 Thank you very much, Eva. It's very nice to be here. You have heard here the series of brilliant lectures. And what I will be talking about today is very, very basic, very elementary, just to set up the scene, set up the language, which is used in non-mechanics. And non-mechanical systems have several faces to them. Much of the interest in non- and micro-mechanics comes from applications. Everybody has a cell phone in his or her pocket. And there is a lot of micro-mechanical systems. Somebody told me about 20 micro-mechanical systems in each cell phone. Non-mechanics probably will allow us to, and is already actually, allowing us making measurements of mass with very high sensitivity force, with extremely high sensitivity, magnetic moments, sub-1 spin resolution. So it's very good for different applications. There are two basic physics aspects of non-mechanics. One is the dynamics itself. Non-mechanical systems are physical systems with complicated properties, interesting and rich. And also, these are unique systems that allow us to ask questions about the behavior of physical systems away from thermal equilibrium. There is much interest in this area, partly coming from biology, and partly primarily coming from this. This is my interest in the first place. Coming from physics, what happens if the system is driven away from equilibrium? There is no thermodynamics. There are no, there is no Boltzmann distribution. What can we say about the dynamics of a system? What can we say in particular about fluctuations in systems away from equilibrium in classical systems, in quantum systems? And to answer this question properly, we need to have systems which are well characterized so that we know what is being measured and what is being calculated and can compare this in a meaningful way. So my plan for the lectures, and the lecture notes are actually on the internet, and I'm not sure I will be following them literally. But my plan today is to set the scene. And in some sense, what I will be talking about has been referred to implicitly or explicitly. I just want to show you how a calculation is done when you are talking about nonlinear dynamics in particular. What is the language for describing nonlinear dynamical systems, nonlinear vibrational systems? What the rotation wave approximation is about? What is the difference between quantum and classical description? And then tomorrow and the day after tomorrow, we will be talking about floquedynamics and non-mechanical systems and other mesoscopic vibrational systems, that is, systems which are sufficiently small so that quantum and classical fluctuations are significant, and at the same time sufficiently large so that you can measure a single system. You don't have to do an ensemble measurement. So we are here in the room. We have pressure. Pressure is an ensemble characteristic of the air in this room. With a non-mechanical system, you can measure a single system, which is at the same time sufficiently big that you can make a single particle measurement. So besides non-mechanical systems, there are modes in the cavities that we have heard about today and, I think, yesterday. There are Josephson junctions, and these are probably the three major types of mesoscopic vibrational systems that are being explored today. Much of what I will be talking about refers to all of them. All of them allow us to peek at what is happening at the microscopic level with systems away from thermal equilibrium. So the first thing about non-equilibrium system is how to drive the system away from thermal equilibrium. And the easiest way to drive the system is to apply force. So we have an oscillator, which is described by this simple dynamical equation. Let's start with the linear oscillator. The mass is set equal to 1, and will always be 1 in what I will be talking about. This is the friction coefficient, and this friction force is linear friction force. That is, the friction force is proportional to the velocity. Omega naught is the frequency, and it is equal to the force. So the answer to this equation is given in terms of the susceptibility. So that is, I solve to find the susceptibility. I say, OK, let's remove the real part here, and I will then restore it. And then I know that the solution of this equation will be q stationary solution. q is equal to chi of omega f e to the minus i omega t. It's a linear equation, and so this is a linear solution. Periodic vibrations add the frequency of the drive. And chi of omega, just plug it in, is what? Is minus omega squared plus omega naught squared minus 2i gamma omega to the power minus 1. I guess it is correct. And if I, instead, put in here a real part, f e to the minus i omega f t, then my vibration become a real part chi of omega f f e to the minus i omega f t. And I know that these are vibrations at the frequency of the drive omega f, so I can write them as a cosine omega f t plus some phase. They're shifted in phase with respect to the driving force. And the amplitude is just the absolute value of this susceptibility. It's f times absolute value chi of omega f. And so this is f divided by omega f squared minus omega naught squared, squared plus 4 gamma squared omega f squared. Probably it is correct. Please interrupt me if I make mistakes. And if I'm going too fast or if I'm going too slow, probably I will not go too slow. And I apologize. Again, this will be very, very introductory, very, very simple. It's to the 1 half. Let me do it like that. Wait a second. Let