15. Unraveling Open System Quantum Dynamics
Ketterle, Wolfgang
In this lecture, the professor discussed motivation for single quantum systems, QMCWF, and models for dephasing.
 The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. MIT.edu. I apologize for the delay. But we are ready for a multimedia show with three different applications. One of them are the clickers. I hope you all took a clicker. It's just two questions today, but really good questions. So today is actually a very nice topic. I think almost every lecture will be a nice topic. But today, it's especially nice. It's called Unraveling Open Quentum System. That may not tell you so much. Well, unraveling means we want to sort of untangle it. We want to understand what is inside. And so we want to develop a better understanding what is inside a density matrix. How can we think about the quantum system in an even more microscopic way than the density matrix tells us? And again, this finds applications in atomic physics because one application of it is spontaneous emission, optical block equations. On the other hand, actually, this class will also teach you something very general about quantum physics. So today's lecture is this part. We'll have three topics. One is I want to provide a motivation for single quantum systems. And you should appreciate this. A single quantum system is not what Schrodinger and Heisenberg told us about. Quantum mechanics was formulated as an ensemble average. But today, we want to talk about is there something in addition about single quantum systems? Part two addresses the derivation of the quantum Monte Carlo wave function method. And this is a formalism which is really simulating, in quite general terms, quantum physics by following, using quantum Monte Carlo simulations, trajectories of individual quantum systems. So here, the individual quantum system, even if you don't have one, becomes a computational tool. And eventually, in part two, I want to give you another example for unraveling open quantum system, which are different models for the phasing. And if you remember that this was on one of your homework assignment, that's correct. I sort of want to here take your homework into the classroom discussion. Well, I think it's nice to start this lecture with a quote by Erwin Schrodinger, where he said, we never experiment with just one electron or atom or molecule. In thought experiments, we sometimes assume that we do. This inevitably entails ridiculous consequences. Well, today, we want to talk about those ridiculous consequences, because many people in our field trap address and observe single electrons and single atoms. Now, this is related to quantum jumps. Quantum jumps has even made it into the popular realm. You can find lots and lots of references on quantum jumping, but I warn you, this is about quantum jumping into different worlds. It's about broadening your horizon. And those authors, those people actually sell you something by claiming that people like Einstein and Stephen Hawking would support the idea that you can jump into different universe, experience another version of yourself, probably a better version of yourself. And this will inspire you to change your life in the world you live. But so this is quantum jumping. I'm not talking about quantum jumping. I will only talk about quantum jumps. So what you see here is really a classic experiment. The one of three experiments which were done almost simultaneously, they've always been racist for the first observation of something. And there was the Hamburg group by Toschek, the Seattle group by Hans Demelt, and the Boulder group by Dave Weinland, who observed the phenomenon of quantum jumps for the first time. Let me just tell you what it is because that's exactly what you want to describe. Here in this situation, you have a system. It's a trapped barium ion with a P and S state. And there is a very fast transition where you can rapidly scatter light and get lots of fluorescence at 493 nanometer. But what happened is every once in a while, the particle has a branching ratio to decay to a D state. And well, D to S, two units of angular momentum, the transition here down to the ground state is very slow. It's slow because it's forbidden. It's even slower because it happens at a longer wavelength of 1,600 nanometer. And this is slow. And it has a lifetime of about 50 seconds. So if you think about it naively, and that's probably the correct approach, you would say the particles in the S state, it's like a light bulb. It scatters many, many photons. But then it branches to the D state. And for 50 seconds, the light bulb is switched off until the particle spontaneously goes down to the S state. And then the light bulb is switched on. And this is exactly what people observed. But I mean, this was 86. That was the time I was a graduate student. And I talked to theorists in the field. And respectable theorists told me that there was a controversy, probably not among experimentalists who use a naive approach, but theorists who believe too much in what they calculate. Because if you solve it quantum mechanically and look what you get, it's a steady state solution, atom in steady state, laser in steady state. And if you calculate the quantum mechanical expectation value, the expectation value is constant as a function of time. So a simple quantum calculation would say the fluorescence is on average, well, just the time average and is constant and is time independent. So there were some serious doubts whether you would observe this or something which looks more like the ensemble average. Well, all the theories soon realized you cannot calculate the steady state average. You have to calculate the correlation function in steady state. And the steady state correlation function has structure exactly at 50 seconds. So in a steady state system, if you're interested in fluctuations and temporal behavior, you can't just calculate quantities. You have to calculate correlation functions of quantities. So anyway, but then, of course, if you now average over many, many realizations, you find exactly an exponential decay with 50 seconds. And this is sort of what you would get from an ensemble average. So this is what quantum jumps are. This is what you should always imagine when we talk about quantum jumps in the rest of this lecture. But I also want to show you a more recent example from search-harroshescope in Paris where they observed quantum jumps recording a single photon in a cavity. So what happens is there's a single photon in this cavity and they can read out whether there is a photon or not by sending root back atoms through. And the single photon is never absorbed, but it causes a phase shift, an AC stark shift, on the root back atom. And then they can read out with the Ramsey interferometer, the AC stark shift. So every atom which passes through the cavity will tell you whether there is a photon or not. And this is shown here. Blue means there is no photon. Red means there is a photon. So here you repeatedly figure out, again, there is no photon, and suddenly there is a quantum jump when the cavity has absorbed a photon or when a thermal photon was released. You may say, what happened here? Well, no measurement is perfect. There is some noise. But by seeing that these are outliers, you clearly see that here is a photon in the cavity and here where the red read out of the atom overwhelms the blue read out. Here you have no photon. And then as time goes by, you can observe the death and birth of microcavity photons. So these are now quantum jumps, not in atomic population, but quantum jumps in the photon number in a cavity. Yeah, time is second. So what is shown here is, I think, what is shown here are 300 milliseconds, and this is a 2.5 second element. What is the density of the line? Each line is a Rötberg atom, which has been read out with a Ramsey interferometer. And I think that means there was a Rötberg atom in state E, which means there is a photon and blue means the blue bar means there was a Rötberg atom in state G. These are two highlighting states of the rubidium atom. Sort of each spike here is one read out of the microchannel plate. So these are detector clicks. You can say yes. You can sort of say blue is yes and red is no. And these are the two clicks. But the separation between two clicks is not uniform. No, because if you have an atomic beam, the atoms come not equally spaced out. There's a randomness when the atom comes. I think it's Poissonian statistics in this experiment. It's a real experiment. It's amazing. Other questions? Timor? Just one more question. So for example, the first bit of blue is all measurements for the same state of the cavity. I don't know if blue means photon or no photon, but if it does mean photon, that means it's the same photon interacting with? Yes. OK. Yes, because I mean, this is something people know that the photon number in this cavity only changes over very long time scales. And you sort of see kind of exactly here how you see here two quantum jumps, one from blue to red and one from red to blue in this cavity. Yes? I don't know if this may be just my ass, it's not very good, but in the bottom graph, or on the bottom graph, in the left corner of the looks like there's some overlap between the dark blue and the dark red. You mean here? Yeah, right there. Is it? Sort of there is a photon in that photon. Sort of simultaneously. I think to the best of my knowledge, I'm really looking at, I've heard wonderful talks about it, you can go back and read the paper, but this probably means n, the atom number in the cavity, and the algorithm they have used to analyze the data says that here the photon number was 0, jump to 1, and jump back again. So you see here, for instance, there is some overlap, but you would still say, since either there is a photon or not a photon, as long as you have a higher number of red clicks and blue clicks, you would say there is a photon. And the little overlap, there is maybe a little bit of time where you don't know whether there is a photon or not. But this is real, because every measurement takes a certain amount of time, since you have a few random clicks. If there are a few random clicks, you do not know yet if the system has jumped, you have to wait a little bit. But this is really also showing how our knowledge about the quantum system comes with time. And therefore, the wave function of the statistical operator changes with time, because wave functions, statistical operator are simply a description of our knowledge about the quantum system. You could have a situation, for instance, you could prepare the cavity that it has an equal probability for 0 and 1, and then what you would read out may just be completely random. So that may be possible, but in that experiment, I think the nature here is that you have most of the time 0 photons because the cavity is cool to very low temperature, and only occasionally does the thermal distribution, the Bose-Einstein distribution of photons in the cavity have a 1. And what we observe are the thermal fluctuations between mainly 0 and occasionally a 1. If you would make the cavity a little bit hotter, it would have maybe half of the time, and 1 and half of the time is 0. But that wouldn't mean that you have completely randomness in your results, because there is a thermal relaxation time over which a photon is created and the photon goes back. The photon would not jump back and forth from the cavity to the walls of the cavity in microseconds. It would probably have an exchange time, and you could sort of measure the thermal correlation time of photons in a cavity. It's a really good experiment. OK, so in that sense, I think I showed you that what Schrödinger said is ridiculous, is real, is the subject of current research, and since for quantum information processing, quantum computers, we need single quantum system is at the heart of quantum information processing. So therefore, we want to now describe single quantum systems. But when I do that, it will become clear at the end of this unit, but I want to give you the warning right away. The way how I describe now a single quantum system may not be unique. There will be, there is actually an infinite number of ways, and I've already mentioned that to you, that there is an infinite number of ways to unravel a density matrix. I mentioned that you can write every density matrix as a probability of pure states, weighted probability of being in pure states, but that always different choices. You can get an infinite number of different combinations of pure states and construct your density matrix. So this was sort of at the level of the density matrix, but similarly, and this will come out of this discussion today, there is also an infinite number of ways to unravel the dynamics, the dynamics involved in the evolution of an open quantum system. So with that, I want to describe the quantum Monte Carlo wave function method. So in the quantum Monte Carlo wave function method, we perform, we don't need an experiment. We don't need sophisticated experiments to observe single quantum system. It's a theoretical method, but we perform thought or Gedanken experiments. So our Gedanken experiments is that we perform measurements. So we assume we have, when we want to describe an ensemble of atoms, we assume we have a single atom. We assume it emits photon, and then we assume that the photon is detected. So the moment the photon is detected, we know, for instance, that the atom is in the ground state. And so our quantum Monte Carlo simulation follows now the trajectory of the single quantum system by saying, I toss a coin, there is a certain probability for detecting the photon. If I detect the photon, the system I know is in the ground state. If I don't detect the photon, it's not in the ground state and such. And this probabilistic approach is just simulated in the quantum Monte Carlo sense. So what is the essence of the quantum Monte Carlo method is that we allow a small time step delta t. It's a real time simulation. And there is a small probability that the system has decayed in that time. So we may now toss a coin, use a random function generator, and ask, has spontaneous emission taken place? And then the two possibilities are yes. And we continue whatever the outcome, whatever the quantum state is. Or we say no. And then with the probability p, we say yes with the probability 1 minus p, we say no. And then we continue the simulation accordingly. So this will become much, much clearer when I write down the formalism for you. But there is one important aspect which I really want to discuss with you. And this is the essence of the method. Namely, how does a wave function change by a measurement? So this is the essential, the only conceptual part for the quantum Monte Carlo wave function. Afterwards, I just show you. Once we have cleared up this conceptual part, the rest is just a few equations. And we immediately show that the quantum Monte Carlo wave function method is equivalent to the optical block equation. But let's really feature and take some time for discussing the concept, namely, the change of a wave function by a quantum measurement. And maybe what this carries home to you is that the nature of an open system with a quantum system which interacts with the environment is actually that the environment permanently performs measurements. And you will actually see that the idea of performing measurements formalized in the quantum Monte Carlo sense gives us an equation for the density matrix which is identical to the optical block equation. So in other words, you can say I'm re-deriving for you master equation now, but from a very, very different perspective, from the perspective of an environment which constantly performs measurement. OK, so let's go through that. How, what happens in a measurement? Well, if our initial state is the ground state, starting very in trivial ways, well, there is no measurement because the particle will stay in the ground state forever. OK, now let's go to the next situation that our atom starts out in excited state. Now, in a time step delta t, we have two possibilities. A photon can be observed or no photon is observed. When the photon is observed, we know for sure that the system is now projected into the ground state. If no photon is observed, well, the system is still in the excited state. OK, so we've discussed what happens when the system is initially in the ground state, when the system is initially in the excited state. But now comes the following question. We prepare the system in a superposition of ground and excited state. And we want to discuss what happens when the photon is observed, what happens when no photon is observed. I think if you observe the photon, it's pretty clear. We have to assume after photon emission, the system is in the ground state. But if no photon is observed, I want to give you three choices and you should address them with a clicker. One choice is that the particle stays in the same state. The second choice is that the particle stays in a superposition state, but the probability to be in the excited state decays, also no photon has been emitted or no photon has been observed. And the third possibility is something else. So in other words, assume you are an observer and you know with 100% detection probability. Let's not discuss technical issues. You have completely surrounded your system with perfect detectors. And you know in the time step delta t, in the time t, no photon was observed. And now you should make a prediction what happens next to the quantum system. And you make the prediction by assuming that the system is described by a wave functional density matrix. And I'm asking you which of the three choices correctly describes what the system will do afterwards, what the system will do next. Is the question very clear? Good. OK. All right. Pretty good. A lot of people, when I taught the class before, said it's a. The wave function is not changing because no photon has been emitted. Now, for those people, but there are only a few among those, I would have now asked the next question. If you have a 50-50 superposition of ground and excited state. And let's say the time for spontaneous emission is 10 nanoseconds. But now you wait one second, and after one second, no photon has been observed. What is your prediction? What in what state is the system at this point? Is the system in the ground state? Or is the system still in a superposition of ground and excited state? Well, since we have the clicker, you can tell me. Is it still a superposition state, or is the system in the ground state? OK. So now you have to help me out, because the people who said B, if they would go back to here, and most people said it is B, if you put in gamma 10 nanoseconds, the inverse of 10 nanoseconds, and the time is one second, the excited state amplitude, the excited state at mixture, is 10 to the minus nothing. So OK, if you want to be a mathematician and say there is epsilon 10 to the minus 10, yes, then you can say there is still a small mixture to the excited state. But by making it so extreme, 10 nanoseconds versus a second, at least in all practical terms, the system is in the ground state. And I think that should teach you what it means if you have a system which is in a superposition of ground and excited state. And after one second, it has not emitted a photon. I mean, then you would say it had its chance. It had plenty of chances to emit, but it decided it doesn't want to emit. And that would mean your 50-50 superposition of ground and excited state means the atom decided that it's in the ground state. You should also realize a 50% superposition of ground and excited state means in 50% of the cases, there will never be a photon emitted. So in other words, what you should realize is if you have a 50-50% superposition of ground and excited state, that the quantum evolution of this system is after a long time, the system is definitely positively, absolutely in the ground state. It's always in the ground state after a long time. In 50% of the cases, with emission of a photon, in 50% of the cases, without emission of a photon. If you have an ensemble of radioactive nuclei, and let's say half of the atoms are in the excited state and can radiate with decay and half are in the ground state, and you send the sample to your friend, and your friend waits many, many, many half times, common sense tells you that all the particles are now in the ground state. Also only half of them have decayed because the other half were in the ground state to begin with. I mean, this