference between the same type particle and mix case or different type particle and mix case. So, analyzing this case, we can argue n-factorial. But this argument is correct for non-interacting particle case. But as far as I see, there was no proof for general cases. If we want to have a proof for this statement, we have to use quasi-static process related to the mixing processes. This is nothing but our construction. Now, I summarize my talk. We construct a quasi-static decomposition process for small thermodynamic systems. And we assume that free energy difference is equal to quasi-static work then we can derive n-factorial factor of the formula. Thank you for your attention. I cannot hear anything. No, we hear you. We hear you. I see some questions in chat, right? Yeah, you want to answer that now. We have two questions in chat. But we can first invite PhD students and early career investigators to ask their questions. Then we can also read the questions from the chat. So, if you have any questions, please write up your question. Okay, so I answer to the questions in chat, right? Okay, so the first question is that if I understand this correctly, the only condition on the particles that they are identical, no other statistical symbol, right? Yes, right. So the point is that the nature of identical nature is expressed in terms of Hamiltonian. So that's all right. And the second question is that is there an experimental procedure to construct the recon potential that avoid measurement? Okay, that is very important question that I back to here. Yes, so this potential actually counts a number of particles. And so far, I don't have any idea of realizing such potential in real experiments. Of course, it's easy to have the numerical simulations. But so this is from the standard viewpoints of interacting particle systems, this is anybody interacting potentials. So such many body interacting potential cannot be designed in real systems. But I think, of course, it is not easy to design the particle interactions, because this is a many body interaction systems, but we may have another idea to realize such potentials. But I don't have any explicit ideas. Yeah, okay, thank you. We have a question by Farshid. So Farshid, please. Yeah, so I just want to make sure I understand the point you were trying to make earlier. Do you mention that if you use Shannon entropy for distinguishable but identical particles? Yes. Do you not get the one over factorial term? Is that correct? Yes. You mentioned the main result, right? Yes. So, of course, yeah. If we use Shannon entropy, we get the homeroid. So this is very simple, just inserting the current ensemble to this expression, then you can get these relations. But this case, we can directly confirm that free energy difference is not equal to process static work. And this process static work was constructed by our method. So I see. Okay, thanks. Okay. So at the moment, we have one more question. And this should be a quick one, please. Oh, it's a bow. So we will have more time later for additional questions and additional discussion. So, again, you can ask your questions or please. Okay, can you hear me, Professor? Can you hear me? Yes. Oh, you just mentioned that the previous words are limited to the non-interacting particle case, including the 2017's work by Masahiro Wuda. But I think that they just assumed the validity of the frustration theorem was absolute irreversibility. And since the theorem also can be applied to indirect particle case, so they showed it. Of course, final conclusion is correct, of course, but the problem is that logic, I mean, detailed argument, right? So as far as I understand, their argument is valid only for the special cases, non-interacting particle cases. Otherwise, they just compare the two different cases, process, but it is not easy to conclude any factorial for the general cases. However, in their article, the claim that they generalize the previous result to in the presence of inter-particle interactions to the case, yeah. From my understanding, the fluctuation theorem can be applied to indirect and particle case. Okay, so you say that, but they didn't give any proof and they actually are honestly writing that this argument is valid for non-interacting cases. Of course, non-interacting cases, we can estimate some modification part and you don't agree that. Okay, I'll check. Thank you very much. Okay, so if you have the, maybe please send some communication to me, right? So maybe we can discuss more. Okay, thank you. Okay, thank you very much. Oh, thank you very much for this great talk, for this great discussion. I hope you will be able to follow the whole session and to join the discussion after that, which