Master class with Andreas Heinrich
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Heinrich, Andreas
We all learn in quantum mechanics lectures how to treat the spin of an electron using the Pauli matrices of an S=1/2 system. However, the magnetic properties of atoms in gas and in particular those in a solid-state environment or in molecules are often much more complex and interesting. We will begin by trying to understand what happens when the spin of a quantum system is larger than S=1/2 at which point ligand fields (crystal fields in solids) become important and lead to important effects such as magnetic anisotropy. We will then move from the treatment of a single spin system to coupled spins. How do you set up spin matrices for such a situation and how do you find solutions to those problems? We will discuss some experimental findings about spin chains on surfaces as studied by STM. If time permits we will try to apply the concepts of coupled spin systems to quantum computation.
 Hi everybody. My name is Andreas and I'd like to tell you a little bit about some of our work on spins on surfaces as you probably know. I also have a talk tomorrow night where I'm going to talk about something quite different. So you won't hear the same thing again. So that's good I guess. And also I want to do something which I've never done before which is to try to teach you that these spin matrices, these poly matrices that you probably hated in quantum mechanics just as much as I hated them at the time, that they are actually pretty useful to calculate things that you can actually measure with a scanning tonic microscope. So that's basically the goal I have. And that might be a little bit boring sometimes because I actually want to use them. I want to show you that you can actually do real calculations with them. So if that's too boring, let me know. And so I don't really have a detailed agenda. I put something together and you have a copy of that. That's not exactly what I'll be using. So you'll see some differences. I'll be happy to provide the exact slides that I'm using if you guys want to have those for later. That's fine. Before we get started though, I would like to take maybe five minutes or so and have each one of you tell me in a few sentences what your background is. Just so I have a little bit of an idea what to expect. So maybe we start over here. Just your name and a couple sentences on what you're working on. Hi everybody, my name is Xenia and I'm from the University of Twenton, and I'm working on the chess transfer through organic molecules. Thank you. Hi, I'm Matthias from the same university and I'm actually trying to make quantum thoughts in silicon-driven nanowires. Hi, I'm Daya from the same university. I am working on electronic quantum inventories. I'm Pessa Pockemer from the University of Amsterdam and I'm working on non-dimensional chemical reactions. Hi, I'm Eningen. I'm from Radbaal University and I'm working on the interaction between oxygen molecules and surface molecules. I'm doing some calculations. Hi, I'm Marc Sudej. I'm from the University of Amsterdam and I'm doing work on quantum reaction in quantum gases. Hi, I'm Eningen Kapitza from the University of Birmingham and I'm working on spin transport in motion metallic also. And I work on coupling with the spin heat transport in gravity and electrical motion. I'm Marc Sudej. I'm also from the University of Birmingham and I work on spin transport in the field and I recently also started a little bit on magnetism. Spin transport is not magnetism? Well, yes. Well, yes, you're right. My name is Jorma, I'm from the Tractinian Earth Dragon and I work on optical gas in the field of ultra-pestuous spin sometimes. My name is Daphne Ruekom. I'm in PUNL. I work on my run-off females in the cookware boxes and today you are actually in my lab. I'm Iman, also from DL and I work on integrated quantum object torsors. I'm Dimak from also from DL and I work on spin qubits and getting more sorry for the subject. Hello, I'm Nick from Leijf, University of Birmingham and I work on spin transport in molecules and then I'm working on the molecular charge in spin transport in qubits and in qubits and then I'm working on the spin transport in the molecules. Hello, I'm Davide Grossi and I'm from the University of Birmingham and I'm working with Tesla. You don't have much to learn here anyhow. Great. So some of you will be really bored if you're working on spin qubits. You probably know more about those spin operators than I do. So you'll really be bored and the rest of you, you know, I see some of you will actually learn something today which is what we want to achieve. Okay, so let me get going. I was making a little bit of fun of the magnetism versus spin, right? But this actually is actually a very, very serious distinction I think between the two and the distinction really lies in whether you think of them as quantum mechanical or as classical objects. But classical here, I mean, and that's probably what you also meant is you use something like a Landau-Liffschild-Gilbert equation or just the direction of the magnetization as the magnetic moment and you can determine that magnetization direction to any precision, right? That's in contrast, of course, to the quantum mechanical description where you use spin operators and you have quantum computing and all these applications in mind. So maybe that's the distinction that you have. This is something that I'm actually very interested in is, you know, if you take a single atom, for example, and you put that on a surface like I'll be using in this discussion today, then we mostly use quantum mechanical concepts. I won't be telling you anything about phase coherent quantum mechanics such as you would need for quantum computation, but we're not going to be trying to get there, but we don't have that with our systems. So it's all going to be the energy eigenstates, essentially, of such systems. On the other hand, if you want to do data storage, then you don't really want to do quantum mechanical systems. You want to have something like a compass needle that's points north or south. That's your magnetic state that you want to use for data storage. And so quantum mechanics is a bad thing for those kind of spin systems. So I'm just in question how you go between, sort of, the million atoms that you have in these things here to the few atoms where you use quantum mechanics and you can ask yourself how that transition region works, which is really what I'm mostly working on. Now, since this is a lecture and I don't really want to talk the whole time, I really need you guys to break this up, right, and ask questions and give me your thoughts on things. So what do we want to learn from such a quantum system? In the end, we want to do basically quantum computation with those. So in the really far future, I'd like to put atoms on surfaces, and I'd like to use a scanning and tunneling microscope as the tool to interact with those guys and do full phase coherent quantum mechanics like some of you are doing if you do qubits in semiconductors, for example. Now, we don't have this yet. For this, you need to know a lot of things and a lot of detail about the interaction between this atom here or the spin, and, for example, the environment that is in there. Now, what we will talk about today to quite some extent is the energy landscape that these spin states here will have. So here, I'm sketching a spin that is larger than the spin one-half, which you probably are using for qubits. So here, it's a spin two, which is actually an iron atom in our case. And this iron atom has five spin states. Those spin states are not degenerate at zero magnetic field, and we will talk about this in quite some detail. I want to show you how we can calculate the so-called zero-fuel splitting from some knowledge about the spin operators. That's one of the goals that I have for this talk, and also how we can then learn something about the dependence of these energy states as a function of, say, an external magnetic field. If we have time, then we can also talk about how we can measure relaxation time. So if you take one of those states and you put it into an excited state, then you might also want to know how long does this spin actually live in this excited state, how fast does it relax back down. That's a T1 relaxation time for those of you working in quantum computing, so not T2, not phase coherent, just the energy relaxation time. And those we can measure with the time resolution down to approximately one nanosecond. If we have time, we can talk about this again. And basically, the technique that I'll be using is spin excitation spectroscopy with STM. So using scanning tunneling microscopy to be sensitive to the magnetic states of those single atoms. Okay, now I promised for some of you who do quantum computing, this is going to be trivial, so you have to be quiet, everybody else. This is sort of the introduction to what is the difference between classical spins, classical magnetization description, and quantum mechanics. So you've probably all learned that electrons have a spin, and you can measure the spin with a Stern-Gerlach experiment. In the Stern-Gerlach experiment, you take a beam of atoms that have a simple magnetic state, just a single electron. Essentially, you can think of this as a single electron flying through space if you want. And you put the single electron in this apparatus, which is basically measuring its magnetization in the z direction, in one particular spatial direction. I call that zero here. And because this is a quantum mechanical thing and it has two states, we see two deflections, the beam deflects into two states with equal opportunity. So you see a signal here, which we call SC plus, and a signal down here, which is SC minus. In practice, this is done with an inhomogeneous magnetic field that deflects the spins depending on the magnetization of their spin and gives rise to this. But don't worry about the details, because we're actually not going to do this, we're just going to do this conceptually here. Okay, so that's not that difficult, right? What you learn from this is that there is such a thing called the spin, and that the spin has two levels that it can obtain. Now, this becomes more interesting if you then use a similar apparatus that you've now rotated by 90 degrees. So now you're measuring the magnetization of this beam of atoms in the z direction. Here I'm blocking out one half of those atoms, so I know that all of these guys are in the SC plus state here. And now I'm sending them through the Stangale experiment here in the x direction, 90 degrees rotated, and I get two beams coming out. I can repeat this now. I block the x direction here, so block out the Sx minus. And so now I have SC plus and Sx plus, and so the question then is what comes out of this apparatus? I've already measured that it is SC plus, I've already measured that it's Sx plus, so if I do the z measurement again, what do I get out? Are you working on quantum computing? What do you get out? Half and half, right? And you probably learned why this is, but I would like to take a moment and maybe take some time with this and show you why this actually works. So let me do this a little bit slower. You have the slide actually in your hand out there. So really what the, I mean the answer for why this is one half and one half is of course because the Stangale experiment in the x direction has destroyed the information from the Stangale experiment in the z direction. These are non-commuting operators we're saying quantum mechanics. But let's look at this a little bit more in detail. So I'll be very sloppy today. The Sx operator is defined as h bar over 2, 0, 1, 1, 0. And we're just going to make this simpler by using just the Pauli matrices here. So I'm leaving out this one half here and I'm also not worried about h bar. Okay, this is just, just don't use this for detailed computations at home or something like that. Now, also what you need to know, the two states that we have, the upstate here is defined as 1, 0 and the downstate is defined as 0, 1. And so you can just, for example, you can apply sigma x, the upstate, right, the 0, 1, 1, 0, times 1, 0. You do this, you get a 1, you get a 0, right? Which just means it's an eigenstate. That can't be right. So see, too fast. Not an eigenstate. So what comes out? 0, 1, right? See, you're paying attention. That's good. So basically applying the sigma x operator on the upstate gives us the downstate, right? But what we're doing in this experiment here is we're then blocking out the Sx minus component here, right? We're blocking out the Sx minus component with basically preparing this guy in the Sx eigenstate, which is, I haven't written this down here, the Sx eigenstate is something like, I don't know, what's the proper terminology here? The sigma x upstate or something like that is something like 1, 1, right? And now we're asking in this last experiment what is the probability of finding this guy in the upstate and the downstate. And this is just of the 1 over square root 2 here. This is 1 over square root 2 times 1, 0, plus 1 over square root 2 times 0, 1. And this is the 1 half that you get, the equal probabilities here, right? Now, then we can also do something maybe a little bit more interesting here, which is, that's what I just told you here. So you can ask yourself what would happen in this experiment here? Basically I've already given you the answer here, I think, right? So the only difference is we're not blocking out the Sx minus here. And so the question is what comes out here, is it still 50%? Right? So we're preparing the system in the SC plus state. We're operating on this with the Stegale experiment in the x direction, but we're not blocking out one of the two beams. And then we're asking the question, what is your state in the z direction? 