Okay, so let's continue, good afternoon everybody. So we stopped before lunch by just writing this full master equation containing the Hamiltonian in the driven scenario plus the dissipation. Now we move quickly to section three, which is to describe the usual and important regime of linearized optomechanics and then we will discuss the application of cooling. So linearized optomechanics, okay. So technically what we do first is to make a transformation of that dynamical master equation which consists in defining a state row by, so basically change the state row in the following way. We just apply a time dependent unitary u to the state row and this unitary is nothing else but this term, okay, e to the i omega l e to the i omega t. This is kind of the energy, it's like the Hamiltonian, it's the time evolution due to the Hamiltonian of the cavity mode but not at the cavity frequency but at the laser frequency. That's just a transformation we do. We will describe all the physics in this frame and this is in quantum optics, sometimes we call like in the rotating frame, okay, these historical names but for us just means we apply this unitary. And then what you can check is that the time evolution of the rotile there now is given by the following master equation. You could put this unitary there in that master equation and see how you rewrite things and without doing any approximation you arrive to this master equation, a Hamiltonian part with a modified Hamiltonian plus the dissipative terms and importantly one can check the dissipative terms remain the same, okay, same as before so you just replace row by row tilde but nothing else whereas the Hamiltonian you need to replace it and the Hamiltonian replaced is basically given by this form u h u dagger minus this term, okay, this just comes, there is no, this is a consequence of you apply this unitary you apply row dot here, you use the product rule, you have three terms and the one with row dot you substitute the master equation then you rearrange things and you can always write it like that. The dissipative term will be the same, the Hamiltonian is modified and actually how it is modified is as follows. So basically this Hamiltonian is nothing else but can be written now like that, okay. The Hamiltonian can be written like that if you apply this transformation. So the only difference is two things, first the frequency of the cavity mode before it was h omega c now is minus h delta 0 where delta 0 I defined as the detuning and I decided to define it like that omega l minus omega c, this is the so-called detuning, okay and this terms comes from this because if you check this, what this is, this gives you a term of the type omega l a dagger a as you can check immediately from the definition of view and then the other important difference is that now you see the time dependence here that we had in the laser has gone away and the reason is because the unitary is such that now u a, u dagger has, but this is now not important so it just goes away. So basically in this rotating frame now the Hamiltonian becomes time independent and this is always something you always want in quantum optics and in quantum physics in general you want to avoid using time dependent quantum Hamiltonians as soon as you have some time dependence in the Hamiltonian you have to think how to make some transformation so that this time dependence goes away, okay this simplifies everything a lot. So this is achieved by applying this unitary which we call going to the rotating frame and note that the detuning we decided to define it like that you could also define it the other way around and then there will be a plus here so it doesn't matter but we decided it like that and then in quantum mechanics we then use the same as we use the following vocabulary we call, we talk about red detuning or blue detuning and what this means is independent of the how you define the tuning physically red detuning we mean that the laser has a frequency or the driving frequency is lower than the cavity frequency namely it is reddish it is more red than the frequency of the cavity okay and blue detune means the driving frequency is higher than the frequency of the resonator it's bluish okay in our definition red detuning means laser is lower than omega c so it's for in our definition that will mean negative detunings and for blue detuning for us means positive detunings that's something you can always choose okay and of course you could also choose a driving resonance so omega l equal to omega c which means the tuning equal to zero okay but so far no approximation good now and as I said before the dissipative terms are just the same so now what we know is that physically this term you see this term is kind of driving the cavity so it will put photons into the cavity of course these photons decay through the dissipation through the dissipator to the environment so on but if you drive sufficiently enough it might be that the steady state is such that you have a steady state of a steady number of photons in the cavity and since there is also photons in the cavity these photons also displays the mechanical motion to a need another position okay and we would like to somehow account for that and actually a trick that we do sorry before before that before that before sorry before that one thing so from these dynamics from this cemetery and this dissipator one can show what is that the following equations whole for the mean value of a okay which you would calculate by the trace of a times rod odd and substitute rod rod odd by this okay and the questions of motion for these are the following this will be the question of motion for a that's a consequence of this master equation and the question of motion for B is and these are the questions of motion for a dry driven optomechanical system in the rotating frame for the mean values of a now precisely you see that the presence of the driving now this is a constant is a source in the mean value in the in this equation so this will for sure populate a okay good so now what we do the next step is to make a change of variables and the change of variables is the following you call the operator a now you call a a tilde times some complex number times identity just decide to make this change of variables I change operators in this way