is what quantum physics tells us. This is what a 50-50 superposition is. But quantitatively, this is, of course, included in this result, which magically almost all of you got right. Namely, there is a superposition of ground and excited state, but the excited state amplitude decays. And to the limit, in the limit that the time is much longer than the decay time, no matter what the coefficient initially was in front of the excited state, this coefficient has decay to 0. Any question about that? Let me just emphasize that, again, and giving you also an example in the reference, a zero measurement modifies the wave function. You can also put it like this. What the wave function is, is the best knowledge you have about the system. The wave function is a way to predict what happens next. And it's the most precise way to predict it. More accurate predictions than the wave function allows cannot be made because of Heisenberg's uncertainty relation. But what happens is, if you have a 50-50% superposition of ground and excited state, and the system has not emitted, then you would say, initially, my guess was half of the atoms are excited, half of the atoms are in the ground state. But if none of them has decayed for a while, you would say, I revise my estimate now. Now I have to assume that more in the ground state. Because having not decayed means there is a higher probability that the system is in the ground state to begin with. So this is how this way of dealing with the wave function automatically adjusts for the knowledge you have gained about the system. Let me give you another example which carries home the same message. And I take it from a pedagogical paper from Dickey, from American Journal of Physics, 49, 926, 1981. And what he discussed is the following. You have a box, and originally, your atomic wave function is completely localized. You oppose the Einstein condensate if you want in this box. Or let's just assume one atom to be more precise. And it's completely delocalized. But now you focus a laser beam into the system. And the laser beam would ionize your atom, and you could count the ion with 100% probability. And for a short moment, you flash on the laser, and the result is, yes, you count. You get a count. In that moment, you would say, my wave function is more localized. I revise my estimate. Well, maybe I shouldn't say ionized. If it's ionized, you destroy the atom. Let's just say you observe fluorescence, so the atom is still alive. But now you have to say the atom is localized in that region. However, if you don't observe anything, you would say, well, now it's actually more probable that the atom is outside, because if it had been in the laser beam with a certain probability, I would have detected a photon. But the non-detection of a photon means that I revise my estimate, and I say it's much more probable that the atom is outside the laser beam than inside the laser beam. So this here is exactly how a non-observation of anything, a zero measurement, modifies your wave function. And the formalism to incorporate that is what we have discussed before. Yes, Cody? So your arguments here so far work the same way as at the wave function, we have just an entirely classical probability, and we're updating our understanding of probability. Do you have anything about the relative base between these two? I want to do that now. This is sort of just to address the basic concepts about what does a measurement mean, what does a zero measurement mean. But now I want to write down everything for you in amplitudes. We want to do exactly the time evolution of the system. Nancy? Is this like inconsistent with energy conservation, like especially the two before this? Because when we are saying that it is in the atmosphere of the ground energy excited state, then we do not observe the flow of the energy in the system. And now we're just going to do that. In a superposition state of ground energy excited state, we are not starting out in an energy eigenstate. So therefore, what we have is not a sharp value of the energy, but an expectation value. And if you repeat the measurement many, many times, the average energy is conserved. And this is exactly what energy conservation in quantum physics says. But if we do not repeat the measurement, then we should not just apply energy conservation. Well, if your particle, if your system is not in an eigenstate of the Hamiltonian, it doesn't have a sharp energy. And that would mean that when you measure the energy now, you have fluctuations. Sometimes you measure higher. Sometimes you measure lower. It's only when you're in an eigenstate that you measure a sharp value. And therefore, you know from the beginning that you will measure an energy diffusion. And sometimes you measure higher. And sometimes you measure lower than the average value. There's nothing wrong about it, but your question is a very good one. You should. I mean, those examples are really, on the one hand, they are trivial, but on the other hand, it's very profound what they tell us about quantum physics and how to apply conservation laws and such. OK, I think let me now formalize exactly how this is done with all the bells and whistles. We assume we have an initial wave function, which is now an arbitrary superposition state of ground and excited state. And we have the environment, which in this case we assume is the zero photon state. It's a vacuum. And we want to exactly solve the Schrodinger equation for that system. But we restrict ourselves to a very small time step. The small time step is smaller than anything else than the natural decay time. Maybe the inverse Rabi frequency if you drive the system. And we are not doing it here, but I will later show you how you can add a laser beam and drive the system. Or it should also be smaller than the inverse detuning. So what is very important here is that we want to deal only with simple cases, namely one photon has been emitted or no photon. So we want to make sure that there is at most one spontaneous emission event during the time delta t. So what I'm writing down now is the total wave function of the system plus the environment. And now we can evolve it. We can ask what is the wave function time later. This can be exactly done by time dependent perturbation theory. And the result is that the system will be in a super position of ground and excited state. And we want to calculate the coefficients alpha prime and beta prime. And we still have the vacuum state. But then we have the possibility that a photon has been emitted. In that case, we are in the ground state. And the direct product with the environment involves now a photon emitted into a certain direction with wave vector k with a polarization epsilon. And we have coefficients beta k epsilon. And we have to sum over all possibilities for the photon to be emitted. So we call this, so the wave function of the total system has now two parts. It's a wave function which I call psi 0. This is a wave function which involves no photon in the environment. And the wave function psi 1 involves one photon in the environment. So this is the ansatz which seems very natural. But you could immediately prove that these are the only possibilities how the system can evolve. And we can verify this by using time dependent perturbation theory. It's actually almost everybody of you has seen it in either in 8.421 or in a more basic course on quantum physics. It's time dependent perturbation theory for the emission of a photon. But usually when you see those treatments, the reservoir, the environment is not treated as explicitly as we do it here. The theory where this is treated exactly in this way how we need it is the Wegener Weisskopf theory, which is nothing else in a perturbative approach. But it's really a perturbation theory not just for the atomic system, how we sometimes present it in a simplified version. It's perturbation theory for the total wave function of the complete system. Timur? You mentioned the time interval, which was in the T, is much smaller than a lot of things. But here we're ignoring, I take it, the counter-rotating terms, in which case we're still at a large enough time step if that's average now. I simply say this because I expect it to be an E with a photon term from the G-dot-tum, so to speak. Yes, well, the counter-rotating term. The first answer I wanted to give you, know this is exact perturbation theory. But I think I get myself in trouble if I would get myself in trouble if I would allow the time step to be extremely short, because then I'm in shorter than 1 over omega. Because during the time 1 over omega, you can say we have counter-rotating terms, or in other words, a particle in the ground state during a time 1 over omega particle in the ground state can emit a photon and reabsorb it. These were those weird diagrams with virtual states, which we discussed earlier in this course. So I think I want to assume here, which I haven't done, that the time step is larger than omega to the minus 1. And as you know, the counter-rotating term has a detuning delta, which is 2 omega. So therefore, it is excluded. Or in other words, when we talk about photons sent into the environment, we want to talk about real photons and not virtual photons. Yes, good point. So if you do a simple lowest order time dependent perturbation theory in the Wegener-Weisdorff approach, you get an exact result for beta prime. You find that beta prime in this superposition is the original amplitude beta, but it has decayed with e to the minus gamma over 2 dt. And since we are only interested in small time steps, we can do a linear expansion of that. The probability dp that a photon has been emitted is the norm of this wave function psi 1. And this using Wegener-Weisdorff perturbation theory is gamma dt times the amplitude squared that the system was excited to begin with. And because of the conservation of the norm, the wave function, the norm of the wave function psi 0, which is the wave function without a photon being emitted, is 1 minus dp. So what I'm telling you here is that this is an exact result of time dependent perturbation theory. Often when textbooks treat spontaneous emission, they are more interested in getting Fermi's golden rule and this rate, but the same approach, if you just write it down, tells you how the amplitude in the excited state evolves. So therefore, what we learn from perturbation theory that alpha prime and beta prime, the coefficients of the wave function without emission of a photon, this evolution of the wave function occurs with a non-Hermitian Hamiltonian. And this non-Hermitian Hamiltonian is our Hamiltonian for the atomic system. But then we have to account for spontaneous decay, and this is done by this non-Hermitian part. So occasionally, you may have heard that people wave their hands and say, your system is described by a Hamiltonian which has an imaginary part for decay, and this is sort of phenomenological. This is definitely not the case, because even if you have a Hamiltonian like this, a pure state would simply decay and remain a pure state. What we are doing here explicitly is we deal with probability in a correct way. Every time the system can branch out into two different possibilities, two different dimensions of the density matrix, one has probability delta p, one has probability 1 minus delta p, and it is only the wave function associated with probability 1 minus delta p, which evolves with this Hamiltonian. So what I'm telling you is the exact solution for the evolution of the total system in terms of a density matrix for the atomic system. And this is exact. So that means we can now write down the just get some extra space. Oops, how does it do? Sorry, I have to insert a new page. At least. OK. OK, so the procedure how we implement this exact solution of perturbation theory is the following. We have a time step delta t. We compute what is the probability that a photon will be emitted. Then we need a random number generator. So we need a number epsilon, which is a random number chosen in the interval 0 and 1. If this random number turns out to be smaller than delta p, then we continue our time evolution on the computer that psi is now in the ground state. And maybe there's a laser beam which excites it again and such, I will add a few bells and whistles later. Otherwise, psi, our wave function, is now the wave function, which is the time evolution with a non-Hermitian operator of the original wave function psi. And since we have a real wave function now with probability 1, we have to renormalize the wave function by this denominator. Then we execute the next time step. This means go to 2 and do the next time step. And then you have so to speak, if you do it many, many times, you get a time-direct trajectory of so to speak one experiment. And then you start again with your system in a wave function psi, and you accumulate a second experiment. And maybe you do it 10,000 times to get enough statistics. And then you sum up everything you want and calculate all the expectation values you want to know about your quantum system. So the claim is that this method called quantum Monte Carlo wave function method is fully equivalent to the optical Bloch equations. And I want to prove it to you by showing that if I take a density matrix, which is an ensemble average over many realizations of those quantum Monte Carlo trajectories, that then this density matrix follows the differential equation, which is the optical Bloch equation. And the proof is shown here. So what I told you is the density matrix after time delta t has now two matrix elements, one with probability delta p. And this is, photon has been emitted, system is in the ground state, with probability 1 minus delta p. We evolve the quantum state with a non-Hermitian Hamiltonian. And then all what is done in the next few steps, we assume that delta t is small. We do a Taylor expansion of the exponent. We neglect quadratic terms in delta t. And then we come to this line. And if we write this as rho of t plus delta t minus the original rho of t, we find that this follows a differential equation, which is exactly the optical Bloch equations. So therefore, what we implemented as a form of doing many, many quantum measurements in the environment is a procedure which is rigorously the same as the optical Bloch equations, which we derived using a master equation approach. OK. This method is very powerful for the following reasons. If you simulate a density matrix with n internal states, or an external or internal states, I should have said with n quantum states, you need n times n matrix elements which can become quite a memory hog for your computer. The wave function at any given moment has only n components. So therefore, there is a computational advantage in using a stochastic wave function approach over simulation of the density matrix. The second thing is that a lot of people, especially experimentalists, like this approach because it reflects directly what they do in the experiment. And I think it's pretty clear that with this approach, you can deal with a great variety of situations. Let me just mention two obvious extensions. One is polarization. If photon is detected, that means your random number produced a number epsilon, which was smaller than delta p. At that point, you can create a second random number which determines if the photon has been detected with sigma plus, sigma minus, or high polarization. Or you can discuss recoil. If the photon is detected, you throw on another random function generator which determines what the k vector of the photon is. So what direction h bar k the photon has taken. And this determines now what is the recoil that change in momentum of your atom. So you see kind of you can start with an atom at zero momentum. You can say it emits a photon. And then your computer always makes a choice based on the random number. And then you say, OK, my photon has now received a recoil kick at 45 degrees. Now your wave function of the photon is such and such. And if you add a laser to the Hamiltonian, the recoil may have double shifted the resonance, but everything is easily taken into account. So you see that this quantum Monte Carlo wave function method can easily be extended to describe external degrees of freedom, multiple laser fields, and all that. Any questions? OK, let me now just generalize the thought. But I think you know already everything about it. We talked about the master equation. We talked about the most general master equation in the Linn platform, where those operators L are the jump operators. And the prominent example for a jump operator was the sigma minus operator, which takes a particle from the excited state to the ground state. And this is a jump operator for spontaneous emission. But you may have many jump operators, maybe spontaneous emission of different photons with different polarization and such. Or the cavity may lose a photon, and then we have a jump operator for the cavity. We went through that. Well, that means now in an exact way that the non-Hemiltonian is the original Hamiltonian with an imaginary part, which comes because of those jump operators. And the general quantum Monte Carlo wave function procedure is that first you ask, has something happened? Has any jump happened? And this gives you the probability delta p. And if no jump has happened, you just do a time evolution with a non-Hemiltonian. But if a jump has happened, then the jump is now the wave function is projected by the jump operator. If you have a sigma minus operator, it takes a particle to the ground state. If you have several operators, a different jump operator may take your particle to another state. And you have then a branching ratio that you know first the jump has happened, and then the question is, which jump has happened? And you just play the probabilistic game. So you see that the quantum Monte Carlo wave function method can immediately be generalized. So let me come back to the special case of spontaneous emission. In that case, we have only one jump operator, which is sigma minus. And the normalization is the square root of gamma. And that actually means that our quantum trajectory, ground state, excited state, is very simple. If we start with a system 100% in the excited state, it's very trivial what happens. Nothing happens. But if then the jump occurs, the particle is in the ground state. Then nothing happens then until infinite time. You then have to start the next trajectory. You start in the excited state. And this time, by chance of the random number generator, the jump happens later. Another time, the jump happens earlier. And what you then have to do is you have to average over all those realizations what is the probability for the particle to be in the excited state. And this is now your estimator for the excited state diagonal matrix element of the density matrix. And if you've written your code correctly, you will find that you get a wonderful exponential decay exactly as you have expected. Questions? Now, yes. Let me now talk about defacing. What I've shown you so far is how we can unravel the density matrix in many microscopic realizations. And I've given you a specific example. We assume probability for spontaneous emission. And with that, we propagate our wave function. But now I come to sort of the nitty-gritty details or the dirty truth that what we have assumed is by no means unique. And I want to explain it to you first by reminding you of a very, very nice homework assignment you have solved. And this was about we have optical Bloch equations for the density matrix. And the solution of the optical Bloch equation is that we have population damping with a damping time t1. And the off diagonal matrix elements are damped with a time t2. Remember, if you have only spontaneous emission, t2 is 2 times t1. But we can have a lot of other processes which can lead to a much, much faster defacing time than a population, than a time of population changes. And in your homework, you have discussed three possibilities. One is, well, spontaneous emission is energy loss. But you have discussed three different possibilities for t2. One, three different possibilities, how the phasing loss of coherence and phase damping can happen. And in this homework assignment, you assumed either that there is a random phase, that the elastic collision which project onto the excited state. And then you got the most crazy and artificial model that every time, randomly, the phase of the excited state flips from plus to minus. So if I would implement that now with a Quento Monte Carlo method, it would have