50, 50? What about the quantum computing does? I don't think it's 50, 50, because the 50, 50 came about because we prepared here an Sx plus state by blocking out the Sx minus state. What we're doing now is we're doing something that is more akin to actual quantum computation. We're operating on a well-prepared state here with this operator, which gives us exactly this here. And then we're asking this state, what is your decomposition in SC plus and SC minus, which is 100% SC minus? Right? So it's not magic. That's really in a nutshell what this whole talk is about. This is probably the only thing that you're really going to learn, that you can use these matrices if you apply them halfway reasonably to learn something about these kind of systems. So we can take our time here. You see the difference between these two experiments? It's just the fact whether you're blocking out this Sx minus here. So this experiment, you don't have to do this at all. This doesn't give you anything, doesn't make any sense in the end. You could just leave this out. That would be exactly the same outcome. This just means that you're preparing your system in the Sx plus state. And then you're asking if you are in the Sx plus state, how much probability do you have to be in the SC plus or in the SC minus state? Which is what this guy asks. In the other hand, if I do this experiment here, then this one really matters. I'm preparing the system in the SC plus state, and now I'm doing something that you would call a quantum computation if you are in this field. You're operating on this state with this operator here, and you're keeping the full result of it and then asking what is the outcome of this. Yeah. This actually means the exact same thing as having no box at all. As having no box that does this experiment in the X direction at all. It does because the outcome of this is the spin down, the flip the spin. But like if you made up a different position of Sx plus and Sx minus, and you made them separately, you would get the result that would be exactly the same as the box. Say that again. If I made an experiment with Sx plus and Sx minus of equal weight, so like in SC, if this guy here, if you express this guy as a superposition of Sx plus and Sx minus, that's going to have equal weight of those two. Right. Well, okay. No, no, no. This is what this is all about. I mean, there's no need to not answer your question. I mean, I get the point that I'm just trying to make a remark that it's very important to understand what the box actually means. Okay. What I mean by those boxes is in the end some way to make these matrices, bring those matrices to life. Right. So this is sort of, I mean, I didn't make these up, right? This is standard textbooks kind of, some of them. Sakurai is my favorite one that actually starts the quantum mechanics with this experiment to show you that classically with the first one. That classically, of course, the outcome is if this is a classical spin, not a quantum mechanical spin, right, you can measure the z component and you can measure the x component. That's fine. Classically, you can determine the magnetization to any precision in all three spatial directions at the same time. That's what we mean by a classical spin. Basically, right? And so the outcome here is that you get SC plus classically, right? And I've used as an example of why quantum mechanics is different from classical mechanics. Okay. Now, I really like your question. I want to make sure that I don't come across as not liking your question here. Okay. So now I want to take maybe 10, 15 minutes to introduce the basic measurement technique that I'm using, which is an STM scanning tunneling microscopy technique based on inelastic electron tunneling spectroscopy, which tells us something about the energy separations between the ground state and the excited states of these spin systems. This is going to be a little bit more like a regular talk. But again, I'd like you to interrupt me and ask questions as we go along. So the experiment we're doing, some of you are doing transport in graphene or other plain out junctions. The sense is related. We run electric current through a tunnel junction where we have a single magnetic atom in the beginning or a chain of atoms on a surface. And we're taking the current from, in the first case, a non-magnetic tip to a non-magnetic sample, but running it through our magnetic atom in the middle. One of you is working on molecular structures, what? One of you two, right? So with the break junction, basically? In a network, you have a sort of gold nanobarber for network, and then I can insert speakers over molecules in the field. And how do you measure the states? In this network, you can stand them on another trans device and then into a complex, you can measure the resistance. Electrical transport going through several molecules at once. This is, in some sense, the single molecule equivalent if you want, right? If you had one of these molecules in between, break junction experiments, for example, can do this to some degree. Okay, so in the simplest case, if we just have a spin one-half, like these guys over here, you apply a magnetic field to a spin one-half. At zero magnetic field, it's degenerate. At finite magnetic field, you have an energy separation. That's the Zeeman energy. Zeeman energy is proportional to the external magnetic field, right? And the property that you can learn here is the so-called G-value, which tells you something about orbital angular momentum and spin angular momentum of the structure that you have here. Now, I'm going to be using only one material here, which is a thin layer of copper nitride, which is prepared on copper. And this is a scanning tunnel in a microscopy image, approximately 100 nanometer by 100 nanometer large or small. And what you see here is in green, the bare surface of copper, the particular spatial direction of copper, 1-0-0 direction. And you can see that that's basically atomically flat and beautiful. Then we prepare on these patches of blue here, which are a single atomic layer of an insulator, copper nitride. That's an area of approximately 10 nanometers by 10 nanometers here, and that's one atomic layer, thick or thin. And then we put magnetic atoms on there, and here says Manganese copper and iron. Yes? So, you can see this picture also telling us something about the height of the copper nitride, so you can collect it in the copper. Right. It does. So, the operating mode here is to keep a constant tunnel current. And so, I'm showing you here the height of the tip as a function of lateral position under the assumption or under the condition that I want the constant current of, say, 1 nanom. And so, if you have an insulator there, then the tip has to be brought in closer over the insulator to keep that 1 nanolamp of current requirement. Yes, and the copper nitride is actually laying on the copper. The copper nitride here physically, actually, this is more likely or more like this. So, you have your copper here, and then the nitrogen is almost sitting inside that copper. So, you can think of, and if there's an end to the copper nitride, right, so there's no nitrogen, and here's nitrogen, see it? And that's basically a planar structure to a certain, the first approximation. So, it's not sitting on top, it's sitting inside the top layer, and it appears lower here, right? It appears like it's a step down because of this requirement to keep a constant current. This we only know from calculations, density functional theory calculations to be precise, but the measurement of the height, we can do that directly here. So, it's about 30 times less current if you keep the same height over the copper nitride than on the bare metal, isn't it? Okay, and so then we can take our STM tip and we can park it on one of these atoms and we can run our current through that. As a function now of the voltage that we apply between the tip and the sample, and this is the conductivity, the IDV here, as a function of this voltage, yes? How do you make this copper nitride? Is this a matter of adding nitrogen? Yeah, so what we have is we have a low temperature scanning tunneling microscope, and attached to that we have room temperature, essentially a simple MBE chamber, metal beam, epitaxy chamber. In the case of nitrogen, so first step is to prepare clean copper. So there are many recipes that other people developed a long time ago, use sputtering to blast off some of the dirt, and then use heating to bring out, you know, make it a clean, flat copper surface. So that's relatively easy to do. And then the nitrogen we have to crack the nitrogen bond. But if you just let gas into a vacuum chamber, you will not make copper nitride because the nitrogen bond is very strong chemically. So you need to break that. We do it very simply by using that same sputter gun that accelerates the nitrogen to something like a kilovolt. That's definitely enough energy, right? That kilovolt of nitrogen then hits the surface and gets implanted somewhere in that material. And if we then heat that up to proper temperatures like 300 degrees C, then this nitrogen comes out and self limits to only a single layer of this copper nitride. So we're using recipes developed by other people mostly to do this. But we can do other materials as well, like MGO, magnesium oxide is one that we're working on right now. So, yeah. So do these nitrogen atoms just arrange themselves in these 10 by 10 projects? They do. So you can see that this is really a fairly regular kind of looking structure, right? They really want to be a certain size. And they also keep, if you see here, they keep only one layer of missing nitrogen here. There's one layer of missing nitrogen. There's one. There's one. So you can get a fairly regular pattern, nano pattern if you want to call it that. And so some people are interested in just using this as a template to grow something else in a regular pattern, for example. For us, that's not that important because we can always find an area where we have copper nitride or not. But this is limited by the strain. The copper nitride has a slightly different lattice constant than the copper. And so it wants to relax that strain. And so if it has space, it will keep some copper in between. If it doesn't have the space, it will leave just one layer, one roll of missing copper. And if you even force more nitrogen in there, you can make a continuous film. But that will also have some defects in it to get some strain relief. So lots of things that we can learn about the growth of this. Yeah. These guys? Those are single atoms that we put there. So every one of these mountains here is a single atom. And in this case, these are manganese copper and iron atoms. Defects are these guys here. So there's some defects in there. But again, it's the beauty of a local probe, right? If you have a defect nearby, just don't look at that guy. Use a different one. So in some sense, that's why I like this technique, because it allows us to make measurements that are extremely well characterized. I know exactly that this is, for example, a manganese atom on the surface. So I can tell DFT people, this is the structure that I have. Calculate for me what does this look like? What does the atomic arrangement look like that I can't see here in detail? So you can do these kind of experiments or these kind of comparisons relatively easily. I mean, if you guys have a bit more technical questions about how this technique works, we can do that too. So I have no agenda in which direction you guys want to take this. OK. Now, so the basic measurement then is we have our SDM tip parked, in this case, on a manganese atom on copper nitride, on copper. And we measure the conductance, conductivity. Yes? Tell you about the no-prosthumance of the manganese atom or iron or whatever. So here you actually see that that's not so easy to tell, right? I mean, if you look at these four guys, they basically look the same. So without prior knowledge, they might have a very slight difference in height, meaning how much current flows through there. But if you don't know a priori who is higher and not, you will not be able to tell. So what we do is, we, of course, before we do something that is complicated like this, we only use one metal atom. So let's say manganese. And we look at the surface. And we look at the spectrum, which I'll show you in the next slide, for all the different bumps, like 100 different ones. And of those, maybe 95 will look the same. And five of them are close enough to defects. They will look different. So we just say, all right. Most likely, this spectrum belongs to a manganese atom. And then you repeat the same thing with another species, and then you can definitely tell them apart. But this is actually not a trivial question at all. You have to do a lot of work to do that. Okay. So back to our manganese atom. We run current through this, and we measure the conductance that flows through this atom here as a function of the voltage that we apply. This is now the basic measurement of inelastic tunneling spectroscopy that I want to explain to you. What you see here is the zero is no voltage applied and corresponds basically to the Fermi energies of tip and sample being at the same energy level, same energy. So that's this point here. We have a constant conductance around the Fermi energy. And then we get a stepwise increase in the conductance to a higher current or higher conductance at a voltage of approximately 1 millivolt here. Yeah. So first question, why is it not symmetric? I would say it's pretty symmetric. Okay. So this is not symmetric. Well, we can talk about that. Yes. And then the second question is... It's a difficult question, though. Okay. Is there any... Okay. The magnetic field perpendicular, is there any particular reason for having that? In this case here, not very much. I'll show you an example where it matters very dramatically which direction you apply the magnetic field. Okay. But to give you a quick answer to your first question is, the difference is that there seems to be a peak here. Right? If you ignore the stepwise increase, if you basically subtract this step from this negative side, you see that there is a peak here, right? There's also this slope here and there is this... So actually there's a peak at the same energy. And this is, to my knowledge, not fully understood yet why there are sometimes peaks and sometimes only steps. It has something to do with higher-order correlation, something like a condo effect happening, if you know what the condo effect is. So higher correlations often... So this is a single... What I'm going to be telling you in the next slide, how this actually works, is a single electron picture and you need more than these single electrons to describe this condo effect. But I cannot tell you why it happens on this side and not on this side. So that's actually a pretty non-trivial question. Not that the other questions are trivial, but that only means that I don't know the answer, really. Okay. Now, how do we know that this is a magnetic excitation? Well, that's also a non-trivial question, right? And we really do this by changing this external magnetic field and you can see that this energy or this voltage where the step occurs goes to lower energies, lower voltages, as we decrease the magnetic field. Now, you see here we have a low magnetic field scanning tunneling microscope. So if you had 33 Tesla here, you would make your step somewhere way out there. Actually, this is well-matched, though, to the measurement temperature of 0.5 degrees Kelvin. So we need the combination of relatively high field and low temperature