 On Friday morning, I will finish today some things. And probably tomorrow, I don't know how far I will go today. Tomorrow maybe I will finish what I want to do today. And then we'll do some maybe some exercises tomorrow, some practice to prepare for the exam. Is there any question about what happened yesterday? Okay. Any question? Okay. Okay. So, yeah, write down all the questions you have and we'll go through all of them either at the end or sometime today or tomorrow. And so now if you have any question about the last course. So let me remember where we were. So what I showed last time is that you can write a field theory for systems of charged ions. And the whole point was that if I have a system of ions, the grand partition function of the system can be written in terms of particle as integral sum over n equals 0 to infinity lambda to the n over factorial n integral product over i dri and e to the minus beta over 2 sum over i dri and j qi, qj, I mean, okay, plus, et cetera, minus beta minus the interaction with the fixed charges, et cetera, right? So what we showed is that what we showed is that you can formulate this z of lambda exactly as a functional integral over all configurations of the field phi. So phi of r is a real field defined at each point in space. And you have e to the minus beta epsilon over 2 integral gradient phi square d3r. So if there is fixed charges here, you would have minus beta sum over i integral dr of qi vc of ri minus r rho f of r, and then minus beta over 2 integral dr dr prime rho f of r vc of r minus r prime rho f of r prime. So this is the electrostatic energy of the system. And the action which comes in here is minus i beta integral d3r rho f of r phi of r. And if you have several species with fugacity lambda i, so this was for one species, but believe me, you can easily show it's just the notation is a bit cumbersome. So it's lambda i integral d3r e to the minus i qi i beta qi phi of r. And this is an exact correspondence. This exact correspondence was obtained by making a change of variable by introducing in here a new variable rho of r equal rho f of r plus sum over i of qi delta f of r minus ri, and introducing and forcing this new variable rho of r by a delta function. And the field phi of r is the conjugate field to rho. So what I showed you is that the electrostatic potential, let's call it psi of r, psi of r is the expectation value of the field i phi of r, the expectation value being defined with this functional integral. What I showed also is that if you look at the concentration, the concentration of ions of type i, so ci of r is equal to lambda i expectation value of e to the minus i beta qi phi of r. Is it clear? I think this is up to where I went yesterday if I remember correctly. And of course, one important thing I need to say is that lambda i is lambda, no, sorry, the lambda i, the fugacity is lambda i which are defined here, are defined in such a way that the bulk, the total number of particles of type i is equal to lambda i d by d lambda i of log z of all the lambda i. This comes from here. And which means if you remember that z of lambda i equals e to the minus beta f of the lambda i, if you divide by v, the volume you have this relation which is that ci equals minus lambda i d by d lambda i of beta f over v. So this is an important relation which will determine lambda i. OK, so how do we calculate now this kind of object? And the standard method, so when you do field theory, there are many methods to calculate things. One of the standard approaches is perturbation theory. So perturbation theory is, as you know, when you have. So first of all, as a preliminary, the only field theory that you can do, that you can calculate exactly, which means the only functional integrals that you can calculate are the Gaussian ones. Yes? Yes. No, but I call it free energy. It's a grand potential, I agree. But for some reason people noted f, but it's omega in principle. It's omega. No, Gibbs free energy, I think, is when you subtract the pressure. One is at fixed volume. One is at fixed pressure. So anyway, it's a notation who cares. I decided to call it f. We are in a free country, so I choose f. It's not, it's the grand potential. I think usually you, the notation, it's not the Gibbs. I think, well, you know, there is, OK, the internal energy is u. Then you have the enthalpy h, which is u plus pv. Then from u is a function, if you look in terms of variables, u is a function of s and v. u is a function of the entropy. If you make a Legendre transform from u to go to temperature, then you go to f of t and v. Now, h, this is a Legendre transform when you add pv. And h is a function of s and p. It's a function of the pressure. And from h you go to g and p. And g is the Gibbs free energy as far as I remember my thermodynamics. And then when you go, so in fact f is a function of t, v and n. n is the number of particles. Yes? You have a question? OK, so just if you define f minus mu n, so if you go from f to f, you do a Legendre transform where you write f of t v n minus mu n. And I think this is omega, no? Omega is the other one, maybe. OK, it's a different name. But then this gives a function which is a function of t, v and mu. It's another function. Maybe I'm mixing up the symbols. OK, anyway, you go from one to the other by a Legendre transform. I didn't explain what's a Legendre transform, but I guess you have seen that in high school. This equation, yes? Or equivalently, this one? No, that's the average Ni. Yes, but what you fix, so you have a system which exchanges particles with the, so for instance, when you have a bath of salt, what is fixed is the average concentration of the salt. So the system can exchange ions with the reservoir. And what is fixed is the average Ni. So when you are in a grand canonical ensemble like this, what is fixed is the fugacity of the ions, so which is the same as the chemical potential, right? Lambda i's are related to the chemical potential like this. So the lambda i's are fixed, but the n can fluctuate, but they have a certain average which is usually fixed, and then they have fluctuations, and one can calculate the fluctuations, but usually the fluctuations are in square root, so they are very small in the thermodynamically. OK, so several approximations are possible. So ways of calculation. So the first one is perturbation theory. So if your theory is not Gaussian, which is in any realistic system, your theory is not Gaussian. So if your theory is not Gaussian, which means quadratic in the fields, then you have to do approximations to calculate things. So one of the possible approximations is perturbation theory. So perturbation theory is when the, if you have the, let me call this the Hamiltonian in this, so if you have h equals h0 plus gh1, if the Hamiltonian can be written as a quadratic part or Gaussian part plus g times kind