and the B I also do the same B tilde plus beta times identity and alpha beta are in principle complex numbers and the motivation is that or what we will then use is that as you will see in a second the mean the mean the absolute value of this alpha and beta will be this will be numbers much larger than one okay will be will be in a scenarios where the absolute value of alpha and absolute value of beta will be larger than one and how do we how do we choose these two numbers so what we do now is the following so if you would not block this change of variables here okay you will see what will happen there will be the following though though so you will get these equations now for for more for these new variables and in particular I will take from all the num all the terms I get here I will focus on the terms that only contain alphas and betas without any mean value of a mean value of B tilde okay so I focus on the following so from here I will choose away any of you want to just say that so I will choose alpha and beta sites that they fulfill the following two equations that zero is equal to and the same for so I can phrase it here like I'm saying I change the the the operator I just if right this operator like that I can always do that no not yet not I've not done any approximation yet this is just a change of variables and I can do that will use this change of variables to make an approximation which will be the linearization but at this point is still I've still not done anything any approximation I'm just making a change of variables and I can always do that no not yet not yet not yet now I've defined a change of variables and I choose alpha and beta such that they fulfill these equations because I want that's my decision okay the motivation a bit is that once we plug this change of variables here you know by forcing alpha and beta to fulfill that a lot of terms will be zero that's convenient okay and I but now just stay with me I just choose off and beta such that these equations are fulfilled so then what happens you can already be is if you now try to solve this linear also the system of equations you see you could basically from here isolate isolate Vita from the second equation as a function of alpha squared plug it here and then you have an equation of third order a cubic equation in alpha okay an equation a cubic an equation which is cubic can in principle have of course three solutions okay and and actually this is very interesting in optomechanics the regime in which you have three solutions is a very well known regime of two mechanics which you try to avoid because the system is what is called by a stable so if you drive in some parameter regime such that these equations have three solutions and the system becomes be stable meaning that as you put light into the into the cavity and then there are two stable positions for the mechanical mode okay and we will always try to avoid that regime one can show you could look now at the properties of this equation that the system has a single solution and hence the system is a stable system has one solution if the detuning is larger than minus square root of 3 over 4 times kappa okay just believe me that so if the detuning is larger than minus the square root of 3 over 4 kappa then there is only one solution okay then there is one solution for alpha one solution for beta so we will focus in that regime we will focus always in the system one solution that becomes the system is a stable in that regime if you are not fulfilling that condition then the system becomes be stable and this was predicted a long time ago and observed experiment we are interested in that regime okay note also that in that stable regime an approximated solution for alpha is this one this is an approximated so you can calculate the exact solution but it doesn't have a very nice closed formula but if if g0 is very small alpha is very well approximated by that which basically is a solution by obtained by neglecting this term here and just solving for alpha okay that's a very good approximation to what alpha is which then implies that the mean value of alpha square is approximated to to this and then from this expression you already see we will be interested in driving strength from recall that omega depends on your power in driving strength so strong that this and sufficiently on resonance such that these divided by this is way larger than one and actually in experiments that's huge that's sometimes 10 to the 8 most of the experiment okay so we will be interested in regimes where this is very large okay for instance sometimes in experiments is even 10 to the 8 okay so it's a huge number hence you see that if this is so huge the mean value of 8 a if we substitute our change of variables it will have this contribution alpha square plus the other terms kind of alpha star a plus alpha a dagger so till this sorry plus so and since this is 10 to the 8 these terms here will be at max 10 to the 4 and this is of the order of 1 so this dominates so this means a number of photons in the cavity if this driving is such that this ratio is of the order means there are kind of 10 to the 8 photons into the cavity so it's highly populated with a lot of electromagnetic field intense intensity okay so in the regime this can be neglected so and then once you have alpha then you can also calculate beta and beta then will also be the shift in the equilibrium position in the sense that now the mean value of x basically is beta plus beta star so the mean value of x now is not zero but the oscillator has moved okay so that's the idea so it was you had first your nice cavity which was in the vacuum okay and then you shine light okay and then you shine light and you populate a lot the cavity and the mirror moves okay and that's the setting we are interested in now strong driving that we put a lot of photons here we are in that scenario now okay and how much alpha and beta is it is obtained by analytically solving or numerically solving this equation and just getting the exact expression for that okay and so far no approximation I just say okay that's what alpha beta is now it will come the approximation which is the so famous linearization because now what I can show I what I can do is now I pluck this change of variables here then alpha and beta are decided are