to make this visible. If you had 33 Tesla, you could do this at 4 Kelvin and you would get similar information. Yes? So the in-voltage is this amplitude of the step corresponding to the electron temperature, actually? The amplitude? The height? No, no, no, no, sorry, not the amplitude. The in-voltage on the next axis. Yes, yes, I'll show you this in the next slide. So the basic information is what is the location of this step as a function of the external magnetic field? So this is what I show you here, the magnetic field, and here is the splitting, basically from 0 to this energy here. And so what you see from this is that's a linear increase corresponding to the Zeeman energy between the two spin states, the ground state and the excited spin state with a slope that you can learn something about this G-value. It's 1.90 in this particular case for manganese on this copper binding site on copper nitride, so that's very specific to the atom and to the local binding site as well. But it does tell you that it's less than 2.00, which is what you would expect for a free electron. Right? That's the Zeeman, that's not the Zeeman, that's the G-value for a free electron, basically. So it's significantly less than that. And that tells you something also in a way that I don't fully understand about orbital angular momentum and spin angular momentum. If you go into atomic physics, if you look up what is the G-value of a certain state in an atom, then this G-value will tell you something about orbital moment and spin moment, in a not very trivial way, basically. Yeah? This basically is lowered because of the environment here. It is lowered because it actually sits on this copper binding site here. It sits like this. I don't know if I have a DFT calculation in here. So it replaces one of the copper atoms that used to be there in this copper nitride. And so it has these nitrogen atoms nearby, and these nitrations are negatively charged. And so it sets up something that we call a crystal field, local electric fields. So this environment here matters very dramatically for these effects. It's not an atom in free space, basically. Okay. But I still haven't explained to you, really, why we got a step and what we can learn about the width of the step. Yes? I really like the fact that you're asking a lot of questions, by the way. Yes? What about the offset? I mean, from your equation, it should go to zero. Yes. So it should go to zero. If you get a splitting at zero magnetic field, you call that zero field splitting. That's something that's well known from electron spin resonance, which is basically the other technique that can measure something like a G-value, right? This is a resonance experiment, usually in gigahertz range microwave fields. And it's, you know, well-established technique over 30, 40 years or something like that. And it can measure G-values much more precisely than we can here, but it needs, like, 10 to the 10 spins to get enough signal to do this, right? And so they coined the term zero field splitting in their measurements. And I will explain to you in quite some detail how this comes about for iron on the same binding site. It's due to the direction dependence of the spin. So if the spin is pointing in the horizontal direction, like here, why did I call this a copper? I was going to call this a manganese. Sorry. This was supposed to be a manganese at me, okay? So if this manganese spin is pointing in the perpendicular direction, it has a certain energy. If it's pointing in an in-plane direction, it might have a different energy. And that's set up by the local symmetry of your binding site here. And that difference in energy, we call that magnetic anisotropy, that is something that gives rise to these finite energy offsets. And I want to take quite a bit of time, actually, to explain how this comes about using something like the spin operators in a few slides, basically. Okay. Sometimes what I do in talks is I say, everybody always says I want questions, right? And then nobody asks questions. So what I do quite frequently is I stop talking. I say, I'm only going to talk for five minutes, and then you're going to have to ask questions. Unfortunately, I don't have to do that with you guys, so keep up the questions. Okay. So that offset is due to magnetic anisotropy. Now, before we go any further, I want to explain to you how this spectroscopy actually works. And so here I made a slide where I have four different cartoons. They are all cartoons, basically, not reality, right? All cartoons by trying to motivate why we see the symmetric stepwise increase in the conductance here. And so here I have the field states on the tip, the field states on the sample. Sample in my terminology, I mean the copper. The copper underneath the spin, basically. And here I show you the energy of the electrons that we have here in relation to the energy splitting between the ground state and the first excited state. So let me just start this. So if we increase the voltage between tip and sample, that shifts the Fermi energies. That's what we mean by applying a voltage between these two electrodes. And nothing interesting happens. We got a constant conductance in DIDV as we increase the voltage until this situation is reached where the maximum energy available is equal to the energy difference between the ground state and this first excited state. All right? When this case is reached, then this electron up here sometimes will give off energy during the tunneling process and come in as a colder electron on the other side. And somebody has to be available to take that energy, because obviously we have conservation of energy. And this is actually possible by this local spin here, the manganese atom, switching from the ground state to an excited state. Yes? How do you know that this is a magnet, with the spin, why couldn't you go to the nitrogen atom, or is it copper atom in this case? In the end, that's a very good question. I mean, of course, we do this experiment also on the bare copper nitride. We don't see anything. We do it with, say, a copper atom instead of in place of the manganese atom. The copper atom happens to be non-magnetic in this case, which is also not a trivial statement. We don't see anything. So we do these kind of control experiments. And then in the end, I will not, this is again a simplified picture, but I can't really tell you this. All I can tell you is that this cluster, whatever it is, that might be the manganese together with the nitricians here, that that together behaves like the spin. So it's not atomic physics at all. It's really this guy together with its local environment that is doing that. Let's see. Okay, so when we have enough energy, we can make a change. This guy can flip. And now what you see is that we have two ways to make it across. We can tunnel elastically and we can tunnel inelastically. So without any proof, you would say that it's much, not much, it's easier for this electron to tunnel than if it cannot make this excitation. It changes the conductance, is all I'm trying to say. And it increases the conductance usually, not always. And then if I further increase the voltage, all the additional electrons can make this excitation and this guy gets kicked more and more often from the ground state to the first excited state. So it's not a peak in the conductance at the voltage corresponding to the Zeeman energy, but it's a stepwise increase. In your picture it's coming back because it's relaxing. Right. So this only works if there is a sufficiently fast relaxation mechanism from the excited state back to the ground state. So I'm basically, this only works if this relaxation mechanism is faster than the average time between tunneling electrons. We usually do this experiment at tunnel currents on the order of one picoamp to one nanoamp. One nanoamp corresponds to 10 to the 10 electrons per second, which is one electron, what is that, every 100 picoseconds. So you need to have relaxation that is on that order to make this possible. I don't have this in this deck of slides, but we can actually get to a regime where the relaxation is not fast enough anymore, and then this breaks down and becomes a bit more interesting in some sense. Okay. So it's symmetric because you can reverse this process. You can apply the voltage the other direction. The same thing should happen to first approximation. And so you should see a stepwise increase on the other side. This, by the way, is the hallmark of inelastic excitations. That is really the difference between zero and a finite voltage or energy that matters, and it's not the absolute value of this energy. That's one way you can distinguish between so-called elastic excitations and inelastic or elastic tunneling processes, changes in the density of states, for example. This is not too dissimilar to superconductor tunneling spectroscopy. Pretty similar in that respect. Okay. Now, your question was earlier, what's the width of this guy here? The width is set by the width of the Fermi energy in the sample and the width of the Fermi energy on the tip. At zero, this would be infinitely sharp. At finite, we actually get a broadening that corresponds to 5.5 times kT because we have two Fermi functions that get broadened here. And that also explains to you why this experiment is experimentally difficult because most people have magnetic fields only on the range of a few Tesla. Very few people have 33 Tesla available. And so you need to make sure that this energy here is bigger than your thermal broadening. So we do this at 0.5 Kelvin and magnetic fields up to 7 Tesla. When I started at IBM, my first task was to build one of these machines. There was no commercial machines available at the time. It took about 10 years or so for commercial machines to become available. Now, they are available if you have a million dollars or something like that to spend. Okay. So what can we do with this? We can do tunneling magneto-resistance. I'll tell you what this is in a moment. We can measure magnetic anisotropy, which is the zero-fuel splitting. I want to spend a lot of time on that. We can measure spin-spin coupling, put two spins next to each other, and learn something about how the combined system works. I want to spend a lot of time on this. And if you guys keep up the questions, which I hope, then that's going to be the end of what we have time for. We can learn something about strong interactions between the spin and the substrate, giving rise to these peaks, which gives you something about the condo effect. I definitely will not talk about this. We can drive the system out of thermal equilibrium, which is basically if the relaxation time is not fast enough. We can learn something about the relaxation time. And we can also measure this relaxation time directly. I do want to show you this at the very end a little bit, because it's kind of a fun technique. Now, so we'll spend most of our time on these first three issues here. Okay. So, Tanneling-Magneto resistance. So, the Nobel Prize in Physics was given to two gentlemen, Albert Ferd and Peter Grunberg. I don't know what's his first name. Peter Grunberg. A few years ago, maybe four years ago or something like that, five years ago? No way. Old people always tell you, you know, it's amazing how time flies, you know. It's scary. So, it was given for the development of transport in solids that is dependent on the magnetization, on the spin direction of magnetic materials, so-called magneto-resistance. Now, this was done at first by growing magnetic layers, thin films, and seeing that if you change the relative orientation of these films, you can get a change in the conductance. This is actually what's used. I still have a computer that has a hard drive here. That's used as the read mechanism in every hard disk drive read head. Actually, we use Tanneling-Magneto resistance. So, we put a insulating barrier between these magnetic materials, and that's how you read the state of your magnetic bit in your hard disk drive. Now, we want to do something similar here on the atomic scale, and let me show you a few experiments how we can do this. This also answers a little bit your question how we know who is who in this case. So, now we have a smaller image here, 6.5 nanometers of copper nitride. That's the blue area, bare copper. That's the green here, and four atoms on the surface, which I've already labeled here as copper, iron, and two manganese atoms. So, we can take this tip, we can park it on the manganese, we do the spectrum that I just explained to you. We get a stepwise increase at finite magnetic field, again 7 tesla here, at approximately 1 millivolt. Now, what we actually learned by doing a lot of experiments actually, is that we can bring this tip into point contact, bring it in very close, and applying a fairly high voltage at the same time. And if we do this right, we can transfer this atom from the copper nitride to the tip. This is actually very simple in this cartoon, but in reality, that's a very difficult experiment to do. Because when we do this, if I go back, when we do this, at this point here, we apply two volts, and approximately 10 microamps of current running through a single atom. So, it's not surprising that you change something. I mean, if you do that, two volts and 10 microamps through a single atom, you should change things. The only thing that's surprising about this is that sometimes, you can actually make this a reproducible process. Now, how do we know this? Again, not a very, very non-trivial question. One thing we know is we can take this tip now, and we can take an image of the same area. And of course, one thing you notice is that this atom is missing. So, we can conclude from this. All right, we can make atoms disappear. All right? How many times does it work? That's a difficult experiment. Anna probably does similar things, so... Yeah, it really depends. It really depends on the shape of the dip. So, if you have a good shape, it can work every time you do it. But if you have a bad shape, it gets bad like hours to get it worked. That would be my answer, too. So, we can easily spend two days where it doesn't work. Ever, basically. We can always make the atoms disappear. They can always disappear. That's easy. But to make it go to the end of the tip, and then in particular to get it back off, that's a difficult experiment. I'll show you on the next slide. Very similar process, basically. Okay, so first thing is this guy is missing. Now, second thing, you also see that these atoms, the remaining atoms, look better. They look more round, right? So, this is a more point-like SDM tip, because in the end, the SDM image is always a convolution of the shape of your tip with the shape of the surface, right? Nature doesn't distinguish between