 OK, good morning, everybody. So let's continue from our discussion yesterday with the last lecture. So today, so we finished starting the discussion about decoherence. So I would like to spend the lecture today also talking about that. So we will go slow. So please ask me if things aren't clear, because that's an important topic, of course, in general. And I will discuss it now for the case of the nanosphere. So recall the scenario that we have is the following. So we have this polarizable object or this massive object that, in our case, it's levitating. And recall here that the quantum system, the degree of freedom that we want to treat quantum mechanically for such an object and, in general, for almost all nanomechanical systems is the center of mass of this object. This is the degree of freedom that we want to treat quantum mechanically, the position. And now, in our scenario, this object is inside a vacuum chamber. And now we want to understand. And then the coherence appears, because this center of mass is the degree of freedom that we treat quantum mechanically, get interacts with other degrees of freedom that are in this piece of reality, let's say. And then, because of this interaction, this degree of freedom gets entangled with other degrees of freedom. And whenever we then don't have access to these additional degrees of freedom, we trace out these degrees of freedom. And then the effective dynamics for the center of mass is then no longer conservative, no longer described by a Hamiltonian, but just opens dynamics, mass per equation, and so on. So of course, the good thing that this object being levitated in high vacuum by definition is already very nicely isolated from all the other degrees of freedom, but still there are some. In particular, what will happen is that if to have this object in vacuum, we need to somehow trap it to suspend it against gravity, a typical way is by shining a laser and using an optical tweezers, as I described in the first lecture. So if we shine a laser, then I'm using photons. These are many degrees of freedom, and some of these photons then interact with the particle in a non-conservative way in the sense that they scatter, and there are photons going away. And these photons that go away provide information about the degree of freedom. I want to treat quantum mechanically, and therefore, this provides decoherence. This would be one source I want to discuss today. This would be light scattering. Or if you want laser light scattering, also related to that, there would be the fact that the walls of your vacuum chamber, they have some temperature. And since they have some temperature, they emit radiation, because they are hot. And these photons that they are emitted from the walls scatter also from the particle and go away. So you could, in principle, also measure these scattered photons emitted by the walls of your vacuum chamber, what we call black body radiation. And they would also provide information about the center of mass. It would be scattering of black body radiation. Scattering of black body radiation. Apart from elastic scattering, it could also happen that these photons are not just elastically scattered, but actually are absorbed inside the particle. Because if the electric constant of the object has an imaginary component, it can also absorb radiation. So some of these photons can be absorbed, could enter inside. And this will also be a source of decoherence, because there will be some recoil and so on. And also, the object has some temperature. The object is levitating. It has some temperature, too. And having some temperature, it means it's emitting radiation. So if we take an infrared picture of all of you, I would see very hot spots, hopefully, for all of you. So that means you are emitting a lot of radiation. So if you would be now in a superposition of being on one corner of the room or on the other one, this radiation would tell me where you are. And therefore, superposition is broken. So that's decoherence due to emission of black-bodied radiation. So therefore, radiation enters into scattering, into absorption, and due to emission. And these are quite unavoidable. So when you think about the Schrodinger-Cat paradox, you should know that it's going to be always very difficult. Because if you take an infrared picture of a dead cat or the cat being alive, it's very, very different. In one case, it's emitting a lot of radiation. The other one doesn't emit. And so even if the box is closed, if this radiation emitted, it heats up the box. And then by just measuring the temperature of the box, I know whether the cat is dead or alive. So it's going to be very difficult to prepare such a superposition. And this we will explain. Then it's also quite unavoidable that we don't have perfect vacuum in this vacuum chamber. It's not absolutely perfect. Therefore, there are some air molecules flying around. And these molecules then from time to time will encounter the particle and scatter away. And they also provide information. Because if I would be able to measure such a gas molecule, I would get some information about where the object is. So this also produces some decoherence. That's scattering of gas molecules. And also another one is that if this object is suspended, even if I suspend it with an extremely clean method that does not create any type of decoherence, the steel, for instance, the whole lab might be shaking a bit. So there are always a bit of vibrations. And if there are a bit of vibrations, the center of mass will also vibrate. And this creates also some noise in the decoherence. This is quite unavoidable. These are the effect of vibrations. Of course, this list could be done, could be very, very long, depending on the experimental situation you have. But I will discuss those ones because they are, I would say, quite unavoidable, even if you do things really, really well. Because of course, there is always temperature around. So this means there is always plug body radiation. There is always a bit of gas. And if I need to use light to trap objects, for sure, there will always be a bit of scattering. I mean, if I have a suspended object, there might be always a bit of vibrations. And I want to quantify that. But of course, this doesn't mean that there are no more. In other experiments, there could be more. For instance, imagine that because I was not careful enough, this particle contains an electrostatic charge inside. Then if the walls are made of metal or so, this charge could move electrons in the surface. And these electrons in the surface would also provide information, would get information about the position of the charge. And therefore, it's great decoherence. That might be many, many things that could happen. But at least I want to discuss those ones. And then the nice thing, if you want, all of these sources would be super well explained according to the laws of quantum mechanics. And this is just a standard quantum mechanics. But the funny thing is that some people have conjecture that on top of all these standard sources of decoherence, there might be additional sources of decoherence, if you want, or there will be mechanisms for which you cannot prepare this system in the quantum regime. Because quantum mechanics breaks down when the mass is very large. And so on. And for instance, we can summarize under the name of these collapse models. The good thing is that these collapse models, the effect these collapse models have to these objects is actually pretty similar to these effects. But just with a different strength. And therefore, if you would like to falsify these models, these