defined at such that all these constant terms can cancel but then I will have two type of terms terms that only contain a till this a till this and terms that contains products of a till this times alpha okay then since alpha is so huge all the terms that contain an alpha dominate versus the terms that do not contain an alpha so then I throw away the terms that do not contain the alphas and then this is the so-called linearization approximation okay so idea I write it so if I substitute my change of variables then I would get all these terms no approximations I put all the terms okay I just put the change of variables and of course I use that alpha and beta are defined such that these terms cancel so that while all these terms are not there anymore because that's my definition of the change of variable so I contain all these terms and then for the beta I also have good now here it comes the famous linearization approximation what I do is you see if I compare this term with this term this term or if I compare this has an alpha this one has a beta and this one has nothing okay so this will be super small for alpha and beta large so what I do is I just throw away this term and I make this approximation that's an approximation and the same here I this term I also neglect that term okay and from the structures of the question you see this term I delta 0 I j 0 is of the same form so I will define a new delta which is delta 0 plus the term here and I here have j 0 times alpha and and also here j 0 times okay so what I do is I now rewrite these approximated equations in the following way we do two things we drop till this and we get and two things have happened and this will be the new equations of motion or the new equations and what I've done is I've defined delta without the sup index 0 as delta 0 plus j 0 beta plus beta star basically that's the detuning now that the cavity has once the mechanical oscillator has displaced has been displaced to the new equilibrium position and I have defined j without the sup index 0 very importantly as this as the j 0 times the modulus of alpha and I said before that modulus of alpha can be of 10 to the 4 so in that way I increase the coupling by four orders of magnitude recall that I said in the beginning of these lectures that G0 was very small now with this trick I increase the coupling by many orders of magnitude between the motion and the light and just what I've done a trick is if alpha is not real which is not real in general I take the modulus and the phase and then I can always absorb this phase into and the operators the operator a okay it's just a minor detail I can really find a that absorbs this phase and for all practical purposes assume alpha is real but anyway this is just so that's what I have done so here it comes these are the linearized equations and they are linear because now a only depends on b whereas before a was depending on the product of b times a yeah this is a nonlinear equation this is a linear equation for the mean values and this this is the famous linearization in optomechanics now what I realize is that actually these equations of motion for the mean values I could obtain from the following Hamiltonian so these equations of motion I can derive from the following effective effective master of effective Hamiltonian okay these these equations of motion can be obtained from this Hamiltonian or from if you want from this from this master equation where the dissipator are the same when you linearize the dissipators remain the same okay same as before same as before and the Hamiltonian is this one and please note the differences in this Hamiltonian first there is no delta sub zero it is delta is a new that you name there is no g not but g which is way larger because you have a modulus of alpha there is no this is a linear Hamiltonian now there are only terms containing two creation or an elation operators okay this is now a plus a dagger times b plus b dagger not a dagger a times b plus b dagger and the driving is out okay so then in optomechanics if you search for optomechanics and you look at the question one depending on the paper equation one is this one or the other one and and this this Hamiltonian here is the so-called single photon or single photon optomechanical Hamiltonian including driving this one is the linearized Hamiltonian okay and and note what we have done we have gone to a rotating frame first we have gone to a displaced frame so we change variable sides that we are in this displaced frame alpha beta then we make an approximation which is to remove all the nonlinear terms and we arrived to the Hamiltonian and master equation okay and basically these requires that these absolute value of alpha and absolute value of beta should be larger than one for the linearization to be okay okay so these only valid for very strong driving okay so if the detuning would be very large this driving would be this alpha would decrease and might not be really valid anymore and so on so one should keep this in mind the cool thing is that now this system as it is written like that this system can be solved exactly analytically in quantum optics that's a very easy master equation why for two reasons this recall so this is what is called a quadratic system because now the Hamiltonian is quadratic it only contains creation and products of up to two creation and relation operators and the dissipative terms come contain jam operators contain jam operators are the operators in front of raw in the written here in the front here and these jam operators are linear in a and a dagger when in the dissipator you have linear terms and in the Hamiltonian up to quadratic the whole system is what is called quadratic and can be solved analytically or exactly what do I mean by that the following so now you could do the following so if you define now define two vectors you find the vector B supes as the vector obtained by the mean value of a the mean value of B the mean value of a dagger and the mean value of B dagger I define this vector this could be complex numbers is a vector of dimension four of complex number and also define this other vector V Q which have more terms it has a square B square AB so basically all the possible products of two annihilation operators a dagger B a dagger square B dagger square a dagger B