the following effect. And I hope you enjoy the graphical representation. You can really think about it that this is what happens, that this is what is inside the density matrix described by this process. So let's assume we have a system which would be in a superposition of crowned and excited state. And that would mean that in the lab frame, the dipole moment would just rotate at the resonance frequency. It rotates, and it is a rotating dipole moment which emits coherent light. But now you assume that you have a random phase. So if you add random phases to it, maybe because you have fluctuating magnetic fields, then the line becomes sort of, that doesn't look random, but you know what I mean, it becomes sort of jagged. The phase distribution is still sort of there, but there is a jitter on top of it. And as a result, the light emitted by this dipole is actually broad. Well, in your second model, you assumed that with a certain randomness, with a certain time constant, there's an elastic collision in the excited state which projects the system into the excited state. At this moment, there is no superposition state anymore, and the dipole moment is 0. So that's what you assume. Or when you have random phase flips, you assume that suddenly this kind of deterministic sine function suddenly jumps corresponding to a minus sign in the excited state. And depending where you are in the cycle, it creates jumps at random places. And the question, of course, is if all those three processes, and that's what you showed in your homework, lead to the same density matrix, which one is correct? Which one is really realized in an experiment on a system which is described by this kind of damping or by the optical Bloch equation? So you can say all or none. Well, you can assume what you want. It doesn't make a difference. And the reason is actually profound. The reason is profound in that sense that the way our goal was to have an open quantum system, which is the atomic system interacting with the reservoir. But the way how we phrase the question, we are only interested in what the atomic system does is, we didn't do extra measurements on the environment. The environment was just a dump for photons, a dump for energy, a dump for whatever we assumed in our defacing mechanism. So let me just take the example of putting photons into the environment. And this is described by a certain damping term. But there is an ambiguity. And this is shown here. If we evolve the system, our atomic system, the density matrix evolves. But the environment also evolves. And for instance, here, the unitary time evolution has taken the excitation from the atom. And we have emitted a photon into the environment. And you remember, we got the optical Bloch equation by doing the partial trace here and just focusing on the atomic part. But now wait a moment. If we trace out the environment, we can trace it out in a different basis set. Remember, I showed you the quantum Monte Carlo wave function. We emit a photon. But we can, for instance, detect the photon in our Gedanken experiment. We can detect it with linear polarization or with circular polarization. If you have a situation where you have an s state which decays to a p state, sigma plus takes you to m equals plus 1, sigma minus takes you to m equals minus 1. So therefore, if a photon is emitted in your quantum Monte Carlo procedure, you would say, now the atom is in the plus 1 state or now the atom is in the minus 1 state. But if you detect linearly polarized light by just putting a polarizer here, and this is a unitary transformation in front of your detector, you would no longer project the atomic system on plus 1 and minus 1. You would project it on m equals 0 or whatever is connected to the linear polarization of the measurement. So therefore, if you simply assume that something has been dumped in the environment and can be used for measurement, there are many ways, many unitary transformations, what you can do to the information, to the energy, to the photons which have been dumped into the environment. And each of them will lead to a very different trajectory in your quantum Monte Carlo system. So therefore, if you just dump the photons and not measure them, you have equal rights to assume that your quantum Monte Carlo wave function jumps to states which correspond to a circular basis or jump to states which correspond to linear basis. And there are many, many possibilities. But each of those possibilities is 100% correct on reveling of the density matrix. And if you just do it right, assume that the measurement is done in a certain basis and you're consistent with that, you will 100% correctly describe the time evolution of your atomic system. Any questions? Timo. Again, with the DG, I'm just curious. So we take it very short, but why don't we run into something like the quantum Tino effect, where if every evolution is initially quadratic, like the problem frequency, you take very short DT, start measuring the system. Shouldn't for a choice of sufficiently short DT, the simulation would give you strength results? That's a very deep question, Timo. And I want to think about it more, but I think I've excluded that by saying that the time step DT is larger than the correlation kind of the environment. So I'm doing some kind of Markov approximation with the environment, whereas the quadratic part of the behavior is a coherent time evolution. For very short time steps, the wave function evolves quadratically. But this part is a coherent evolution, and that's related to the fact that an atom can emit a photon. But in the very first moment, before the vacuum has transported away, it can take the photon back. And the result of that is that at very short times, the exponential decays doesn't start out exponentially. It starts out a little bit flatter. We get the exponential decay. We get firmness, golden rule. And we get optical Bloch equations only if your time step is larger than the correlation time of the reservoir. And the fact, you remember when we derived the master equation, we had to say we do a step which is sufficiently small for the dynamics of the atomic system, but sufficiently large that we are not getting into any memory effects of the reservoir. And we've done the same assumption here. Other questions? Thank you. I'm just confused. We are now always saying the environment measures the system in regular time intervals. That's a T which are large compared to the correlation time. It's smaller compared to the evolution of the atom. But now I wonder, isn't that exactly the Markov approximation we've made when we derived the master equation? And I was just wondering, if we just may need to derive it by making the master equation. Maybe we don't actually need the idea of constant imaginary. Maybe a physical interpretation of the Markov approximation? Oh, yes, exactly, Niky. I think you said it really very, very nicely. We derived a master equation just by assuming that the system takes a photon and past the correlation time, the photon is taken for good by the environment. And with the Markov approximation, we formalized that there is no memory effect. The photon is not stored like in a cavity. The photon has disappeared. And by saying that this time is very short, this was the Markov approximation in the derivation of the master equation. But that also means if the photon has disappeared, has separated from the atomic system, we are now free to make a measurement. And of course, it shouldn't matter whether we make the measurement or not. But what I was able to do today is I was making now the assumption, let's assume we make a measurement. We measure all the photons which have been emitted. And the measurement is probabilistic. And we fold this probabilistic evolution into our quantum system. And what we obtained was exactly the same time evolution for the density matrix as we got from the master equation. So in other words, this should tell you that interactions with an open quantum system where irreversibly, energy, photons, angular momentum, or whatever flows into the environment, once it has flown into the environment, you can measure it. And you get all the stochastics from the measurement. But even if you don't measure it, you get exactly the same stochastics as if you had measured it. It's all the same. Jenny. I want to think about it. It seems sort of weird to me that we're making these measurements or lack of measurements, these interactions at regular intervals. It doesn't really matter. All we have to make is you could actually make delta t a random variable. It wouldn't change anything. The only thing we have to make sure is we have to make the time interval short enough that we don't have maybe two photons emitted in that time. We just want to make sure that it's a simple probabilistic branching yes or no. And that's the only requirement here. So this is a formula that's a phase constant. You say now it's wave function involved, and then it collides. And this is where the measurement comes in. So now, say equivalently, it's the same wave function involved in the collection. When you make the measurement, you say that operator involves. So these are possible people. So they've all some kind of formalism like this, but using operator. Because here, even though the measurement is operator, it is still described as state. I haven't seen it, Mark, but I'm absolutely certain that you could describe the same physics in the Heisenberg picture where the time evolution is with operators. Because what we have here is we have a time step delta t where the system evolves as an isolated quantum system, and then we measure again. And so I think if you would use operator equation, you would have an evolution of the operator with the same Hamiltonian, which is non-Hermitian Hamiltonian. And then the measurement would do some form of projection. I have to think about it what the projection would be in terms of operators. But my gut feeling is you can always take the transformation where you put the time dependence into the operators and not into the wave function. On the other hand, I have some little bit misgivings about that because the Quendom-Onde Carlo wave function was really developed in order not to deal with matrices, not to deal with something such as dimension n times n if n is the number of components of the wave function. It was specifically formulated to have the simpler description with n coefficients for the wave function. But conceptually, I think it is evolution with a Hamiltonian measurement. Evolution with a Hamiltonian measurement. And I assume this could also be done with operators. Thanks. I don't know how important is the randomness of the epsilon that we are choosing because no computer is actually generating random numbers? Well, I think how random a random number has to be is really dealt with in computer science and mathematics. But to our results, we always get this exponential behavior of random numbers. I don't know the answer. I mean, here, conceptually, it should be a completely random number. And I think there are algorithms which produce pseudo-random numbers. But if the pseudo is close enough, if there is a recurrence time of your random number, if your random number is your series of random number, repeat itself after several billion random numbers, this gives you a small arrow bar. And if you only do a limited sampling and limited number of time trajectories, I don't think it matters. Of course, if you want to have ultimate precision, then also the precision of the random number may enter through the vector at some point. I'm going to talk a little bit about the three different equations of the equations of the photons. But if we know the photons in the equations, then we can decide which one is the precision. Correct. If we would know more about the environment, if we would say we measure the photons in a certain basis, or when energy is dumped in the environment by a defacing mechanism, we look at the environment. Then, of course, we would add extra information to it. And then certain quantum trajectories would reflect the extra knowledge we have. And of course, we would then describe everything in these spaces. On the other hand, what you should learn from this is once the information, once the photon has escaped, it doesn't matter for the time evolution of your atomic system in which basis you measure the photon. So the evolution of the atomic system itself, at least when you average, is unaffected by these basis. However, if you make a coincidence measurement that you say your atoms which has emitted is still flying through your vacuum chamber, and now you figure out the photon is emitted as circularly polarized light, then, of course, you know that this atom which is still available has emitted circularly polarized light. And then you have actually obtained extra information about your atom. In other words, the photon and the atom was entangled like a bell pair. And if you now do a measurement on one part of the bell pair, you know more about the other part. But this is just how quantum physics works. If you don't use this information for anything, then you could have as well not measured the polarization. And it would have no effect on the atomic density matrix. OK, that's it for today. We have class at the usual time on Wednesday.
 The following content is provided under a Creative  Commons license.  Your support will help MIT OpenCourseWare continue to  offer high quality educational resources for free.  To make a donation or view additional materials from  hundreds of MIT courses, visit MIT OpenCourseWare at  ocw.mit.edu.  MIT.edu.  I apologize for the delay.  But we are ready for a multimedia show with three  different applications.  One of them are the clickers.  I hope you all took a clicker.  It's just two questions today, but really good questions.  So today is actually a very nice topic.  I think almost every lecture will be a nice topic.  But today, it's especially nice.  It's called Unraveling Open Quentum System.  That may not tell you so much.  Well, unraveling means we want to sort of untangle it.  We want to understand what is inside.  And so we want to develop a better understanding what is  inside a density matrix.  How can we think about the quantum system in an even  more microscopic way than the density matrix tells us?  And again, this finds applications in atomic  physics because one application of it is spontaneous  emission, optical block equations.  On the other hand, actually, this class will also teach  you something very general about quantum physics.  So today's lecture is this part.  We'll have three topics.  One is I want to provide a motivation for single quantum  systems.  And you should appreciate this.  A single quantum system is not what Schrodinger and Heisenberg  told us about.  Quantum mechanics was formulated as an ensemble average.  But today, we want to talk about is there something in  addition about single quantum systems?  Part two addresses the derivation of the quantum  Monte Carlo wave function method.  And this is a formalism which is really simulating, in quite  general terms, quantum physics by following, using  quantum Monte Carlo simulations, trajectories of  individual quantum systems.  So here, the individual quantum system, even if you don't  have one, becomes a computational tool.  And eventually, in part two, I want to give you another  example for unraveling open quantum system, which are  different models for the phasing.  And if you remember that this was on one of your homework  assignment, that's correct.  I sort of want to here take your homework into the  classroom discussion.  Well, I think it's nice to start this lecture with a quote  by Erwin Schrodinger, where he said, we never experiment  with just one electron or atom or molecule.  In thought experiments, we sometimes assume that we do.  This inevitably entails ridiculous consequences.  Well, today, we want to talk about those ridiculous  consequences, because many people in our field trap  address and observe single electrons and single atoms.  Now, this is related to quantum jumps.  Quantum jumps has even made it into the popular realm.  You can find lots and lots of references on  quantum jumping, but I warn you, this is about quantum  jumping into different worlds.  It's about broadening your horizon.  And those authors, those people actually sell you  something by claiming that people like Einstein and  Stephen Hawking would support the idea that you can jump  into different universe, experience another version  of yourself, probably a better version of yourself.  And this will inspire you to change your life in  the world you live.  But so this is quantum jumping.  I'm not talking about quantum jumping.  I will only talk about quantum jumps.  So what you see here is really a classic experiment.  The one of three experiments which were done almost  simultaneously, they've always been racist for the first  observation of something.  And there was the Hamburg group by Toschek, the Seattle  group by Hans Demelt, and the Boulder group by Dave  Weinland, who observed the phenomenon of quantum jumps  for the first time.  Let me just tell you what it is because that's exactly what  you want to describe.  Here in this situation, you have a system.  It's a trapped barium ion with a P and S state.  And there is a very fast transition where you can  rapidly scatter light and get lots of fluorescence at 493  nanometer.  But what happened is every once in a while, the particle  has a branching ratio to decay to a D state.  And well, D to S, two units of angular momentum, the  transition here down to the ground state is very slow.  It's slow because it's forbidden.  It's even slower because it happens at a longer wavelength  of 1,600 nanometer.  And this is slow.  And it has a lifetime of about 50 seconds.  So if you think about it naively, and that's probably  the correct approach, you would say the particles in the  S state, it's like a light bulb.  It scatters many, many photons.  But then it branches to the D state.  And for 50 seconds, the light bulb is switched off until  the particle spontaneously goes down to the S state.  And then the light bulb is switched on.  And this is exactly what people observed.  But I mean, this was 86.  That was the time I was a graduate student.  And I talked to theorists in the field.  And respectable theorists told me that there was a  controversy, probably not among experimentalists who use  a naive approach, but theorists who believe too much in what  they calculate.  Because if you solve it quantum mechanically and look what  you get, it's a steady state solution, atom in steady state,  laser in steady state.  And if you calculate the quantum mechanical expectation  value, the expectation value is constant as a function of  time.  So a simple quantum calculation would say the fluorescence  is on average, well, just the time average and is constant  and is time independent.  So there were some serious doubts whether you would  observe this or something which looks more like the  ensemble average.  Well, all the theories soon realized you cannot calculate  the steady state average.  You have to calculate the correlation function in steady  state.  And the steady state correlation function has  structure exactly at 50 seconds.  So in a steady state system, if you're interested in  fluctuations and temporal behavior, you can't just  calculate quantities.  You have to calculate correlation functions  of quantities.  So anyway, but then, of course, if you now average over  many, many realizations, you find exactly an  exponential decay with 50 seconds.  And this is sort of what you would get from an  ensemble average.  So this is what quantum jumps are.  This is what you should always imagine when we talk about  quantum jumps in the rest of this lecture.  But I also want to show you a more recent example from  search-harroshescope in Paris where they observed  quantum jumps recording a single photon in a cavity.  So what happens is there's a single photon in this cavity  and they can read out whether there is a photon or not by  sending root back atoms through.  And the single photon is never absorbed, but it causes a  phase shift, an AC stark shift, on the root back atom.  And then they can read out with the Ramsey interferometer,  the AC stark shift.  So every atom which passes through the cavity will tell  you whether there is a photon or not.  And this is shown here.  Blue means there is no photon.  Red means there is a photon.  