tip and surface. So, if you have a very sharp atom on the surface, then really what you're seeing is the shape of your tip. The shape of the last few atoms of your tip, right? So, that's what this one is. Not a very nice image. You see this atom is slightly deformed, because our tip has two atoms that are close to the very end instead of one, right? But here you see it's better. So, maybe two points for this mechanism, right? The atom is gone, and the atom image looks sharper. But then we can also take this tip now and measure a spectrum on the remaining manganese atom. And what you see then is the following. So, here you see an inelastic tunneling spectrum measured on the same manganese atom, but now with a modified tip. And what you have here, again, at minus 1 millivolt, you have a step, and at plus 1 millivolt, you have a step, but the height of these steps is different, right? This step is about half as tall as this step over here. So, the amplitude has changed, the energy has stayed the same. Okay, so then how do we do the reverse? We, again, we take this tip and we bring it back down to the surface, point contact, several microamps of current, and now we apply a polarity difference, opposite polarity, and then sometimes this works. This is a difficult process, actually, to get this back off. And, you know, I work for a company, so we actually do not spend a lot of time trying to understand why this works, right? We don't have time for that in the end, because I can't, you know, I don't have a small group. We always have to focus on the highlights basically, we can't focus on this, because in the end, who cares, right? If it works, it works. My bosses don't care. I can guarantee you that. Okay, and so then we can take this empty tip and we can image again, and you can see we can go back to exactly the same shape of this tip without the manganese atom on it, but this manganese atom has moved from here to here, right? So we can move this manganese atom by a few lattice sites, actually two or three in this case, by using this manipulation process, right? Now, if you have a tip that works, the biggest structure we've built with total precision was 96 atoms. With the same tip, we could do this 96 atoms, 96 times in a row, without messing up, basically. So if it works, it works. Okay, so go back to the same atom, we can measure its spectrum again, of course we go back to exactly the same spectrum that we had to begin with, because we have exactly the same tip and exactly the same thing on the surface, and that's again, it's the beauty of this, right? If you know it to this precision, what you have, everything has to be exactly the same. We can pick up a copper atom, which was the guy up here, so we now have a copper at the end. You see that the spectrum is essentially exactly the same as what I call the empty tip here. It's not 100% the same. If you look very carefully, you can see that there are slight differences. I don't know if you can see the brown curve underneath there, that's the old one, right? Because the density of states changes a little bit between having a copper atom there and not having a copper atom there, right? But to first approximation, the height of these steps is again identical. And go back to the empty one, back to manganese, we can do this again many, many times. Okay, so how does this work? One problem that we have, or one complication that I haven't mentioned to you yet, is that the manganese atom is not a spin-1-1-2 particle. Manganese, if you look at the periodic table, is in the middle of the D-shells, of the 3D shell. So if you just look up at the atom periodic table, you have five electrons in the D-shell, giving rise to a spin of 5 half and to zero orbital angular momentum. That's what the periodic table tells you, basically. So if you have a 5 half atom, you should have six excitations or six energy levels in the Zeeman energy diagram, but we only see one excitation. That's kind of weird, right? So we have some selection rules at play here, and I want to explain that to you. Now, the selection rule stems from the fact that we're actually using electrons for this excitation mechanism, and it's actually not so difficult to understand. So the electrons, of course, are spin-1-1-2 particles. The tunneling electron is a single electron, spin-1-1-2, and we have to conserve not only energy, but we also have to conserve spin angular momentum. Those are two, I mean, not spin, but angular momentum has to be conserved, unless you produce orbital angular momentum, the spin has to be conserved, basically, in this tunneling process. So really what we can do here is we can exchange spin between the manganese atom and the tunneling electron, and so really this is an exchange interaction between these spins, and so that means that we can only go from the ground state to this first excited state. The electron here, the tunneling electron, just does not have enough orbital moment to get you from 5-1-2 to 1-1-2, basically. Can you, if you do it fast enough, can you then excite the excited atom and excite the excited atom? Yes, yes. So if we don't give it enough time to relax back down from this state to the ground state before the next electron comes about, right, then we can drive it from this state to this excited state, right? So sequential electrons will do that, basically. Practically also working, so... Yes, yes, yes. And, well, I don't think we will have time, but we did this experiment a couple of years ago, and this tells you something about how fast this relaxation back down to the ground state is, because we can make this same measurement as a function of the tip sample distance. So if you say at 1 millivolt, you know, at some height I have 1 nanohamb of current, and if you bring your tip in closer, at 1 millivolt you have more current, or if you pull your tip back, at 1 millivolt you have less current, right? So that's something that people that are doing transport in, say, planar tunnel junctions, they can't do that, right? You basically have a tunnel junction that has a certain resistance, given by the thickness of your insulator, and so that's it. So at 1 volt, you always have the same current, basically. We can have this freedom by changing this distance. Okay, so why does this give rise to this asymmetric spectrum here? It's not so difficult to understand if you allow me to make a simple simplification here. If you think that this manganese atom, together with an external magnetic field, on this tip, you know, it's sitting on the metal here, so let me make this simpler. Let me ignore quantum mechanics here, and let me use quantum mechanics here, right? They're both atoms. That's not a trivial statement, right? This is a pretty drastic approximation. But if you think about how, what you learn about spin polarized currents, usually in solid state physics, then what you mean by having a spin polarized material, like cobalt, for example, what you learn is that you have a different number of electrons of one spin flavor at the Fermi energy than the other spin flavor. But usually you say you have more majority spins than minority spins at the Fermi energy, and so your transport will be spin dependent, spin polarized. Now, if we say that this tip behaves like this, so I have fewer or minus one-half electrons and more plus one-half electrons, then you can understand why the spectrum becomes asymmetric. Now, we have elastic current going from tip to sample in this window here. We have elastic current going in the opposite direction. I'm just changing the polarity here of the bias voltage. More interestingly, if I go in the inelastic window here, I have to start as a minus one-half electron because I want to transfer angular momentum here. I have to start as a minus one-half electron, oops, and I want to end up as a plus one-half electron. So in this direction, I can do this with a certain probability. In this direction, I can do this with fewer electrons because I have less initial states available here. And the beauty of this is that this allows us to measure this degree of spin polarization here, usually defined as the density at plus voltage and minus voltage, minus this, divided by the average of the two. So we can assign the spin polarization directly to this asymmetric spectrum. The degree of asymmetry tells us what the spin polarization is, giving this a simple formula here. And in this case, this comes out to be approximately 0.3. Spin polarization 0.3 means that you have twice as many electrons of one spin flavor than the other spin flavor, just using this formula here. Now, this is interesting because, A, it tells us that it degrees well, that we put a magnetic atom on this tip and we apply a magnetic field to magnetize this atom, basically. That makes a spin polarized tunnel junction. And it allows us to quantitatively measure what the spin polarization is. This is very difficult in practice in other materials. If you ask somebody, some of you are doing spin transport, right? How do you know what the polarization of your, you use cobalt electrodes? How do you know what the spin polarization is of your cobalt? It's very difficult to measure, right? It's very difficult to measure. I mean, people have done it by measuring, using a superconductor and pushing magnetic material into the superconductor, using those kind of devices, but it's very difficult to measure. So here we have a way to measure this for these simple tips, basically. Yeah? Maybe I missed it. Why don't you have from minus or half to minus or half? Because the selection rule is basically that we start with the atom in the ground state and we need to transfer a certain direction of angular momentum to this, right? If you would transfer the opposite amount, you'd go from this state down one state, but you don't have a state below that, basically. You were first. My question would be, is spin polarization, is it typical for manganese or can you also, well, change it? Should I get another value there? It's somewhat typical, but not reproducible in the sense that every time you pick up a manganese atom, you will get exactly that spin polarization. Of course, it depends on magnetic field over temperature, right? At zero magnetic field, you will not get any spin polarization to first approximation, right? You need to polarize, you need to magnetize, quote, unquote, the single atom here, Curie-Weiss law, basically, right? So just B over T. So in the saturated regime, so if you apply more than one Tesla at 0.5 Kelvin, you are in the saturated regime, and then I would say it's typically in that range. If you pick up more magnetic atoms, you can make it bigger or smaller than this, right? And it does not depend very dramatically on whether it's manganese or iron. Can you, because we are coming from one manganese to the other one, can you have them aligned in anti-barrel, for example, because now if you have to apply a certain saturated field, all your manganese atoms are in the same direction, right? Not in the ground state, basically. In the ground state, they will be aligned, and you get a certain current, right? So the answer to that is really no, in the sense of having a steady state, you know, tunnel junction where you can do this real experiment where you change one orientation versus the other. We can do a time dependent, right? If we kick this guy out of the ground state, then this will be rotated, if you think about it classically, and your current will change for a short while until it relaxes back down to the ground state. Can you, for example, just pick up another manganese atom trying to change the, for example, field, or change the total spanning for that guy? In practice, the answer is no. You know, but what other people have done is built structures of, say, five or ten atoms on the surface, a big blob of iron atoms, for example, and then those will be a steady magnetic state in a certain direction that's paramagnetically coupled to each other, and then you could have this. If you could transfer that to the tip, then the answer would be yes, but that doesn't actually work that way. So the subject is also the same, and that depends on the substrate? That depends extremely heavily on the local environment around these atoms. Could it be different from the one in the tip? It's very, very, very different, guaranteed to be different. Yes. And we don't know it, right? I told you so, you know, big word that we know everything about the structure on the surface. We don't really know much about the structure of the tip, right? So the tip is always a black box for us, in some sense. I'm interested in what your next slide is. How do you know my next slide? Yeah, you're a student. Oh, okay. So I understand the picture that the high-intensity amplitude is observed by the density of states, but the topic also seems to indicate that there's an energy difference. So it takes more energy to reach the mindful mouthpiece. This energy difference here? Yeah. It's, yeah, it's, it's, this is supposed to be a picture. I agree with you. This is not, not the greatest picture. Let's do this over here. So if you think about what happens when you apply a magnetic field to a normal metal, right? You have your parabola of states, basically, right? Energy versus K, for example. Well, I don't know if K is good here. And now, and this is spin up and this is spin down. You take a normal metal, not a ferromagnet, but a normal metal, and you apply a magnetic field to this. You get the Zeeman energy of the electrons in the metal, basically. And in some sense, what you do, the first approximation is you just shift one of these parabolas down and keep the other one at the same, at the same place. And you could do this change, you could change the Fermi energies or you can change the bottom of the space here, right? Now, what happens, of course, is that the Fermi energy will stay constant because you can get spin, the spin is going to, I mean, the Fermi energy is going to be the same, basically. You're not going to build up a voltage inside your material by applying a magnet to a normal metal, I mean, applying a magnetic field to a normal metal, right? And so, I just draw this, basically, as a shift in this parabola. So, I shift one of these parabolas here by the Zeeman energy. So, this shift there would be the Zeeman energy, right? So, this distance here would be the Zeeman energy in the metal, basically, for the quasi-particles. Don't take that too seriously. I mean, this is really meant to be a total cartoon, right? If you don't like this parabola, I'm perfectly okay with this. But just say that you have a difference of the number of electrons at the Fermi energy of the two different spin flavors, right? That's really all that's supposed to mean. Maybe that's taken the cartoon