dagger we should not miss anyone a V dagger a dagger a and V dagger B this vector which contains 1 2 3 4 5 6 7 8 9 10 mention 10 okay so these are all basically all the first moments and all the second moments now because the system is quadratic if you now calculate the equations of motion for these mean values you obtain the following you obtain that actually time derivative of these components can be obtained as a matrix MS times this vector BS so basically the first moments are closed so the differential equation for these four moments only depend on their first moments we saw we see actually this first equation is this one you see the mean the time derivative of the mean value of a time derivative of the value of B only depends on the mean value of a and B and B is V daggers and so on but only on the first moments so and the question of motion for the quadratic more so the quadratic terms also is closed and it has always this form okay you'll have a matrix times the same vector of the variables plus another vector with con with terms but the important thing is that the system is closed okay if the system would not be quadratic when you calculate the time derivative of a term of second order you would get terms of three order of order three and then you should take the derivative of the term of order three and you would get terms of order four and so on and the system would never close okay and that's why you cannot solve exactly whereas when the Hamiltonian is quadratic and the dissipator is linear this equation differential equations closed close and if they close you can solve them okay and this is actually super nice and because actually this linearized optomechanics still contains a lot of physics such as cooling and everything okay so for instance now once you have these differential equations I recall you that now the time evolution of the first order moments can be calculated by just doing this I will define and the terms for Bq as a function of time okay where where these are the columns the so this lambda matrices the columns are the eigenvectors are eigenvectors of the matrix MS or the matrix MQ and the D is a diagonal matrix with eigenvalues and here in this notation the BSS is defined as basically the steady state of the quadratic okay this is all definitions just for you to know that the only exercise you need to do is carefully obtain these matrices M and Q and A which you can do if you do this calculation by the master equation a bit slowly on a paper with pencil then you will get these matrices this matrix is four times four this one is ten times ten but then you could put it on Mathematica have a little code where given the physical parameters boom you just solve find eigenvectors eigenvalues of this matrix and construct the solution as a function of time okay and here they are very interesting note that some components of this vector are for instance how many phonons do I have as a function of time how many photons do I have as a function of time and I could plot the dynamics of the system for any value of the parameters dumping everything there okay so I really encourage you to if you are interested to actually do this exercise and to have this code for you you will be able to understand and learn a lot of physics because at the end even though we are talking about optomechanics these are just the dynamics of two coupled harmonic oscillators with dissipation and in many situations in physics you end up having that setting of two harmonic oscillators coupled with dissipation okay so there are already a lot of nice physics that you can do that first of all recall that for the system to be stable so that these solutions do not exploit as a function of time you already see you want that these eigenvalues should have a real part that is a negative if the real part of the eigenvalues are positive once we both as a function of time this thing could explode and it does happen the system can become unstable meaning that the system would just grow and of course physically this means that at some point you will not be linear anymore you will explore non-linearities and so on but the system can become unstable so you have to make sure if you make such a program to check always that the system is stable which basically means that the real part of these eigenvalues are smaller than zero so that's it so this master equation contains all a lot of interesting physics the physics you can analyze actually exactly without further approximation by just doing this exercise and in the last 10 15 minutes I will show you I have a little program where we do that and we can immediately show some examples but before that let me discuss two things so first of all one thing regarding the coherent dynamics in optomechanics there are two regimes either in which you are red the tune meaning you send photons that have energy smaller than the cavity which that means that the tuning so there is the red the tune scenario which means delta is of the order of minus omega m so basically you shine photons that have a frequency smaller than the cavity frequency and the and the difference is comparable to the mechanical frequency which was megahertz when this happens then one obtains the so-called beam splitter type of interaction and in that setting you can do what is called the rotating wave approximation which then allows me to describe the effective dynamics you can see that the dynamics basically are very well described by this Hamiltonian so the same as before by dropping two terms okay you see in the interaction I have four terms but if the detuning is basically very similar to minus omega m these two oscillators have the same frequency and if they have the same frequency you see that here there are terms that would kind of not conserve energy this would be terms that they generate one excitation in the phone on a one in the photo and if these two modes have the same frequency this doesn't conserve energy so to first order you can throw this away okay and then you only keep the terms that conserve energy but these the rotating wave approximation is only valid if the coupling is sufficiently small okay valid if coupling is much smaller than omega m if the coupling starts to be comparable to omega m the rotating wave approximation is not valid okay and what this term the does