of interaction, non-Gaussian part, if g is small, you can do an expansion of all quantities in powers of g, and this is perturbation theory. Now, in this case, there is no a priori small parameter, and the way to do is to do what is called the saddle point approximation or the loop expansion, and this is what I will start to discuss here. OK, so who has ever heard about loop expansion, saddle point approximation? Yes? Oh, everybody knows. OK, so I will review it rapidly. It's very good. Excellent. OK, so the loop expansion or the saddle point approximation is an approximation when you have an integral like this, let's say, consider I'll do it for one dimension, integral dx e to the minus l f of x. So, of course, this, as it is written here, it's not a one-dimensional integral, it's a multidimensional, infinite-dimensional integral, but the method is really the same. So essentially what you want to know is what happens when l goes to infinity. So when l goes to infinity, if your function e to the minus l f of x can look something like this, I don't know, it goes to zero at infinity, when l becomes larger, the relative weight of the various, of the various minima are, you see, let's say if you have a maximum, which means a minimum of f at some point, it's going to be more and more dominant because the correction of the, OK. So let me do it in the following way. I will explain the way to do it, and then I will explain what is the idea behind. So first you look for the minimum of f of x. So call it x zero. So f of x zero is such that df by dx at x zero equals zero. So it's one of these points where the function, sorry, it's the function f of x, it's a minimum. I plotted here the exponential. So the minimum would be something like, I don't know, so it can be, let's say, this point, for instance. So then what you do is you expand f of x around this point x zero. So if you expand f of x around, so it's f of x equals f of x zero, so I'll write it f zero. Right? When I put an index zero, it means that the function or the derivatives, everything is calculated at zero. Then there is a term plus x minus x zero f prime zero. But this term doesn't exist because you are at the minimum of the function. So this is zero. Then you have plus one half f second zero x minus x zero square. Let me write the next one. f one two three over six x minus x zero cube plus et cetera. So now if I rewrite the integral i of l and I replace f of x by this expansion around x zero in here, so the first term, there is a first term which comes out which is e to the minus l f zero. And then let me call x minus x zero. Let me call it. So then it's integral du e to the minus one half f second zero u square times l minus l. And the rest is sum n equals three to infinity f n zero over factorial n u to the n. Right? This is the generic part here. And of course, because I assume that we are at the minimum of f of x, this implies that f second zero is positive. So this term is really like this, a convex. OK. So then I do a rescaling and I will write z equals square root equals u square root of l f second zero. So then this will be equal to e to the minus l f zero. So du will be integral dz divided by square root of l f second zero. And then here I will have e to the minus z square over that. Right? The change of variable, the rescaling was done in such a way that I get z square over two. And then minus l minus, I'll write it as sum n equal three to infinity f second f zero over factorial n. And u to the n is z to the n over f second zero to the n over two. And here I have a factor l. So there is a factor l minus one minus n over two. And now if I look at the expansion when l goes to infinity, so I am doing this kind of expansion. So if l goes to infinity, you see that the term which are here, since n is larger than three, they go to zero. So I can do an expansion. I can expand this in powers of one over. So this is really powers of one over square root of l, because it starts for n equals three. It's one minus three half, so it's one over square root of l. The next order term, so if I write the terms here in this expansion, so the term of order zero is e. It would be e to the minus z square over two minus one over square root of l f zero of order three by f second zero to the three half z cubed minus, et cetera. The next one would be minus one over l f zero of order four and I forgot the factor of six. And here would be 24, et cetera. And this is to be integrated d z. So then what I am just saying is that when l goes to infinity, I can re-expand this in powers of l. And this is just e to the minus z square over two times one minus one over square root of l, et cetera. You do the expansion of all the rest, order by order in powers of one over square root of l. And that will give you an expansion of i of l in the form, so i of l finally will be e to the minus l f zero divided by square root of l f second zero times. So the integral of e to the minus z square over two, it's a square root of pi, so it's going to be one minus something divided by square root of l minus or plus something of order one over l plus, et cetera. So if you do this kind of expansion, you have the dominant term. And this I can re-exponentiate because l goes to infinity, so the one over square root of l goes to zero. So this goes like e to the minus l f zero. Then there is minus one half log l, which is this one over square root of l. Then there is a term minus one half log f second zero. And then there, if I re-expand this, I get plus order one over square root of l plus all the other terms. So this is really an expansion. It's an asymptotic expansion. So what's interesting is, of course, that i of l is like a partition function. It's an integral. What you're interested in is always the log of the partition function, which is the free energy. So when you take the log of i of l, this saddle point method gives you automatically an asymptotic expansion for the free energy in the powers of l for large l. So the dominant term will be proportional to l. Then there is a trivial correction, which doesn't depend on anything which is log l. But then you have a constant term. So there is a dominant term for the l. Then a constant term. And then corrections of order one over square root of l, which you can expand to any order you want, by some kind of perturbation theory. Yes? Yes. This exponential, because l going to infinity, this goes to zero. So all these terms become small when l goes to infinity. So it's one minus f zero z cubed. Of course the integral, and then you do the integral, the next term will be minus one over l. But you see terms of order one over l, you get one from here from the first order expansion, and you get one from the square of the expansion of this one. So you have two contributions for the term one over l. Yes? And then you do the integral over z. But then the integral