So here you repeatedly figure out, again, there is no  photon, and suddenly there is a quantum jump when the cavity  has absorbed a photon or when a thermal photon was released.  You may say, what happened here?  Well, no measurement is perfect.  There is some noise.  But by seeing that these are outliers, you clearly see that  here is a photon in the cavity and here where the red read  out of the atom overwhelms the blue read out.  Here you have no photon.  And then as time goes by, you can observe the death and birth  of microcavity photons.  So these are now quantum jumps, not in atomic population,  but quantum jumps in the photon number in a cavity.  Yeah, time is second.  So what is shown here is, I think, what is shown here are  300 milliseconds, and this is a 2.5 second element.  What is the density of the line?  Each line is a Rötberg atom, which has been read out with  a Ramsey interferometer.  And I think that means there was a Rötberg atom in state  E, which means there is a photon and blue means the  blue bar means there was a Rötberg atom in state G.  These are two highlighting states of the rubidium atom.  Sort of each spike here is one read out of the microchannel  plate.  So these are detector clicks.  You can say yes.  You can sort of say blue is yes and red is no.  And these are the two clicks.  But the separation between two clicks is not uniform.  No, because if you have an atomic beam, the atoms come  not equally spaced out.  There's a randomness when the atom comes.  I think it's Poissonian statistics in this experiment.  It's a real experiment.  It's amazing.  Other questions?  Timor?  Just one more question.  So for example, the first bit of blue is all measurements for  the same state of the cavity.  I don't know if blue means photon or no photon, but if it  does mean photon, that means it's the same photon interacting  with?  Yes.  OK.  Yes, because I mean, this is something people know that the  photon number in this cavity only changes over very long  time scales.  And you sort of see kind of exactly here how you see  here two quantum jumps, one from blue to red and one from  red to blue in this cavity.  Yes?  I don't know if this may be just my ass, it's not very good,  but in the bottom graph, or on the bottom graph, in the  left corner of the looks like there's some overlap between  the dark blue and the dark red.  You mean here?  Yeah, right there.  Is it?  Sort of there is a photon in that photon.  Sort of simultaneously.  I think to the best of my knowledge, I'm really looking  at, I've heard wonderful talks about it, you can go back and  read the paper, but this probably means n, the atom  number in the cavity, and the algorithm they have used to  analyze the data says that here the photon number was 0,  jump to 1, and jump back again.  So you see here, for instance, there is some overlap, but  you would still say, since either there is a photon or not  a photon, as long as you have a higher number of red clicks  and blue clicks, you would say there is a photon.  And the little overlap, there is maybe a little bit of time  where you don't know whether there is a photon or not.  But this is real, because every measurement takes a certain  amount of time, since you have a few random clicks.  If there are a few random clicks, you do not know yet  if the system has jumped, you have to wait a little bit.  But this is really also showing how our knowledge about the  quantum system comes with time.  And therefore, the wave function of the statistical  operator changes with time, because wave functions,  statistical operator are simply a description of our  knowledge about the quantum system.  You could have a situation, for instance, you could  prepare the cavity that it has an equal probability for 0  and 1, and then what you would read out may just be  completely random.  So that may be possible, but in that experiment, I think the  nature here is that you have most of the time 0 photons  because the cavity is cool to very low temperature, and  only occasionally does the thermal distribution, the  Bose-Einstein distribution of photons in the cavity have a  1.  And what we observe are the thermal fluctuations between  mainly 0 and occasionally a 1.  If you would make the cavity a little bit hotter, it would  have maybe half of the time, and 1 and half of the time is 0.  But that wouldn't mean that you have completely randomness in  your results, because there is a thermal relaxation time over  which a photon is created and the photon goes back.  The photon would not jump back and forth from the cavity to  the walls of the cavity in microseconds.  It would probably have an exchange time, and you could  sort of measure the thermal correlation time of  photons in a cavity.  It's a really good experiment.  OK, so in that sense, I think I showed you that what  Schrödinger said is ridiculous, is real, is the  subject of current research, and since for quantum  information processing, quantum computers, we need single  quantum system is at the heart of  quantum information processing.  So therefore, we want to now describe single quantum systems.  But when I do that, it will become clear at the end of  this unit, but I want to give you the warning right away.  The way how I describe now a single quantum system may not  be unique.  There will be, there is actually an infinite number of  ways, and I've already mentioned that to you, that  there is an infinite number of ways to unravel a density  matrix.  I mentioned that you can write every density matrix as a  probability of pure states, weighted probability of being  in pure states, but that always different choices.  You can get an infinite number of different combinations of  pure states and construct your density matrix.  So this was sort of at the level of the density matrix,  but similarly, and this will come out of this discussion  today, there is also an infinite number of ways to  unravel the dynamics, the dynamics involved in the  evolution of an open quantum system.  So with that, I want to describe the quantum Monte Carlo  wave function method.  So in the quantum Monte Carlo wave function method, we  perform, we don't need an experiment.  We don't need sophisticated experiments to observe single  quantum system.  It's a theoretical method, but we perform thought or  Gedanken experiments.  So our Gedanken experiments is that we perform measurements.  So we assume we have, when we want to describe an ensemble  of atoms, we assume we have a single atom.  We assume it emits photon, and then we assume that the  photon is detected.  So the moment the photon is detected, we know, for instance,  that the atom is in the ground state.  And so our quantum Monte Carlo simulation follows now the  trajectory of the single quantum system by saying, I  toss a coin, there is a certain probability for  detecting the photon.  If I detect the photon, the system I know is in the ground  state.  If I don't detect the photon, it's not in the ground state  and such.  And this probabilistic approach is just simulated in the  quantum Monte Carlo sense.  So what is the essence of the quantum Monte Carlo method is  that we allow a small time step delta t. It's a real time  simulation.  And there is a small probability that the system  has decayed in that time.  So we may now toss a coin, use a random function  generator, and ask, has spontaneous emission taken  place?  And then the two possibilities are yes.  And we continue whatever the outcome, whatever the quantum  state is.  Or we say no.  And then with the probability p, we say yes with the  probability 1 minus p, we say no.  And then we continue the simulation accordingly.  So this will become much, much clearer when I write down  the formalism for you.  But there is one important aspect which I really want to  discuss with you.  And this is the essence of the method.  Namely, how does a wave function change by a measurement?  So this is the essential, the only conceptual part for the  quantum Monte Carlo wave function.  Afterwards, I just show you.  Once we have cleared up this conceptual part, the rest is  just a few equations.  And we immediately show that the quantum Monte Carlo wave  function method is equivalent to the optical block  equation.  But let's really feature and take some time for discussing  the concept, namely, the change of a wave function  by a quantum measurement.  And maybe what this carries home to you is that the  nature of an open system with a quantum system which interacts  with the environment is actually that the environment  permanently performs measurements.  And you will actually see that the idea of performing  measurements formalized in the quantum  Monte Carlo sense gives us an equation for the density  matrix which is identical to the optical block equation.  So in other words, you can say I'm re-deriving for you master  equation now, but from a very, very different perspective,  from the perspective of an environment which constantly  performs measurement.  OK, so let's go through that.  How, what happens in a measurement?  Well, if our initial state is the ground state, starting  very in trivial ways, well, there is no measurement  because the particle will stay in the ground state forever.  OK, now let's go to the next situation that our atom starts  out in excited state.  Now, in a time step delta t, we have two possibilities.  A photon can be observed or no photon is observed.  When the photon is observed, we know for sure that the  system is now projected into the ground state.  If no photon is observed, well, the system is still in the  excited state.  OK, so we've discussed what happens when the system is  initially in the ground state, when the system is initially  in the excited state.  But now comes the following question.  We prepare the system in a superposition of  ground and excited state.  And we want to discuss what happens when the photon is  observed, what happens when no photon is observed.  I think if you observe the photon, it's pretty clear.  We have to assume after photon emission, the system is in  the ground state.  But if no photon is observed, I want to give you three choices  and you should address them with a clicker.  One choice is that the particle stays in the same state.  The second choice is that the particle stays in a  superposition state, but the probability to be in the  excited state decays, also no photon has been emitted or no  photon has been observed.  And the third possibility is something else.  So in other words, assume you are an observer and you know  with 100% detection probability.  Let's not discuss technical issues.  You have completely surrounded your system with  perfect detectors.  And you know in the time step delta t, in the time t, no  photon was observed.  And now you should make a prediction what happens next  to the quantum system.  And you make the prediction by assuming that the system is  described by a wave functional density matrix.  And I'm asking you which of the three choices correctly  describes what the system will do afterwards, what the system  will do next.  Is the question very clear?  Good.  OK.  All right.  Pretty good.  A lot of people, when I taught the class before, said it's  a.  The wave function is not changing because no photon has  been emitted.  Now, for those people, but there are only a few among  those, I would have now asked the next question.  If you have a 50-50 superposition of ground and  excited state.  And let's say the time for spontaneous emission is 10  nanoseconds.  But now you wait one second, and after one second, no  photon has been observed.  What is your prediction?  What in what state is the system at this point?  Is the system in the ground state?  Or is the system still in a superposition of ground and  excited state?  Well, since we have the clicker, you can tell me.  Is it still a superposition state, or is the system in  the ground state?  OK.  So now you have to help me out, because the people who said  B, if they would go back to here, and most people said it  is B, if you put in gamma 10 nanoseconds, the inverse of  10 nanoseconds, and the time is one second, the excited  state amplitude, the excited state at mixture, is 10 to  the minus nothing.  So OK, if you want to be a mathematician and say there is  epsilon 10 to the minus 10, yes, then you can say there is  still a small mixture to the excited state.  But by making it so extreme, 10 nanoseconds versus a second,  at least in all practical terms, the system is in the  ground state.  And I think that should teach you what it means if you have  a system which is in a superposition of ground and  excited state.  And after one second, it has not emitted a photon.  I mean, then you would say it had its chance.  It had plenty of chances to emit, but it decided it  doesn't want to emit.  And that would mean your 50-50 superposition of ground and  excited state means the atom decided that it's in the  ground state.  You should also realize a 50% superposition of ground and  excited state means in 50% of the cases, there will never  be a photon emitted.  So in other words, what you should realize is if you  have a 50-50% superposition of ground and excited state,  that the quantum evolution of this system is after a long  time, the system is definitely positively, absolutely in  the ground state.  It's always in the ground state after a long time.  In 50% of the cases, with emission of a photon, in 50%  of the cases, without emission of a photon.  If you have an ensemble of radioactive nuclei, and let's  say half of the atoms are in the excited state and can  radiate with decay and half are in the ground state, and you  send the sample to your friend, and your friend waits  many, many, many half times, common sense tells you that  all the particles are now in the ground state.  Also only half of them have decayed because the other  half were in the ground state to begin with.  I mean, this is what quantum physics tells us.  This is what a 50-50 superposition is.  But quantitatively, this is, of course, included in this  result, which magically almost all of you got right.  Namely, there is a superposition of ground and  excited state, but the excited state amplitude decays.  And to the limit, in the limit that the time is much  longer than the decay time, no matter what the coefficient  initially was in front of the excited state, this coefficient  has decay to 0.  Any question about that?  Let me just emphasize that, again, and giving you also an  example in the reference, a zero measurement modifies the  wave function.  You can also put it like this.  What the wave function is, is the best knowledge you have  about the system.  The wave function is a way to predict what happens next.  And it's the most precise way to predict it.  More accurate predictions than the wave function allows  cannot be made because of Heisenberg's uncertainty  relation.  But what happens is, if you have a 50-50% superposition of  ground and excited state, and the system has not emitted,  then you would say, initially, my guess was half of the  atoms are excited, half of the atoms are in the ground state.  But if none of them has decayed for a while, you would say,  I revise my estimate now.  Now I have to assume that more in the ground state.  Because having not decayed means there is a higher  probability that the system is in the ground state to  begin with.  So this is how this way of dealing with the wave function  automatically adjusts for the knowledge you have gained  about the system.  Let me give you another example which carries home the same  message.  And I take it from a pedagogical paper from Dickey, from  American Journal of Physics, 49, 926, 1981.  And what he discussed is the following.  You have a box, and originally, your atomic wave  function is completely localized.  You oppose the Einstein condensate if you want in  this box.  Or let's just assume one atom to be more precise.  And it's completely delocalized.  But now you focus a laser beam into the system.  And the laser beam would ionize your atom, and you  could count the ion with 100% probability.  And for a short moment, you flash on the laser, and the  result is, yes, you count.  You get a count.  In that moment, you would say, my wave function is more  localized.  I revise my estimate.  Well, maybe I shouldn't say ionized.  If it's ionized, you destroy the atom.  Let's just say you observe fluorescence, so the atom is  still alive.  But now you have to say the atom is  localized in that region.  However, if you don't observe anything, you would say, well,  now it's actually more probable that the atom is  outside, because if it had been in the laser beam with a  certain probability, I would have detected a photon.  But the non-detection of a photon means that I revise my  estimate, and I say it's much more probable that the atom is  outside the laser beam than inside the laser beam.  So this here is exactly how a non-observation of anything,  a zero measurement, modifies your wave function.  And the formalism to incorporate that is what we have  discussed before.  Yes, Cody?  So your arguments here so far work the same way as at  the wave function, we have just an entirely classical  probability, and we're updating our understanding of  probability.  Do you have anything about the relative base between these two?  I want to do that now.  This is sort of just to address the basic concepts about  what does a measurement mean, what does a zero  measurement mean.  But now I want to write down everything for you in  amplitudes.  We want to do exactly the time evolution of the system.  Nancy?  Is this like inconsistent with energy conservation, like  especially the two before this?  Because when we are saying that it is in the atmosphere of  the ground energy excited state, then we do not observe  the flow of the energy in the system.  And now we're just going to do that.  In a superposition state of ground energy excited state,  we are not starting out in an energy eigenstate.  So therefore, what we have is not a sharp value of the  energy, but an expectation value.  And if you repeat the measurement many, many  times, the average energy is conserved.  And this is exactly what energy conservation in  quantum physics says.  But if we do not repeat the measurement, then we should  not just apply energy conservation.  Well, if your particle, if your system is not in an eigenstate  of the Hamiltonian, it doesn't have a sharp energy.  And that would mean that when you measure the energy now,  you have fluctuations.  Sometimes you measure higher.  Sometimes you measure lower.  It's only when you're in an eigenstate that you measure a  sharp value.  And therefore, you know from the beginning that you will  measure an energy diffusion.  And sometimes you measure higher.  And sometimes you measure lower than the average value.  There's nothing wrong about it, but your question is a very  good one.  You should.  I mean, those examples are really, on the one hand, they  are trivial, but on the other hand, it's very profound what  they tell us about quantum physics and how to apply  conservation laws and such.  OK, I think let me now formalize exactly how this is done  with all the bells and whistles.  We assume we have an initial wave function, which is now an  arbitrary superposition state of ground and excited state.  And we have the environment, which in this case we assume  is the zero photon state.  It's a vacuum.  And we want to exactly solve the Schrodinger equation for  that system.  But we restrict ourselves to a very small time step.  The small time step is smaller than anything else than the  natural decay time.  Maybe the inverse Rabi frequency if you drive the  system.  And we are not doing it here, but I will later show you how  you can add a laser beam and drive the system.  Or it should also be smaller than the inverse detuning.  So what is very important here is that we want to deal only  with simple cases, namely one photon has been  emitted or no photon.  So we want to make sure that there is at most one  spontaneous emission event during the time delta t.  So what I'm writing down now is the total wave function of  the system plus the environment.  And now we can evolve it.  We can ask what is the wave function time later.  This can be exactly done by time dependent  perturbation theory.  And the result is that the system will be in a super  position of ground and excited state.  And we want to calculate the coefficients alpha prime and  beta prime.  And we still have the vacuum state.  But then we have the possibility that a photon has  been emitted.  In that case, we are in the ground state.  