this beam splitter this is just a is a term that exchange energy between one mode and the other okay and let me show you that let's see if this works so this is a little code that basically does this okay it just solves these equations for the given parameters so what I do now is I put the two so here I put the two this should be the two frequencies okay I put them equal okay so when I call omega omega cavity but would be minus the detuning okay and I put these two equal then I put some couplings J0 and I put a very small dissipation and initially I just put 10 to the 1000 excitations into the mechanical mode and zero excitations of the cavity mode then if I put I calculate the time evolution of the number of excitations in each mode you just see some super nice and stable oscillations and this is calculated using the rotating wave approximation and without the rotating wave approximation then you see I start with 1000 excitations in the mechanical mode and then after a G time I just exchange energy from one to the other okay now for instance you could say that this for a very low dissipation if now you increase a bit kappa okay which is a decay of the photons then you would see that these oscillations are just dumped okay you exchange energy between the two but the photons you know you put the 1000 phonons you put them into photons but the photons can start to decay disappear from the cavity so all these energies being lost through the cavity okay and if you would increase even more okay now you just done very fast so if instead you keep the cavity very clean but you put dissipation on the mechanical motion then what you get is you also the oscillations get dumped but now the energy is no loss because recall that the mechanical the bath of the mechanical bath has some temperature so you can put energy from the bath into the system so that's why it doesn't go to zero it goes to the steady state which I chose a number such that is the same as the initial so if I increase the damping a lot you know just goes to the steady state okay so both the photon and the cavity goes to the same number of excitations as the bath okay and you see in all these cases both the rotating wave approximation and then is valid so if I put again low dissipation in both okay and now what I do is I increase a bit the coupling now the coupling was the frequency was one and the coupling was 0.1 so the coupling was 10% of the frequency if I increase the coupling for instance truth yeah sorry to 30% you see it still looks again but you see the oscillations start to look a bit different they are not perfectly this is this is when the rotating wave approximation starts to fail this is the real stuff this one is not completely correct and if you increase a bit more this is very sensitive actually if you put already 40% you see it's completely different so then the rotating wave approximation starts to fail it's not correct to neglect this seemingly non conserving conserving energy terms because the coupling is so large that even these modes are not well defined without the coupling and actually there is a if you put really close to zero to half the frequency the system gets very very different and actually if you would put five this is a well-known instability then the system isn't stable so actually you start to put a lot of energy to the system again it gets crazy okay so this is just for for the beam splitter type of interactions let's go back to the small couple okay good so that's for when the when you are red the tune where you send photons that have a frequency smaller than the cavity frequency the other regime that I just briefly mentioned is what happens if you are blue the tune you are blue the tuned if you are blue the tune which means Delta is comparable to omega m but positive so you send photons that have an energy higher than the cavity photons then the Hamiltonian in the rotating wave approximation it contains the other two terms okay and this term this type of interaction is called a two-mode squeezing interaction because what it does it the system is unstable you keep putting energy into the system but in a way that you create what is called a two-mode squeeze state between the mechanical mode and the light which basically it generates a highly entangled a highly entangled state between the motion and the mechanics okay so this is also a very interesting regime but now what I want to do for this last I will ask you more 10 extra minutes to discuss a bit more of ground state cooling okay and the sign of G is not important it doesn't change the physics so I kept track to as it is different now but no no no as far as I know because everything will go with G square and the sign here is not so important okay so now the last thing I want to discuss also qualitatively and with the with the simulation is the ground state cooling but basically we have all the ingredients okay so how how the situation is is the following so we have this mechanical mode of frequency omega m the we have this mechanical mode optical which in the rotating frame has a frequency of minus delta but recall that originally has a frequency omega c and now in the driven regime they are linearly coupled by G then omega m is coupled to a bath of temperature T and the cavity mode also to a bath of the same temperature okay and the couplings are the following so the energy and so the mechanical mode loses energy into the bath with a rate gamma times n bar omega m plus one and this and it gains energy with a rate gamma n bar times of omega m and the key is that the optical mode only loses energy through a rate kappa and actually this is in this setting this is actually very interesting no so we have the bath of the two systems is the same it has the same temperature so what and of course cooling means that you are flowing energy from this mechanical mode to here so how come this happens if this temperature is the same okay the reason is that the temperature is the same but the entropy of the mode is very different because we call that because of this you see here there is only one arrow so there is a net flow of energy going here and fundamentally that happens because since the cavity mode has a frequency way larger than the mechanical mode for the same temperature the mean the thermal mean number occupation at