over z, so this one will give zero, because it's z cubed e to the minus z square over two. But now if I take the term of order one over l, so there is a term like f zero four over two, 24 f zero second square z to the four. And then there is a term plus one over two, the square of this term, which is of order z to the six, which comes from the square of this. So now I give you a funny result, which is not funny, but it's the manifestation of a very important theorem for Gaussian integrals, which is called the weak theorem. And the weak theorem, OK, so I don't know where to write it. I want to keep this. I don't need any more. And so for this simple case, let me show you our dx, let's say e to the minus x square over two. So I normalize it. And x, so if I take x to some power, if the power is odd, it will give zero. So if I take x to the two n, this is weak theorem. And the way to do it is the following. I represent this as a vertex with two n points like this. And then you do all the possible contractions two by two by pairs in all possible ways. So this will generate, so this is how you construct Feynman diagrams. OK, I just do it rapidly, but this is weak theorem. So, for instance, if I have something like this, which are numbered one, two, three, four, this will generate this or this or even one and three. So the rule is that this integral here is the number of ways you can do complete pairings of this diagram on itself. So for instance one, the one here, you have two n minus one possibilities to pair. So let's say you pair it like this. Then you have two n minus two. So you take the next one, you have two n minus three. Have you ever heard about weak theorem in field theory? Yes? Yes? Good. So weak theorem tells you that when you have a Gaussian integral or in field theory, if you have a Gaussian or quadratic action, the expectation value of a product of fields is the sum over all set of complete contractions, pairwise contractions of the operators. Yes? Sorry? Yes? OK, so it's a minimal point. I will show why it's a subtle point, because, OK, I'll come to that precisely now. OK, so just to finish this, once you do this expansion, this simple integral and in the case of field theory, the weak theorem tells you how to calculate this expectation value of z cubed, z four, z six, et cetera, to all orders. And in the case of field theory, it's a bit more complicated. But given the time, I cannot, I mean, so here z cubed will give zero, z four will give factor of three, z six will give 15. So you can calculate order by order to any order you want, to any precision you want. This integral, when L goes to infinity. OK, so I will come to your question of subtle point in a minute when I will do it for the, so the idea is simple, and I will not do it systematically, in principle, we can do it systematically, and you can do it for the case of the field theory of charge ions. So what you do, so let me give you some vocabulary. This is called, this is the subtle point value, and it turns out it's called the mean field value. We will see that it corresponds to the mean field in the case of, in the case of field theory. This is the first correction, so this term, which is this one, this is called the quadratic fluctuations. Why is it called the quadratic fluctuations? It's because it comes from this term. It comes from the integral of this term, and then all the rest, all these corrections are order loop corrections. So it's one loop. This is one loop, when it's one over L squared, it's two loops, et cetera. OK, so this is just vocabulary. Yeah, you had the question, yes? When I define what? Yes? No, it's positive. It's positive because I said it's a minimum, so if it's a minimum, F zero is, F second is positive. So, OK, so no, precisely this is where one point I wanted to say. So, if you have two minima, let's say X zero and X prime zero, OK? So, let's assume that you have, so I erase all these. The idea is clear, right? You just do the expansion. OK, so if you have two minima, X zero and X prime zero. So, for both minima, of course, F prime is zero. Such that, let's say, F of X zero is smaller than X prime, rather than F of X prime zero. OK, so then, as we saw, the dominant, OK, I erased it unfortunately, but up to small corrections, the corresponding expansion would give you here E to the minus L F zero. And here would be E to the minus L F prime, F of X prime zero, OK? Now, if F of X zero is smaller than this, then this is larger than this. So, the contribution of this is larger than this. Not only that, but if I look at the ratio of the two, so the ratio, so if you want, when L goes to infinity, if I expand it around here, I would get E to the minus L F zero. And this, if I expand it around here, I would get E to the minus L F of X prime zero. If I look at the ratio of these two numbers, I would get E to the minus L F zero minus F of X prime zero. And since this is positive in the limit when L goes to infinity, this goes to zero. So, which means that the relative contribution of the other saddle, so you're right, in principle, you should write that I of L, I could imagine expanding around X zero and around, I could imagine writing that it's E to the minus L F zero plus E to the minus L F of X prime zero, et cetera. But what I'm saying is that the dominant one is the one with lowest value of F because relative to the other one is exponentially small compared to. So, if the minima are well separated, you should expand around one minimum and do the correction around minimum, one minimum. Now, if what can happen and what happens in many, in quite many situations, not here but in many problems, it can happen that you have an infinite number of minima which are degenerate. In this case, you have to sum over L minima and expand a little bit around each minima and the situation is more complicated. But here we will stick to one minimum and expand around one minima. So, is the idea clear? No question? Okay. So, the strategy is simple. Look for a minimum of the action of this and then expand around it. So, what we will do is we will expand just to quadratic order. In other words, we will write, we will look, we will look at what we will do. So, I will write, I will look for a phi zero which is a minimum of this function. Right? So, first of all, I will write it, I will introduce a fake parameter L. So, write Z, I forget the lambda i, so Z of L is integral d phi e to the minus L. Times beta epsilon over 2 integral gradient phi square dr plus i beta integral dr rho f phi minus sum over i lambda i integral dr e to the minus i beta qi. Okay? So, I just put the L in front and I will do like if L goes to infinity, like if L is very large. Okay? So, first thing to do is to look for the minimum phi zero. So, this is in a functional integral. It's always of the form exponential of something and there's something called the