predictions, what you should do is an experiment where all these standard sources of decoherence are really minimized, such that the dominating source of decoherence would be this exotic one. And then if you do the experiment and you see quantum effects, you will have falsified these collapse models. It will be always very tough to corroborate or to prove that if you see decoherence, that's because of collapse models. Because there might be many other standard sources that you didn't control. But if you see quantum effects in a regime where all this is minimized and collapse models should be there, and you still see quantum effects, then you for sure falsified them. But this we will be able also to incorporate into the analysis in the same fashion. So if you want to know, so I will make a summary of how to analyze these things, but some details. I mean, these things are known for many decades. But we wrote a paper where we summarized some of these things and with the same notation that we use in the lecture. So if you want to go for further details, you can look into this paper, where this thing is discussing the context of levitating objects. And then there you will find many references to go into these things in more detail, OK? Good. So what I said yesterday is that all this type of decoherence can be very well cast into this type of master equation, where you say that the density matrix describing the state of the degree of freedom we want to treat quantum mechanically. In this case, the center of mass, for instance, along one axis, x, can be expressed in this way. The density matrix in the position basis for the degree of freedom I'm interested in, the position along x, can be described by the Schrodinger equation plus a decoherence term of that form. And I said yesterday that this decoherence rate is actually might depend on the distance of, so it might depend on which of the diagonal terms I'm looking at in the following way. So this guy actually is a function of x minus x prime in the following way. I mean, this is an approximation, which is actually a model that I use or that people use to be able to characterize all these sources of decoherence pretty well. And the good thing is that by definition then we will be able to somehow parametrize all these sources of decoherence just with two parameters. This is a small gamma that is a frequency, and this A that is a landscape. And this is really a very cheap way to model many different things, actually. And of course, this is an approximation, but it's a very good one. And these two parameters are defined here. And then because of the form of this function, there are clearly two regimes. Whenever I'm looking at correlations which are much smaller than this lens scale A, whenever I'm looking at these correlations that are much smaller than 2A, then this function looks like that. So the decoherence rate grows with the correlation distance. And this constant that appears here, we'll redefine it as a new one that is going to be very important. Let's call it capital lambda. So in that regime, decoherence then is described just by a single parameter, this capital lambda, this localization parameter. And if you are in the opposite regime where you look at correlations that have a lens scale much larger than 2A, then the decoherence rate is actually independent of x minus x prime, and it's just given by gamma. So there are these two regimes. We call this one the long wavelength regime, and this one the short wavelength. And I gave you some intuition that this lens scale A has to do with the de Broglie wavelength of the particles that are coming and scattering from the object. And this small gamma has to do with the flux of how many particles are coming per second. So now let's discuss how all these sources of decoherence, how do you model them, or what are the values of these two parameters that you need to characterize these sources of decoherence. And then we will also see, look at some numbers to have a feeling. OK, questions about this? By the way, note that this parameter capital lambda has units of frequency divided meter squared. A is a lens scale, gamma is a frequency. Good. So then let us go to these standard sources of decoherence. So let us start with the light scattering. So imagine that I now shine a laser light to the nanosphere. Then there is some scattering, and this produces decoherence. This actually we have already discussed in the first lectures. And what we saw is actually that you remember. So if you go back to the discussion, you will see that what we obtained is that first of all, the parameter 2A is going to be related to the wavelength of the laser, because that's the wavelength of the photons that are in the laser. And of course, for optical lasers, this is one micrometer. So provided I'm interested, and of course, the correlations of a nanosphere will be typically much more than a micrometer. For instance, if I'm trying to cool the particle to the ground state, if I would be in the ground state, the correlations of the wave function are of the order of the zero point motion. So recall that if you were able to cool the nanosphere into the ground state or an harmonic trap of omega t, the wave function is just a Gaussian wave function with a width given by the length scale h bar divided 2m omega t. And this is a good number to remember. So if you use a trap frequency of 10 to the 6 hertz, so megahertz, and then you assume now the mass is n times an atomic mass unit. OK? And being the number of atoms, then this length scale is always of the order of 100 nanometers divided the square root of n. That's typically the length, the width of a trapped atom in a potential of 10 to the 6 hertz. Now if you trap a much more massive atom that contains actually a mass that is n times atomic mass unique, the wave function in the ground state gets shrinked to a length scale given by that. And then, for instance, since I typically have it in a nanosphere, if you now you take a nanosphere of 100 nanometers, you typically have 10 to the 7, 10 to the 8 atoms. This means I get four orders of magnitude here. So I have 10 to the minus 7 on top divided 10 to the 4. I get 10 to the minus 11, which is a typical number for the zero point motion for a nanosphere, 10 to the minus 11 meters, OK? So smaller than the size of an atom that has an Armstrong. So this means the wave function in the ground state has correlations of the length scale of the zero point motion, which is 10 to the minus 11 meters. And of course, then 10 to the minus 11 meters is much smaller than the wavelength of a laser, which is 10 to the minus 6. So for sure, the light scattering is described by decoherence in that limit, in the long wavelength limit, OK? So since then, x0, so the zero point motion, is much smaller than the wavelength of the laser. For sure, I'm in the long wavelength limit. And then I know that decoherence then scales like that with this factor capital lambda. And this is, so we showed, and actually we showed in the first lecture or in previous lectures, I told you that the light scattering produces a master equation of that form. There was the Hamiltonian part. And then there was this guy here. You remember, if you look back at your notes, you will see that we wrote this master equation for the motion where we had this gamma m, OK? This guy now can be rewritten in this form, OK? Can be rewritten in terms of this localization parameter lambda, OK? Where now, in our case, the lambda for light scattering then is just given, I write it here, is just given scales with the polarizability square, the intensity of the