too far. I see what you mean. I had one question about, so you don't say you don't know what the surface tip structure of the tip is. But you also said when you're probing the resonance, you're basically probing the state. Yes, you're imaging the tip to some degree. So, we learned something about it. And you could, in principle, do sort of a deconvolution and take the image that you get, maybe for different atoms, right? If you have one atom that's very tall, one atom that looks very short, you know, you could learn something. We haven't done this in practice. I don't know. Has anybody done that? Probably not. I mean, it's also about structure. You need sort of a good structure for your tip to make this process work. So, hand-waving, yes. Hand-waving, you know, you learn something about, all right, this tip might work just because it looks like this. And then it works half the time or something like that, right? I mean, so it's not, that's not really scientific. No, it's not. No. I'm sure that, I mean, it is. In the end, it is. But we haven't put the time and effort in that. Okay. Now, I think we already went through this. But as I said in the beginning, I do want to spend a little bit of time on these operators trying to explain to you now how we can understand these experiments a little bit more, you know, from a theory point. Now, what I'm going to be telling you sort of in the next few sections is not something that a theorist would actually tell you, because it's not really theory. This is trivial for a theorist, right? But it's just a little bit beyond what normal experimentalists learn about spin operators. And so I think that's why it might be worthwhile for some of you. Okay. So, first step here is how can we model this, basically? So, we start with our spin manganese being in the ground state, M equals plus 5 half. And we ask if I have a tunneling electron that starts as a spin-down electron, what can it do? It can flip and become a spin-up electron, but we have to conserve angular momentum, and that means that the manganese has to go to an excited spin state, right? That is the inelastic tunneling channel. But you can also do the following process. You can keep this down electron as a down electron. And the manganese, of course, stays as M equals plus 5 half and remains an M equals plus 5 half. That's a perfectly allowed transport process as well. You could take an up electron and start in the ground state and keep an up electron and still be in the ground state of the manganese. And the fourth channel, you start as an up electron. I think you asked that. Somebody asked the question. And we start in the M equals plus 5 half. We flip this to a down electron and go to an M equals plus 7 half. That state doesn't exist. So, this probability is zero, basically, right? So, we have three different probabilities here. And can we learn something about what happens, basically? Can we do some modeling here? And through quite a bit of work, what we actually learned is that it's actually not that difficult. So, the tunneling electron, which is the state that is either up or down, and I'm not using full quantum mechanics here, so there's no, you know, coherent manipulation or something like that. So, this is either up or down the tunneling electron in the copper of the tip, or in the tip and in the copper in the metal, right? So, this is, this is say the tip here and this is, this is in the metal underneath. So, that's the tunneling electron operators. And then the operators corresponding to the local atom, the manganese with its local environment here. So, that's, and if for those of you who are familiar with this, this is an exchange interaction. You know, S1 dot S2 times usually a parameter like J, that's called the Heisenberg exchange interaction, and tells you something about exchange of electrons, interactions between electrons. And I want to spend a little bit of time understanding what these operators actually mean. Now, we have here something a little bit more complicated. We have an additional potential scattering term. And I can't explain to you why this comes about. Theorists will explain to you why this comes about, okay? But I know that we need this parameter to do good modeling, right? So, I'll just use it here. You can grill me later if you want on why this works. Is it zero field or what is it? Do you fit it for this to something? Yes. So, we fit, there's, there's, this holds at any magnetic field. And we basically fit it, you know, for example, this u parameter is a single parameter, one number basically is a real number. And we fit this to a certain set of measurements, for example, as a function of magnetic field. But we want this parameter to be constant, independent. It should be a parameter that only depends on having manganese in this particular configuration. Okay, this means u is x basically. And then everything else should come out of that basically. Okay, so this term here, sigma dot s is actually a pretty difficult animal, if you look at this a little bit more carefully. Because this dot s, what that really means, this is just a definition of what sigma dot s means. It means sigma x sx plus sigma y sy plus sigma z sc. All of these are operators in this sense. Right? That's a pretty non-trivial, trivial term if you really want to calculate that. Right? And let me see, I think I have this written down here in a little bit slower, because I want to take some time here and bring this to life a little bit. So, where is the term? So, okay, so what we have there is, in the end, let me just write this down. So we have four cases here. I don't think this is in the slides. I'm not sure. But I'll give those to you. Actually, I don't think I have this in the slides. So you can either get my handwritten notes or if you care about this, you probably have to write this down. So, the sigma initial of the tunneling electron and the sigma final, so it can be from up to up, from up to down, from down to up, and from down to down. Those are the four cases that we sort of evaluate as we do these simulations. And again, I'm ignoring quantum coherent kind of states here. The electron in the tip could be in a simple position of states, but I'm not dealing with this at all here. So just sort of classical magnetization states. So let's just take an example here. So one of these probabilities here is up electron as final, sigma plus as minus, sigma minus as plus, sigma z as z plus u up as initial. All right? So if I want to calculate what's the probability of this process happening, the electron starts as an up electron and ends as an up electron. This is just one of the four cases that I have to calculate basically. So that's this case over here. Then you'll see that what happens with, you guys are probably familiar with sigma plus and sigma minus, right? That's sigma x and sigma y. Those are raising and lowering operators, just like you learned in basic quantum mechanics. So, but you know that if I operate sigma minus on the upstate, I get the downstate. And then I multiply or I check the downstate overlap with an upstate is zero. If I operate sigma plus on an upstate, I get zero. So that tells you that these two terms don't play any role in this particular case. So I can simplify this by saying this is s sigma z sc plus u up as initial. Absolute value squared. So you can do sort of these simple terms. And so you learn that if you don't change the spin of a tunneling electron, the plus and minus terms don't play a role. On the other hand, if I do the opposite, right, if I do change the spin to down s final, sigma plus s minus plus sigma z sc. This by the way is the part where you can yell at me and say this is too slow if you want. I can do worse handwriting, but not much. If you want to calculate this one here, now you have to find the term where the up electron changes to a down electron by applying these sigmas, right? And that is which one? That one, right? So now I have down s final sigma minus s plus up as initial. Absolute value squared. If you want to calculate basically what are the occupations of these six m states I have for my manganese, right, in the finite magnetic field, you have to keep track of these processes and do something like a bookkeeping of which state are you in. Maybe you start in the ground state, you run an electron through, with a certain probability it goes in the first excited state. Maybe now you do a relaxation mechanism or not, right? And then you keep track of this and you calculate what is the current as a function of the occupation of these states, following these rules here basically. On the other hand I can simplify this even further because I know that sigma minus operated on up and down excluding an h bar or something like that. I don't know the h bars, I don't have these written down here, right? That is s final, oops, s plus s initial, right? I've basically applied the sigma minus here to this upstate. I get, you know, excluding factors of two or something like that, I don't have all these factors written down and an h bar probably, right? You get the downstate here and then you can evaluate this and you get a certain value out, right? And so you are left with this process here. So basically what this tells you is by doing some bookkeeping, you can keep track of the tunneling electron, what the tunneling electron is doing and you're left with some operators that depend on the manganese atom that is sitting on the surface, right? And that's why in this description here I call this the elastic tunneling. That's basically the tunneling where the tunneling electron does not change. If it starts as an up, it ends as an up. If it starts as a down, it ends as a down. And the inelastic one where it actually does a flip basically, right? So the inelastic one have the x and y which is the same as plus and minus operators and the other one has this sigma z in it. I don't have this written down any better than that, right? I mean I think you get the idea how you can do this kind of modeling. This is basically an exchange direction for a certain amount of plug. We really don't make any, yeah, we treat this in a very funny way if you think about this. Are you a quantum dot person? In a quantum dot you actually think about the electron hopping on the dot and hopping back off, right? Here I'm not doing this basically. The electron is zapping through there and it leaves behind an excitation or not. And I'm asking, this is more like a scattering process in that sense. I don't really tell you that the electron lives on this atom here for 10 femtoseconds or something. I have no idea, right? I can tell you if the electron goes through and it makes an excitation in the spin, how long does the spin live? I can measure that. But I can't tell you how long this electron that actually went through there lives on this atom. I can't even tell you if this is an allowed question to ask. I know this is a good hand-waving. I really don't know the answer to that question. It bugs me. I don't really know how this transport process really works, obviously. But we do know that it actually looks like it does this exchange interaction, obviously. That's really what is happening from experimental measurements. Is this something, I mean, I guess you can sort of fit that effect with the gate and this effect of exchange interaction? Yeah, that's in the end somehow what this U parameter does. It's the ratio of this exchange interaction to a potential scattering. It's only one parameter that matters. It's not two parameters. On the strength of the tunnel current? No, it does not. Yeah, I think the answer to that is that the tunneling process is always slow in comparison to the transition time of this electron. A while ago, just before you guys were born almost now, but before you guys started physics, basically, people were wondering how fast does an electron tunnel? What's the time corresponding to a tunneling process? And I don't really know what came out of this discussion, but that's at least 15, 20 years ago, that was an unsettled problem. And I don't know if this is settled by now or not. Can you actually assign a time to such a tunneling process? I don't know. Anybody know? I read about this a while ago. It depends on the transfer of the wave function and the overlap. That's how we can calculate, basically, how much current will flow. That's the overlap of the wave functions. But it doesn't really tell you anything about the time that takes that tunneling process. It's a really non-trivial question, I think. So, you know, this to excluding a factor of one half or something like that, oops, sorry, you can actually equate sigma x sx with sigma plus s minus and so on. That's true to a factor of one half or something like that. So that's the same thing, basically. Okay, so I'll tell you a couple slides how we can measure this, this u, basically. And then I want to go into this zero field splitting. So with a non-spin polarized tip, you know, we get our manganese spectrum that we talked about quite a bit now. And here we have elastic tunneling and inelastic tunneling, and the inelastic tunnelings are the same because we have a non-spin polarized tip. If we take a spin polarized tip, we have more inelastic current in this channel than in this channel, right, that's just the polarity of the bias voltage. And sometimes, I think one of you asked, can we make different tips? By picking up more atoms, yes we can. And sometimes we can even make inverse spin polarized tips. If we pick up a lot of atoms, maybe we'll make an anti-farrer magnet on the tip, who knows? We don't know. But sometimes we actually see that the opposite argument holds, and so, you know, basically what this means is that the end atom on the tip is anti-aligned to the external magnetic field. And maybe that was the question, can we make this? Sometimes we can, but we can't do this reproducibly. I can't say, all right, if I pick up 10 manganese atoms, I will make this. It just happens to be sometimes the case. Okay, now this U, not from this measurement, but from something similar, let me just skip that part, we can calculate this U parameter, the quantity that this belongs to, okay? And for the manganese atom, the dashed line here is the value where you get the maximum tiling magneto-resistance. And that I will show you in a calculation in a couple slides. So for manganese, the manganese atom looks like it wants to be fully spin-polarized, basically. Whereas an iron atom, that line here is 1.0 in our parameters, right here is 1.25. That's just because the iron is not spin-5 half, it's spin-2, right? So we didn't use a proper, you know, that, I mean, the difference between 1 and 1.25 is just the magnitude of the spin, right? That's just the way that, I'm not a theorist. A