action. Okay? So, this action, which I can call S or f of phi. So, the first thing to do is to write, to look for phi zero of R such that delta f of phi zero by delta phi zero of R equals zero. Yes? No, but L is not the length. L is just a parameter. It's an artificial parameter which I will imagine that it is large and at the end I will put L equals one because the real quantity, that's the real thing that you want to calculate. So, by the way, I didn't comment what is the physical meaning of this saddle point expansion or this kind of thing. The meaning is that you look for the, so this is like a probability, right, in the partition function because you remember that the Boltzmann weight would be, the Boltzmann weight is just e to the minus, f of x divided by i of L. I could have called it Z. So, you see that the Boltzmann weight, that's the probability to find the particle at point x. You see that if I use the expansion that I was showing, this means that the Boltzmann weight has a form typically e to the minus L. So, this is f zero e to the minus L over two x minus x zero square and divided by, so it turns out that this is e to the minus L f zero and then it's a square root of two pi, there is f second zero. So, these two disappear. So, you see that when you do this kind of approximation, it means that you assume that x is going to fluctuate around the maximum value and at x equal x zero, it's x equal x zero is really the value for which the probability is highest, which means it's the most probable value of x. So, here if we do this kind of approach, the phi zero that you will get is the configuration of field which has the highest probability to be seen. So, it's the, right? If we minimize, then you see that the Boltzmann weight of the field that you will get is maximal. So, it's the most probable field. That's the whole idea of this saddle point method. It's like in perturbation theory. When you do perturbation theory, it's a parameter of expansion. So, you do a formal expansion. It's like if you do a series, you write exponential x equals one plus x plus x square over two, et cetera. When you do this, you assume that x is small. But in the end, you can put x equal whatever number you want if you want to calculate. If you want an approximation to e to the x, you can calculate this with x equal one. But the expansion is an approximation which is valid in principle for small x. That's okay. You got the idea? So, it's an asymptotic expansion. It's not a power expansion because you have, you have, it's not an entire function. It's asymptotic. It has no radius of convergence whatsoever or something like that. It's just valid when L goes, so it's an expansion when L goes to infinity. You can write as many terms as you want in this expansion. And then, once you have written it, it's really like this, you can calculate with any value of L. This is very common in physics. You, all these large L, large N, they're all kind of approximations. All these expansions are done like that. And that's the general philosophy. Okay. So, the first thing we do is to try to look for the configurations which will maximize the Boltzmann weight of the system. So, the most probable, so the idea is you look for the most probable configuration and then you look at small fluctuations around this configuration. This is exactly what you have here when you do this expansion. X zero is the most probable configuration because if X equals X zero, it is like this. And then you have quadratic Gaussian fluctuations around this minimum. Okay. So, the first thing to do is to calculate the functional derivative of this action with respect to phi. So, you all remember, so you had a tutorial on functional derivatives. So, if I take the functional derivative of this, so the functional derivative of the first term is minus Laplacian phi. And I call it phi zero. You remember that the functional derivative of gradient phi square is minus Laplacian phi, right? With a factor of two. So, it's two Laplacian phi. If you don't remember, I will do it. But for the moment, I will assume that you know it. Then functional derivative here will give plus i beta rho f because it's a linear term. And then minus sum over i lambda i, then functional derivative of this with respect to phi. So, there is plus i lambda i beta lambda i qi e to the minus i beta qi phi equals zero. And so, I can erase the beta. And so, the equation is just minus epsilon Laplacian phi zero equals minus i rho f minus i sum over i lambda i qi. OK, so it looks like phi is a purely imaginary field. So, I will write psi zero equals i phi zero. It's my right. I decide to. So, if I do this, so which means I multiply this equation by i, then this equation in terms of this field psi zero is minus epsilon Laplacian psi zero equals rho f plus sum over i lambda i qi e to the minus beta qi psi. OK, so this looks identical to the Poisson-Voltzmann equation that we had before that we had derived in the beginning. The only difference with the Poisson-Voltzmann is that in Poisson-Voltzmann here we had the bulk concentration. Here is the fugacity in originality. So, we will see how to reconcile this because remember that the lambda i's have to be defined. The lambda i's are defined by the equation c i equals lambda i d by d lambda i of minus beta f over v. But now, as we see, so this z of f, I'll write it as e. So, we saw that when you write the expansion, this integral scales like e to the L times something. So, I write it as e to the minus beta L f. So, when I write this expansion, you see that f is a function of L, right? F has an expansion in powers of L when L goes to infinity. And therefore, you see that the lambdas, this is an equation for lambda, so it means that lambda has an expansion in powers of L, right? Because f is a function of L, you expand it in powers of L. C is a number. It's given. So, if you want to satisfy this equation, it means that lambda is a, it can be a, so what I'm saying is that f of L will have the form f zero plus one over L f one plus et cetera, right? Because as we saw when you expand by the saddle point method, the free energy has an expansion in powers of one over L. So, if you put this in here, it means necessarily that you will have lambda i equals lambda i zero plus one over L lambda i one plus et cetera. And all this is to be determined. You have to make this equation exact, order by order in powers of one over L. So, what we will show, and it's fairly trivial, is that to order zero, so in this equation, you have to write this equation, then you calculate f at order zero, and this will give you that the lambda i, so the lambda i which should come here because you are at order zero in powers of L, the lambda i zero is just C i, the concentration, the bulk concentration of the system. I will show you that in a minute, but so the idea is that everything is expanded to a certain