me do it like that. I want to have epsilon squared. I want to have f squared. Thank you, Elsa. So this is what I want to have. And I will reproduce this expression here as f squared over omega f squared minus omega naught squared. Let me simplify it a little bit before I do it. Because what I will be talking about for much of this class today and tomorrow is the situation where the driving is almost resonant. So resonant driving omega f minus omega naught is much less than omega naught. That is the frequency of the drive is close to the eigen frequency of the oscillator. And I will use delta omega as omega f minus omega naught. And then I can rewrite this equation in the following form that a squared is equal to f squared. So look what I do. I write this as omega f minus omega naught squared times omega f plus omega naught squared. Now, this difference is small. I keep it. This is almost twice omega f squared. And therefore, I will write it. I have then this 4 omega f squared here and 4 omega f squared here. So I will write it as f squared over 4 omega f squared. And here I have what is left, delta omega squared plus gamma squared. Or I will make that simpler. So for now, I will write it as omega f minus omega naught squared plus gamma squared. Now, this is almost Lorentz, and sometimes you heard today that this was called Lorentz and complex Lorentz. If you want to find the absorption coefficient, the absorption coefficient is the power absorbed by the field qf cosine omega ft. And averaging means that I average the work of the force. This is the work done by the force. I average it over the period. And this is proportional to f squared omega f imaginary part of chi over omega f. Again, if you use this expression, to find q dot, you have to differentiate this over time. You will have omega f and i. Now, you take the real part. Because it is i, it becomes imaginary part of chi. So absorption coefficient is expressed in terms of the imaginary part of the susceptibility. And this imaginary part of the susceptibility, as you see, this is the formula that is Camillear. That imaginary part of susceptibility is gamma over gamma squared plus omega f minus omega naught squared, when omega f is close to omega naught. And this is this Lorentz and function that you all familiar with. So this is often how you do measurements. You apply a drive to a non-mechanical system, and you measure its response. And you change the frequency of the drive, and you, again, measure the response. And you find that this response is described by this Lorentz and function, which tells you what the frequency of the non-mechanical system is, and what is its decay rate. And this is at the very naive level, where we don't talk about the coherence and other effects, which can, of course, incorporate. So this was a linear system. What will happen if the system is nonlinear? Again, this is something we can play with. I will not use this. And the best thing about nonlinear systems is that you don't have a luxury of using complex variables. You better use real forces and be careful with what you are doing. You can use complex language, of course, but be careful. So what is the simplest kind of nonlinearity that you can encounter? So you have a nonlinear system, and you bend it. For example, this is a nanotube. You bend it. And a linear assumption is that the restoring force is proportional to the amplitude of your vibration. This is the restoring force from the tension industry. Now, we are here in Trieste. It's on one side of the peninsula. If we go straight west to the opposite side of the peninsula, we will find ourselves in Pisa. And about 350 years ago, Galileo was there, and he was watching the pendulum. And he found that the pendulum performs periodic vibrations. And this is how clocks became practical. This is the period of vibrations of a pendulum that Galileo found. What you were told in your first physics class was that these vibrations are independent of their amplitude, which, of course, is not true. They are dependent on the amplitude. Galileo just was watching chandelier in this cathedral, and the vibration amplitude was very small. Compared to the length of the chandelier. The fact that vibration amplitude depends on the vibration frequency depends on amplitude. It's more or less obvious if you think of a pendulum. And if you make the whole turn up, then the pendulum just freezes over there. So clearly, the vibration period at this point is infinite. So here it is one frequency. Here this frequency is zero. So the frequency clearly depends on the vibration amplitude. And it took about 250 years between Galileo's invention and Jacobi found a way to describe the properties of the vibrations of the pendulum and introduced Jacobi functions. These elliptic integrals and elliptic functions, which are called Jacobi elliptic functions, describe vibrations of the pendulum. So these vibrations of a pendulum are not sinusoidal in time, the independent time in terms of Jacobi elliptic functions. What a simple, very simple model that captures the non-linearity is the model which keeps the first non-trivial, non-linear term in the restoring force. And if I look at the string, I understand that if I move the string straight, if I move it up and down, the string doesn't have any difference if this bending is the same. So the non-linear term in the displacement of the string must be proportional to q cubed, where q is the displacement. Because if it were proportional to q