 And then it's open bar. Questions. I will come back to exercises, whatever, to get you ready for a successful exam tomorrow. So I remind you that we wrote for a system of ions in solution. So if you have, let's say, a certain number of species with the fugacity lambda i and charges qi, the partition function can be written as a functional integral d phi e to the minus beta epsilon over 2 integral dr gradient phi of r square minus i beta rho f of r phi of r plus sum over i lambda i integral dr e to the minus i beta qi phi of r. So this was what we have painfully shown two classes ago, which I've been discussing yesterday. So let me specialize to simplify a little bit the notation to the case of a salt 1, 1, which means that there is q plus equals, so q minus, there are two kinds of charges, q minus and q plus. So q minus is minus q plus and q plus equals e, one electronic charge to simplify. And I will study, to simplify also, I will study a bulk solution. So it's a salt and bulk for simplification. So rho f equals, so of course the concentration c, 0 of the salt, it's a neutral object. So the concentration of the plus and the minus charges is the same. So it's the same lambda for both. So this term reduces to lambda e to the minus i beta e phi plus e to the plus i beta e phi. And so the partition function can be written as an integral over all field configurations phi as e to the minus beta epsilon over 2 integral dr gradient phi square plus 2 lambda integral dr cosine beta e phi. And it's three dimensional integrals or any dimensional depending where you are. By the way, there is something I forgot to tell you, which is an interesting thing that if the system is in one dimension, then the problem is like quantum mechanics. If you, for those of you who have studied finite path integrals, it's just a note in passing. I will not ask you that in the exam. But you see, if you are in 1D, if r is 1D, so all the integrals here are sum, let's say, between 0 and l if l is the size of the system. So this term is something I sum from 0 to r to l integral dx d phi by dx square. And then you have some potential term, v of phi of x. And this exponential of something like that, it's like a Feynman-Path integral. And therefore, it represents the matrix element of imaginary time, evolution operator, quantum mechanical operation. And this allows to solve all these models in 1D by just solving the corresponding Schrodinger equation. This is a remark in passing. So as I said before, the standard way to evaluate this kind of integral is by doing a loop expansion. So you write that this is an integral d phi. So introduce this fake artificial parameter e to the minus l. So l is a parameter, and we will look for asymptotic expansion, beta epsilon over 2 integral gradient phi square minus 2 lambda integral cosine beta e phi. And of course, this is to be taken at l equals 1 at the end. But before taking l equals 1, you do an asymptotic expansion for l going to infinity. So you get a free energy, as we saw, that you can expand in powers of 1 over l. And this free energy, this is e to the minus beta l times some f0 plus a correction term, which goes like 1 over l f1 plus et cetera. And the systematic loop expansion will give you all the various terms of this expansion in powers of 1 over l. So the way to do, as we saw, is first you look for the saddle point. So if I call this object s of phi, it's a certain function of phi, the saddle point is defined by delta s by delta phi 0 of r equals 0. And so here the functional derivative of this with respect to phi will give you, so the functional derivative of this term is minus beta, the factor 2 disappears. So minus beta epsilon Laplacian phi 0. So it's plus 2 lambda beta e sine beta e phi 0 equals 0. And as you remember, the real electrostatic potential is psi 0 i phi 0. So in terms of psi 0, if you replace phi 0 by minus i psi 0, you get the equation, which is that minus epsilon Laplacian psi 0 equals, so Laplacian 1 plus 2 lambda e sine beta e psi 0. Right, this is just a trivial thing. Anyway, it's trivial in the sense that if you're in the bulk, as we are here, the solution, the obvious solution to this equation, you have translational invariance because the system is homogeneous. And so the solution is just psi 0 equals 0, or phi 0 equals 0. So in a salt solution, an infinite salt solution, the electrostatic potential is 0. Everything is 0. And to 0's order, I just replace here phi by phi 0. So to 0's order, I get that z is e to the minus l times minus 2 lambda. So it's plus 2l lambda times the volume. And so this is e to the minus beta x0. So I forgot also to write something, which is that the lambda is the fugacity of the system. And it is determined by the fact that the total number of particles is determined by lambda, so it's the total average number of particles. And it's given by n average is equal to lambda by d lambda of log z. Now, this is the total number of charges. So here, if I have n plus ions, n minus ions, with n plus equals n minus, so this is 2. So it's really 2n of each ion species, because this is really the total number of ions, which you have here, because you have the same lambda for each species of ions. So this is really 2n ions. So here, if you see this equation, and this is e to the minus beta l f0, because the first order term is here, so you see that f0 over v, or minus better log z, is equal to 2 lambda v, which implies that 2n equals lambda d by d lambda log z equals 2 lambda v, which means that lambda equals n over v, which is just c0, the concentration of one of each species of ions. So you see that to lowest order, the fugacity is just the concentration of the ions. And that's very general. If you have many species like this, at any to lowest order, lambda i under 0 is just the concentration, the bulk concentration of the ions number i. Sorry? You put l equal to 1 for n. Not really. It's because I am doing the expansion in powers of l. So the first order term is e to the minus beta l f0. And