order in one over L, and you have to be consistent in all the orders in one over L in the expansion, okay? Yes? It was the expectation, but today it's just i times five. So, yes? Because no, I'll come back to this. It's just, yes, so when you have, okay, first of all, let me show you one thing. When you have, when you do the saddle point, and this is where the, so it's related to this question of saddle point. So, in principle, we saw that what you're doing is that you're integrating L from minus infinity to plus infinity, say, e to the minus L f of x. Now, if you find, okay, so the integral is like this. Now, assume that everything goes to infinity, to zero at infinity, and everything is well behaved. So then you know that you can deform the contour of f integration as you want in the complex plane, as long as you don't cross any pole of the function f of x. As long as you don't cross any singularity of the function f of x, you can go wherever you want. And now, of course, if you have a saddle point, which is somewhere in the complex plane, it's perfectly legal to deform your contour and do the expansion around this point, because if you do this, if you get a better saddle point, which means a better minimum of the function, a better extremum of the function along this path, then it will be a better approximation to this integral. So, what I am saying is that I call it saddle point, and it is because, essentially, the saddle point that you choose seems to tell you that it's not on the real axis, but there is a, because on the real axis, you will not find any solution. There is no minimum of this. But if you deform the phi's and allow them to go in the complex plane, you find that there is one saddle point which is purely imaginary here. And now it's called a saddle point because you know that when you have a function, a complex function in the plane, it cannot have a minimum. If it's an analytic function, it cannot have a minimum. It can only have saddle points, right? That's why it's called a saddle point method. And in principle, what you're supposed to do is you know when you have a saddle point like this, you look at the second derivatives, and you have regions, two perpendicular regions. In certain areas, the function increases, and in other areas, the function is bound. So in this region, the function increases, and in this region, it's a maximum, and this dictates the way the path is. If you have a saddle point like this, you are not allowed to go through it because you would have to go over large regions to go back to infinity because the path has to be fixed at infinity. So you can deform it like this. Is it clear? More or less? Right? You have a path from minus infinity to it plus infinity. You can deform it in the complex plane as long as you don't hit any singularities of the function. And you can select any saddle point provided it has this shape, namely that it allows you to go back to infinity while not increasing. If you have a saddle point like this, if you go through it, if you cross, here it's increasing. A saddle point, it looks like a parabola, it's like this. So if you follow this direction, you go up, and you cannot satisfy that you go to zero. So in other words, this is it. And here you can look and you can show that the saddle point phi zero, so it's the saddle point, it's not the integral, is pure imaginary. And therefore, that's how you get this equation. Is it approximately, yes? Sorry? Can you talk a little bit lower? Because I cannot hear her. Yes? Why not? You mean for this equation or in general? Well, the saddle point approximation is just looking which part, which field here gives you the most contribution, the biggest contribution to this integral. It turns out that in this case it's on the imaginary axis, but it could be in the real axis. I mean, if I give you, there are some cases, so in this case it's pure imaginary axis, but in some cases it can be in the complex plane or it can be in the real axis. It depends. If it's on the real axis, in general, you don't have to move your contour. You don't have to do, you just take it as it is. In this case it turns out that you have to do this change. And this change is okay because it shows then that the electrostatic potential, if you remember, I erased it but it was there, the electrostatic potential psi i is the expectation value of i phi i. Sorry, not phi i. Psi of r is the expectation value of i phi of r. That's what I showed last time. And it turns out that when you are at the saddle point approximation, so i of l, we saw that the p of x in the saddle point approximation goes like e to the minus one-half x minus x zero square over times f second zero divided by square root of two pi f second zero. And there is a factor of l. So if you calculate expectation value of x at this approximation level, it's just x zero. You just do the integral. You shift x by x zero on the rest of the integral, right? If I calculate integral dx x p of x, of course, if I go to higher order in l, I will have corrections. But at this order, this is just integral. So I write x minus x zero equals u. So it's integral du of x zero plus u. And then e to the minus l over two u square f second zero divided by this. So the first term is x zero, which is constant. And the second one is zero, which means that at the saddle point approximation, the expectation value of any quantity is equal to the value at the saddle point. So expectation value of i phi of r in the saddle point approximation is just i phi zero, where phi zero is the saddle point. Yes? Ah, yes, yes. Sorry. Yes, yes. It's just to confuse you a little bit more. Yeah, otherwise it's too simple. OK, you very fast, you run out of notations. Very often people call this s. S as action. OK, I don't know. We can call it, let's call it s of phi. OK, let me keep f because you wrote already f. So I mean there shouldn't be too many ambiguities. Actually, you see that at the saddle point, these two will be the same, right? Because f of at phi zero will be the free energy at the lowest order, right? To lowest order, if you remember, it's f of x zero at the value. So f of phi zero is really the f at lowest order. So it's not exactly completely stupid. It's just a little bit stupid. OK, so this is the equation. So essentially you have what happens is that the Poisson-Boltzmann equation and the Poisson-Boltzmann electrostatic potential appears to be the configuration of the field, which has maximal contribution to the free energy to the Boltzmann weight. So it's the contribution, which is dominant, let's say. And then what you want to do is to look for fluctuations around this dominant configuration. OK, let me just do one more thing. And then I will give examples. The one more thing is that let's do a case, for instance, assume that there is no external charges. So it's just a bulk solution of iron. Example, take