light, and then this pre-factor. There is this pre-factor always polarizability, the intensity of the laser, the wavelength of the laser to a power of 5, and just natural constants, OK? So this you can already put. So then if you would have now, recall that this is the localization parameters, which has units of frequency divided meter squared. So for instance, now if you have a particle that you want to cool to the ground state, the heating rate is going to be this times the zero point motion squared, because that's the correlation that I'm looking at, OK? So if I'm trying to cool down this particle to the ground state, I will have a cooling rate. But the heating rate is going to be this gamma m, which is then the zero point motion squared times this gamma laser, OK? And now you can already put numbers. So you could play with this expression and see for a nanosphere, if I use a laser beam of some power, then you can already put numbers and get the heating rate in terms of hertz. And then you should compare this heating rate in hertz with the cooling rate that can be achieved with high band cooling, OK? You would need for sure the cooling rate to be larger than that, otherwise experiments could not cool these particles to the ground state, OK? Good. The good thing is that now whenever we have scattering of light, basically the localization parameter always has this form, polarizability squared, the intensity of the light, and the length scale of the electromagnetic field, and this constant, OK? So hence, we can already predict what's going to be the decoherence rate due to scattering of not now laser light, but of the radiation emitted by the walls, OK? You will see now how we will make, so of course, this can be derived very rigorously, but I will give you now a hand-waving way to already get the scaling, which is a really nice thing. So by the way, just one comment. So this is always the dominating source of decoherence in these levitating experiments in high vacuum. Whenever they use a laser, you scatter so many photons that this is by far the main source of decoherence, OK? So whenever I have a laser on, that's the main source. But now you could say, OK, imagine I have the laser on, I have the particle trap, and now I use sideband cooling. I cool it to the ground state, and I keep the cooling so that the particle is really in the regime, and then suddenly I switch off the laser. I take out the laser, OK? So now this main source of decoherence is not there anymore, and still the particle is in a quantum regime. And now this thing would evolve, in particular, as the center of mass will just expand coherently. But now, while it is expanding, OK, there is no decoherence due to the laser light, but of course, there are the other sources of decoherence. And these are, of course, then very relevant. So in particular, let us analyze now this decoherence due to the scattering of the photons emitted by the walls of the vacuum chamber. This is then scattering of a blackbody radiation, OK? So the first thing to know is what is the wavelength of these photons, to know whether we are in the long wavelength limit or the short wavelength limit. So therefore, the first thing we need to know is the thermal photons. What is the wavelength they have? Because this wall has some temperature T, imagine the back. This is a room temperature experiment, so this vacuum chamber is at 300 Kelvin. And therefore, this is emitting photons of 300 Kelvin. So then the thermal photons have a wavelength that is given. So I will all do orders of magnitude, so not the precise factors, but if you write the length scale with the energy of these photons, kBT, HRC, and I now put numbers at 300 Kelvin, well, or if you want. This number is always the following. If 1 millimeter divided the temperature in units of Kelvin. So if I would have one Kelvin photon, it will have a length scale of 1 millimeter. It will have a wavelength of 1 millimeter. So if I have 300 Kelvin, I lose two orders of magnitude. So I have 10 to the minus 5, so 10 micrometers. So room temperature photons are as long as 10 micrometers, so even larger than the laser photon. Therefore, for sure, I'm in the long wavelength limit, too. So this means for sure, then the 2A, it's going to be this guy, which is for sure much larger than the zero point motion. Yeah, and also, then recall that this is the zero point motion if the particle is in the trap. If I switch off the trap, the wave packet would start to grow, grow, grow, grow, grow, grow, grow. Provided the size of the wave packet is smaller than 10 micrometers, for sure, I can still describe the coherence due to this long wavelength limit. Meaning that the scattering of a thermal photon does not provide maximal information about the position, does not provide what is called which path information, because it's so broad that I don't know, you know. So imagine, so I have a superposition where you're, so you are in a state where you are on the left or on the right of the room, and then on the thermal photons, imagine has a wavelength much larger than this room. So then even if I scatter from you and I get this photon back, it's going to be difficult for me to know whether it comes from your left or from your right. And that's that situation. So with one photon, I get a bit of information, but not a lot. So that's why I'm still in that limit. Okay, then let's predict what is gonna be the decoherence rate. So we will be in the long wavelength limit, so we need just this parameter lambda to characterize this source of decoherence. And we can predict it as follows. So let's make a back of the envelope estimation. So as I told you before, so that's always this form. So I have the polarizability, the intensity of the radiation and the wavelength. So I can already predict that what I will have for this scattering of blackbody photons is gonna be the polarizability evaluated at the wavelength, at the thermal wavelength. So the refraction index at the thermal wavelength, okay? So times the typical magnitude for the thermal field times omega t divided c. So the thermal, the thermal, sorry, this should be t, five, one, okay. Okay, now let's write this in a more convenient way. So the omega t for a thermal photon, of course, is just kBt divided h bar, you will be of that order. And the typical electric field for a thermal radiation. So you can extract like that. So you write the energy, kBt, and then you know that using an electric field, I can always write an energy scale by putting an epsilon, putting the e squared and I'm multiplying by a volume because you know the volume integral of e squared times epsilon null is an energy. And then the volume I put here is, of course, again, lambda t to a power of three, okay? So you can always do it like that, c omega t to a power of three, a second. So then I'm putting this together and now writing this omega t in terms of temperatures, okay? What I get is the following. I basically will get that also writing then recalling that alpha goes with, recall epsilon null and the volume of the particle and the refraction index of minus one, refraction index plus one. Then this thing already, you see that it scales like the following. So then you had already this amazing scaling with the temperature. So this decoherence rate goes with the radius of the particle to a power of six, which is a reminiscence of the Rayleigh cross section, scattering cross section of a dipole, which always goes like R to the six and then this super high power on the temperature of the wall, okay, then t to the nine, okay? And then this is the refraction index evaluated at the