here I have l also, so I don't put l equals 1 yet. I identify order by order. So this was simple enough. And now I go to the Debye-Huckel theory. Or let me not call it Debye-Huckel for the moment. So it's quadratic fluctuations. So quadratic fluctuation is, you write that phi of r is phi zero of r. And phi zero of r, we saw that phi zero is zero, so I will not write it. Plus 1 over square root of l psi of r. And psi is called the fluctuating field. So this is zero. The saddle point value of phi of psi is zero. And therefore, we can write the correction term. So in the exponent, since phi zero is zero, then the partition function reduces too. So there was this term when I put phi zero equals zero. So it's integral dix. So z is the, yes? Because phi zero is the solution of this equation, and we solve it as phi zero equals zero. So I have my saddle point, and then I look at fluctuations around the saddle point. So I write that, I make just, this is a change of variable, right? Phi is the maximally, the field which has the maximal weight in your exponent. And then you look at small fluctuations around that field, and you expand in powers of these fluctuations. And the fact that it's small fluctuations is reflected by the fact that you put a factor one over square root of l. Yes? Yes? Psi zero is i phi zero. This is because we saw that the electrostatic potential is related to the integration field by, in all this formalism, the real electrostatic potential, which I write psi, is expectation value of i phi. No? Because you see, if you multiply both sides by i, I multiply this equation by i. So i phi zero is psi zero, and i sine sine beta e phi is sinh beta e phi. That's because it's one over two i e to the i beta e phi zero minus pi. So you multiply by i, this disappears, and this is psi zero minus psi zero. So z, so the first, there is, so z is integral d psi of e to the minus l beta epsilon over two integral gradient psi square minus two lambda integral cosine beta e psi, so psi over l over square root of l, because phi I write as psi over square root of l, and then you expand. So I'm looking at the correction, the quadratic correction, that's the term of order one over l. So this is already of order one over l, so if I want this term to order one over l, I expand to second order, and the cosine to order one over l here will be one minus beta e square psi square over two l. So the one with the e to the minus two lambda will give me the term e to the two lambda l v. It's the contribution of the one. If I put one here, when you integrate the one, it gives the volume, so the volume times lambda l. So this is the zero order contribution that we got before, and then, so I put approximately, because I stopped to second order, then I have integral d psi of e to the minus. So I have a factor of l here which cancels this one, and this one also, because I have, by definition, I have expanded to order one over l. So my quadratic form here, which goes like psi square, has a factor one over l in front. So the result is e to the minus beta, so beta epsilon over two integral gradient psi square minus two lambda, right? I have minus minus minus, so three times minus minus psi, minus two lambda beta e square over two integral psi square. And this I rewrite as integral d psi e to the minus beta epsilon over two integral dr of gradient psi square plus kappa d square psi square, where kappa d is the Dubai length, I mean the Dubai parameter, which you read of here. So kappa d square is equal to two lambda beta e square. No, sorry, I made a mistake. It's beta square, beta square. But since I factorize beta epsilon, so it's two lambda beta e square over l. So, yes? Sorry? Just this. Yes, I mentioned coming back configuration corresponds to the minimum. But we always look at the extreme amount and then we proceed. Yes, but you see the fact that it's a maximum or a minimum, you will see by looking at the fluctuation because the fluctuation is like the second derivative of the function. So if the quadratic form, which is here, is definitely positive, it means that you are at a minimum. And of course, when you look at it like this, when you go to Fourier transform, you see that this is k square plus kappa square. So this is what I will do now. And this tells you that, indeed, the solution psi zero equals zero was a minimum. Okay? Okay, so now we have to calculate this, right? We have this interesting property that the partition function z is e to the lambda lv times integral d psi e to the minus beta epsilon over 2 integral d3r. And I write it a slightly different way, psi of r minus laplacian plus kappa d square psi of r. Why do I write it like this? Because the only difference with respect to here is that the gradient psi you integrate once by part and this integral gradient psi square is equal to minus integral psi laplacian psi. Okay? Now, if you remember what I showed you about Gaussian integrals before, so Gaussian integrals, you will have that product over i d xi e to the minus one half xi aij xj. So I did it with a plus bi xi summation over repeated indices. This is a constant, 2pi to the n over 2 over square root of determinant of a times e to the one half bi aij minus one bj. So in the continuum, so we have something which is very similar to that, but in the continuum, right? It's like a quadratic form in the psi variables. So if you discretize this, it's really a form like this. So the result is generalized at the following way, that integral d xi of r e to the minus one half xi of r of r. So it's an integral d r d r prime xi of r a of r r prime minus or common, it doesn't matter, xi of r prime. And this is the only thing that I would use. So this b is zero. So this is proportional up to an infinite factor which is here to e to the minus one half log determinant of a. We saw this already last time where determinant of a is the so-called Fredon determinant of the operator a. So the, right? And this, so if you remember that for any operator determinant of a is product