rho f equals zero. It's a bulk electrolyte. So an electrolyte is just a solution of ions of all kinds, whatever you want. So if rho f is equal to zero, right, then the solution is psi zero equals zero. No protest. You just put it in. So in fact, what you get is zero equals sum over i lambda i qi. And because of charge neutrality, sum over i of lambda i qi is zero. So if you have charge neutrality, which you always have, and if you have no rho f, the solution is psi zero equals zero. So if psi zero is equal to zero, which means phi zero equals zero, if I look here, you see that z, so to lowest order, will be equal to, so this is zero, this is zero. And then I have just e to the lambda sum over i, sorry, e to the l sum over i lambda i times v, plus corrections if I had done the expansion to higher order. But here, right? Yes? No, no. X zero, what plays the role of x zero is psi zero. Initially I had this integral dx e to the minus l f of x. So this is integral d phi of x, phi of r, e to the minus l f of phi. So looking for a minimum x zero is equivalent to looking for phi zero. So phi zero is the saddle point. It's not a point. It's not the x zero. It's a function which minimizes the f of phi. So I'm looking for this function which will give you the maximal weight here. Is it clear? So then the point is that this, you expand as e to the minus l f of phi zero plus some correction which I will calculate after. But the lowest, to lowest order in plus correction of order one over l. So there is a correction of order one plus et cetera. So in this class I will calculate the f of phi zero and the f one. So what I'm saying is that the f of phi zero is just f where you replace phi zero by its value at the saddle point. So the value at the saddle point is zero. You plug phi zero equals zero here. So this is zero. There was no external field. So you get this. And this is the integral of r when phi is equal to zero. It's just the volume. So this is what you get. And therefore you see that beta f and this z of l is e to the minus beta l f of l. So this tells you that to order zero f zero, beta f zero of l. So beta f zero is the order one is equal to minus sum over i lambda i times v. And therefore you see that the equation for the particle number, this equation, to all, yes? The second, what? I mean, because if you assume that the, yes? And then the sum over all the species is zero k. Yes. Then there's the second derivative of phi zero, right? Sorry. So this equation, if I put zero here, I just verify. I just check that if I put phi zero equals zero, it works. No? So what else do you want? The second derivative of function is zero, right? The second derivative? On the left of this equation there's second derivative, right? The Laplacian. The Laplacian is zero. If psi is zero, the Laplacian is zero. I don't understand if you are starting assuming rho f equals zero or phi equals zero. No. I assume rho f equals zero. So there are no external charges. So it's just a bulk solution of ions. So if I have a bulk solution of ions without, by just translation and invariance, I can be sure that psi is going to be independent of r in space, right? Yes. There is no breaking of translational symmetry in the system. If I don't have fixed charges, everything is translationally invariant. So psi should be independent of r. So Laplacian of psi will be zero. And then because of invariance, so psi equals zero is a, actually psi can be any value you want. Any constant works, right? Because if I have any constant, this will give me zero. And this, if it's a constant, it comes out. No, it doesn't work because of the qi. So it has to be zero, right? Because of translation and invariance. Psi is constant. And because of this, the only solution is psi zero equals zero. How, what? How did the volume come out? How the volume, because look here. I have integrals everywhere. So if I put phi equals zero, so this term, if I put phi equals zero doesn't contribute. There was no rho f. So here I have integral dr of one. That's the volume. Yes? Absolutely. So what I'm saying is that from this analogy, the order when I, you see, when I was doing this integral, the expansion was the order zero of f of, so if this is i of l, if you want, was lf of x zero. So the lowest order term is f, the value of the function, take at the saddle point. So the value of the function taken at the saddle point, that's this. So what I'm saying is that at lowest order, this is e to the minus beta l times f zero plus one over lf one plus et cetera. It will be like that. So what I'm saying is that f zero is given by this. And furthermore, it will be the value of lambda i zero because lambda itself will depend on l. I'm not sure we can see that today. So maybe in the beginning next time. But the lambda, since lambda is determined by this equation, if we expand f in powers of one over l, the lambdas have to be expanded also in powers of one over l so as to satisfy this equation which doesn't depend on l. So they have to be adjusted order by order in powers of one over l. It's a little bit like a renormalization group, in fact, where you request that certain quantities are independent of the cutoff or things like that. It's exactly the same procedure. Okay, is it clear or more or less? I mean, it's not, it's rigorous. It's not just vague. It's really, if you write an expansion of f, well, actually the strategy would be the following. You do an expansion of, you give yourself a certain lambda and you do an expansion of f as a function of lambda. So f would be f zero, not as a function of lambda as a function of l. So plus one over l, f one plus one over l squared, f two, et cetera, with fixed lambda. And then I say that I should have c i equals lambda i by d lambda i of minus beta f zero plus f one over l plus et cetera, divided by v. So you see that this will give you a function of lambda, which depends on l. And you can re-expand the lambda as a function of l. And you will have an expansion of lambda as a function of l, like this. Yes? Yes? Yes. Yes, so I was going to do it here. So by doing this, by replacing, I get that beta f zero is minus lambda i v, which is just this term. Right? Because I put phi equal zero. So I get minus sum over i lambda i times v. So then I do lambda i d by d lambda i of minus. So I get that c i. At this order, I should have c i equals lambda i d by d lambda i of minus beta f over v. And minus beta f over v is sum over i of lambda i. And sum over i, like the derivative is one. So you get really this. Exactly. So if rho f is not equal to zero, if rho f is localized in a certain space, so if you look asymptotically very far from rho f, so rho f has to be finite. So the distribution is always the same. It was like the question of the normalization. So if you have rho f, which is a finite object, if you have charge neutrality, then at large distances phi goes