squared, then the restoring force would have the same component for up and down displacement. But it may not have the same component for up and down displacement. It has to change sign, because these displacement are the same. And therefore, this is the first non-linear term in the restoring force for a system with symmetry. In this case, it is inversion symmetry. And so what I do, I add here a non-linear term. Now, if I forget about this part, this model is called Duffing-Duffing Oscillator. This non-linearity is called the Duffing non-linearity. Or sometimes it is called from optics. It is care non-linearity. And this term is used more and more frequently in nanomechanics. Although the term comes from optics. And in optics, if you have a non-linear optical medium, the polarization of the medium is, well, we are used to having kappa e. But then there is another term. And this is called care non-linearity in optics, where e is the electric field and p is the polarization. So as you see, this term has the same structure as this term. And therefore, the Duffing non-linearity is very frequently called care non-linearity. Now, let's talk about this role of the Duffing or care non-linearity. And to begin with, I consider an oscillator without friction force. So just a few oscillators. This is a motion of a particle in a potential well. This is, by the way, just the simplest term that you have if you expand the restoring force on a pendulum. If you look at the pendulum, as I said, the frequency depends on amplitude. The restoring force is not proportional to the amplitude. OK, well, let's keep the lowest of this term in this dependence of the restoring force on the amplitude. This is the lowest of the term. And I can also think about this system as a particle oscillating in a potential well. But this potential is non-parabolic. So the energy of the system is conserved if there is no friction force. And it is 1 half p squared plus 1 half on a non-squared q squared plus 1 quarter gamma q to the force. And the equations of motion are Hamiltonian equation q dot is equal to p. And if I use here, now this e and this e are from different alphabets. This is the electric field. This is energy. So let me remove one of the alphabets to avoid confusion. If I call this Hamiltonian, then the Hamiltonian equation of motion q dot is dh dp. And p dot is minus dh dq. So q dot is p. p dot is dh dq is minus omega naught squared q minus gamma qq. If I plug this into this, if I just differentiate it again, p dot is q double dot. I reproduce this equation. So these are Hamiltonian equations of motion. But I know that this motion of the particle are just vibrations. So I understand that the vibration period depends on amplitude. How to find it? Well, very simple. Vibration period is integral dq over q dot calculated along the orbit. So I start with this point q. I move to this point where I stop. And then I come back. I calculate how much time it has taken me to make this vibration. I find the vibration period. And now I will plug this in as integral dq over square root of. So I want to express q dot. From this equation, I understand that this is q dot squared, right? p is q dot. So I can express this q dot in terms of energy and this part. So let's see if I get it right to e minus omega naught squared q squared minus 1 half gamma q to the force. And I understand that this vibration back and forth, the period is just twice, the vibration one way. So I will put here factor of 2, integral from q minimum to q maximum where q minimum is this position and q maximum is this position. Well, so can I calculate it? Well, Jacobo did it. Can I use? Now, here is my question for you. So suppose, as I said, my gamma is small. My nonlinearity is small. My nonlinearity parameter is gamma. Can I expand this function in gamma? Did anybody try? Just as it stands, the answer is yes, but I have to be smart. But I have to do it carefully. I have to do it carefully because I have to be careful about these limits of integration. So first of all, without this term, this integral, you know this integral is independent of E, because I can write it as a vibration that time of third, or so I think I will use this tree. But when I take gamma q to the force, I have to take into account that if I do the expansion, I will have a singularity here. I have to be more careful. I have to take into account some shift. And yes, the answer is yes. You can do the expansion. You just have to know what you are doing. And if you know what you are doing, then what you will have is 2 pi over omega naught. That part I know, because this is the period of the harmonic oscillator. Then what I will have 1 minus, OK, what can I possibly have here? I have to have a correction which is proportional to gamma. I have to have a correction which will be proportional to E. Then I have to maintain dimension. And to maintain dimension, I think I have to have omega naught cubed. I'm not sure. No, you want omega naught to the force. Of course, sorry. And then there is a coefficient here. And this coefficient is equal to 3 quarter. Now, how do you know the sign? Well, if gamma is positive, then the potential becomes steeper as I squeeze it. So the period should go down. The frequency should go up. And so this is what I have. So this is my expression. And I can also simplify it slightly. And what I want to do is to express this in terms of the amplitude. And first of all, I will write that omega of E is 2 pi over t, which is now a function of E. And this is omega naught. And I have to inverse it. And this is plus 3 water gamma E over omega naught to the force. Now, this is small correction. What I want to do is to express E in terms of the vibration amplitude. So if I have a harmonic oscillator, then my vibrations q are A cosine omega naught t for a harmonic oscillator. Therefore, the energy is 1 half q dot squared plus 1 half omega naught squared q squared. And this is, as you see, 1 half omega naught squared A squared. So I can also rewrite this in terms of the amplitude. 