the at the cavity mode at the cavity frequencies way lower smaller than than than this one so basically the cavity mode has lower entropy of basically is in the ground state is in the vacuum and these in the rate equations this means you see there is no energy coming up from here so there is if the couplings are correct and the parameters are correct there is a net flow of energy going to the right okay not also that these two modes originally they have very different frequency omega m and omega c omega m is 10 to the 6 cavity frequency could be 10 to the 15 and they have to talk to each other and to oscillators if they have very very different frequency they don't talk to each other but now by the driving we adjust this frequency such that this the tuning now in the rotating frame has a similar frequency of omega m that's why they talk okay that's just from the dynamical equations but what is the mechanism of cooling let's first also I mean this is a bit from the equations but let me also give you a bit of intuition of what is going on okay so the idea is you send photons and these photons for cooling should be read the tune should have a frequency smaller than omega c then by being smaller than omega c they can only enter into the cavity by gaining energies from the phone okay so basically you send photons that have less energy than the cavity photons they only enter inside because they absorb one phone and once they are inside they don't go they don't emit back into the phone on but at some point they decay to the bath then this energy is lost okay and in that way you can create a process where you are extracting phonons that's a kind of a similar picture of this one and in a more diagrammatic scheme so of course one can develop the theory from the master equation do all of these and understand this perfectly all the physics that include in these dynamical equations so you could just explore yourself once you have the program but I just tell you how the things work so if I make this diagram okay so this I define energy levels the first index is how many photons do I have in the cavity zero photons in the cavity one photon in the cavity the second one is how many phonons do I have I have and phonons and phonons and here n plus one phonons zero photons okay so therefore the energy between these two levels is always the same is h bar omega c and the energy between these two levels is h bar omega m and of course omega c is way larger than omega m so that's correct now the idea is that what we we shine a laser that has a frequency omega l which is lower than omega c you see this would be omega c it's smaller than omega c and how much smaller precisely so small that this difference that is missing is omega m the detuning I said we tune it such that is omega m so what happens by shining this this this frequency is the now two things happening this frequency is resonant to this interaction so this you enhance that interaction a lot so this transition amplitude which is the one in which you annihilate one photon by putting one photon okay this was this beam splitter interaction this one is resonant and very strong whereas to go to the other side to basically excite the photon and excite the phonon this is very much of resonant this is very far the tune okay so then what happens is you start to oscillate like that and if the cavity is lossy as soon as you excite the photon the photon decays it goes away okay and then you go to here so you basically do this transition and then you go down and then since the laser is still on you go to this one where now you have one n minus two you excite this one and then you go down again and in this way you just pump you go to the left and at some point you will hit the zero zero which is the ground state and then you cannot go more to the left okay so this is what is called the sideband cooling this is a very this is the sideband cooling and what it requires it requires that the mechanical frequency should be larger than kappa okay because in a sense this makes that these two transitions are very well distinguishable okay because if kappa would be very large this line width would be so large that there would be no difference to these difference in energy levels okay if kappa the line width of this level is very large here and here will look the same so you will not distinguish these two transitions okay so you require this condition omega m larger than kappa for effective cooling and also and also for and this cooling scheme is efficient efficient for a small coupling which means that the coupling should be smaller than kappa okay which means that you don't want that this thing is oscillating a lot you want that as soon as it goes up boom in the case okay and if you fulfill these two conditions then one can show so basically if you fulfill this you can see that the effective cooling rate how much energy you are removing from the oscillator as a function of time this effective cooling rate is basically g square over kappa over 2 this is the rate at which you are flowing energy from here to there yeah okay this is the cooling rate so this is the flow of energy from the mechanical mode here and then to cool you should make sure that this cooling rate is higher than the heating rate that comes from the bath so then what you need is cooling rate cooling rate larger than the heating rate and the heating rate of the mechanical mode is gamma times m bar so this basically the requires that g square over kappa over 2 is larger than gamma over 2 n bar which then this is typically defined as like that as this condition so you need that the what is called the cooperativity g squared divided kappa gamma times 4 is larger than m bar of them of the mechanical bath if this is fulfilled then you will definitely cool okay and I'm just coding results you have to believe me on that again everything is encoded in these equations and you could do some theory to obtain that analytically all these conditions but if you fulfill that then the final occupation number the final temperature the final occupation number for the mechanical mode is basically n bar omega m plus c so divided by c plus this fundamental limit kappa omega m square okay this is in these assumptions then if your cooperativity is large enough such that this ratio is smaller than one