to zero. So you solve this equation asymptotically at large distances. And you will see that at large distances, psi goes to zero. And then when you plug it back here, you see that the region where psi is non-zero, you see when you have a neutral solution, a solution which is electrostatic charge neutral, and you have finite charged objects, then you have a Dubai length. And the field everywhere, the electrostatic potential will be non-zero, only up to a certain distance lambda d Dubai length around there. Beyond that, essentially the field goes to zero exponentially fast. Yes. OK, so now in case, of course, in the case where the charges rho f are not zero, you have to solve the full Poisson-Boltzmann equation. And that's it. One thing I forgot to show you is one example. What happens in the case of a one-one salt? So one-one salt, you have two, you have plus and minus components, symmetric. So the partition functions z is integral d phi e to the minus beta epsilon over two integral gradient phi square minus i beta integral rho f phi dr. And then here, if I have a salt, so the two components will have the same chemical potential because of charge neutrality. So you'll have lambda integral d three r e to the minus i beta q phi plus e to the plus i beta q phi. And q equals e or whatever. So this is just two cosine. Just to mention, this is a very common field theory, which is called the Sein-Gordon field theory. If you have no external potential, but it doesn't matter, this is a source field. So this is called Sein-Gordon. And the Sein-Gordon theory is very well known and very studied, particularly because in two dimensions, it's an integrable system and it has exact solutions, complete exact solutions in two dimensions, but not in three. I don't know if you have ever studied or heard about Sein-Gordon theory, but OK. So once we have the, and by the way, if you write the saddle point, yes? Because in 2D, this operator is marginal. The anomalous dimension is zero. So I mean the canonical dimension is zero. So it's complicated, too. OK, yes. So now, once you have done the mean field, the average, so it's called mean field because in fact it's not a mean field, it's the most probable solution. So you replace the whole integral by the value at one field. So now let's look what happens for the fluctuations. OK, so the fluctuations, and it is called the quadratic fluctuations. So I will write back the equation here. So it's minus epsilon Laplacian psi zero equals rho f plus sum over i. So at this, so lambda i qi and psi zero equals i phi zero. OK, so the next step. So the partition function Z of L, as we saw, is an integral d phi of e to the minus L minus beta o. OK, I call this what, this was f. No, I don't remember, was it f of phi? So let me call it s of phi because it's, no, because of the beta I didn't, I don't want to put the beta. OK, it's just to confuse you a little bit. So it's e to the minus L s of phi. So now what I want to do is expand around phi zero. So I write phi equals phi zero plus. And if you remember, I made a change of variable because in the integral i of L it was integral e to the minus L f of x zero plus L times x minus x, so sorry, minus L x zero square over 2 f second zero. So the variable really should be, I want to define u equals x equal square root of L x minus x zero. I want to include the square root of L here so that the expansion is more natural. So I will write phi equal phi zero plus a certain field psi divided by square root of L. OK, it's my choice. So, and I will do the expansion. In principle you're supposed to do it to any order, but then the calculations are more and more complicated. So I will do it to second order. To second order I write that s of phi is equal to s of phi zero plus psi over square root of L. And I do the functional expansion to order two. So it will be s of x zero plus s of phi zero, sorry, plus one over square root of L integral dr psi of r delta s by delta phi zero of r plus one over 2L integral dr dr prime psi of r delta 2s by delta x phi zero of r delta phi zero of r prime psi of r prime plus et cetera. Do you understand this expansion, this kind of expansion? This is the generalization of Taylor expansion for function. It's a functional Taylor expansion, right? So it's like when you have, if you have a function s of many fields, phi one, phi n, which are the values of the field phi zero at all the point in space, you see it's the value at the point here plus sum over i of phi d by d phi i et cetera. So this is really the simple generalization of the expansion of a function around a certain field phi zero. So it's phi zero of r, of course, plus psi of r. And psi of r is called a fluctuation field, yes? Where did I put it? I didn't put one yet. Where did I? No, at the end when you want to evaluate quantities and compare with, and see what is the value, what is the approximate value of the quantity, you put L equals one. So I do the expansion in numbers of L to any order. And then if I want to see the approximation, to see actual values, if I want to calculate really the free energy, at that point I have to put L equals one. But I don't, I mean it's an expansion. It's like if I expand a function, I can calculate all the coefficients one after one. And if I want to calculate the function at x equals one, at the end I will put x equals one. But to calculate things, you see what I, yes? This one? Yes, it's, you, it's, you, in this kind of writing, you have to imagine r. It's like an index. Okay, forget the phi zero. So this, this is what, you see, phi of r is really like the value of phi, if you, if you had a discretized, if the space was discretized, phi of r is really phi, it's the collection phi at point one, at point two, phi at all points on the lattice. That's what means phi of r. It's a field. So a field is defined by the value of the function at all points in space. So you can imagine you have a lattice, something like that. Now, if I have a function s, which is a function of phi one, phi n, which is the equivalent of this phi of r, if I want to expand around some phi zero, I will write that it's phi one zero plus psi one, phi two zero plus psi two, right? If I have a certain configuration of my field, it will correspond to phi one zero, phi two zero, phi n zero. This is my field, phi zero. It's defined in each point of space. So I have this. So then this is like a function of many variables, and I do just a Taylor expansion. So the first term will be s of phi one zero, the phi two zero, plus sum over i of psi i ds by d phi i zero, which is exactly this term. So the sums become integrals. And then the second order term will be the cross term sum over i and j of psi i d two s by d phi i zero, d phi j zero, psi j, et cetera. And then, of course, you replace all the sums by integrals. Yes? Sorry, can you? It's the same. I mean, you do the function of