thermal wavelength, okay? And actually this expression up to factors is exactly the one that you get if you do this calculation really carefully, okay? And the right expressions are given in the reference I told you before. But from here it's actually interesting to put now some numbers. So, and this is the localization parameter, okay? So if now you would have your particle trap in the ground state and inside the wall that is emitting photons and scattering, the decoherence rate is gonna be then this lambda black body times the zero point motion square. If instead of being in the ground state, you have a larger wave packet, then here you would change the zero point motion by the lens scale of your wave packet or your super position state. And now you can also again put numbers into that. And I think it's nice to do it. So, let me show these numbers here. Yeah. So if you use as I said before, for instance, in SI units, you put the zero point motion to be of 10 to the minus 12, which is a good number. You use the C 10 to the eight. You use the radius of the sphere to be 100 nanometers. And then you recall that KB is 10 to the minus 23 in SS units and H bar 10 to the minus 34. And then you take N of the order of two, the refraction index at the thermal wavelength, okay? You put these numbers and then you automatically can get that this decoherence rate once the particle is in the ground or is near to the ground state. So this black body zero point motion square is of the order of, let's do it carefully, but you will get this of the order of 10 to the minus 27 times the temperature of the vacuum chamber in kelvin's to a power of nine hertz. So this means that if I have the, if the vacuum chamber is at the, it would be at 1000 kelvin's, I would get the coherence rate of one hertz. So heating rate of one hertz is quite small. This is just to show you that if you want to cool to the ground state, the heating due to the black body radiation from the wall is completely negligible, okay? But of course, note that there is a dependence here with the zero point motion. So if instead of being in the ground state I'm trying to prepare a wave packet of 100 nanometers, for instance, really delocalize the center of mass over 100 nanometers, this I would go instead of 10 to the minus 12, I would go to 10 to the minus seven. So these are five orders of magnitude which being squared are 10 orders of magnitude. Then I would get 10 to the minus 17, okay? And this means that then if I would be a hundred, I would try to do this expansion on a chamber at 1000 kelvin's, then I would get the coherence rate of 10 to the 10 hertz. So be careful. So that's why the scattering of black body radiation is negligible. So the effect is negligible once you want to cool to the ground state, okay? No problem. But if you then switch off the trap and let the wave packet expand, then this actually is the dominating limiting factor because of that, this is scaling with a land scale and this enormous dependence on the temperature, okay? Then you could say, okay, good. Then what I do is I just don't do a room temperature experiment. I just put this vacuum chamber in a deal fridge such that I have this temperature then instead of being one kelvin, this is actually 10 milli kelvin's, okay? It's 10 to the minus two. So I get 10 to the minus 18 here, okay? Then I can prepare really large superposition, okay? That would be a possibility, but this temperature is gonna be very cold. But what happens about the internal temperature of the sphere? You are shining lasers. This particle is absorbing these photons and it's really hot. As I told you, already in experiments, they measure that this particle has a temperature of 1,000 kelvin. And even if it's environment, if the walls are at milli kelvin's, if you are shining a laser, you are absorbing this heat and the cooling process, so the thermalization with the wall, it's not happening because it takes ages and you are just putting heat into the particle. So this will still be at 1,000 kelvin's and therefore will emit radiation. So let's estimate what is the decoherence into that. This would be the third one. Now emission, now let's talk about emission, emission of black-body radiation. In this case, I cannot, so you have to believe me. I will write you what is the expression. Again, the thermal photons you are emitting have a very long wavelength, so still you will have that two way which is the wavelength of the photons. It's still much larger than the relevant length scale. It's still in the problem, so we are in the long wavelength limit. And therefore the decoherence is characterized by a localization parameter which in this form, in this case, for the emission, just believe me, this has this expression, it is scaling. So the main difference is that now this doesn't depend on the real part of the polarizability but on the imaginary part of the polarizability because if I'm emitting photons, it means I'm losing energy from the particle and this also means I can absorb these photons so this can only happen if my electric constant is imaginary. So this depends on the imaginary part of the refraction index at the thermal wavelength and then the scalings that appear, this goes with the volume of the sphere and then it depends on the internal temperature to a power of six. So because you might all remember that a thermal, so an object, we always learn in basic physics that it emits thermal radiation with a power that is proportional to the surface, right? But this is only true if the object is larger than the wavelength of the photons that it emits, okay? This is okay if the object is much larger than the wavelength of the photons which is okay with a large object. As soon as the object is actually smaller than the thermal wavelength, it doesn't emit proportionally to the surface but to the volume, okay? And that's why here it appears the volume, okay? Not the surface, just a detail. And then there is this dependence on the internal temperature, okay? And there are some prefactors here that are now not relevant. But and therefore, since for an optically manipulated sphere, this internal temperature is of the order of maybe even several hundreds of Kelvin, close to 1,000 Kelvin. Even if I would do this experiment in a cryostat, if I use light, this object is that hot and it's emitting photons and then you could also put numbers and you will see that again, for if the particle is in the trap and it's really, you are trying to cool to the ground state, the heating rate which is this got land emission times a zero point motion. Also for 1,000 Kelvin is completely negligible. This heating rate is not important. However, if now I'm trying to expand the wave packet to really large length scales for matter-waving interferometry, then the decoherence rate is actually this times the length scale of the wave packet, sigma, which can be many orders of money to larger than the zero point motion. Then this is the dominating, is a super limiting decoherence factor for matter-waving experiments of hot particles. Okay, and in the reference I gave you, you can see quantitative statements about that, okay? But for instance, yeah. But for instance, I can already tell you that for a room temperature nanoparticle, you could not expand in free space for timescales comparable to few milliseconds. After few milliseconds, you would have destroyed any coherence in the wave packet. Okay, even if you are in high vacuum in a very cold environment, because the object is hot, okay? Because it's, because this source of decoherence is due to the emission of photons. And the electric object