1 plus 3 8 A squared over omega naught squared. So this is the vibration frequency of the Duffin oscillator. This vibration frequency for weak nonlinearity depends on the amplitude, as does the vibration frequency of a pendulum. Again, this is the case of small nonlinearity. Small nonlinearity means not that just gamma is a small parameter. It means that this correction is small compared to 1. That is the energy is small, or the amplitude is sufficient for small. For strong nonlinearity, for large amplitudes, nonlinearity is always strong. OK, so what now happens is the following phenomenon. So I no longer need this, and I no longer need this. I can think of what happens when I have a nonlinear resonator and I drive it by a periodic force. So now I have this equation. Well, I have the vibration frequency as a function of squared amplitude, and I drive here amplitude squared. For sufficiently small amplitude, it's a linear function. It starts at omega naught and goes up. And this is not up to scale. So the nonlinearity is comparatively weak. I'm talking about comparatively small change of frequency. Now I drive my system at some frequency omega f. Intuitively, I can imagine that what can happen is that my system will have small vibration amplitudes like this, of course, vibrations excited by the drive. Now for this drive, for this vibration amplitude, the vibration frequency is largely detuned from the drive frequency. So I'm not having very good resonance. So my amplitude is small. It's a self-consistent regime. But what can also happen is that if for some reason my amplitude is large somewhere here, then the vibration frequency is very close to the drive frequency. So I have a very good resonance. And therefore my amplitude is large. So I can imagine for that for the same driving field, for the same amplitude of the driving force, for the same frequency of the driving force, I can have two self-consistent regimes, one with large amplitude and another with one with small amplitude and the other with a large amplitude. Now, formally, it looks in the following way, which actually somewhat surprisingly is correct. Or maybe not surprising. So what I did, I said that my amplitude depends on frequency. So what I have to do in this expression for the amplitude of force vibrations, which we had before, I will just add here 3 8 gamma A squared over omega 0 squared. So now I have an equation for the squared amplitude. So let me rewrite it because I know that sometimes it is confusing. A squared times omega F minus omega 0 plus minus, it's minus of course, minus C8 gamma A squared over omega 0 squared squared plus gamma squared is equal to F squared over 4 omega F squared. So I just multiplied A squared by this denominator. And voila, I have an answer for the equation for A squared. What is the order of this equation? OK, let's count. 1, who is voting for 1? Let's vote. Nobody. 2, 3. OK, so who votes for 3? OK, well, it's the majority. So we agree on 3 on the fact that this is a cubic equation. A squared, A squared squared becomes A squared squared times A squared. That's A squared to the power C. So a cubic equation, how many real roots can a cubic equation have? Let's again vote, right? Or we should not. So a cubic equation can have three roots or one real root. So now we have a situation where we can have three real roots or one real root. So let's try to draw the dependence of the amplitude of force vibrations now on the amplitude F. So let's see what we have. A squared versus F squared. So let's start with small F squared. What is the dependence of A squared on F squared if F squared is small? If F squared is small, we drive the system weakly. It should be linear, right? Because the amplitude is just a linear oscillator, amplitude is small. The oscillator is linear. So it starts linearly. But then, of course, it does not continue linearly. It goes somehow like that. And then we understand that there is a region where there may be three solutions. So there is a region where there are three solutions. So there is this region and something like that. So what I'm drawing here now, and we will go over this more carefully how it works out, is that in this region, there are three solutions of this cubic equation. And I drew two of them with solid lines and the third with a dashed line. The full curve looks like that. So what happens here? The amplitude becomes so big that this solution, even though you are far from resonance, it doesn't matter. The amplitude is big anyway. So this solution does not exist. We have only one solution of force vibrations. So here I clearly have one solution. Here I also clearly have another solution. Here I have three solutions. And as I explained, I expect self-consistent solutions with small and large amplitude. I may have another one. Now it turns out that this solution is unstable. And we will talk a little bit more about stable and unstable later today. But let me just