and omega m is larger than kappa then this can be smaller than one and then we say that we just by shining a laser we wait and the center of mass of this mechanical mode goes to the quantum ground state these are experiments that have been done this last year's okay I know this is now a lot of conditions but again if you build this little program which is not so hard you can study this by yourself and I show you an example and with that we will conclude the lecture today so this is what I'm doing here I put so let's so these are first of all I hear these two numbers means I tune the detuning to be completely read the tune so equal to minus omega m and then I put a sufficiently small coupling you see G is small compared to so I'm sorry and I'm using a couple of zero point one so you see kappa is smaller than omega m so I'm fulfilling I'm fulfilling this condition I'm also fulfilling this one so then if I plot the number of phonons as a function of time I just get this type this is phonon in a log scale as a function of time then you see I start from I put initially 10,000 phonons okay and then I just wait and it's the decaying exponential is a function of time and the dash line here is you know just one so I really go to the ground state okay so if I would not do that correctly for instance if my kappa if my yeah or if I put too much mechanical damping so for instance mechanical damping is 10 to the minus 6 then for instance I don't cool exactly to to one I cool a bit too too high okay this is the blue line so I don't cool below one so but now with a bit of care you can choose parameters such that you end up really cooling to the ground state okay and basically all these conditions are fulfilled so for instance in that particular case the condition that is not fulfilled is this one this one is not fulfilled hence I'm limited by that if for instance I don't if my kappa is not so good so for instance let's let's reduce again the mechanical damping it is the situation of before that I was ground state cooling but now say my kappa is not so great okay so it's not it's too large compared to the mechanical frequency for instance is 1.1 a bit slow sorry see I'm cooling not so effectively and things like that and as I said before this is something that making this little program for you probably is maximum or full day or less okay just to write the questions carefully check that there are no mistakes obtain the matrix and then make a little program that they are going to license so given the input parameters which again the input parameters are only the two frequencies g kappa gamma and initial conditions and temperature you just get numbers into these matrices you find eigenvectors eigenvalues and then you just solve that and plot this is a function of same that's what I'm doing here okay and then you could really explore all these regimes and understand and check with the papers that there are some theory papers where they show you how to describe all these limits here do you have questions I rushed a bit at the very end but but the main goal of this lecture was that that you really understand how you derive from the very beginning this final simple linear Hamiltonian plus dissipation from which you can basically study a lot all the physics by yourself exactly without further approximations okay


Great. All right. So thank you very much for the opportunity to present this paper and I'm doing this on behalf of myself and Professor Charles Aka whom I co-authored this paper with. And unfortunately I wasn't able to join you physically but I've been following some of the presentations in the different sessions. So the paper I'm presenting on is Domestic Savings in Sub-Saharan Africa, the case of Ghana. And my presentation is going to be following this outline so I'll provide some brief introduction and background to the study. And then I will share what's the main research objectives where the data and then their profile estimations. I'll discuss some of the results that we found and I'll finalize my presentation by providing a summary and conclusions and also sharing some recommendations based on our study. So it's already been established from the previous speakers how important savings is when it comes to capital accumulation and investment and the important role this actually plays in driving economic growth particularly in the long run. So because of the importance of savings, most of the macroeconomic growth models that we know of have continuously encouraged countries particularly for developing countries to be able to increase savings and investment in order to drive their growth and development agenda. Now this has been done several times like you know over the long period most Sub-Saharan African countries have struggled over the period to be able to mobilize resources from domestic resources to be able to finance their growth and development agenda. So if you go back in time using data from the 1970s to the 1990s for instance the data shows that for this particular sub-region there's been consistent declines in private savings over time going from 11% to about 9% in this period. And using more recent data over the two decade period from 2000 to 2020 we still see a similar kind of declines in savings from about 30% to about 20%. So what this means is that in view of the low savings that these countries are recording most of these countries actually... Hello? We can't hear you. Alright, great. I got some feedback so I wanted to be sure that I wasn't speaking to myself. So in view of the low savings domestic resource mobilizations that these countries have faced over time most of these countries have had to rely very heavily on foreign aid and also extend our sources of finance to be able to drive their development agenda. In the case of Ghana there's been a recent call to actually have development that goes beyond aid. So we have a very popular mantra in Ghana that talks about growth beyond aid which has brought this new focus and renewed interest in mobilizing resources on the domestic market to be able to drive the growth and development agenda that we seek to how. Now what does the literature say about the determinants of domestic savings? And this has been very well documented in the literature. So we do have income related variables and the previous speakers