can only emit radiation if the electric constant is imaginary. Because of the, if you want, it's because of the fluctuation dissipation theorem. If it emits, it's because its polarizability is fluctuating and this requires an imaginary part of the electric constant. So emission has to do with absorption. And you can only emit if you can absorb. And you can only absorb if the imaginary part is non-zero. So if you want the microscopic picture of why we are all emitting radiation is because we have charges inside any object that are thermally fluctuating. And because of the thermal fluctuations, it emits radiation, okay? And these fluctuations and these thermal fluctuations are then, we describe them in electromagnetism by putting an imaginary part of the electric constant. Good, okay, just a comment. If this is the decoherence rate due to the emission, the decoherence rate due to the absorption, absorption is actually the same. The only thing that changes is the temperature. This would be the internal temperature. If you consider the decoherence due to absorption, the only thing you need to change is the temperature. Now it's not gonna be relevant here, the internal temperature, but the external temperature, okay? External, external, okay? Because now in the emission, you are emitting photons that have a thermal wavelength even by the temperature of the nanosphere. For the absorption decoherence, you are absorbing photons that have a wavelength even by the thermal, by the temperature of the environment, which is different, okay? This could be the cryogenic temperature. That's why if I have this levitating sphere in a cryostat, the scattering of black body photons is negligible because it depends on the photon on the temperature of the environmental photons, which are really, really cold. The absorption is also negligible because it has to do with the external temperature, but the decoherence due to emission is, it depends on the temperature of the body, which is really hot, okay? And then that's the levitating factor. So to then overcome that, then the message is, okay, we need to use cold objects, okay? Because if we use also cold objects, this decoherence rate will be very much suppressed. And this seems difficult to be done with optically manipulated the electric spheres because they will always be hot because they absorb light. So a perfectly transparent object that doesn't exist. Then the strategy that some people suggested is then to say, okay, then let's not manipulate these objects with light, but with static fields, magnetic fields that do not hit the bulk of objects. And in that context, one could use then magnets or superconducting microspheres. Then, so in our group, we work on doing some proposals to do this matter wave experiments, but not with optically manipulated objects, but with magnetically levitated, magnetically manipulated either superconducting spheres or magnets because with the main motivation that I can use an object whose bulk temperature is really small, it actually can be cryogenic too, okay? And then you suppress also these sorts of decoherence with the price of having to manipulate everything with magnetic fields and being in cryogenic environment, okay? And this is actually the main motivation. Good. So we covered already then the decoherence due to the scattering of lasers. So if I use a laser, we discuss the decoherence due to the black body radiation, either the one emitted by the sphere or the one emitted by the walls of the vacuum chamber. The next one is what happens with gas molecules. Questions, please interrupt me, yeah? Do you have any questions or no? Yeah. In general, when you are talking to the question is, those of you are always saying about the localized particles in a way, but are you using this classical and it is a unit of reflection? Yeah. So it seems like it kind of breaks down to any of the units of reflection which is the localized particles. So which units of reflection is the decoherence or the classical? Yeah. Yeah. So here what is important is to think about always what is the degrees of freedom I'm treating quantumly, okay? So here the degree of freedom that I say is the quantum system is the center of mass position. Then the refraction index here appears because the center of mass is associated to an object, okay? And these objects can interact with other degrees of freedom, in particular the degrees of freedom of the electromagnetic field, okay? And the way these degrees of freedom of the electromagnetic field are connected with the center of mass, the degrees of freedom, is by electromagnetic interaction with the body. And the electromagnetic interaction with the body has to do with polarizability and that's where the refraction index enters, okay? I don't know if I answered to your question, but so in this sense, even if the center of mass is the localized, I don't care in every, so if I have a superposition of being left or right, in every branch of the superposition, I have a body with a refraction index. And of course, the refraction index is the same whether the particle is in the left or on the right. Otherwise, this would create decoherence. If you would do an experiment where for instance, imagine I do a matter with interferometry, experiment where I send a particle going through the left of the room or through the right of the room. Imagine that on the left of the room, the temperature is higher, okay? So then when the particle goes to the left, the temperature is higher, so the refraction index, if you want changes, whereas when it's on the right, not. This will provide which path information and would create decoherence, too. So these types of things would be relevant. But provided the refraction index doesn't change in all the branches of the superposition, and this is just a parameter. Any other question? Okay. Then now for the scattering of air molecules or gas molecules depending on the environment, of course. First thing to again always do is, okay, what's the wavelength of these molecules? What is the de Broglie wavelength? And then the de Broglie wavelength of these gas molecules can be estimated, I mean, or it's even by the Planck's constant divided the momentum that these particles have. If these particles are in thermal equilibrium, so have some temperature, I can know what is the thermal momentum they have. So this is, of course, just, let me put it over here. So the two pi's are so I don't care now, okay? This is just orders of magnitude. So then I just have, so this is mass times the energy, so the square root, this gives me a momentum, okay? And this will be the mass of a typical gas molecule. Let's consider, for instance, a typical mass of a molecule in air, okay? And this is the thermal energy. So if I use, for instance, the mass of the typical air molecule to be 30 atomic mass units, which is quite okay. Then this wavelength is of the order of 0.5, okay, 0.15 nanometers divided the temperature in Kelvin. This means that gas molecules have a very short wavelength. Actually, for a room temperature air molecule, this is like 10 to the minus two nanometers, so 10 to the minus 11. So comparable to the zero point motion, okay? So that's why here we have to be careful. So depending in which regime you are, so for instance, the typical scenario is if you are cooling the particle to the ground state, then typically the zero point motion, so the zero point motion will be smaller than the wavelength of the gas. So for describing heating rates when you are trying to cool to the ground state, typically you can use the long wavelength limit, but if you are trying to do matter