And the direct product with the environment involves now a  photon emitted into a certain direction with wave vector k  with a polarization epsilon.  And we have coefficients beta k epsilon.  And we have to sum over all possibilities for the photon  to be emitted.  So we call this, so the wave function of the total system  has now two parts.  It's a wave function which I call psi 0.  This is a wave function which involves no photon in the  environment.  And the wave function psi 1 involves one photon in the  environment.  So this is the ansatz which seems very natural.  But you could immediately prove that these are the only  possibilities how the system can evolve.  And we can verify this by using time dependent  perturbation theory.  It's actually almost everybody of you has seen it in either  in 8.421 or in a more basic course on quantum physics.  It's time dependent perturbation theory for the  emission of a photon.  But usually when you see those treatments, the reservoir, the  environment is not treated as explicitly as we do it here.  The theory where this is treated exactly in this way how  we need it is the Wegener Weisskopf theory, which is  nothing else in a perturbative approach.  But it's really a perturbation theory not just for the atomic  system, how we sometimes present it in a simplified  version.  It's perturbation theory for the total wave function of the  complete system.  Timur?  You mentioned the time interval, which was in the T, is  much smaller than a lot of things.  But here we're ignoring, I take it, the counter-rotating  terms, in which case we're still at a large enough time  step if that's average now.  I simply say this because I expect it to be an E with a  photon term from the G-dot-tum, so to speak.  Yes, well, the counter-rotating term.  The first answer I wanted to give you, know this is exact  perturbation theory.  But I think I get myself in trouble if I would get myself  in trouble if I would allow the time step to be extremely  short, because then I'm in shorter than 1 over omega.  Because during the time 1 over omega, you can say we have  counter-rotating terms, or in other words, a particle in  the ground state during a time 1 over omega particle in the  ground state can emit a photon and reabsorb it.  These were those weird diagrams with virtual states, which  we discussed earlier in this course.  So I think I want to assume here, which I haven't done,  that the time step is larger than omega to the minus 1.  And as you know, the counter-rotating term has a  detuning delta, which is 2 omega.  So therefore, it is excluded.  Or in other words, when we talk about photons sent into the  environment, we want to talk about real photons and not  virtual photons.  Yes, good point.  So if you do a simple lowest order time dependent  perturbation theory in the Wegener-Weisdorff approach,  you get an exact result for beta prime.  You find that beta prime in this superposition is the  original amplitude beta, but it has decayed with e to the  minus gamma over 2 dt.  And since we are only interested in small time steps,  we can do a linear expansion of that.  The probability dp that a photon has been emitted is the  norm of this wave function psi 1.  And this using Wegener-Weisdorff perturbation theory is  gamma dt times the amplitude squared that the system was  excited to begin with.  And because of the conservation of the norm, the wave  function, the norm of the wave function psi 0, which is the  wave function without a photon being emitted, is 1 minus dp.  So what I'm telling you here is that this is an exact  result of time dependent perturbation theory.  Often when textbooks treat spontaneous emission, they are  more interested in getting Fermi's golden rule and this  rate, but the same approach, if you just write it down, tells  you how the amplitude in the excited state evolves.  So therefore, what we learn from perturbation theory that  alpha prime and beta prime, the coefficients of the wave  function without emission of a photon, this evolution of the  wave function occurs with a non-Hermitian Hamiltonian.  And this non-Hermitian Hamiltonian is our Hamiltonian  for the atomic system.  But then we have to account for spontaneous decay, and this  is done by this non-Hermitian part.  So occasionally, you may have heard that people wave their  hands and say, your system is described by a Hamiltonian  which has an imaginary part for decay, and this is sort of  phenomenological.  This is definitely not the case, because even if you have a  Hamiltonian like this, a pure state would simply decay and  remain a pure state.  What we are doing here explicitly is we deal with  probability in a correct way.  Every time the system can branch out into two different  possibilities, two different dimensions of the density  matrix, one has probability delta p, one has probability 1  minus delta p, and it is only the wave function associated  with probability 1 minus delta p, which evolves with this  Hamiltonian.  So what I'm telling you is the exact solution for the  evolution of the total system in terms of a density matrix  for the atomic system.  And this is exact.  So that means we can now write down the just get some  extra space.  Oops, how does it do?  Sorry, I have to insert a new page.  At least.  OK.  OK, so the procedure how we implement this exact solution  of perturbation theory is the following.  We have a time step delta t.  We compute what is the probability that a photon  will be emitted.  Then we need a random number generator.  So we need a number epsilon, which is a random number  chosen in the interval 0 and 1.  If this random number turns out to be smaller than delta p,  then we continue our time evolution on the computer  that psi is now in the ground state.  And maybe there's a laser beam which excites it again and  such, I will add a few bells and whistles later.  Otherwise, psi, our wave function, is now the wave  function, which is the time evolution with a non-Hermitian  operator of the original wave function psi.  And since we have a real wave function now with probability  1, we have to renormalize the wave function by this  denominator.  Then we execute the next time step.  This means go to 2 and do the next time step.  And then you have so to speak, if you do it many, many times,  you get a time-direct trajectory of so to speak one  experiment.  And then you start again with your system in a wave function  psi, and you accumulate a second experiment.  And maybe you do it 10,000 times to get enough statistics.  And then you sum up everything you want and calculate all  the expectation values you want to know about  your quantum system.  So the claim is that this method called quantum Monte  Carlo wave function method is fully equivalent to the  optical Bloch equations.  And I want to prove it to you by showing that if I take a  density matrix, which is an ensemble average over many  realizations of those quantum Monte Carlo trajectories, that  then this density matrix follows the differential  equation, which is the optical Bloch equation.  And the proof is shown here.  So what I told you is the density matrix after time  delta t has now two matrix elements, one with probability  delta p.  And this is, photon has been emitted, system is in the  ground state, with probability 1 minus delta p.  We evolve the quantum state with a non-Hermitian Hamiltonian.  And then all what is done in the next few steps, we assume  that delta t is small.  We do a Taylor expansion of the exponent.  We neglect quadratic terms in delta t.  And then we come to this line.  And if we write this as rho of t plus delta t minus the  original rho of t, we find that this follows a differential  equation, which is exactly the optical Bloch equations.  So therefore, what we implemented as a form of  doing many, many quantum measurements in the environment  is a procedure which is rigorously the same as the  optical Bloch equations, which we derived using a master  equation approach.  OK.  This method is very powerful for the following reasons.  If you simulate a density matrix with n internal states,  or an external or internal states, I should have said  with n quantum states, you need n times n matrix elements  which can become quite a memory  hog for your computer.  The wave function at any given moment has only n components.  So therefore, there is a computational advantage in  using a stochastic wave function approach over  simulation of the density matrix.  The second thing is that a lot of people, especially  experimentalists, like this approach because it reflects  directly what they do in the experiment.  And I think it's pretty clear that with this approach, you  can deal with a great variety of situations.  Let me just mention two obvious extensions.  One is polarization.  If photon is detected, that means your random number  produced a number epsilon, which was smaller than delta p.  At that point, you can create a second random number which  determines if the photon has been detected with sigma plus,  sigma minus, or high polarization.  Or you can discuss recoil.  If the photon is detected, you throw on another random  function generator which determines what the k vector  of the photon is.  So what direction h bar k the photon has taken.  And this determines now what is the recoil that  change in momentum of your atom.  So you see kind of you can start with an atom at zero  momentum.  You can say it emits a photon.  And then your computer always makes a choice based on the  random number.  And then you say, OK, my photon has now received a recoil  kick at 45 degrees.  Now your wave function of the photon is such and such.  And if you add a laser to the Hamiltonian, the recoil may  have double shifted the resonance, but everything is  easily taken into account.  So you see that this quantum Monte Carlo wave function  method can easily be extended to describe external degrees of  freedom, multiple laser fields, and all that.  Any questions?  OK, let me now just generalize the thought.  But I think you know already everything about it.  We talked about the master equation.  We talked about the most general master equation in the  Linn platform, where those operators L are the jump  operators.  And the prominent example for a jump operator was the sigma  minus operator, which takes a particle from the excited  state to the ground state.  And this is a jump operator for spontaneous emission.  But you may have many jump operators, maybe spontaneous  emission of different photons with different  polarization and such.  Or the cavity may lose a photon, and then we have a jump  operator for the cavity.  We went through that.  Well, that means now in an exact way that the non-Hemiltonian  is the original Hamiltonian with an imaginary part, which  comes because of those jump operators.  And the general quantum Monte Carlo wave function procedure  is that first you ask, has something happened?  Has any jump happened?  And this gives you the probability delta p.  And if no jump has happened, you just do a time evolution  with a non-Hemiltonian.  But if a jump has happened, then the jump is now  the wave function is projected by the jump operator.  If you have a sigma minus operator, it takes a particle  to the ground state.  If you have several operators, a different jump operator may  take your particle to another state.  And you have then a branching ratio that you know first  the jump has happened, and then the question is, which jump  has happened?  And you just play the probabilistic game.  So you see that the quantum Monte Carlo wave function  method can immediately be generalized.  So let me come back to the special case of spontaneous  emission.  In that case, we have only one jump operator, which  is sigma minus.  And the normalization is the square root of gamma.  And that actually means that our quantum trajectory,  ground state, excited state, is very simple.  If we start with a system 100% in the excited state,  it's very trivial what happens.  Nothing happens.  But if then the jump occurs, the particle  is in the ground state.  Then nothing happens then until infinite time.  You then have to start the next trajectory.  You start in the excited state.  And this time, by chance of the random number generator,  the jump happens later.  Another time, the jump happens earlier.  And what you then have to do is you  have to average over all those realizations  what is the probability for the particle  to be in the excited state.  And this is now your estimator for the excited state  diagonal matrix element of the density matrix.  And if you've written your code correctly,  you will find that you get a wonderful exponential decay  exactly as you have expected.  Questions?  Now, yes.  Let me now talk about defacing.  What I've shown you so far is how  we can unravel the density matrix  in many microscopic realizations.  And I've given you a specific example.  We assume probability for spontaneous emission.  And with that, we propagate our wave function.  But now I come to sort of the nitty-gritty details  or the dirty truth that what we have assumed  is by no means unique.  And I want to explain it to you first  by reminding you of a very, very nice homework assignment  you have solved.  And this was about we have optical Bloch equations  for the density matrix.  And the solution of the optical Bloch equation  is that we have population damping with a damping time t1.  And the off diagonal matrix elements  are damped with a time t2.  Remember, if you have only spontaneous emission,  t2 is 2 times t1.  But we can have a lot of other processes  which can lead to a much, much faster defacing time  than a population, than a time of population changes.  And in your homework, you have discussed three possibilities.  One is, well, spontaneous emission is energy loss.  But you have discussed three different possibilities for t2.  One, three different possibilities,  how the phasing loss of coherence and phase  damping can happen.  And in this homework assignment, you  assumed either that there is a random phase,  that the elastic collision which  project onto the excited state.  And then you got the most crazy and artificial model  that every time, randomly, the phase of the excited state  flips from plus to minus.  So if I would implement that now with a Quento Monte Carlo  method, it would have the following effect.  And I hope you enjoy the graphical representation.  You can really think about it that this  is what happens, that this is what  is inside the density matrix described by this process.  So let's assume we have a system which  would be in a superposition of crowned and excited state.  And that would mean that in the lab frame,  the dipole moment would just rotate  at the resonance frequency.  It rotates, and it is a rotating dipole moment which  emits coherent light.  But now you assume that you have a random phase.  So if you add random phases to it,  maybe because you have fluctuating magnetic fields,  then the line becomes sort of, that doesn't look random,  but you know what I mean, it becomes sort of jagged.  The phase distribution is still sort of there,  but there is a jitter on top of it.  And as a result, the light emitted by this dipole  is actually broad.  Well, in your second model, you assumed  that with a certain randomness, with a certain time constant,  there's an elastic collision in the excited state  which projects the system into the excited state.  At this moment, there is no superposition state anymore,  and the dipole moment is 0.  So that's what you assume.  Or when you have random phase flips,  you assume that suddenly this kind of deterministic sine  function suddenly jumps corresponding to a minus  sign in the excited state.  And depending where you are in the cycle,  it creates jumps at random places.  And the question, of course, is if all those three processes,  and that's what you showed in your homework,  lead to the same density matrix, which one is correct?  Which one is really realized in an experiment  on a system which is described by this kind of damping  or by the optical Bloch equation?  So you can say all or none.  Well, you can assume what you want.  It doesn't make a difference.  And the reason is actually profound.  The reason is profound in that sense  that the way our goal was to have an open quantum system, which  is the atomic system interacting with the reservoir.  But the way how we phrase the question,  we are only interested in what the atomic system does is,  we didn't do extra measurements on the environment.  The environment was just a dump for photons,  a dump for energy, a dump for whatever  we assumed in our defacing mechanism.  So let me just take the example of putting photons  into the environment.  And this is described by a certain damping term.  But there is an ambiguity.  And this is shown here.  If we evolve the system, our atomic system,  the density matrix evolves.  But the environment also evolves.  And for instance, here, the unitary time evolution  has taken the excitation from the atom.  And we have emitted a photon into the environment.  And you remember, we got the optical Bloch equation  by doing the partial trace here and just focusing  on the atomic part.  But now wait a moment.  If we trace out the environment,  we can trace it out in a different basis set.  Remember, I showed you the quantum Monte Carlo wave  function.  We emit a photon.  But we can, for instance, detect the photon  in our Gedanken experiment.  We can detect it with linear polarization  or with circular polarization.  If you have a situation where you have an s state which  decays to a p state, sigma plus takes you to m equals plus 1,  sigma minus takes you to m equals minus 1.  So therefore, if a photon is emitted  in your quantum Monte Carlo procedure,  you would say, now the atom is in the plus 1 state  or now the atom is in the minus 1 state.  But if you detect linearly polarized light  by just putting a polarizer here,  and this is a unitary transformation  in front of your detector, you would no longer  project the atomic system on plus 1 and minus 1.  You would project it on m equals 0 or whatever  is connected to the linear polarization of the measurement.  So therefore, if you simply assume  that something has been dumped in the environment  and can be used for measurement, there  are many ways, many unitary transformations,  what you can do to the information, to the energy,  to the photons which have been dumped into the environment.  And each of them will lead to a very different trajectory  in your quantum Monte Carlo system.  So therefore, if you just dump the photons and not measure  them, you have equal rights to assume  that your quantum Monte Carlo wave function jumps  to states which correspond to a circular basis  or jump to states which correspond to linear basis.  And there are many, many possibilities.  But each of those possibilities is 100% correct  on reveling of the density matrix.  And if you just do it right, assume  that the measurement is done in a certain basis  and you're consistent with that, you will 100% correctly  describe the time evolution of your atomic system.  Any questions?  Timo.  Again, with the DG, I'm just curious.  So we take it very short, but why don't we  run into something like the quantum Tino effect, where  if every evolution is initially quadratic,  like the problem frequency, you take very short DT,  start measuring the system.  Shouldn't for a choice of sufficiently short DT,  the simulation would give you strength results?  That's a very deep question, Timo.  And I want to think about it more,  but I think I've excluded that by saying  that the time step DT is larger than the correlation  kind of the environment.  So I'm doing some kind of Markov approximation  with the environment, whereas the quadratic part  of the behavior is a coherent time evolution.  For very short time steps, the wave function  evolves quadratically.  But this part is a coherent evolution,  and that's related to the fact that an atom can emit a photon.  But in the very first moment, before the vacuum has  transported away, it can take the photon back.  And the result of that is that at very short times,  the exponential decays doesn't start out exponentially.  It starts out a little bit flatter.  We get the exponential decay.  We get firmness, golden rule.  And we get optical Bloch equations  only if your time step is larger than the correlation  time of the reservoir.  And the fact, you remember when we  derived the master equation, we had to say we do a step which  is sufficiently small for the dynamics of the atomic system,  but sufficiently large that we are not  getting into any memory effects of the reservoir.  