remind you, again, talking about the pendulum. I was saying that the pendulum can have this position when it is pointing upward. Well, in principle, it can. In practice, it can't because it is unstable. A little deviation from this position draws it down. And so this solution, it exists. It's a real root of this equation. It's just an unstable root. And this is why I draw it with a dashed line. So in this region, I have three solutions. These points play a special role, and we will talk about them. But before we do it, and in line with what I'm going to talk about, about quantum and classical language of the oscillator, let's think for a second about the quantum feature of the unharmonic oscillator that I'm talking about. So again, the Hamiltonian that I had before here was this. Now, what is the Hamiltonian of the quantum oscillator? Well, quantum Hamiltonian is usually written in terms of the raising and lowering operators, ladder operators. And you are familiar with this. So a dagger a, and q is a square root of h over 2 omega naught a plus a dagger. And the p is, respectively, minus i omega naught square root of h over 2 omega naught a minus a dagger. And quantum Hamiltonian is something that you clearly have seen, h bar omega naught a dagger a. I can put here plus 1 half. I don't want to, but I can. This is for a harmonic oscillator, you know. Now, I say that this nonlinearity is weak. I want to find the correction to the spectrum for weak nonlinearity. So what is the leading order correction to the energy for weak nonlinearity? How do I calculate it? It's just the diagonal matrix element of the perturbation. So the eigenstates of this Hamiltonian are called fox states, n, a dagger a, n is n, n. This is you have seen. Do you see what I'm writing here? I don't know. Tell me where I should draw the line. OK, you don't. You don't see it. Chairs can be removed. Microphones can be removed. OK, so what I was writing is that these operators have this property that a dagger a times the eigenstate of the harmonic oscillator is n times this eigenstate. This is an eigenstate of a dagger a. This is the Hamiltonian of the harmonic oscillator. If I, so this is the Hamiltonian of this part, if I want to calculate the correction to the energy from this part, then I say that my E n is n hn to the leading order is h bar omega naught n plus 1 half from this part plus 1 quarter gamma n q to the fourth n. The diagonal matrix element. This is the first order correction to the energy. And take my word for it. How to calculate it? You have to plug in this expression for a plus a dagger, raise it to the fourth power, and just count what you have. So this will be h bar omega naught n plus 1 half plus h bar v n times n plus 1 half. I think, of course, 1 plus 1. Right, right. It's 1 half h bar v, and v is 3 quarter h bar gamma. Probably over omega naught squared, I think, I'm not sure. So this is just a very simple calculation. Plug this in and calculate the diagonal matrix element. So what happens now with these energy levels? What is the quantum analog of the dependence of the vibration frequency on their amplitude or their energy? It means quantum mechanically that the energy levels of the oscillator become non-equidistant. So for a harmonic oscillator, you have energy levels separated by h bar omega naught, h bar omega naught, h bar omega naught. It's just linear in n. When you have n squared, there will be plus v plus. So how to calculate it? If n is 0, and this n is 1. So the difference is n 1, 1 half is just v. Here n is 2, so 2 times 3 is 6 over 2, 3 minus 2. So this is 2v. This is plus 3v, and so on. So the frequencies become different. This is the analog of the dependence of the classical oscillator frequency on energy. For the quantum oscillator, the transition frequency, the spacing between the energy level, depends on the energy on the level number. So this is the quantum picture of a nonlinear oscillator. This is the classical picture of the nonlinear oscillator, and the expected response of the nonlinear oscillator to the external drive. Now, the next question that I want to ask is, well, this was kind of a hand-waving argument, right? I want to do it seriously. And to do it seriously means that I don't want to write an equation like that and claim that this is what happens. I want to start from scratch and describe the dynamics of the oscillator. To do it, and we will do it both quantum mechanically and classically, we make a transformation from the chord. So what we will be doing is the following. We say that our oscillator has high Q factor. That is, the decay rate is much less than omega naught. The detuning delta omega, which is omega f minus omega naught, is also small. And the nonlinearity is small. So what does it mean that the nonlinearity is small if I look on this equation? It means that this term is, in some sense, smaller than this term. That is, gamma Q squared, this is the typical shift of the frequency of the oscillator due to the nonlinearity, is much less than omega naught. What I have in this case is the separation of time scales. I have vibrations of the oscillator, which are excited by the external field with frequency close to the oscillator eigenfrequency. And these vibrations are fast. And I have some slow change of the vibration amplitude and of the phase of the vibrations. This change is controlled by the decay rate, by the frequency detuning, and by the nonlinearity. And therefore, I want to look at these slow