derivative with respect to phi and you replace by phi zero, or you replace first by phi zero and you take the, it's the same. Yes? So you choose it, so in principle you take s of phi, you do the derivative, and you put phi zero. Yes? Yeah. Or you write it as a function of phi zero, you do the derivative, and then you implement phi zero. Yeah, so I mean, it's the same. So the only important thing is that this quantity is zero. This quantity is zero because the phi zero, by definition, is the field which minimizes the action s of phi. Right? That's how we construct phi zero is the one which minimizes s of phi. So s of phi is equal to s of phi zero plus one over two L integral dr dr prime psi of r delta to s by delta phi of r delta phi of r prime psi of r prime plus etc. Actually, yes? Why what? It is because by definition, by definition phi zero is the field which minimizes the action. That's how it is constructed. So delta s by delta phi zero is by definition zero. That's how we choose it. So actually, instead of doing all these second derivatives things and etc., I can write directly that z of L, if I make the change of variable that I was saying which is here, so it's integral d. So I write it as integral d psi now because I have this change of variable. So obviously, this is a constant. It's just like a shift x zero. So from this, I have d phi called d psi. Right? It's like a normal integral. It's a multidimensional, infinitely multidimensional integral. So it's integral d psi of e to the minus L times beta epsilon over 2 integral gradient phi zero plus gradient psi square plus i beta integral dr rho f phi zero plus psi minus sum over i lambda i integral dr e to the minus i beta q i phi zero plus psi. So I just do the change of variable and then I will expand to second order because it will be maybe simpler than that. So this, if I expand this, I have a first term which is gradient phi zero square plus two gradient phi zero gradient psi plus psi, this is this. Then here I have a term, so beta epsilon over 2 integral of this plus i beta integral rho f phi zero plus psi and here minus sum over i lambda i integral dr. And then here I expand again to second order in psi. So if I expand to second order in psi, I have e to the minus i beta q i phi zero times 1 minus i beta q i psi minus beta square q i square over 2 psi square. So now what I say is that everything, so because phi zero is the minimum or the extremum of the action, all the terms of order one in psi, the sum of all terms of order one in psi are zero. It's just this, because it's the minimum. So since I have chosen by definition phi zero to be the minimum of the extremum of the action, the first order term in psi should be zero. So eventually what one gets is that z of L is equal to e to the minus. So I will collect the terms in pure phi zero, e to the minus L beta epsilon over 2 integral gradient phi zero square plus i beta integral rho f phi zero minus sum over i lambda i integral dr e to the minus i beta q i phi zero. So this is the first term. So the terms in psi are zero. The term linear in psi don't contribute. They are zero. And then I have integral d psi of e to the minus. So actually I made the mistake. It's psi over square root of L. The change of variable, I forgot the square root of L. So here was a square root of L and here is a square root of L, which means in the expansion I have a square root of L here, L here, square root of L, and here square root of L and L. The change of variable is here. So since there is a factor of L, you see that all the quadratic in psi carry a factor of 1 over L, and therefore we have e to the minus L times 1 over L. So there is no more factor of L. So I have beta epsilon over 2 integral gradient psi square. This term doesn't count. And then plus sum over i lambda i beta square qi square over 2 psi square. And this you recognize something which we saw in the beginning, which is the Bifurcal theory. So z of L equals e to the minus L beta epsilon over 2 integral gradient phi 0 square plus i beta integral rho f phi 0 minus sum over i lambda i integral e to the minus i beta qi phi. And then integral, so this is the contribution of the saddle point. So that's the dominant contribution. And this is the contribution of the quadratic fluctuations around the saddle point, right? It's small fluctuations around the field phi 0, which, and it's given by integral d psi of e to the minus beta epsilon over 2. And you can see that it's just the integral d3r of gradient psi square plus kappa d square psi square, where kappa d square is the Bifurcal constant given by sum over i of lambda i beta qi square over epsilon. Question? Yes? It's because when you do the expansion, the coefficient, you write s phi 0 plus psi over square root of L. So it's an expansion around phi 0. Therefore, the coefficient of psi of all the psi of the linear part in psi, the coefficient is just delta s delta phi 0, which is 0. In which term? The second term. This one? No, there is no contribution. No, there is. No, there is. Here? Yeah. But this is linear in psi. So in fact, no, it's not this term is 0. It's the sum of all the linear terms in psi. So the sum, let me see if I have some. OK, the sum of linear terms. So you have one linear term in psi here. You have one linear term here, and you have one linear term here. So the sum of all these linear terms in psi, the coefficient of the psi is 0. And you can see that easily to phi 0. And you see the coefficient of psi here, if you integrate by part, it will be minus Laplace and phi 0. Here, it's rho f. And here, it's this. And you see that you recognize it's exactly the coefficient of psi will be minus Laplace and phi 0 plus i. OK, so with some beta, with some epsilon minus plus i beta rho f plus or minus plus i beta lambda i e to the minus beta q i phi 0. And this is by definition the Poisson-Boltzmann equation that we had. So this is 0. So the coefficient of psi is precisely the equation that you put equal to 0. So you don't have even to worry. It's like this because you chose the phi 0 so that it minimizes the. OK, so the conclusion for today and next time, I don't know what I will do if we want to do some. OK, so the conclusion is that when you go to, when you do the saddle point for this functional integral, the lowest order term is the Poisson-Boltzmann theory. And the first order, the quadratic correction to Poisson-Boltzmann theory. So the fluctuations are described by the Debye-Huckel theory. This is the Debye-Huckel free energy. And, OK, and next time we will see that this is just e to the minus one half log of the determinant of the operator minus Laplacian square plus kappa d square. OK, so I will stop for today. So tomorrow I will try to maybe sketch the calculation of this quantity and I will try to answer questions. So if you have questions, prepare questions. If things are not clear.