And we've done the same assumption here.  Other questions?  Thank you.  I'm just confused.  We are now always saying the environment  measures the system in regular time intervals.  That's a T which are large compared to the correlation time.  It's smaller compared to the evolution of the atom.  But now I wonder, isn't that exactly the Markov approximation  we've made when we derived the master equation?  And I was just wondering, if we just  may need to derive it by making the master equation.  Maybe we don't actually need the idea of constant imaginary.  Maybe a physical interpretation of the Markov approximation?  Oh, yes, exactly, Niky.  I think you said it really very, very nicely.  We derived a master equation just by assuming  that the system takes a photon and past the correlation time,  the photon is taken for good by the environment.  And with the Markov approximation,  we formalized that there is no memory effect.  The photon is not stored like in a cavity.  The photon has disappeared.  And by saying that this time is very short,  this was the Markov approximation in the derivation  of the master equation.  But that also means if the photon has disappeared,  has separated from the atomic system,  we are now free to make a measurement.  And of course, it shouldn't matter  whether we make the measurement or not.  But what I was able to do today is  I was making now the assumption, let's  assume we make a measurement.  We measure all the photons which have been emitted.  And the measurement is probabilistic.  And we fold this probabilistic evolution  into our quantum system.  And what we obtained was exactly the same time evolution  for the density matrix as we got from the master equation.  So in other words, this should tell you  that interactions with an open quantum system  where irreversibly, energy, photons, angular momentum,  or whatever flows into the environment,  once it has flown into the environment,  you can measure it.  And you get all the stochastics from the measurement.  But even if you don't measure it,  you get exactly the same stochastics  as if you had measured it.  It's all the same.  Jenny.  I want to think about it.  It seems sort of weird to me that we're  making these measurements or lack of measurements,  these interactions at regular intervals.  It doesn't really matter.  All we have to make is you could actually  make delta t a random variable.  It wouldn't change anything.  The only thing we have to make sure  is we have to make the time interval short enough  that we don't have maybe two photons emitted in that time.  We just want to make sure that it's  a simple probabilistic branching yes or no.  And that's the only requirement here.  So this is a formula that's a phase constant.  You say now it's wave function involved, and then it collides.  And this is where the measurement comes in.  So now, say equivalently, it's the same wave function  involved in the collection.  When you make the measurement, you say that operator involves.  So these are possible people.  So they've all some kind of formalism like this,  but using operator.  Because here, even though the measurement is operator,  it is still described as state.  I haven't seen it, Mark, but I'm absolutely certain  that you could describe the same physics in the Heisenberg  picture where the time evolution is with operators.  Because what we have here is we have a time step delta t  where the system evolves as an isolated quantum system,  and then we measure again.  And so I think if you would use operator equation,  you would have an evolution of the operator  with the same Hamiltonian, which is non-Hermitian Hamiltonian.  And then the measurement would do some form of projection.  I have to think about it what the projection would  be in terms of operators.  But my gut feeling is you can always  take the transformation where you put the time dependence  into the operators and not into the wave function.  On the other hand, I have some little bit misgivings  about that because the Quendom-Onde Carlo wave  function was really developed in order  not to deal with matrices, not to deal with something  such as dimension n times n if n is the number of components  of the wave function.  It was specifically formulated to have the simpler description  with n coefficients for the wave function.  But conceptually, I think it is evolution  with a Hamiltonian measurement.  Evolution with a Hamiltonian measurement.  And I assume this could also be done with operators.  Thanks.  I don't know how important is the randomness of the epsilon  that we are choosing because no computer is actually  generating random numbers?  Well, I think how random a random number has to be  is really dealt with in computer science and mathematics.  But to our results, we always get this exponential behavior  of random numbers.  I don't know the answer.  I mean, here, conceptually, it should be a completely random  number.  And I think there are algorithms which produce pseudo-random  numbers.  But if the pseudo is close enough,  if there is a recurrence time of your random number,  if your random number is your series of random number,  repeat itself after several billion random numbers,  this gives you a small arrow bar.  And if you only do a limited sampling and limited number  of time trajectories, I don't think it matters.  Of course, if you want to have ultimate precision,  then also the precision of the random number  may enter through the vector at some point.  I'm going to talk a little bit about the  three different equations of the equations of the photons.  But if we know the photons in the equations,  then we can decide which one is the precision.  Correct.  If we would know more about the environment,  if we would say we measure the photons in a certain basis,  or when energy is dumped in the environment  by a defacing mechanism, we look at the environment.  Then, of course, we would add extra information to it.  And then certain quantum trajectories  would reflect the extra knowledge we have.  And of course, we would then describe everything  in these spaces.  On the other hand, what you should learn from this  is once the information, once the photon has escaped,  it doesn't matter for the time evolution of your atomic system  in which basis you measure the photon.  So the evolution of the atomic system itself,  at least when you average, is unaffected by these basis.  However, if you make a coincidence measurement  that you say your atoms which has emitted  is still flying through your vacuum chamber,  and now you figure out the photon is emitted  as circularly polarized light, then, of course,  you know that this atom which is still available  has emitted circularly polarized light.  And then you have actually obtained extra information  about your atom.  In other words, the photon and the atom  was entangled like a bell pair.  And if you now do a measurement on one part of the bell pair,  you know more about the other part.  But this is just how quantum physics works.  If you don't use this information for anything,  then you could have as well not measured the polarization.  And it would have no effect on the atomic density matrix.  OK, that's it for today.  We have class at the usual time on Wednesday.
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. MIT.edu. I apologize for the delay. But we are ready for a multimedia show with three different applications. One of them are the clickers. I hope you all took a clicker. It's just two questions today, but really good questions. So today is actually a very nice topic. I think almost every lecture will be a nice topic. But today, it's especially nice. It's called Unraveling Open Quentum System. That may not tell you so much. Well, unraveling means we want to sort of untangle it. We want to understand what is inside. And so we want to develop a better understanding what is inside a density matrix. How can we think about the quantum system in an even more microscopic way than the density matrix tells us? And again, this finds applications in atomic physics because one application of it is spontaneous emission, optical block equations. On the other hand, actually, this class will also teach you something very general about quantum physics. So today's lecture is this part. We'll have three topics. One is I want to provide a motivation for single quantum systems. And you should appreciate this. A single quantum system is not what Schrodinger and Heisenberg told us about. Quantum mechanics was formulated as an ensemble average. But today, we want to talk about is there something in addition about single quantum systems? Part two addresses the derivation of the quantum Monte Carlo wave function method. And this is a formalism which is really simulating, in quite general terms, quantum physics by following, using quantum Monte Carlo simulations, trajectories of individual quantum systems. So here, the individual quantum system, even if you don't have one, becomes a computational tool. And eventually, in part two, I want to give you another example for unraveling open quantum system, which are different models for the phasing. And if you remember that this was on one of your homework assignment, that's correct. I sort of want to here take your homework into the classroom discussion. Well, I think it's nice to start this lecture with a quote by Erwin Schrodinger, where he said, we never experiment with just one electron or atom or molecule. In thought experiments, we sometimes assume that we do. This inevitably entails ridiculous consequences. Well, today, we want to talk about those ridiculous consequences, because many people in our field trap address and observe single electrons and single atoms. Now, this is related to quantum jumps. Quantum jumps has even made it into the popular realm. You can find lots and lots of references on quantum jumping, but I warn you, this is about quantum jumping into different worlds. It's about broadening your horizon. And those authors, those people actually sell you something by claiming that people like Einstein and Stephen Hawking would support the idea that you can jump into different universe, experience another version of yourself, probably a better version of yourself. And this will inspire you to change your life in the world you live. But so this is quantum jumping. I'm not talking about quantum jumping. I will only talk about quantum jumps. So what you see here is really a classic experiment. The one of three experiments which were done almost simultaneously, they've always been racist for the first observation of something. And there was the Hamburg group by Toschek, the Seattle group by Hans Demelt, and the Boulder group by Dave Weinland, who observed the phenomenon of quantum jumps for the first time. Let me just tell you what it is because that's exactly what you want to describe. Here in this situation, you have a system. It's a trapped barium ion with a P and S state. And there is a very fast transition where you can rapidly scatter light and get lots of fluorescence at 493 nanometer. But what happened is every once in a while, the particle has a branching ratio to decay to a D state. And well, D to S, two units of angular momentum, the transition here down to the ground state is very slow. It's slow because it's forbidden. It's even slower because it happens at a longer wavelength of 1,600 nanometer. And this is slow. And it has a lifetime of about 50 seconds. So if you think about it naively, and that's probably the correct approach, you would say the particles in the S state, it's like a light bulb. It scatters many, many photons. But then it branches to the D state. And for 50 seconds, the light bulb is switched off until the particle spontaneously goes down to the S state. And then the light bulb is switched on. And this is exactly what people observed. But I mean, this was 86. That was the time I was a graduate student. And I talked to theorists in the field. And respectable theorists told me that there was a controversy, probably not among experimentalists who use a naive approach, but theorists who believe too much in what they calculate. Because if you solve it quantum mechanically and look what you get, it's a steady state solution, atom in steady state, laser in steady state. And if you calculate the quantum mechanical expectation value, the expectation value is constant as a function of time. So a simple quantum calculation would say the fluorescence is on average, well, just the time average and is constant and is time independent. So there were some serious doubts whether you would observe this or something which looks more like the ensemble average. Well, all the theories soon realized you cannot calculate the steady state average. You have to calculate the correlation function in steady state. And the steady state correlation function has structure exactly at 50 seconds. So in a steady state system, if you're interested in fluctuations and temporal behavior, you can't just calculate quantities. You have to calculate correlation functions of quantities. So anyway, but then, of course, if you now average over many, many realizations, you find exactly an exponential decay with 50 seconds. And this is sort of what you would get from an ensemble average. So this is what quantum jumps are. This is what you should always imagine when we talk about quantum jumps in the rest of this lecture. But I also want to show you a more recent example from search-harroshescope in Paris where they observed quantum jumps recording a single photon in a cavity. So what happens is there's a single photon in this cavity and they can read out whether there is a photon or not by sending root back atoms through. And the single photon is never absorbed, but it causes a phase shift, an AC stark shift, on the root back atom. And then they can read out with the Ramsey interferometer, the AC stark shift. So every atom which passes through the cavity will tell you whether there is a photon or not. And this is shown here. Blue means there is no photon. Red means there is a photon. So here you repeatedly figure out, again, there is no photon, and suddenly there is a quantum jump when the cavity has absorbed a photon or when a thermal photon was released. You may say, what happened here? Well, no measurement is perfect. There is some noise. But by seeing that these are outliers, you clearly see that here is a photon in the cavity and here where the red read out of the atom overwhelms the blue read out. Here you have no photon. And then as time goes by, you can observe the death and birth of microcavity photons. So these are now quantum jumps, not in atomic population, but quantum jumps in the photon number in a cavity. Yeah, time is second. So what is shown here is, I think, what is shown here are 300 milliseconds, and this is a 2.5 second element. What is the density of the line? Each line is a Rötberg atom, which has been read out with a Ramsey interferometer. And I think that means there was a Rötberg atom in state E, which means there is a photon and blue means the blue bar means there was a Rötberg atom in state G. These are two highlighting states of the rubidium atom. Sort of each spike here is one read out of the microchannel plate. So these are detector clicks. You can say yes. You can sort of say blue is yes and red is no. And these are the two clicks. But the separation between two clicks is not uniform. No, because if you have an atomic beam, the atoms come not equally spaced out. There's a randomness when the atom comes. I think it's Poissonian statistics in this experiment. It's a real experiment. It's amazing. Other questions? Timor? Just one more question. So for example, the first bit of blue is all measurements for the same state of the cavity. I don't know if blue means photon or no photon, but if it does mean photon, that means it's the same photon interacting with? Yes. OK. Yes, because I mean, this is something people know that the photon number in this cavity only changes over very long time scales. And you sort of see kind of exactly here how you see here two quantum jumps, one from blue to red and one from red to blue in this cavity. Yes? I don't know if this may be just my ass, it's not very good, but in the bottom graph, or on the bottom graph, in the left corner of the looks like there's some overlap between the dark blue and the dark red. You mean here? Yeah, right there. Is it? Sort of there is a photon in that photon. Sort of simultaneously. I think to the best of my knowledge, I'm really looking at, I've heard wonderful talks about it, you can go back and read the paper, but this probably means n, the atom number in the cavity, and the algorithm they have used to analyze the data says that here the photon number was 0, jump to 1, and jump back again. So you see here, for instance, there is some overlap, but you would still say, since either there is a photon or not a photon, as long as you have a higher number of red clicks and blue clicks, you would say there is a photon. And the little overlap, there is maybe a little bit of time where you don't know whether there is a photon or not. But this is real, because every measurement takes a certain amount of time, since you have a few random clicks. If there are a few random clicks, you do not know yet if the system has jumped, you have to wait a little bit. But this is really also showing how our knowledge about the quantum system comes with time. And therefore, the wave function of the statistical operator changes with time, because wave functions, statistical operator are simply a description of our knowledge about the quantum system. You could have a situation, for instance, you could prepare the cavity that it has an equal probability for 0 and 1, and then what you would read out may just be completely random. So that may be possible, but in that experiment, I think the nature here is that you have most of the time 0 photons because the cavity is cool to very low temperature, and only occasionally does the thermal distribution, the Bose-Einstein distribution of photons in the cavity have a 1. And what we observe are the thermal fluctuations between mainly 0 and occasionally a 1. If you would make the cavity a little bit hotter, it would have maybe half of the time, and 1 and half of the time is 0. But that wouldn't mean that you have completely randomness in your results, because there is a thermal relaxation time over which a photon is created and the photon goes back. The photon would not jump back and forth from the cavity to the walls of the cavity in microseconds. It would probably have an exchange time, and you could sort of measure the thermal correlation time of photons in a cavity. It's a really good experiment. OK, so in that sense, I think I showed you that what Schrödinger said is ridiculous, is real, is the subject of current research, and since for quantum information processing, quantum computers, we need single quantum system is at the heart of quantum information processing. So therefore, we want to now describe single quantum systems. But when I do that, it will become clear at the end of this unit, but I want to give you the warning right away. The way how I describe now a single quantum system may not be unique. There will be, there is actually an infinite number of ways, and I've already mentioned that to you, that there is an infinite number of ways to unravel a density matrix. I mentioned that you can write every density matrix as a probability of pure states, weighted probability of being in pure states, but that always different choices. You can get an infinite number of different combinations of pure states and construct your density matrix. So this was sort of at the level of the density matrix, but similarly, and this will come out of this discussion today, there is also an infinite number of ways to unravel the dynamics, the dynamics involved in the evolution of an open quantum system. So with that, I want to describe the quantum Monte Carlo wave function method. So in the quantum Monte Carlo wave function method, we perform, we don't need an experiment. We don't need sophisticated experiments to observe single quantum system. It's a theoretical method, but we perform thought or Gedanken experiments. So our Gedanken experiments is that we perform measurements. So we assume we have, when we want to describe an ensemble of atoms, we assume we have a single atom. We assume it emits photon, and then we assume that the photon is detected. So the moment the photon is