variables. They will control my fast vibrations. And incidentally, you see already from here why nonlinearity and when nonlinearity becomes strong when it is weak. So by all measures, this is the condition of weak nonlinearity. The nonlinear shift of the frequency is small compared to the frequency. Almost to the frequency has changed a little bit. But I don't have to compare this with the eigenfrequency. I have to compare this with the frequency detuning and with the decay rate. And if they are all small, then my small nonlinearity leads to strong effects. And one of these strong effects is the bistability of the vibrations that I was talking about before. Incidentally, if this condition is not met, if my nonlinearity is really strong, then the vibrations of the oscillator becomes very complicated. It displays an oscillator, displays what is called dynamical chaos. That is the trajectories of motion diverge. And the oscillator dynamic becomes very complicated. The attractors become chaotic attractors. And it's a very, very complicated dynamics. A lot of interesting things about dynamics far from some of the equilibrium can be learned without driving the oscillator very hard. They can be learned already in this regime. And this is what I'm going to tell you about. So how to describe it? That's, again, what I will be talking about is the language in which my linear vibrations are described. So I have to change from my fast oscillating coordinate and fast oscillating momentum to something that does not oscillate fast. And this something, this term has been used here already several times. It's called quadrature. And what I will do is I will write a canonical transformation. u cosine phi plus p sine phi. And p is minus q cos minus p sine phi plus p cosine phi. So what is written is I have p and q. And as I wrote it, I just rotated q and p to some different frame. So this is angle phi. This is q. And this is p. So this is a canonical transformation. It's just rotation. What I want to do is to make this phase run. This is the change to the rotating frame. So I had small q and p. I switched to large q, capital Q's and p's. And this q's and p's, this frame rotates. Now what I will do is to make life simpler, I would put here c and put here c omega f. Is this transformation still canonical? So what is the property of a canonical transformation? Quantum mechanical. Quantum mechanical is probably easier because classical canonical information actually more fancy than quantum. And you probably haven't seen them much. So what is, if I did not put this coefficient, what is the commutation relation between capital P and capital Q? Forget about this coefficients. What is the commutation relation between p and q? Minus i h bar, right? Now I rotated the system. What is the commutation relation when I rotated the system? It's minus i h bar. The system doesn't care. I look at it at this angle, or I drifted my head. So it will go also minus i h bar. This is the condition that the transformation is canonical. My commutation relations don't change. Now if I put here constant c and c omega f, will this commutation relation change? Of course it will. It's no longer canonical transformation. So what I did, I first was nice. I made a canonical transformation. And then I decided that I don't want to be nice. I changed, rescaled it because it will be more convenient for me because I will then switch the dimensionless variables. And of course, the commutation relation is no longer i h bar. But of course, what has changed is just that there is extra disk efficiency. So my commutation relation is minus i lambda. And this lambda is h bar omega f c squared. I will put this somewhere here. Lambda is h bar omega f c squared. Just this disk efficient. Is it correct? No, it's not correct. To make it correct, I have to put here h bar divided, right? Because I stretched it. So my uncertainty relation has to be squeezed back. So I have to put here divided by h bar omega f c squared. Now it is correct. It's like you measure volume in centimeters. And then you change your unit to meters. The volume meters 100 times bigger than a centimeter. So in meters, it will be 10 to the volume 10 to the 6 smaller. So you divide, not multiply. OK, so this is the canonical transformation that I'm making. And with your permission, I will erase this part. I don't need this. Right, I will need this. Now I want to derive, of course, equations of motion for my oscillator in this coordinate. And this is what we will be doing today. So again, I rewrite the Hamiltonian. We had the Hamiltonian in variable p and q. And I now change variables to capital Q and p. And again, if this were a canonical transformation, I would have equation of motion, just Hamiltonian equation. I would have q dot is dh over dp. And p dot is minus dh over dq. This is not a canonical transformation. So my h will be rescaled. And there will be a coefficient here. We will figure this out. Now in the presence of dissipation, I know that in the presence of dissipation, my this equation had term minus 2 gamma p. Now I will have here some dissipative correction from this term of dissipation. So I know the structure of the equation that I expect to obtain. And now I want to obtain that. So the way to do it is to do it. I will just make a short. It's good to do a long calculation before dinner. Then you are sufficiently angry at the dinner. And you don't eat much. Or vice versa. So I have two equations. I have equation q dot is equal to