wave interferometry, then typically the size of the wave packet is actually larger than the wavelength of gas molecules. And this means, and that's why we had the discussion in the very beginning, that in that regime you need to use the short wavelength limit, okay? And this makes sense because if I have a superposition, if I have the localized sphere over a length scale larger than this, which is in principle doable, then a single scattering event will provide maximal information, okay? That's why, because if I have a superposition of having the sphere here or there and with a distance of one micrometer, then as soon as one air molecule touches the sphere, I could measure this air molecule and I would know exactly whether the particle was on the right or on the left, okay? So then the decoherence rate in that case is not, is basically given by the probability to scatter a single particle, okay? So in that limit the rate I will get for the decoherence, which is this gamma, it just has to do with the probability to scatter a single particle. Do you see the difference? With the photons, that was not the case. If I scatter a single photon, I get a bit of information about what the particle is but not complete. So then the decoherence rate has more to do with how many particles, how many photons do I scatter per second, basically? Whereas here it's just, what is the probability to scatter a single particle? Okay, it's slightly different. That's the meaning of this short wavelength limit or the long wavelength limit. Whereas if the particle is still in the ground state, single scattering events do not provide complete information about the position of the particle. So single scattering events cannot resolve the zero point motion. And therefore single scattering events still does not provide maximum decoherence. And that's why here we are still in the long wavelength limit. So let's now get an estimation of the other parameter I need to have to characterize decoherence. Recall that I told you in this model of position localization I need two parameters to characterize the decoherence. One is the A parameter and the other is the gamma. The A is always given by the de Broglie wavelength. So for air molecules I already have the parameter, is this one. So I now just need the gamma. Need to obtain the gamma to be able to characterize decoherence with two gas molecules. And this can be done also, can be done rigorously, but in this lecture I just do it a bit intuitively. So first of all, let me tell you already that this gamma air can be written like that and it's a pre-factor. This is the thermal velocity of the air molecules, which is of the order of... This is actually the rate you get and this depends just on the radius of the sphere. That's the geometrical cross-section, which makes sense. The thermal and then the linear momentum of the air molecules for a given temperature and the pressure, okay? That has to do with the pressure in the vacuum chamber. And this gives you a frequency. That's the rate. And recall that in the short wavelength limit, this means that all of the Avanol terms in the master equation will decay exponentially with this rate. All of them decay exponentially with the rate that is proportional to the pressure and to the size of the particle, okay? So let us try to guess this. So where does this come from? And yeah, and this can be done as follows. Let's try to... So as I told you, this has to do with the probability to suffer a single scattering event. Or if you want this, is the number of scattering events per second. Okay, so let's try to guess what is this number. So imagine that you are now in a region where there is some density of air molecules given by rho, okay? So there is this density, okay? And now I have a particle. And now let's go to the reference frame where the particle is actually, and these air molecules are all moving with B bar, okay? With the mean velocity. But now let's go to the reference frame with a moving reference frame with a sphere moving at velocity B bar, okay? So all the air molecules are on average, are basically almost not moving. So then the number of particles, yeah. So the volume density of air molecules covered by the particle in one second, okay? It's gonna be, even it will give you an estimation of how many particles are you scattering, okay? One second, this moves by a, so the particle is displaced by a distance B bar. And all the particles in this volume are for sure scattering the sphere, a scattering from the sphere. And therefore, basically the estimation of this gamma air, how many particles I'm scattering in one second is just given by the volume occupied by the sphere, okay? So this is the volume that the sphere occupies in one second, okay? Because the sphere has a surface, r squared, and it's moving with a velocity B bar. So in one second, basically I cover all this volume given by s squared B prime. And all the molecules that are inside here, okay? The density of molecules is given by the density, so these are all the particles I will scatter. And then the only thing I need to do once I have that is I use the ideal gas formula. So I know that the pressure I can write in terms of the density of air molecules times KBT. And then I already have that, is I can write like D KBT divided. And then I just write KBT as mass times velocity squared. And I obtain exactly the same scaling that I have here. Okay, this is just to justify that this decoherence rate has to do with how many particles I'm scattering per second, okay? Up to a pre-factor, okay? So again, let's put numbers now into that expression. So in SI units, I'm recalling that one millibar is 10 to the minus two Pascal's, I'm sorry, 10 to the two. Then if I take a sphere of 100 nanometers and I consider the typical mass of the gas particles to be 10 times the atomic mass unit, so 10 to the minus 26 in SI units. And I consider this to be in a room temperature environment. So temperatures of the order of 100 calories. Then I automatically get the decoherence rate for such an object. Or how many particles do we scatter per second is 10 to the 12 times the pressure in millibars. And all of these it hurts. Okay, so this means that if I have a sphere of 100 nanometers in an environment of one millibar, I scatter 10 to the 12 particles, 10 to the 12 air molecules per second. Therefore, to do matter wave interferometry, for sure I need to go to quite high vacuum. For instance, ultra high or very, how it's called, yeah, high vacuum. Ultra high vacuum is what, 10 to the minus 11 millibars. I'm not an experimentalist, but you should tell me. So I guess 10 to the minus 11, 10 to the minus 12 millibars. So this means that even in ultra high vacuum, I will scatter a single air molecule per second. So this means that if I want to do matter wave interferometry, I want to let the wave packet expand. At max, I have one second of time. Because after one second, I will scatter a single air molecule and I will destroy the wave packet. Okay, so, and this is for sure a necessary condition if you want to really prepare large superpositions of massive objects, you really need to be in a very good vacuum. So, as far as I know, if you would take such a vacuum chamber, you would prepare it in vacuum, like maybe 10 to the minus eight, 10 to the minus nine millibars. And now you would put this vacuum chamber inside the deal fridge to 10 millikelvings. There are indirect measurements that tell you that the vacuum there would be really, really large. So approaching 10 to the minus 16, 10 to the minus 17 millibars. Because there were some experiments in the Gabrielle's