detected, we know, for instance, that the atom is in the ground state. And so our quantum Monte Carlo simulation follows now the trajectory of the single quantum system by saying, I toss a coin, there is a certain probability for detecting the photon. If I detect the photon, the system I know is in the ground state. If I don't detect the photon, it's not in the ground state and such. And this probabilistic approach is just simulated in the quantum Monte Carlo sense. So what is the essence of the quantum Monte Carlo method is that we allow a small time step delta t. It's a real time simulation. And there is a small probability that the system has decayed in that time. So we may now toss a coin, use a random function generator, and ask, has spontaneous emission taken place? And then the two possibilities are yes. And we continue whatever the outcome, whatever the quantum state is. Or we say no. And then with the probability p, we say yes with the probability 1 minus p, we say no. And then we continue the simulation accordingly. So this will become much, much clearer when I write down the formalism for you. But there is one important aspect which I really want to discuss with you. And this is the essence of the method. Namely, how does a wave function change by a measurement? So this is the essential, the only conceptual part for the quantum Monte Carlo wave function. Afterwards, I just show you. Once we have cleared up this conceptual part, the rest is just a few equations. And we immediately show that the quantum Monte Carlo wave function method is equivalent to the optical block equation. But let's really feature and take some time for discussing the concept, namely, the change of a wave function by a quantum measurement. And maybe what this carries home to you is that the nature of an open system with a quantum system which interacts with the environment is actually that the environment permanently performs measurements. And you will actually see that the idea of performing measurements formalized in the quantum Monte Carlo sense gives us an equation for the density matrix which is identical to the optical block equation. So in other words, you can say I'm re-deriving for you master equation now, but from a very, very different perspective, from the perspective of an environment which constantly performs measurement. OK, so let's go through that. How, what happens in a measurement? Well, if our initial state is the ground state, starting very in trivial ways, well, there is no measurement because the particle will stay in the ground state forever. OK, now let's go to the next situation that our atom starts out in excited state. Now, in a time step delta t, we have two possibilities. A photon can be observed or no photon is observed. When the photon is observed, we know for sure that the system is now projected into the ground state. If no photon is observed, well, the system is still in the excited state. OK, so we've discussed what happens when the system is initially in the ground state, when the system is initially in the excited state. But now comes the following question. We prepare the system in a superposition of ground and excited state. And we want to discuss what happens when the photon is observed, what happens when no photon is observed. I think if you observe the photon, it's pretty clear. We have to assume after photon emission, the system is in the ground state. But if no photon is observed, I want to give you three choices and you should address them with a clicker. One choice is that the particle stays in the same state. The second choice is that the particle stays in a superposition state, but the probability to be in the excited state decays, also no photon has been emitted or no photon has been observed. And the third possibility is something else. So in other words, assume you are an observer and you know with 100% detection probability. Let's not discuss technical issues. You have completely surrounded your system with perfect detectors. And you know in the time step delta t, in the time t, no photon was observed. And now you should make a prediction what happens next to the quantum system. And you make the prediction by assuming that the system is described by a wave functional density matrix. And I'm asking you which of the three choices correctly describes what the system will do afterwards, what the system will do next. Is the question very clear? Good. OK. All right. Pretty good. A lot of people, when I taught the class before, said it's a. The wave function is not changing because no photon has been emitted. Now, for those people, but there are only a few among those, I would have now asked the next question. If you have a 50-50 superposition of ground and excited state. And let's say the time for spontaneous emission is 10 nanoseconds. But now you wait one second, and after one second, no photon has been observed. What is your prediction? What in what state is the system at this point? Is the system in the ground state? Or is the system still in a superposition of ground and excited state? Well, since we have the clicker, you can tell me. Is it still a superposition state, or is the system in the ground state? OK. So now you have to help me out, because the people who said B, if they would go back to here, and most people said it is B, if you put in gamma 10 nanoseconds, the inverse of 10 nanoseconds, and the time is one second, the excited state amplitude, the excited state at mixture, is 10 to the minus nothing. So OK, if you want to be a mathematician and say there is epsilon 10 to the minus 10, yes, then you can say there is still a small mixture to the excited state. But by making it so extreme, 10 nanoseconds versus a second, at least in all practical terms, the system is in the ground state. And I think that should teach you what it means if you have a system which is in a superposition of ground and excited state. And after one second, it has not emitted a photon. I mean, then you would say it had its chance. It had plenty of chances to emit, but it decided it doesn't want to emit. And that would mean your 50-50 superposition of ground and excited state means the atom decided that it's in the ground state. You should also realize a 50% superposition of ground and excited state means in 50% of the cases, there will never be a photon emitted. So in other words, what you should realize is if you have a 50-50% superposition of ground and excited state, that the quantum evolution of this system is after a long time, the system is definitely positively, absolutely in the ground state. It's always in the ground state after a long time. In 50% of the cases, with emission of a photon, in 50% of the cases, without emission of a photon. If you have an ensemble of radioactive nuclei, and let's say half of the atoms are in the excited state and can radiate with decay and half are in the ground state, and you send the sample to your friend, and your friend waits many, many, many half times, common sense tells you that all the particles are now in the ground state. Also only half of them have decayed because the other half were in the ground state to begin with. I mean, this is what quantum physics tells us. This is what a 50-50 superposition is. But quantitatively, this is, of course, included in this result, which magically almost all of you got right. Namely, there is a superposition of ground and excited state, but the excited state amplitude decays. And to the limit, in the limit that the time is much longer than the decay time, no matter what the coefficient initially was in front of the excited state, this coefficient has decay to 0. Any question about that? Let me just emphasize that, again, and giving you also an example in the reference, a zero measurement modifies the wave function. You can also put it like this. What the wave function is, is the best knowledge you have about the system. The wave function is a way to predict what happens next. And it's the most precise way to predict it. More accurate predictions than the wave function allows cannot be made because of Heisenberg's uncertainty relation. But what happens is, if you have a 50-50% superposition of ground and excited state, and the system has not emitted, then you would say, initially, my guess was half of the atoms are excited, half of the atoms are in the ground state. But if none of them has decayed for a while, you would say, I revise my estimate now. Now I have to assume that more in the ground state. Because having not decayed means there is a higher probability that the system is in the ground state to begin with. So this is how this way of dealing with the wave function automatically adjusts for the knowledge you have gained about the system. Let me give you another example which carries home the same message. And I take it from a pedagogical paper from Dickey, from American Journal of Physics, 49, 926, 1981. And what he discussed is the following. You have a box, and originally, your atomic wave function is completely localized. You oppose the Einstein condensate if you want in this box. Or let's just assume one atom to be more precise. And it's completely delocalized. But now you focus a laser beam into the system. And the laser beam would ionize your atom, and you could count the ion with 100% probability. And for a short moment, you flash on the laser, and the result is, yes, you count. You get a count. In that moment, you would say, my wave function is more localized. I revise my estimate. Well, maybe I shouldn't say ionized. If it's ionized, you destroy the atom. Let's just say you observe fluorescence, so the atom is still alive. But now you have to say the atom is localized in that region. However, if you don't observe anything, you would say, well, now it's actually more probable that the atom is outside, because if it had been in the laser beam with a certain probability, I would have detected a photon. But the non-detection of a photon means that I revise my estimate, and I say it's much more probable that the atom is outside the laser beam than inside the laser beam. So this here is exactly how a non-observation of anything, a zero measurement, modifies your wave function. And the formalism to incorporate that is what we have discussed before. Yes, Cody? So your arguments here so far work the same way as at the wave function, we have just an entirely classical probability, and we're updating our understanding of probability. Do you have anything about the relative base between these two? I want to do that now. This is sort of just to address the basic concepts about what does a measurement mean, what does a zero measurement mean. But now I want to write down everything for you in amplitudes. We want to do exactly the time evolution of the system. Nancy? Is this like inconsistent with energy conservation, like especially the two before this? Because when we are saying that it is in the atmosphere of the ground energy excited state, then we do not observe the flow of the energy in the system. And now we're just going to do that. In a superposition state of ground energy excited state, we are not starting out in an energy eigenstate. So therefore, what we have is not a sharp value of the energy, but an expectation value. And if you repeat the measurement many, many times, the average energy is conserved. And this is exactly what energy conservation in quantum physics says. But if we do not repeat the measurement, then we should not just apply energy conservation. Well, if your particle, if your system is not in an eigenstate of the Hamiltonian, it doesn't have a sharp energy. And that would mean that when you measure the energy now, you have fluctuations. Sometimes you measure higher. Sometimes you measure lower. It's only when you're in an eigenstate that you measure a sharp value. And therefore, you know from the beginning that you will measure an energy diffusion. And sometimes you measure higher. And sometimes you measure lower than the average value. There's nothing wrong about it, but your question is a very good one. You should. I mean, those examples are really, on the one hand, they are trivial, but on the other hand, it's very profound what they tell us about quantum physics and how to apply conservation laws and such. OK, I think let me now formalize exactly how this is done with all the bells and whistles. We assume we have an initial wave function, which is now an arbitrary superposition state of ground and excited state. And we have the environment, which in this case we assume is the zero photon state. It's a vacuum. And we want to exactly solve the Schrodinger equation for that system. But we restrict ourselves to a very small time step. The small time step is smaller than anything else than the natural decay time. Maybe the inverse Rabi frequency if you drive the system. And we are not doing it here, but I will later show you how you can add a laser beam and drive the system. Or it should also be smaller than the inverse detuning. So what is very important here is that we want to deal only with simple cases, namely one photon has been emitted or no photon. So we want to make sure that there is at most one spontaneous emission event during the time delta t. So what I'm writing down now is the total wave function of the system plus the environment. And now we can evolve it. We can ask what is the wave function time later. This can be exactly done by time dependent perturbation theory. And the result is that the system will be in a super position of ground and excited state. And we want to calculate the coefficients alpha prime and beta prime. And we still have the vacuum state. But then we have the possibility that a photon has been emitted. In that case, we are in the ground state. And the direct product with the environment involves now a photon emitted into a certain direction with wave vector k with a polarization epsilon. And we have coefficients beta k epsilon. And we have to sum over all possibilities for the photon to be emitted. So we call this, so the wave function of the total system has now two parts. It's a wave function which I call psi 0. This is a wave function which involves no photon in the environment. And the wave function psi 1 involves one photon in the environment. So this is the ansatz which seems very natural. But you could immediately prove that these are the only possibilities how the system can evolve. And we can verify this by using time dependent perturbation theory. It's actually almost everybody of you has seen it in either in 8.421 or in a more basic course on quantum physics. It's time dependent perturbation theory for the emission of a photon. But usually when you see those treatments, the reservoir, the environment is not treated as explicitly as we do it here. The theory where this is treated exactly in this way how we need it is the Wegener Weisskopf theory, which is nothing else in a perturbative approach. But it's really a perturbation theory not just for the atomic system, how we sometimes present it in a simplified version. It's perturbation theory for the total wave function of the complete system. Timur? You mentioned the time interval, which was in the T, is much smaller than a lot of things. But here we're ignoring, I take it, the counter-rotating terms, in which case we're still at a large enough time step if that's average now. I simply say this because I expect it to be an E with a photon term from the G-dot-tum, so to speak. Yes, well, the counter-rotating term. The first answer I wanted to give you, know this is exact perturbation theory. But I think I get myself in trouble if I would get myself in trouble if I would allow the time step to be extremely short, because then I'm in shorter than 1 over omega. Because during the time 1 over omega, you can say we have counter-rotating terms, or in other words, a particle in the ground state during a time 1 over omega particle in the ground state can emit a photon and reabsorb it. These were those weird diagrams with virtual states, which we discussed earlier in this course. So I think I want to assume here, which I haven't done, that the time step is larger than omega to the minus 1. And as you know, the counter-rotating term has a detuning delta, which is 2 omega. So therefore, it is excluded. Or in other words, when we talk about photons sent into the environment, we want to talk about real photons and not virtual photons. Yes, good point. So if you do a simple lowest order time dependent perturbation theory in the Wegener-Weisdorff approach, you get an exact result for beta prime. You find that beta prime in this superposition is the original amplitude beta, but it has decayed with e to the minus gamma over 2 dt. And since we are only interested in small time steps, we can do a linear expansion of that. The probability dp that a photon has been emitted is the norm of this wave function psi 1. And this using Wegener-Weisdorff perturbation theory is gamma dt times the amplitude squared that the system was excited to begin with. And because of the conservation of the norm, the wave function, the norm of the wave function psi 0, which is the wave function without a photon being emitted, is 1 minus dp. So what I'm telling you here is that this is an exact result of time dependent perturbation theory. Often when textbooks treat spontaneous emission, they are more interested in getting Fermi's golden rule and this rate, but the same approach, if you just write it down, tells you how the amplitude in the excited state evolves. So therefore, what we learn from perturbation theory that alpha prime and beta prime, the coefficients of the wave function without emission of a photon, this evolution of the wave function occurs with a non-Hermitian Hamiltonian. And this non-Hermitian Hamiltonian is our Hamiltonian for the atomic system. But then we have to account for spontaneous decay, and this is done by this non-Hermitian part. So occasionally, you may have heard that people wave their hands and say, your system is described by a Hamiltonian which has an imaginary part for decay, and this is sort of phenomenological. This is definitely not the case, because even if you have a Hamiltonian like this, a pure state would simply decay and remain a pure state. What we are doing here explicitly is we deal with probability in a correct way. Every time the system can branch out into two different possibilities, two different dimensions of the density matrix, one has probability delta p, one has probability 1 minus delta p, and it is only the wave function associated with probability 1 minus delta p, which evolves with this Hamiltonian. So what I'm telling you is the exact solution for the evolution of the total system in terms of a density matrix for the atomic system. And this is exact. So that means we can now write down the just get some extra space. Oops, how does it do? Sorry, I have to insert a new page. At least. OK. OK, so the procedure how we implement this exact solution of perturbation theory is the following. We have a time step delta t. We compute what is the probability that a photon will be emitted. Then we need a random number generator. So we need a number epsilon, which is a random number chosen in the interval 0 and 1. If this random number turns out to be smaller than delta p, then we continue our time evolution on the computer that psi is now in the ground state. And maybe there's a laser beam which excites it again and such, I will add a few bells and whistles later. Otherwise, psi, our wave function, is now the wave function, which is the time evolution with a non-Hermitian operator of the original wave function psi. And since we have a real wave function now with probability 1, we have to renormalize the wave function by this denominator. Then we execute the next time step. This means go to 2 and do the next time step. And then you have so to speak, if you do it many, many times, you get a time-direct trajectory of so to speak one experiment. And then you start again with your system in a wave function psi, and you accumulate a second experiment. And maybe you do it 10,000 times to get enough statistics. And then you sum up everything you want and calculate all the expectation values you want to know about your quantum system. So the claim is that this method called quantum Monte Carlo wave function method is fully equivalent to the optical Bloch equations. And I want to prove it to you by showing that if I take a density matrix, which is an ensemble average over many realizations of those quantum Monte Carlo trajectories, that then this density matrix follows the differential equation, which