p. And I have equation p dot is equal to minus 2 gamma p minus omega naught squared q minus gamma q plus f cosine omega fp. And that's it. So let me try to see what follows from this equation. The interesting thing and an observation for a long time of reviewing papers. I cannot tell you how many times I have seen this transformation done wrongly. So be careful. You have two variables q and p, small q and small p. You change to two variables, capital Q and capital P. You have two equations. You have two variables. You change to two variables. They are not independent. So let's do the first equation first. So I have to differentiate this expression to have this expression. So my q dot is equal. If I differentiate this, c q dot cosine phi plus p dot sine phi minus. Now I have to differentiate phi. So minus c omega f, derivative of phi is omega f, q dot sine phi minus p cosine phi. And this is equal to this line. This is how I define this transformation. So what I get instantly is that this expression is actually this expression. Therefore, this transformation means that q dot cosine phi plus p dot sine phi is 0. These variables are related, as q and p were related. And therefore, when people are telling you that you have to have this q is smooth and this p is smooth and these are fast oscillating terms shifted by phase, how can they be equal to 0? We will see this is one of the pitfalls of the rotating wave approximation, because the next thing that we will do is the rotating wave approximation. And we have to do it with our eyes kept open. So let me try to do the rotating wave approximation a long expression here. And you will tell me if you can see it. If you cannot see it, I will suffer. Maybe I will try this line here. So I have to plug to differentiate this and plug it into this equation, which is a long thing to do. But what we will have is I differentiate p dot. I look at the left-hand side. Minus c omega f q dot sine phi minus p dot cosine phi. This was easy. Now I differentiate phi minus c omega f squared q cosine phi plus p sine phi. And this was just this term, right? Well, it's not as bad as it looks, because now this is equal to minus 2 gamma p. I have the expression for p. So it is plus 2 gamma c omega f q sine minus p cosine. At some point, I will stop writing phi, because we know that all the sines and cosines are sine of the consines of phi. Well, I can write it like this. Let's now look at this term. Minus omega naught squared c q cosine phi plus p sine phi. And for now, let's just forget for a second about the nominating. We'll come to the nominating later. Plus f cosine phi plus gamma. Well, I can make life much easier looking at this expression. If I notice that, first of all, I can divide by c omega f. So I divide by c omega f. I divide by c omega f. I have omega f here, c omega f squared. So the first thing that I notice is that this term looks very much like this term. Don't they? What is the difference? Omega f squared and omega naught squared, right? But they are close to each other. So this is what this whole thing is about, that I choose this. I consider the case where omega f is close to omega naught. And then these two big terms almost cancel each other. So if I move this term around here, what I will have, and keep in mind that omega f minus omega naught is built to omega, and omega f squared minus omega naught squared is 2 omega f delta omega. What I have is 2 c omega f delta omega times q cosine phi plus p sine phi. This makes life much easier, because now I will try to divide by c omega f. So what I will have here minus q dot sine phi plus p cosine phi is equal to, again, 2 gamma q sine phi minus p cosine phi plus 2 delta omega q cosine phi plus p sine phi m plus f over c omega f cosine phi. Can you read this line, or it's already too low? Because I will be using it in a minute. God, that's hard. You were concerned that I will not use this whole box. I'm almost there. OK, so what I will do now, I will multiply this equation by sine phi and this equation by cosine and add them together. So this equation plus this equation. The fate of the first term is dramatic, right? It drops out. I have q cosine sine, and I have minus q dot sine cosine. The fate of this term is actually much better, because I have p dot cosine squared phi plus p dot sine squared phi. So this sum becomes p dot. This is a nice left-hand side, right? Now is equal to, let's see what I have here. Let's start with 2 gamma. And then I'm done with this equation, by the way. So I just deal with this equation as it stands. It's 2 gamma q sine phi cosine phi minus p cosine squared phi plus 2 delta omega q cosine squared phi plus p sine phi cosine phi plus f over c omega f cosine squared phi plus 0 linearity. Well, and at this point, at this point, I will do the rotating wave approximation. And here is something that people usually put into their back pocket. So is this term bigger than this term? Let's vote, or you are too tired to raise a hand. So I give you three choices. Bigger, smaller, or the same? So who is for bigger? Who is for smaller? OK, and who is for the same? Well, of course, they are the same. I am clearly in the minority, but of course they are the same. There is no reason to say that this term, unless q is much less than p, but these terms are the same. So what's the difference? The difference is you don't agree? Let's calculate them. Let's take phi equal to pi over 4. What is sine pi over 4? 