over alpha of lambda alpha, where lambda alpha are the eigenvalues of the operator a, this is e to the sum over alpha of log lambda alpha. So of course this assumes that a is definite positive, which means that all the eigenvalues, in order for this integral to make sense, all the eigenvalues of a should be positive, strictly positive. It can be zero, but then you have to be careful. But here we'll see that there is no problem. So, and this is okay. It's e to the sum over alpha log lambda alpha, which you can write as exponential trace log of the operator a. So this is an identity that one uses all the time in field theory. Determinant of operator a is the exponential of trace log a. Is this familiar to anybody? Yes? Okay. Sorry. Okay, so just to finish this, so you see that this is the operator a, which you can write as a of r minus r prime is, so it's minus Laplacian plus kappa d square delta of r minus r prime, right? You put it in. That's what you have. And therefore the eigenvalues of these operators are obtained by Fourier transforming. Yes? You have a question? No, you're just, okay. Just putting your sleeves up. Okay. That's an excellent question. Thank you. So when you have any operator, which is any operator of r r prime, if it's a function of r minus r prime, its eigenvalues are just the Fourier transforms of a of r minus r prime. So the eigenvalues of the operator a, and that's obvious. I mean, you just write if you have integral d. The eigenvalues are by definition a of r minus r prime. Let's say x of r prime equals, how can I call it, x of r. You can check easily that if you take x of r to be plane waves, so it means x of r equals e to the ikr. You can see that indeed you have integral, so let's call it xk. Then you have e equals a tilde of k. Just replace and it's an exercise. It takes two seconds. So what I want to say is that the eigenvalues of the operator a are just a twinkle, a tilde of k, where k is the Fourier transform component of this. And this is just k squared plus k d squared. So the result to which I wanted to get is that z is essentially e to the lambda lv e to the minus one half. So this is a result which is quite important. So it's e to the minus one half sum over all the eigenvalues k of log of k squared plus kappa squared. And when you take the continuous limit of these, this is just e to the lambda lv minus v over 2 integral d3k over 2 pi cubed log k squared plus kappa squared. It's a bit fast, but whenever you have an integral, something like this, this functional integral here is just e to the minus one half trace log. The operator which is here minus Laplacian plus kappa d squared. And the trace log of this kappa d squared is just v times the integral d3k over 2 pi cubed of the log of the eigenvalue which is k squared plus kappa d squared. OK. It's maybe not completely trivial, but OK. Just so why did I want to show you this? Because of course the k range from minus infinity to plus infinity. And the fact that k ranges from minus infinity to plus infinity makes this integral divergent. This is the first example of divergences that you have at one loop. So these quadratic corrections are called one loop corrections. So the one loop correction is divergent if you take k. So you have this integral. You have to calculate something like integral d3k over 2 pi cubed log k squared plus kappa d squared. And this, if you go to spherical coordinates, you will find integral from zero to infinity dk k squared over 2 pi cubed log k squared plus kappa d squared. So of course this integral is widely divergent. So the way out of this is by introducing what's called a short cutoff or UV cutoff, ultraviolet cutoff, which is lambda. So what I do is I put lambda here. And lambda is finite. Lambda is a momentum. So it's 2 pi over a distance a. And a is the, in fact, it's the smallest distance at which you can imagine that particle can come close together. It's a short distance beyond which you don't know what happens, but essentially the ions cannot come closer than a. So when you have a field theory which is renormalizable, what you can show is that eventually the limit a going to zero exists, but here it's not the case. Here the theory depends on a. So you have to choose a. And everything will depend on one parameter, which is this cutoff a. So then you can do the calculations. And I will not do it because everything takes longer than what I think. I had planned to do it. So I will stop here. I can give you the result, actually. Yes. So the result is that beta f over v. So to order zero it's minus 2c. That's order zero. And the next order correction is plus 1 over l. Lambda cube over 12 pi square times minus kappa d over lambda square times 1 minus kappa d over lambda times minus 1 lambda over kappa d plus log of 1 plus kappa d over lambda square. OK. So this is the final expression. So this is the zero order correction. And this is the 1 over l correction to the free energy, which means that this is the one loop for the quadratic calculations, corrections to the free energy, as a function of lambda. So you don't have to write it just to show you. It's a bit lengthy calculation, but it's possible to do it. OK. So I'll stop here for the course. I don't want to go beyond because otherwise we will not have time to come back to all this. So is there any question on this or on there? Yes. There is still the l in the extension. Which l? Yes, but now, so once you have done this, there is a correction of order 1 over l square, of course. So now you put l equals 1. So if you put l equals 1, this is an approximation to the free energy of the system to quadratic order. Does this free energy count? It's the fluctuations around the mean field. So this is essentially the screening, the effect. So it's the fluctuations in the screening, the fluctuations of densities, of concentrations of everything in the system. What can I tell you more? It captures the local inhomogeneities of the electrostatic potential of the concentrations of everything.