group in Harvard, where they did, well, where they had some measurements done in a cryostatic high vacuum and the measurements somehow were required. So the results they were getting, where a necessary condition was that the vacuum chamber there was at 10 to the minus 17 millibars. Of course, there are, as far as I know, no ways to measure such a pressure. So you can only measure it indirectly and these measurements show that, but this would be then the ultimate limit. But for sure, that's an important limitation, okay? Recall also that furthermore, this decoherence rate goes with a radius of square. So if instead of 100 nanometers, you want to use a particle of 10 micrometers, then you gain two more orders of magnitude, which are four orders of magnitude, so that's even more demanding. So what I like from these discussions is that sometimes we think about fundamental questions like does quantum mechanics permit me to prepare a, could I prepare a soccer ball into a superposition state of its position being separated by two meters? Does quantum mechanics allow that from the fundamental point of view? That's an interesting question, but from almost the same fundamental point of view, these type of analysis already show that this is impossible according to quantum mechanics, but of course, let's say because in practice, you cannot have perfect vacuum. And even if you have a bit of pressure, this creates a lot of decoherence, okay? Then we don't know, this we should explore, but the problem is that, yeah. So even as far as I know, I think in interstellar regions, the vacuum I think is 10 to the minus 17, 10 to the minus 18. We don't have a, so then we don't have a theory, a theoretical, a well-founded theory that from first principles predicts that, principle according to quantum mechanics, of course, yes, but quantum mechanics doesn't account for gravitational effects, and we don't have a unified theory that accounts for general relativity and quantum mechanics. And of course, if you have a very massive object, this is also distorting space-time, so then we don't know whether we can prepare how to describe superpositions of space-time, and so on, so on. Principle, the theory, we don't know, yeah. But still in practice, it's good to know these numbers because maybe, you know, these also put some almost fundamental limits of the parametric regime that could be explored, even in super-challenging experimental conditions. Good. Just, yeah, one further comment that I want to do with air molecules. Again, recall, this is, this decoherence rate appears here and in a situation of high vacuum, okay? And this should, and then this should be related, but it's different to what we say. What is the Q factor of a levitated sphere due to damping because of gas, okay? It's slightly different, and I want to comment about that because there are always confusions about this. Gamma air should not be confused with gas damping. And the idea is the following. So what we understand from gas damping is if you have a trap nanosphere in the presence of gas, you can describe its motion with a Langevin equation. It's kind of the typical Brownian motion equation. I have a random fluctuating force and this random fluctuating force. If the baromenic thermal equilibrium has to have an associated damping, okay? That is given by eta. And for instance, for a white force noise, like in Brownian motion, typically, this stochastic force has some correlations given by the fluctuation dissipation theorem be related in that way. For white noise, the fluctuations for a force that averages to zero, but has correlations that depend on the temperature and are connected with the damping by the fluctuation dissipation theorem. Good. Then one could show that this viscosity that appears here, that is what we call the damping, can be expressed in this form. I don't show it now because time is running out and I want to explain something else, but this damping rate is related to the gamma air that I derived before, what I call the decoherence rate, but very importantly with this pre-factor. So this is also a decoherence rate. This is a frequency, but it's related to the decoherence rate by this pre-factor. The mass of air of an air molecule divided the mass of the sphere and this is of course much, much smaller than one. So the damping is much smaller than the gamma air. And why I'm saying that? Because typically, when you will see talks by experimental is in levitating spheres, they will talk about the Q factor due to damping of air molecules and typically what they will do is they will show this Q that is the trapped frequency of the nanosphere divided eta, the air damping. And this is a function of pressure and they make these nice plots to see how Q grows by decreasing the pressure. So this is the Qs that can immediately high vacuum get to 10 to the 12, really high Qs for levitating particles. But this doesn't mean one should not make the confusion to say, ah, from this Q, I obtain this rate and this is my decoherence rate. That would be a mistake. The decoherence rate is the gamma and the gamma has this pre-factor, which is much larger. Okay? So this actually is a similar kind of mistake that you should not do also with clamp oscillators. With clamp oscillators, there is what we call the damping rate, gamma bar. Okay? And then there is the decoherence rate. So this is the damping rate. But then you have the decoherence rate. The decoherence rate is or the heating rate is actually n bar times gamma. Okay? It's always a typical confusion. So because sometimes the Q factors, some people might write it as the mechanical frequency divided by gamma. And from here you could derive gamma and say, oh, that's the decoherence rate. No, the decoherence rate has to be multiplied by n bar. And n bar recall that is always of the order of kT divided by h bar omega multiplied by gamma. So actually the Q, so the decoherence rate, which is this, can be written as this is just kBT h bar omega times gamma. And then omega times, so gamma over omega is 1 over Q. So the decoherence rate, therefore, is always kBT. Divided h bar Q. Okay? It is, of course, related, but there is always this suggestion. So that's for clump oscillators. For the nanospheres regarding air molecules, it's not the same, but somehow there is always this confusion too. One thing is the viscosity due to the gas that creates some dumping. And the other one is the decoherence rate that even in the absence of viscosity, there is a still decoherence every two times you can scatter a single air molecule. And this has to do with what I said about position localization decoherence, that this is, if you want in this scenario, where I have a really high vacuum, I have a Brownian particle that still didn't go to equilibrium. So it still is far from getting to the steady state. But before getting the steady state, it doesn't mean you don't scatter air molecules. Of course, you scatter them and they provide which part information. And that is the question about this. Good. Then the last source of decoherence I wanted to discuss is vibrations. So here the idea is very simple. So imagine I have my particle trap. But now the center of the trap is actually vibrating. Or there is some force acting into this particle that is vibrating. It's just shaking the center of the trap. Then what basically happens is that now my Hamiltonian is time dependent because I have the one that I had before plus a fluctuating force. Just a force that is shifting the center of the trap. In particular, if my potential was fluctuating, it means that the center of the potential now is