1 over square root of 2 cosine, right? So they are clearly the same, right? You will not get me on that one. No, they are the same. The difference is that, remember that phi is omega f t. And I want my p to be, I want to find in p a part which is not oscillating fast. And here is what you were saying. This term sine phi cosine phi is 1 half sine 2 phi. It oscillates like crazy. Whereas cosine squared is 1 half plus 1 half cosine 2 phi. This oscillates like crazy, but this does not. Therefore, if I now say I want to keep in this right-hand side those terms which are smooth, which don't oscillate like crazy, then I will drop this term. Then I will drop this term. And what I will have is p dot is equal to this is 1 half minus gamma p plus delta omega q minus f over 2 c omega f plus gamma. Why can I drop these terms? So these terms are not small. But if I imagine that I'm looking at this equation and I see that p dot is changing on timescale 1 over gamma or maybe 1 over delta omega, it's changing slowly. And if I integrate this term over time, the time over which this change, this will change is 1 over gamma 1 over delta omega for a long time. These terms were far from oscillating. If I integrate them, they cancel out. So if I say that my q is constant and my q varies on this time slow time, if I integrate this term, it's the integral of sine 2 omega ft. Integral of sine 2 omega ft is very small. So this is the idea of the rotating wave approximation. You say that you have terms which, in p and q, there are parts which smoothly vary in time. And there are parts which oscillate. The oscillating parts are small. Their derivatives are not small. So imagine that I have p is p0 plus p1 sine 2 omega ft. I say that p1 is much less than p0. But p dot is p0 dot plus 2 omega f p1 cosine omega ft 2 omega ft. This term, even where p1 is small, when I multiply it by large number 2 omega f, this term becomes large. So the idea of the rotating wave approximation is to separate terms which vary smoothly in time from small terms which are nevertheless possible oscillating. If you look at the derivatives, they are the same. If you look at the terms themselves, they are different. So this is what underlies the rotating wave approximation. And this is what we are doing here. And so I want to finish this equation. So what I have is this equation for p dot. Now I could do the same thing for the equation that I had before here. And I had an equation for p dot and q dot. And I could multiply them by sine and cosine instead of cosine and sine. And then the equation that I would have got would be q dot is equal to minus gamma q minus delta omega p and plus gamma, blah, blah, blah. There will be no term proportional to f because I multiply this by sine phi. And sine phi times cosine phi is oscillating fast. So this is what I have. So what happened here compared to what we had before is the following. I look at this expression. And I see that this form, this part, has a Hamiltonian form with this h tilde equal to minus 1 half delta omega p squared plus q squared. Then if I differentiate in q dot over p, I will have minus delta omega p. If I differentiate over in p dot over q, I will have plus delta omega q. So it's a Hamiltonian or harmonic oscillator with unit frequency delta omega in the rotating frame. Now I have also the term from the drive. And the term from the drive is only in p. So this will be plus f over 2c omega f. And I don't know why it is plus, why it is minus. It must be plus. You didn't correct me. I was making mistakes. Q minus 50 seconds. OK, this is without the non-linearity. And now I have to add non-linearity to these equations, right? So how to do it? You remember I had this non-linear restoring force, which was gamma q cubed, which was gamma c cubed, q cubed cosine cubed phi plus 3p squared q sine squared phi cosine phi plus 3p cubed squared sine phi cosine squared phi plus p cubed sine cubed phi. And you remember what I was doing with this equation. I plugged in q and p, and I was multiplying them in turn by sine phi cosine phi and adding and subtracting. And so I have to do the same with this term. So what will happen? So for example, if I multiply this term by cosine phi and average out over the period, I will have cosine to the fourth phi. What is it if I average it over the period? I'm sure you remember the number. It is 3h. With this term, I will have sine squared phi cosine squared phi. And if you also remember, this is 1h. Here, I will have sine phi cosine cubed phi. What does it average to? Zero, right? Sine is odd. Cosine is even. And this will also average down to zero. Therefore, in equation for p dot, I will have in my Hamiltonian plus gamma with a parameter times what will survive 2 mega F. q cubed times 3h times q cubed plus p squared cubed. And I understand that I have to have a Hamiltonian, which tells me that what I will have in my Hamiltonian is, I wrote it in the wrong place, this is what I have here. I have here gamma c squared over omega F q times 3h mm down plus p squared q. And therefore, in my Hamiltonian, I have to have a combination which will be symmetric, which will be q squared plus p squared squared. And how this combination works out, I understand that if I got it in p and I know the structure of the equation like that, this is what I have to have. And therefore, I have the equations for q dot and p dot. I know that this dissipative term looks like minus gamma q and minus gamma p. And I have complete classical description of the system. Thank you.