have already alluded to some of these important determinants. So we have GDP per capita, we have economic growth rates, we have interest rates which have all been well documented. And we also have macroeconomic variables like fiscal policies, monetary policies and most importantly for developing countries that rely mainly on primary commodities. We have that important macroeconomic volatilities and the external shocks that it comes with. So you have a situation where you have unfavorable terms of trade and that actually works its way into adversely affecting savings in these countries. In more recent times we've had growing literature and in the previous session before the lunchtime there was a whole session on like institutions, political institutions and environments and how this is very important for domestic resource mobilizations. You can think of things like if your financial sector or your legal system is not really strong then people are not able to carry out financial contracts and this creates a lot of lack of confidence in the system. So there's also things relating to democracy, things relating to corruption which all have important implications on domestic savings. So these are all things that have been well documented in the literature. Now although we have learned a lot when it comes to these determinants there's still some gap in the existing literature. And the first one has to do with the fact that most of the existing studies that we have have focused more on a block of countries at a go. So you look at studies that focus on sub-Saharan Africa, you look at studies on domestic resource mobilization that focuses on the Sardic regions. So what this class is that we lose a lot of the heterogeneities or the nuances of individual countries and we are not able to see for a particular country what is actually driving these savings behavior. And aside that many of the studies have also focused on more short-run relationships which also limits our understanding of what works or what the relationships are which are important for long-term policymaking for domestic resource mobilization. So this takes us to the main research objective here which is to be able to determine the long-term drivers of savings in Ghana and also to be able to establish if indeed there are long-term relationships between savings and then the variables that we think are relevant in the Ghanaian context. Now how do we implement this, you know, achieve this research objective? We use data from the 1980s to 2020 and then we use and this particular period was chosen because it has very important like it captures a lot of the major policies and programs that have been implemented over the time. And we believe that focusing on this particular period would give us like a good understanding of what is changing and how what is changing is actually affecting savings in our context. So some of these important policies include the economic recovery program in the 1980s and then the structural adjustment programs. And this was the very turbulent times in the history of our country. And then the 2000s became more of the stabilizing period where we have the growth and poverty reduction strategies one and two. And aside these policies, we also have the monetary, we've had quite a number of monetary and financial reforms because if we look back at the crisis period, a lot of repressive policies were implemented. And that's actually discouraged by national deepening and savings and some of these policies included. Anyway, so I think given the time I'm just going to highlight the main findings from our study. So basically we found that in our case there was no long run relationship between private savings and the variables that were considering. But in the short run we did find the impact or the effect of the significant effects of GDP per capita, which initially, which was negative, but then the square was a positive effect. So our interpretation of that is that at the low levels of income, beyond a certain threshold, we start to see that possibility between income and savings. So that means that at some point people, households or economic agents would have to deal with their basic needs first before if their incomes increase up to a certain point, then they get to increase savings. Then we also saw that there was a significant effect on the lack of money, which is like the proxy for monetary policy, which then also tells us that policy makers would have to be at the back of their minds that whatever policies that they are implementing today has a persistent effect in the future. So you find people's behavior being based on what they know to be the case in the past. So these are the main findings that we had and we thought that the policy implications there would be that we should continue as a country to pursue policies that would increase households income so that once they are able to deal with their basic needs, we actually see that beyond that threshold there will be increases in savings. So I think I would end here so that I can give other people the chance to respond to the discussions. I sincerely apologize that you were not able to see my full presentation. Thank you. Thank you, Monica.


This cow over my shoulder is busy eating its lunch and a little while later it's going to be eating it again That's because it's an animal we call a ruminant Cattle along with deer sheep goats elk and antelope are all categorized as ruminants ruminants digest their food in multiple steps First they plant material and then later regurgitate that material as cudd This cudd is chewed again for better digestion This process is called ruminating Ruminants have four different stomachs and each stomach performs a different purpose in helping to break down and digest food microbes in the first chamber of their stomachs allow the animals to eat cellulose that is found in plant cells and Turn it into nutrients they can use Ruminating also allows an animal to go out in the open eat a large amount of food quickly and Then go off somewhere safe where it can spend time Digesting its food out of danger from predators Interestingly to ruminate can mean to thoughtfully ponder or to meditate on some topic From the University of Wyoming Cooperative Extension Service. I'm Barton Stam Exploring the nature of Wyoming