theorist would do this better, and this would be, you know, always 1 or something like that. But you can see that the iron atom does not correspond to this maximum value of U, and so this U really has some physical meaning. It tells you something about how this tumbling process works, I think. But what it tells me, I don't know. So it really we treat it as a fit parameter, basically. A fit parameter. But only one parameter, right? If you take manganese on this binding site on carbonitride, you have a U of 1.25. You take iron on the same binding site, you have U of 1, basically. Okay, so we can determine, all you can learn from this, we can determine this value. And the histogram here is taking different atoms, measuring this, you know, about 100 times for different atoms, doing all this fit procedure, and you can see that it does vary a little bit. Who knows why? But on average, this is what comes out. I want to spend a moment on these two, both of them. So I'll have to figure out how to do this. But, okay, so magnetic anisotropy. We talked a little bit about that zero field splitting, right, for the iron atom. At zero magnetic field, there seemed to be some energy required to make an excitation. And I want to tell you in a little bit more detail how this works. And I'm going to use a different atom where this is much more dramatic than the iron. Now, what this all comes down to is, of course, if you have only the Zeeman Hamiltonian, at zero magnetic field, all the states have to be degenerate. This is proportional to B, so at zero field, B is zero. If you take the spin and you break the rotational symmetry of space, and you can do this either by putting this spin on a surface like I'm doing here, or by putting the spin in a molecule, or doing anything else that breaks the rotational symmetry of space, then in general, the energy of the eigenstates will depend on the direction that the spin is pointing in. It is allowed to depend on this direction, basically, right, because the local environment is making the directions different. And this gives rise to two effects. One, it gives rise to the zero field splitting. And I will use this kind of picture to describe the zero field splitting. This is based, again, on these quantum models, and it's called a spin Hamiltonian, which is used all the time in electron spin resonance. This is how people describe electron spin resonance. But it's not so well used in normal solid state physics, I would say. So molecular magnets people, chemistry people, they use this all the time, basically, as their picture. Now, here's the first term that plays any role in this. dSc2. So now the z is not the direction of the magnetic field. It's the direction given by, for example, the surface, perpendicular to the surface. In the manganese case, I applied the magnetic field perpendicular to the surface, because that's the easiest case, because the spin wants to point perpendicular to the surface. Now, what happens here, it's an sc2 term. So you see that if I apply my magnetic field along the same spatial direction, then this is a term that is b times sz. And it's sc and sc2, they are commuting operators. So that means that the energies just add up, basically. That's this picture over here. So you have a certain energy at zero field, and on top of that, you add the linear term of the Zeeman splitting. If I apply this magnetic field in a different direction, let's say x, right? Now I have sx here and sc2 here. Now, sx and sc2 do not commute, and then the world is more complicated. So as I increase this term here, the Zeeman term, I'll get some complicated dependence of the energy levels on the magnitude of this magnetic field in this direction. So having magnetic anisotropy in my pictures gives rise to zero field splitting, and it gives rise to a direction dependence of my magnetic field. And they are very closely related to each other. Now, let's do this experiment. We take an iron atom on the same binding site on copper nitride. So all I've really done is I replaced the manganese atom with an iron atom. And I take the spectrum here at zero magnetic field and at low temperature, and you see something quite dramatically different. Remember, the manganese atom had just a tiny little feature at zero magnetic field, a tiny little dip, basically. This guy, and this is plus minus 10 millivolts. So this is 10 times bigger energy scale, basically, and it has three excitations at zero magnetic field. One at 0.2 millivolts, one at 4 millivolts, and one at 6 millivolts. So in some sense, the zero field splitting here is dramatically bigger than in manganese, about 500 times bigger to be precise. And so this is just because we're using a different atom here. So definitely this guy has zero field splitting. To be more basic question, how do you, it's probably a story, where does the choice come from for iron and manganese? Why those guys? Or why do I know? No, no, why those guys? Why those guys? Well, they are the 3D transition metals. Those are used in magnetism all the time. That's not really a good answer, but that's why we did it, basically. And so we've looked at most of the 3D transition metals, and I'm just showing you those two cases here today. If you use F electrons, you should be able to see something similar, but the F electrons are much closer confined to the core of the atom. And so it's not clear to me if you can actually see this current or not. So there's a lot of experiments basically to do. But for us, it's just that's the way we chose it to be for no particular reason. Okay, so we definitely have zero field splitting. What about direction dependence? So let's apply the magnetic field in plane here, along this particular direction in plane. And what you see here is the zero Tesla field, the spectrum. I just showed you that before. And now one Tesla, three Tesla, five Tesla, and seven Tesla. And what you see is that the 4 millivolt zero field mode goes up in energy as an increased magnetic field, and the 6 millivolt also goes up in energy. Right? And also the middle guy here disappears pretty rapidly as we increase the magnetic field along this spatial direction. If I apply the magnetic field in a perpendicular direction, also an in-plane magnetic field, but 90 degrees rotated, with respect to this iron atom and this particular spatial direction here, you see, of course, at zero Tesla, they're the same, have to be the same. And then we increase the field. Well, the 4 millivolt mode stays roughly constant. The 6 millivolt mode goes down in energy. And the central mode seems to be sticking around much longer. Right? So quite different dependence here. Yes? So you understand what the two forward arrows, what are the top two are creating? This one, with the model I'm showing you in the next couple slides, this would be the location of this excitation at seven Tesla. So you should be seeing something there if it was visible. But it's not the transition, or you just cannot measure it? The transition amplitude is zero, basically. It's the original. We just don't see it. I'll show you this in some quantitative detail, actually, which is one of the goals I have. Okay. And the third direction out of plane magnetic field is yet another dependence. Okay. So by now you know what I do. I plot the center of the step, right? That's here as a function of the external magnetic field. That's basically extracting the information from this IETS measurement, inelastic tunneling spectroscopy measurement. And then I can get a very good fit to this. With a simple Hamiltonian that has only a few parameters. Okay. So that's the fit that I get when I use this Hamiltonian. So I have a Zeeman energy, not shown here. Then I have this dS e squared term. So that's a certain spatial direction that I have a so-called easy axis magnetic anisotropy here. That's this term here. And then I have a transverse magnetic anisotropy. Now, if you look at these terms, again you have S c and S x and S y. So again we have non-commuting operators. So you have to do the calculation. You can't just simply see what this does basically. Right? That's kind of the outline of my talk in the end. You can actually do this by using these spin operators in a fairly simple way that you can all do without, you know, go into quantum mechanics two or three courses or something like that. Right? So, now the parameters that we measure from this, this fit we get by getting d equals minus 1.5 millivolts and e of 0.3 millivolts. Right? Just take those as fit parameters. This, yeah. So do they correspond to something physical? So can I? Yes. So this minus 1.5, will you tell me? Yes. So this d equals minus 1.5 together with the fact that this is the simple direction, meaning the direction where all the energies are just adding up together. Right? This is much more complicated than this in some sense. It tells you that this is the preferred direction of the spin. So this spin wants to point in plane, whereas the manganese spin that we talked about before wanted to point out of plane, but very weakly. It has very weak preference to point out of plane, where this parameter for manganese is 50 times smaller. Right? So the magnitude of this parameter tells you how much does it want to point along a certain spatial direction. And 1.5 is very, very big. The cobalt atom in the bulk of cobalt, for example, would have a, you know, d equivalent that is about 500 times smaller than this. So because it has this, this has these local nitrogen atoms nearby, these guys here, they set all the energy scales for this magnetic interaction here. They play a very important role. Okay. So for the next 20 minutes or so, I want to take time to show you how this kind of model actually works. Yes. Yes. Yes. Yes. It's completely set by the local environment, basically. And also depends on the atom itself. Yes. Manganese is very different from iron, very different from cobalt, basically. So is there a way that you can measure like a normal spring atom by... That's why I chose the manganese to begin with. So if you exclude that tiny little wibble at zero magnetic field, right, that was a small fraction of the Zeeman energy. Right? And this here, here the zero field splitting is a large fraction of the Zeeman energy, right? So this manganese is the closest I have to a free spin. All right. Basically. You know, it's... why is this? Because manganese is an S equals 5 half system and that's a half filled D shell. And a half filled D shell to a very good approximation is spherical symmetric, just like a fully filled shell. And so that's why half filled shells are very, you know, less susceptible to those kind of influences. Okay. So how can we understand how this stuff works? So, of course, you know that if I just have the Zeeman energy and I have zero magnetic field applied, I don't get any excitation, right? I have only a zero energy excitation. Nothing to learn from that for this. So, and I have five energy states, right? All degenerate here. And so if they're all degenerate, I can label them any way you want. Ignore this for right now. It doesn't make much sense. If I look at finite magnetic field, this is something you learned in quantum mechanics. What you learned in quantum mechanics is that the magnetic field sets the quantization axis. Right? That's what's typically taught. And so that's what you see here. So if you apply a magnetic field along a certain direction, let's say, SZ, but doesn't really matter which direction that is, then this direction is a good basis to use. And so what we do is we use the M eigenvalue of the SZ operator along that particular spatial direction, which is set by the operator choosing a certain direction of the magnetic field. And so that's what these are here. So M equals plus two, plus one, zero, minus one, and minus two. And those are very good. Those are good eigenstates here, right? So the eigenstate corresponding to the ground state is 100 percent M equals plus two. The ground state corresponding to the first excited state of 0.9 millivolts is 100 percent M equals plus one. This is just repeating your basic quantum mechanics just written down maybe a little bit differently. Right? You all know this, basically. OK. But now let's take a moment and look at this term, dSZ square. OK? And so I want to write this down a little bit for you to not go too fast here. So first off, we have an S equals two system, which I haven't proven to you, but let's assume that that's the case here. So how do you write the SZ operator? Right? Remember for an S equals one-half, it was just minus one and one, basically, or minus one-half and one-one, plus one-half, if you want to be precise. So here, the SZ operator is two, one, zero, minus one, minus two, with everything else being zero. Right? And if you want to know what SZ square is, you can do the matrix product if you want. Put another matrix next to it, do the matrix multiplication, and what you'll see is that this is four, one, zero, one, four. And everything else is zero. Right? This happens to be the case because it's a diagonal matrix. Easy, basically. Just for shits and giggles, what if we want to use this and we want to know what's the energy SZ square, DSE square, if I'm in the plus two state? Right? I have to convert this to a vector because I'm using a matrix here. A vector that corresponds to the plus two that's written up here, that's one, zero, that's the first line up here. Right? That's one, zero, zero, zero, zero. Okay? So I multiply, let me make that simpler here, and let's say DSE square of plus two. All right? So you get my D here, and SE square is this matrix, and I have, what did I say, plus two, one, zero, zero. Right? Do the matrix multiplication here, and you get four times one, zero, zero, zero, zero. So this is an eigenvector of this operator with eigenvalue four. Right? Do the same for the minus, M equals minus two state. You also get four. Right? So you have a two-folded generacy of these states. Nothing magic about this. Simple matrix multiplication, basically. Yes. Yes. Now, let me take this one step further. What do I do with the, where is it? There. Okay? So this is really simple. Now, let's ask ourselves, for example, what is the current that I get for an up electron if I don't change the spin in comparison to the current that I get for a down electron when I do the same thing? What? This is something like, I'm not looking at the inelastic components, I'm just looking at the elastic components. Right? So those are four of the tunneling processes that we have to worry about if we want to calculate the current here. Now, we can do this, right? SZ operated on plus two gives us two times plus two. Right? Let me make this slowly. Plus two gives us two sigma Z. A little bit tricky because I have the U in there. I can't