fluctuating with a random variable. This is an stochastic variable. This means that automatically the stochastic force that is acting into the particle is just minus m. Second. Minus, of course, m omega t squared. So basically now there is a fluctuating force acting into the particle that is given by the vibrations. This is a random variable that has units of meters. So it's a length scale. It's the vibrations. And now the force is proportional to the mass, of course, and to the trap frequency because that's depending on the slope of the potential. I'm talking about that because for a vibrating particle, for levitating particles, this is actually quite relevant because the force scales with the mass. So this means that we will see in a second that the fact vibrations have into a levitating particles scales with the mass. So this means that if people are carefully... So if people trapping an atom are a bit careful to have vibrations low, experimentally trapping a nanosphere have to be m times more careful. And since a nanosphere has of the order of several millions, almost billions of atoms, this means the effect of vibration is a million times or a billion times stronger when you trap a nanosphere. So this is actually relevant. And in particular, if you want to trap microspheres in a cryostat, microsphere has 10 to the... 12, 10 to the 13 atoms. And in a cryostat, vibrations are always a bit more challenging, of course, than in room temperature experiments. So this is relevant. And this is relevant now for experimentally trying to levitate superconducting microspheres in cryostats. And let me be a bit more quantitative about that. So what you can see is that quite... quite nicely, the effect of vibrations, again, can be described as decoherence to a wavelength limit. I cannot justify now intuitively why, but just believe me. And the localization parameter that appears here, lambda, is just the following. So it's going to be m to the power of 4. The trap... No, sorry. It's going to be this... The localization parameter is going to be this one. So the mass squared, the trap frequency is h bar. And this function here is the power spectral density of the vibrations. So that's evaluated at the trap frequency. So the power spectral density function of the vibrations at the trap frequency is basically the Fourier transform. Sorry. The time integral. And this has the units of meter squared divided by hertz. The quantity that they tell you about vibrations. So if they say I have good vibration isolation, they always give you what's this number in meters divided square root of hertz. And therefore, if you want to cool the particle to the ground state, the heating rate due to vibrations is going to be this lambda times the 0 point motion squared. So the heating rate, then it's going to be this lambda 0 point motion squared. So then I have omega t to the power of 4 2 times h bar squared. The 0 point motion squared, which is h bar divided 2 times the mass omega to the power of 3 and the power spectral density of the omega t. So therefore I have m omega 3 divided 2 times h bar and this. And that's the units are okay because that's the meter squared divided by frequency. So then I have mass frequency squared length scale squared, which is divided by h bar, which is a frequency. Okay? And this is really the heating rate that you have to fight if you want to cool this object to the ground state. And now you can plug numbers. I don't know if I plug them here, so we can improvise, but yeah. Let's do that. So for instance, so the mass I write like the mass of the particle is going to be how many atoms do I have by the atomic mass unit? Which is 10 to the minus 27. Then I have h bar 10 to the minus 54 and the frequency which is 10 to the 6. So I have 10 to the 18. Let's see if I mess up now. So I will have 10n 10 to the 34 10 to the 18. Yeah. And this is the vibration which I can write as meters divided square root of 1 to a power of 2. And this is the number I would need. And then you can already see that so this I might have messed up. So, okay, let's not analyze numbers because I will mess up now. But just do it carefully at home. And you will see that this number is actually very demanding for the surprisingly demanding for the vibration isolation you require to cool these nanoparticles to the ground state. Okay, because typical numbers for these vibrations as far as I know so for instance in a cryostat it should be fairly easy to get this vibration isolation at megahertz of the order of let's say 10 to the minus 10 meters per square root of hertz. Okay. This means that the center of the trap is vibrating by 0.1 nanometers with an amplitude of 0.1 nanometers and once and so this will be squared this is 10 to the minus 20 and you have to compensate to this big number that appears here. Okay, so this is maybe not sufficient. So and of course the best values would be like LIGO type of values which is 10 to the minus 21 but the range you have. Okay, and then if the mass is very big so if you try to levitate a very big mass you will not be able to compensate so this hitting rate would be actually larger than the cooling rate. So then due to vibrations you would not be able to cool to the ground state. So time is over right Florian? Yeah, so just maybe two minutes to conclude so I was a bit slower than what I expected but I hope was okay. So now what you can see in the reference I gave you is that these famous collapse models can be expressed in the same fashion. So you will get now a localization parameter for a given collapse model such as the continuous-spontaneous localization model that was developed here among other people by Professor Girardi and others there would be a localization parameter for this collapse model what is called the continuous-spontaneous model there would be also a localization parameter for the collapse model conjectured by B.O.C. and 10 rows which has to do with the conflict between quantum mechanics and in general relativity and there is an expression for this and then what is very nice to do is to say if I want to falsify these models I need to make sure that my standard sources of decoherence the ones that I described today provide decoherence which is weaker than this decoherence created by this model and then you can see how challenging it is to falsify these models it is definitely challenging but not impossible so and that's what some of these papers analyze to see that if you have a sufficiently large object of the order of these nanospheres a bit bigger and sufficiently large super positions larger localizations in very nice environments such as high vacuum low temperatures both of the of the environment and of the sphere low vibrations and so it could be possible to be in regimes where this would be by far the dominating source of decoherence and therefore if you make like a double slit experiment and you see interference fringes these collapse models could be falsified if you do the experiment and you don't see fringes okay that might be interesting it could also be that there is some source of decoherence standard one that you didn't control so as I said very difficult to say if you don't see fringes that this is because quantum mechanics breaks down but it's definitely as always in science easier to falsify conjectures and therefore if you see you can falsify them and this is somehow discussed in this reference that I gave you and with many other references that are related to this one and that they are cited so I apologize that I didn't have time to further discuss this but I hope it was okay so thank you very much for your attention