is the optical Bloch equation. And the proof is shown here. So what I told you is the density matrix after time delta t has now two matrix elements, one with probability delta p. And this is, photon has been emitted, system is in the ground state, with probability 1 minus delta p. We evolve the quantum state with a non-Hermitian Hamiltonian. And then all what is done in the next few steps, we assume that delta t is small. We do a Taylor expansion of the exponent. We neglect quadratic terms in delta t. And then we come to this line. And if we write this as rho of t plus delta t minus the original rho of t, we find that this follows a differential equation, which is exactly the optical Bloch equations. So therefore, what we implemented as a form of doing many, many quantum measurements in the environment is a procedure which is rigorously the same as the optical Bloch equations, which we derived using a master equation approach. OK. This method is very powerful for the following reasons. If you simulate a density matrix with n internal states, or an external or internal states, I should have said with n quantum states, you need n times n matrix elements which can become quite a memory hog for your computer. The wave function at any given moment has only n components. So therefore, there is a computational advantage in using a stochastic wave function approach over simulation of the density matrix. The second thing is that a lot of people, especially experimentalists, like this approach because it reflects directly what they do in the experiment. And I think it's pretty clear that with this approach, you can deal with a great variety of situations. Let me just mention two obvious extensions. One is polarization. If photon is detected, that means your random number produced a number epsilon, which was smaller than delta p. At that point, you can create a second random number which determines if the photon has been detected with sigma plus, sigma minus, or high polarization. Or you can discuss recoil. If the photon is detected, you throw on another random function generator which determines what the k vector of the photon is. So what direction h bar k the photon has taken. And this determines now what is the recoil that change in momentum of your atom. So you see kind of you can start with an atom at zero momentum. You can say it emits a photon. And then your computer always makes a choice based on the random number. And then you say, OK, my photon has now received a recoil kick at 45 degrees. Now your wave function of the photon is such and such. And if you add a laser to the Hamiltonian, the recoil may have double shifted the resonance, but everything is easily taken into account. So you see that this quantum Monte Carlo wave function method can easily be extended to describe external degrees of freedom, multiple laser fields, and all that. Any questions? OK, let me now just generalize the thought. But I think you know already everything about it. We talked about the master equation. We talked about the most general master equation in the Linn platform, where those operators L are the jump operators. And the prominent example for a jump operator was the sigma minus operator, which takes a particle from the excited state to the ground state. And this is a jump operator for spontaneous emission. But you may have many jump operators, maybe spontaneous emission of different photons with different polarization and such. Or the cavity may lose a photon, and then we have a jump operator for the cavity. We went through that. Well, that means now in an exact way that the non-Hemiltonian is the original Hamiltonian with an imaginary part, which comes because of those jump operators. And the general quantum Monte Carlo wave function procedure is that first you ask, has something happened? Has any jump happened? And this gives you the probability delta p. And if no jump has happened, you just do a time evolution with a non-Hemiltonian. But if a jump has happened, then the jump is now the wave function is projected by the jump operator. If you have a sigma minus operator, it takes a particle to the ground state. If you have several operators, a different jump operator may take your particle to another state. And you have then a branching ratio that you know first the jump has happened, and then the question is, which jump has happened? And you just play the probabilistic game. So you see that the quantum Monte Carlo wave function method can immediately be generalized. So let me come back to the special case of spontaneous emission. In that case, we have only one jump operator, which is sigma minus. And the normalization is the square root of gamma. And that actually means that our quantum trajectory, ground state, excited state, is very simple. If we start with a system 100% in the excited state, it's very trivial what happens. Nothing happens. But if then the jump occurs, the particle is in the ground state. Then nothing happens then until infinite time. You then have to start the next trajectory. You start in the excited state. And this time, by chance of the random number generator, the jump happens later. Another time, the jump happens earlier. And what you then have to do is you have to average over all those realizations what is the probability for the particle to be in the excited state. And this is now your estimator for the excited state diagonal matrix element of the density matrix. And if you've written your code correctly, you will find that you get a wonderful exponential decay exactly as you have expected. Questions? Now, yes. Let me now talk about defacing. What I've shown you so far is how we can unravel the density matrix in many microscopic realizations. And I've given you a specific example. We assume probability for spontaneous emission. And with that, we propagate our wave function. But now I come to sort of the nitty-gritty details or the dirty truth that what we have assumed is by no means unique. And I want to explain it to you first by reminding you of a very, very nice homework assignment you have solved. And this was about we have optical Bloch equations for the density matrix. And the solution of the optical Bloch equation is that we have population damping with a damping time t1. And the off diagonal matrix elements are damped with a time t2. Remember, if you have only spontaneous emission, t2 is 2 times t1. But we can have a lot of other processes which can lead to a much, much faster defacing time than a population, than a time of population changes. And in your homework, you have discussed three possibilities. One is, well, spontaneous emission is energy loss. But you have discussed three different possibilities for t2. One, three different possibilities, how the phasing loss of coherence and phase damping can happen. And in this homework assignment, you assumed either that there is a random phase, that the elastic collision which project onto the excited state. And then you got the most crazy and artificial model that every time, randomly, the phase of the excited state flips from plus to minus. So if I would implement that now with a Quento Monte Carlo method, it would have the following effect. And I hope you enjoy the graphical representation. You can really think about it that this is what happens, that this is what is inside the density matrix described by this process. So let's assume we have a system which would be in a superposition of crowned and excited state. And that would mean that in the lab frame, the dipole moment would just rotate at the resonance frequency. It rotates, and it is a rotating dipole moment which emits coherent light. But now you assume that you have a random phase. So if you add random phases to it, maybe because you have fluctuating magnetic fields, then the line becomes sort of, that doesn't look random, but you know what I mean, it becomes sort of jagged. The phase distribution is still sort of there, but there is a jitter on top of it. And as a result, the light emitted by this dipole is actually broad. Well, in your second model, you assumed that with a certain randomness, with a certain time constant, there's an elastic collision in the excited state which projects the system into the excited state. At this moment, there is no superposition state anymore, and the dipole moment is 0. So that's what you assume. Or when you have random phase flips, you assume that suddenly this kind of deterministic sine function suddenly jumps corresponding to a minus sign in the excited state. And depending where you are in the cycle, it creates jumps at random places. And the question, of course, is if all those three processes, and that's what you showed in your homework, lead to the same density matrix, which one is correct? Which one is really realized in an experiment on a system which is described by this kind of damping or by the optical Bloch equation? So you can say all or none. Well, you can assume what you want. It doesn't make a difference. And the reason is actually profound. The reason is profound in that sense that the way our goal was to have an open quantum system, which is the atomic system interacting with the reservoir. But the way how we phrase the question, we are only interested in what the atomic system does is, we didn't do extra measurements on the environment. The environment was just a dump for photons, a dump for energy, a dump for whatever we assumed in our defacing mechanism. So let me just take the example of putting photons into the environment. And this is described by a certain damping term. But there is an ambiguity. And this is shown here. If we evolve the system, our atomic system, the density matrix evolves. But the environment also evolves. And for instance, here, the unitary time evolution has taken the excitation from the atom. And we have emitted a photon into the environment. And you remember, we got the optical Bloch equation by doing the partial trace here and just focusing on the atomic part. But now wait a moment. If we trace out the environment, we can trace it out in a different basis set. Remember, I showed you the quantum Monte Carlo wave function. We emit a photon. But we can, for instance, detect the photon in our Gedanken experiment. We can detect it with linear polarization or with circular polarization. If you have a situation where you have an s state which decays to a p state, sigma plus takes you to m equals plus 1, sigma minus takes you to m equals minus 1. So therefore, if a photon is emitted in your quantum Monte Carlo procedure, you would say, now the atom is in the plus 1 state or now the atom is in the minus 1 state. But if you detect linearly polarized light by just putting a polarizer here, and this is a unitary transformation in front of your detector, you would no longer project the atomic system on plus 1 and minus 1. You would project it on m equals 0 or whatever is connected to the linear polarization of the measurement. So therefore, if you simply assume that something has been dumped in the environment and can be used for measurement, there are many ways, many unitary transformations, what you can do to the information, to the energy, to the photons which have been dumped into the environment. And each of them will lead to a very different trajectory in your quantum Monte Carlo system. So therefore, if you just dump the photons and not measure them, you have equal rights to assume that your quantum Monte Carlo wave function jumps to states which correspond to a circular basis or jump to states which correspond to linear basis. And there are many, many possibilities. But each of those possibilities is 100% correct on reveling of the density matrix. And if you just do it right, assume that the measurement is done in a certain basis and you're consistent with that, you will 100% correctly describe the time evolution of your atomic system. Any questions? Timo. Again, with the DG, I'm just curious. So we take it very short, but why don't we run into something like the quantum Tino effect, where if every evolution is initially quadratic, like the problem frequency, you take very short DT, start measuring the system. Shouldn't for a choice of sufficiently short DT, the simulation would give you strength results? That's a very deep question, Timo. And I want to think about it more, but I think I've excluded that by saying that the time step DT is larger than the correlation kind of the environment. So I'm doing some kind of Markov approximation with the environment, whereas the quadratic part of the behavior is a coherent time evolution. For very short time steps, the wave function evolves quadratically. But this part is a coherent evolution, and that's related to the fact that an atom can emit a photon. But in the very first moment, before the vacuum has transported away, it can take the photon back. And the result of that is that at very short times, the exponential decays doesn't start out exponentially. It starts out a little bit flatter. We get the exponential decay. We get firmness, golden rule. And we get optical Bloch equations only if your time step is larger than the correlation time of the reservoir. And the fact, you remember when we derived the master equation, we had to say we do a step which is sufficiently small for the dynamics of the atomic system, but sufficiently large that we are not getting into any memory effects of the reservoir. And we've done the same assumption here. Other questions? Thank you. I'm just confused. We are now always saying the environment measures the system in regular time intervals. That's a T which are large compared to the correlation time. It's smaller compared to the evolution of the atom. But now I wonder, isn't that exactly the Markov approximation we've made when we derived the master equation? And I was just wondering, if we just may need to derive it by making the master equation. Maybe we don't actually need the idea of constant imaginary. Maybe a physical interpretation of the Markov approximation? Oh, yes, exactly, Niky. I think you said it really very, very nicely. We derived a master equation just by assuming that the system takes a photon and past the correlation time, the photon is taken for good by the environment. And with the Markov approximation, we formalized that there is no memory effect. The photon is not stored like in a cavity. The photon has disappeared. And by saying that this time is very short, this was the Markov approximation in the derivation of the master equation. But that also means if the photon has disappeared, has separated from the atomic system, we are now free to make a measurement. And of course, it shouldn't matter whether we make the measurement or not. But what I was able to do today is I was making now the assumption, let's assume we make a measurement. We measure all the photons which have been emitted. And the measurement is probabilistic. And we fold this probabilistic evolution into our quantum system. And what we obtained was exactly the same time evolution for the density matrix as we got from the master equation. So in other words, this should tell you that interactions with an open quantum system where irreversibly, energy, photons, angular momentum, or whatever flows into the environment, once it has flown into the environment, you can measure it. And you get all the stochastics from the measurement. But even if you don't measure it, you get exactly the same stochastics as if you had measured it. It's all the same. Jenny. I want to think about it. It seems sort of weird to me that we're making these measurements or lack of measurements, these interactions at regular intervals. It doesn't really matter. All we have to make is you could actually make delta t a random variable. It wouldn't change anything. The only thing we have to make sure is we have to make the time interval short enough that we don't have maybe two photons emitted in that time. We just want to make sure that it's a simple probabilistic branching yes or no. And that's the only requirement here. So this is a formula that's a phase constant. You say now it's wave function involved, and then it collides. And this is where the measurement comes in. So now, say equivalently, it's the same wave function involved in the collection. When you make the measurement, you say that operator involves. So these are possible people. So they've all some kind of formalism like this, but using operator. Because here, even though the measurement is operator, it is still described as state. I haven't seen it, Mark, but I'm absolutely certain that you could describe the same physics in the Heisenberg picture where the time evolution is with operators. Because what we have here is we have a time step delta t where the system evolves as an isolated quantum system, and then we measure again. And so I think if you would use operator equation, you would have an evolution of the operator with the same Hamiltonian, which is non-Hermitian Hamiltonian. And then the measurement would do some form of projection. I have to think about it what the projection would be in terms of operators. But my gut feeling is you can always take the transformation where you put the time dependence into the operators and not into the wave function. On the other hand, I have some little bit misgivings about that because the Quendom-Onde Carlo wave function was really developed in order not to deal with matrices, not to deal with something such as dimension n times n if n is the number of components of the wave function. It was specifically formulated to have the simpler description with n coefficients for the wave function. But conceptually, I think it is evolution with a Hamiltonian measurement. Evolution with a Hamiltonian measurement. And I assume this could also be done with operators. Thanks. I don't know how important is the randomness of the epsilon that we are choosing because no computer is actually generating random numbers? Well, I think how random a random number has to be is really dealt with in computer science and mathematics. But to our results, we always get this exponential behavior of random numbers. I don't know the answer. I mean, here, conceptually, it should be a completely random number. And I think there are algorithms which produce pseudo-random numbers. But if the pseudo is close enough, if there is a recurrence time of your random number, if your random number is your series of random number, repeat itself after several billion random numbers, this gives you a small arrow bar. And if you only do a limited sampling and limited number of time trajectories, I don't think it matters. Of course, if you want to have ultimate precision, then also the precision of the random number may enter through the vector at some point. I'm going to talk a little bit about the three different equations of the equations of the photons. But if we know the photons in the equations, then we can decide which one is the precision. Correct. If we would know more about the environment, if we would say we measure the photons in a certain basis, or when energy is dumped in the environment by a defacing mechanism, we look at the environment. Then, of course, we would add extra information to it. And then certain quantum trajectories would reflect the extra knowledge we have. And of course, we would then describe everything in these spaces. On the other hand, what you should learn from this is once the information, once the photon has escaped, it doesn't matter for the time evolution of your atomic system in which basis you measure the photon. So the evolution of the atomic system itself, at least when you average, is unaffected by these basis. However, if you make a coincidence measurement that you say your atoms which has emitted is still flying through your vacuum chamber, and now you figure out the photon is emitted as circularly polarized light, then, of course, you know that this atom which is still available has emitted circularly polarized light. And then you have actually obtained extra information about your atom. In other words, the photon and the atom was entangled like a bell pair. And if you now do a measurement on one part of the bell pair, you know more about the other part. But this is just how quantum physics works. If you don't use this information for anything, then you could have as well not measured the polarization. And it would have no effect on the atomic density matrix. OK, that's it for today. We have class at the usual time on Wednesday.
single quantum systems. And you should appreciate system, which are different models for the start this lecture with a quote by Erwin