quite do that so slowly. Let me operate both of them. So if I operate the SC on the upstate, I get two times the upstate. If I operate the sigma Z on the upstate, I get one half times the upstate. So I get one half times two. Right? One half times two plus U times up plus two absolute value square. Right? Nothing magic, this is just matrix multiplications or basic quantum mechanics. Right? Now, so this positive gives you, if you do one step further here, gives you something like, right, this is now just numbers. So I can take those numbers out and this operated on this is one. Right? So I get out of this one plus U in brackets square. Do the same over here. Right? Now I have SC on plus two, this plus two. Sigma Z on the down is minus one half. Right? So I get here minus one plus U square. Right? So and remember that I told you for iron, the U was 0.75 in these arbitrary units which I'm using here consistently for this U. And I said that the maximum value for spin polarization for iron is one. And what this means is that if you had a U of one, that means that if you have an up electron that's tunneling, you get a certain current. If you have a down electron that's tunneling, minus one plus one is zero. You have 100% spin polarized current. Right? This channel is exactly zero if you have the maximum value of spin polarization. Now it's not exactly because it's 0.75, but you can see that there's quite a dramatic difference between an up electron tunneling through this iron atom and a down electron tunneling through this. Right? And I'm not, and again, I'm not using any magic here. This is just simple matrix physics basically. Okay. So that's what we learned from this, but going back to our measurement, we only get one excitation at zero magnetic field and not three. So this is not the full story. Right? We had one excitation at one millivolts, at 0.2 millivolts, one at four millivolts and one at six millivolts. And so that part we just did. What transitions are allowed, that's that. Okay. Now, if we include this transverse magnetic anisotropy term, right, then what you see is that you get a much more rich picture of what the quantum states are here. The ground state is now a mixture of m equals plus two and m equals minus two. That means that this iron atom in the ground state is pointing its spin along one particular spatial direction equally up and down. Quantum coherence of a position of these two states, but if you're pointing at the same time up and down, you have no net magnetic moment left. Right? They cancel out. So in other words, this iron atom at zero magnetic field is non-magnetic. It's canceling its own spin. It's only possible because you're doing it quantum mechanically, right? Classically it's either going to point this way or that way. This comes out if we believe this Hamiltonian just from diagonalizing this matrix. If you look at the first excited state here at 0.17 millielectron volts, that's also a superposition of plus two and minus two, but with a different phase here, with a minus one. Right? Just comes out of the matrix. That's just what comes out if you diagonalize this matrix basically. And so on. Right? So this is, again, not too difficult if you believe in this kind of matrix stuff or in this kind of quantum mechanics. Now let me see what I was going to show you here. Okay. Let me not write this down because I do want to show you the second part too here. Now I showed you the SC operator, right? That's the two, one, zero, minus one, minus two. You can also look up what the SX and the SY operators are. Right? So let's have zero on the diagonal and have some finite values on the one off diagonal. This is just looking up in the textbook basically what is SX and SY for these guys. Now you can matrix multiply, which I actually did, you know, paper and pencil kind of guide that I am. You can just multiply it by itself and you get this matrix out for SX square and this matrix out for SY square. They're not diagonal, so it's not trivial. You can just square these components here and get your matrix out, right? That doesn't work. We have to actually do the matrix multiplication here. Then we can look at this term as actually E times SX square minus SY square. So we look SX square minus SY square has only six components in that matrix here. And so then our full Hamiltonian excluding magnetic field is DSC square plus E SX square minus SY square. This term here gives you the diagonal elements and this guy here gives you these off diagonal elements. Right? You can just do that. That's not difficult. You can just calculate this and then you can diagonalize this matrix that I did not do myself, right? That's beyond my powers of matrix algebra or something like that. But you get out that this is the ground state. If you take this and put in the D and the E value that we measured basically and this is the first excited state of this. All right? So this is again not particularly difficult. But what I want to do now for a second here is I want to show you what happens when I operate the S plus operator here on this guy. Remember that one of the terms that we had is S plus operated on the ground state and then we can ask what that gets. So S plus operated on the ground state. Whoops. Ground state. Now the S plus operator is related to the SX, right? And the SY. It's a simple operator really. 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0. All right? This is just again looking up what does the operator for S equals 2 look like. And now we want to operate this on the ground state. Now the ground state has some components of plus 2, some are 0, and some are minus 2. So that's the plus 2 component here. So that's the 0.69 something rather. 0.70, 0, minus 0.17, 0, 0.70. Right? This is just another way for writing this. Really the same thing. This is just how you translate this thing to a matrix. And then, right, I mean you know how to do this, right? You go this one times this one times this one times this one times this one gives you 0 in the first one. Right? This one times this one times this one gives you a finite value. This one, this one, gives you a finite value of minus 0.42. The next one, ding, ding, ding, you do that and so on. Just basic matrix multiplication and you get 0, 1.400. Right? Nothing magic about it. And what this, another way to write that is, that is basically times minus 0.42 times plus 1 plus 1.40 times minus 1. Just the same thing basically. Now the next state that I have here, one higher than this is, I go back to this in a second. So the next higher state, I want to skip that, is, I don't have this written down here. If I go back, if you go back to this one here, the, so this is the ground state, first excited state, the second excited state. And this is a state that has plus 1 and minus 1 components in it. Right? And remember for our overlap, we always have final and then the s plus here, initial absolute value square. And so you can project, this one is, we just calculated this. You can project this on the final state and you see that that has a lot of overlap with the final state that has plus 1 and minus 1 components. Yeah? What's that again? Why is it not normalized? Why is it not normalized? Because I'm very sloppy. Very sloppy when it comes to H bars and normalization and so on. Right. This one probably has a factor of square root something missing in normalization. Right? I just sort of try and do, highlight kind of the ways that this works, basically. Not to be used for actual computations. Right? Now, this is what we get out of this and if you want to do this overlap here, you just get the plus 1 component overlaps with this plus 1 here, 0.71 of this and this one over here and you get a finite value coming out. Right? So, that means with the operator S plus, we can go from the ground state to the second excited state. We can also go to this excited state with these operators. Also a mixture of plus 1 and minus 1 and the last one has only a component of m equals 0 and we can't reach that at all with our interaction Hamiltonian. Right? We don't have a way to make it to m equals plus 2 with only 1 S plus from the ground state. Right? So, in a nutshell, we can calculate what these transition intensities are and so you can see that forward. So, from the ground state to this one, we have a large overlap if we operate with the SC operator, the elastic part which I called elastic. It's not elastic because it's changing the energy but it is non-spin flip basically and we can also get to this first excited, second excited state here because here we have a delta m of plus 1 and minus 1 or the other way around, sorry for that and we can get equally to this guy here. Now, they have a different amplitude. Why is the one in blue smaller than the one in green? Right? If we had a little bit more time, I could just write this down. It's just the fact that here we have a certain combination of these numbers and you get this together with this phase here, multiply this and you'll see that one is smaller than the other one by a factor of 3. Right? So, all of this detail, all of this, you can get from this, from this spin Hamiltonian which has a few parameters, the D and the E and the direction of the easy axis and the sigma dot s plus u, right, which has one parameter. All of this, the height of these things and the energy falls out of this space here. What does not fall out is the fact that there's a peak here somewhere, right? It's not just a step, it's a peak also and that goes back to this maybe condo effect kind of thing which I don't fully understand. What? I don't understand, not fully, I mean, I don't understand it, not even halfway, basically. Okay? So, in a nutshell, using relatively simple matrix stuff you can look up in a simple textbook, basically, you can get this full understanding relatively easily. Okay? Let me skip this part here. Because one of you guys asked earlier, how do we know the differences? Why? I showed you the manganese and the iron, I mean, they're very different. You can definitely tell them apart. A cobalt atom also very different, right? So once you know who is who, you can actually tell them apart relatively easily and the copper atom is just flat because it's a spin zero in this case. Okay? So, I'm sorry I'm going a little bit faster here because it's 20 minutes to beer time. And I don't want to, but I do want to tell you a little bit about what you do when you have more than one spin. And then we'll call it quits. Okay, so single manganese atom, you know how this works by now. Two manganese atoms completely different. Like this guy, this spectrum here really has nothing to do with this spectrum. Right? Yes? Doesn't matter, actually. You can put the tip here, here or in the middle, doesn't matter. It's delocalized because the atoms are fairly close together and we really can't individually tunnel in only one. They really look like a cigar. Maybe I have an image, I don't know. I'm sure I have an image later. You'll see it's a cigar, so we don't really, if we build them with a little bit further spacing then we can actually tell which atom we're tunneling into and then it can depend on which atom it is, which is what Anna is working on, actually. And delft, not in my lab. Okay, three atoms, very different again. Right? Sort of has a little feature in the middle, similar to the single one, and has some big energy stuff, maybe similar to the dimer. And four atoms, different again. Right? So, spectroscopically, if we know what we're talking about, they're very different. If we do our experiment carefully, we know exactly that there's four atoms and not three. Right? And that's an experiment. If you try to do this with other techniques, other than STM, it's very difficult. Right? If you want to make a spin chain, and you want to, you know, there's some materials that behave like spin chains, but that chain has a certain length. And if you want to make that length one less or one more, you know, you have to talk very hard to your chemistry friends or materials friends to make that material. It's very difficult to do. Right? So, we can do this here because of this combination of atom manipulation and local spectroscopy, basically. Okay? So, what can we understand about this? Going back to the dimer here, at first observation is this is zero field, four Tesla and seven Tesla. First observation, around the energy where the single manganese atom does all of its physics, all of its spin physics, namely, you know, plus minus one millivolts, nothing happens in the dimer. Right? Looks like it's non-magnetic and like the iron atom, basically. It's at zero Tesla. Now if we look at the higher energy, higher magnetic field, sorry, at seven Tesla, it looks like we have actually three steps. You know, this one here at five millivolts, this one at six, and this one at 6.5 maybe. Right? So, observation, I mean, no interpretation of this at this point. And so, we can plot what is the center of this step and this step and this step as a function of magnetic field. And we can do this maybe down to three Tesla and you can see that we have one step that stays constant, one step that goes up with magnetic field and one that goes down. Right? And that, again, should look familiar to you. That looks like an S equals one system. An S equals one has quantum numbers M equals plus one, zero, and minus one and they would behave exactly like that. So what's happening here? Well, not too surprising in the end. We have two very strongly coupled magnetic atoms here. They happen to couple anti-ferromagnetically such that you make a spin singlet. A spin singlet is really a non-magnetic state. A spin singlet is a state where you have a total spin of zero. That doesn't mean that the manganese itself is spin zero. You can take two electrons, right? First electron is up, second one is down, minus, second electron is down, first one is up. That's a singlet state for two electrons. And that has no net spin. Total spin of the state is zero. Right? So that's what the ground state is like. It's a spin zero state. And then the first excited state is an S total equals one made out of two S equals five half manganese atoms by making magic combinations of spin states basically. And then you can get three quantum numbers and they all have allowed transitions according to our spin interaction Hamiltonian. Pretty simple, right? So we use the Heisenberg interaction, J S1 dot S2, as a way to model this kind of system. This is simple. Heisenberg interaction is S total is S1 plus S2. Right? That's just definition of S total operator. S total square is S1 square plus S2 square plus two times S1 dot S2 because the two are commute with each other. And that's your Heisenberg. These guys are constant and so you really can say that the S total square operator is the same as the Heisenberg interaction for two spins excluding the factor of two. So this S1 dot S2 is the same as S total basically here for two spins. And so we have anti-firm magnetic coupling. Ground state is S total equals zero. First excited state is S total equals one with an energy difference of J. And so this is a way to measure this J, 6 millivolts. So we can quantitatively determine the strength of this interaction again by just making this inelastic tunneling spectroscopy. And so with that, I'd like to very briefly show you how we can actually calculate things when you have more than one spin because that's probably something that everybody in this room who doesn't do quantum computation has never seen. How do you write a spin matrix for a spin that is composed of two spins? The quantum people here I'm sure have seen this, but the other guys probably haven't seen this. So let me go back to the x operator, sigma x for a single spin one half particle is zero, zero, one, one. Now how do you do this if you have two spins? How do you write this down? How do you even do the bookkeeping basically? I'm not talking science here. The science, that's all the Heisenberg interaction has to offer. But how do you write this down? How do you diagonalize a matrix that has more than one spin in it? How do you do the bookkeeping? How do you set up this matrix basically? So if we take two spins, well one spin sorry, we have two states that we keep track of, up and down. We call the up one one zero, we call the down one zero one in this picture. If we take two, two spins this is one spin, two spins, how do we do this? Two spins, our states are basically first one is up, second one is up, that's now one zero zero zero. How do I write this best? Up, down, zero one zero zero. Up, zero zero one zero. Down down, zero zero zero one. So with two spins you have four basis states that you have to keep track of. In general the number of these states is two to the power of n, which is why quantum computation works. It's exponential in the number of spins so your Hilbert space becomes big very fast. But two we can still write the matrices down, that's simple. So how do you write an SX operator in this? First off what we know, what we want and the moment the, that was my time to end, so give me a couple minutes. So if I want to flip the first one and I don't want to flip the second one, I can write this as a sigma operator on the first guy and identity on the second guy and I know that my outcome of this should be first one flipped, second one not flipped because the sigma X is a flip operator. But how do I make this matrix here? And that's actually not that difficult. If you keep this, this definition basically, this is not science, this is just definition and bookkeeping. So how do you make this operator here? So basically what you want to do, the sigma X is this guy here. So you set up your matrix zero, identity, identity zero. And by this I mean that you get your four dimensional matrix here by saying this is zero, zero, zero, zero. This guy is zero, zero, zero, zero. And then I have an identity matrix here, identity matrix there. Just bookkeeping, no science in this. Just the way that quantum computing people basically do the bookkeeping and that's how you calculate a spin chain if you want to set up the SX plus SX square plus SY square plus SC square for chain of two spins of spin one half basically. So let's just take a quick look what this does if I operate on one, one. One, one is one, zero, zero. So one, zero, zero. First line is zero, second line is zero, third line I get this one times this one is one, fourth line is zero. Look at this again, that's just down, up. Final thing, what if we want to operate on the other guy? We want to flip the second one. Then I do identity times, oops, times sigma X and my matrix looks first I do the identity. So I get sigma X, sigma X, zero, zero. And then I fill this into my full dimensions, sigma X is zero, one, one, zero. This is zero, zero, zero, this is zero. And then I get another sigma X here, zero, one, one, zero. Right? That's different than this matrix here because somehow the matrix has to do the bookkeeping for you, which operator you're actually operating on, which space you're operating on. Does that work? Well, let's do the same thing. We operate on one, one, one, on up, up, which is one, zero, zero, zero. And we get this guy is zero, this one is one, this one is zero, this one is zero. If you look that guy up, that is the second one there which is up, down. Right? Okay, that's not really telling you any science, this is just telling you how the bookkeeping works and with that I'm almost finished. So I brought this down in your slides here so you can look at this a little bit more. I was going to have you do a few exercises too but that's not going to work. We do the Heisenberg interaction, okay? So, finally, we do this, of course, with Mathematica or Matlab or something like that, set up these matrices for a spin chain up to length six of magnitude, you know, s equals five half. You get a very big matrix very quickly, like a million by a million or so matrix, right? Something like that, I don't know. Very big matrices very quickly. We know the j from the dimer, from the trimer, we can actually determine that the magnitude of the spin of the individual is five half, right? Using the j from the dimer, we can look at, all right, the best agreement is with s equals five half and that's it, that's all the parameters that there are, right? Only the j and the s for the individual one. We're using the simplest, not only nearest neighbor Heisenberg interaction here basically. And then for the even chain, you can see that this fits very well and for the odd chain, also fits very well, right? So if you want to do this, you know, you have to know how you translate this into matrices and how you diagonalize these matrices, right? This is sort of the beginning of how you set up these matrices, okay? And with that, I think I would like to stop and just in a summary, I guess I tried to teach you two things. One is that you can learn something about these quantum mechanical systems of spins on surfaces using scanning tunneling microscopy, right? And we in particular use inelastic tunneling spectroscopy to learn something here. And second, that you can actually model these things with fairly simple basic physical models if you are not afraid to use these spin operators and spin matrices, right? So those are kind of the two messages. So thank you guys very much for coming. And I promise tomorrow will be completely different. So I think we have about 10 minutes or so, actually five minutes. So, you know, please shoot. So if you want to do phase coherent measurements, that's one of the many parts that I didn't have time. There's a lot more stuff in the slides that you have. So what do you need to do phase coherence? You need something that keeps the clock for you, right? You need usually you use a time-oscillating magnetic field or time-oscillating electric field, something that oscillates in sync with the energy spacing of the energy levels of the system here. And that can give you coherent oscillations between the ground state and that first excited state, right? So what we need is two things. One, we need to make a system where the relaxation times are long enough that you could have any chance of doing that. If you do this on copper nitride, the relaxation times, T1 relaxation times are on the order of a nanosecond. Way too fast for anything phase coherent to work, right? So that's one. Make them longer. Maybe make them a microsecond or something like that. And the second one is include something like a time-dependent field that sort of sets the clock basically. So we are trying to do both of these processes, both of these steps. How do we want to extend T1? The T1 is currently limited by the fact that this copper nitride is so thin. So we have, for example, the iron atom here and the relaxation time. So if I put the iron atom into an excited state, this relaxation process is completely dominated by an electron from the copper coming in and going back out. So we need to make this insulator thicker. So that's how we think we can get rid of this relaxation mechanism. Then you will have other relaxation mechanisms, but you have an atomically small system. So nuclear spins will not be as bad as they are, for example, in gallium arsenide quantum dots or something like that. But we don't know if this works. We have never seen anything that is T2 phase coherent. Is there any prediction for the T2? Well, you know what kind of good theorist I am now, right? So you're not going to get a good answer from me for that. I don't know. Nobody knows. I mean, there's many examples where point defects in a material can have very long relaxation times. Like the fashionable one today is the NV center in diamond that has relaxation times that can be up to milliseconds or longer, T2 and T1. Phosphorus in silicon is another one. Relaxation times can be very long. So I don't think there is a very general reason to expect them to be long or not. You have to really work hard on your local symmetry and so on. The beauty that we have is you sort of got the feeling for we have quite a bit of flexibility, right? So we can build different atoms, different surfaces, different structures. So the phase space of possible experiments is huge. But we have a problem with these conduction electrons, which we have to solve for sure. So one step at a time, I would say. What's the big downside of increasing these insulators? Things become more difficult. Yeah, I mean, part of your SEM, the general curve. That's right. So you need to find a balance. Of course, we want to maybe able to tunnel through this. So we need to find a balance where we can still get enough current through this to do our measurement, but where this process here becomes long enough that it doesn't really matter. And so what we're doing is, on copper nitrate, we can't make this any thicker. It's just self-limiting to one layer. Not in the neck, but that's the way nature works in this case. And so for other materials like magnesium oxide, we can make them one, two, three, four, five layers thick. And then we're in the process, basically, of measuring these things that I told you here as a function of the thickness of these materials. Sort of the space of experiments that I showed you is maybe five years of work. So this is not fast, basically. But if you change instead of copper, you use HOPG for a five, for example. Would you expect changing things? Oh, it certainly will change everything. I mean, I think you got a feeling for that the local environment plays a very important role. HOPG is probably hexagonal, if I'm not mistaken. So that's very interesting because you can set up frustrated spin systems and whatnot. I haven't done it, so I don't know if that would actually work. The magnetic anisotropy will also be different because the symmetries are different. You probably don't get charges nearby, like the nitrogen atoms. So magnetic anisotropy will probably be much less. So you can see I can make speculations, but one has to try it. You can, I mean, HOPG is definitely conductive enough that one can work with this, with STM. I have another question. Sure. I don't know if there is any experiment with the spin-polarized STM working on the spin valve, a planar planar, but to see all the spin accumulation in the channel and how it behaves and a probe spinalization of systems like this. There is a vast amount of stuff known about spin-polarized STM. I mean, the most prominent group in that respect is in Germany and Hamburg. We've done quite a bit of work with spin-polarized as well. So add another layer of spin-polarized stuff on top of this. Now to actually see a spin valve. So basically you want to make a planar junction. So the problem will be that you don't have two tips. So if you use your tip for the injection, you can't measure away from the tip what is the contact. Yeah, so you'd have to do that. So you'd have to make a lateral structure and that has not been done as far as I know. Anytime you go for this level of more complexity, it becomes very difficult. This experiments work beautifully in UHV clean samples, right? No dirt at all. That's what we can do with STM. As soon as you do lithography or any stuff like that, the world becomes much more difficult for us. That's probably the best answer to that. One slide further. Let's see. Oh, there you go. Did you program it or did you help to program this? This one here? This was, this is the original IBM that was built by Don Eigler, the guy I inherited the lab from. That was the first atom manipulation, 1989, 1990. Tomorrow I'll talk exclusively about atom manipulation and how we can build structures and have fun with that. No spectra. No inelastic tunneling spectroscopy at all. Just pictures, that's right. Just pretty pictures. What does IBM expect from you? Science papers. And even more important than that, New York Times articles or mentioning of IBM as a leader in nanoscience. If we can portray to the academic community and even more importantly to the general, to the shareholders basically, that IBM is a leader in nanoscience and that this stuff might have applications to the future of IBM's business, then we win. So tomorrow I'll show you two examples where we use atom manipulation to do computation and where we use atom manipulation to do data storage. And that, I think, you know, that's why IBM funds this kind of work. That's also why we can't focus on figuring out exactly what happens in this manipulation process. IBM doesn't care about that. And I have to be careful. I have to do stuff that is interesting to the company. So even though I don't work on anything that is product related, nobody at IBM expect any of the stuff I do to make an actual product, I still have to be able to demonstrate faithfully that the work I do has net benefit to the company. So in principle, you work in the marketing department? Yes. Yes. I often say that I actually work for the marketing department. Yes. Of course, the marketing department doesn't pay for it, but they should. I mean, maybe some of you saw the movie that we made, The Boy and his Adam. That was actually paid by the marketing department. That was approximately a million-dollar kind of budget to promote this and produce this and all that stuff. Because it's a marketing firm that's involved in it. They have a lot of... It's not IBMs. I mean, this is... You know, we use a commercial marketing firm in Manhattan. And you know, I don't know how much... I don't know the number, basically, right? But it's definitely more than $100,000. They were involved in this for about a half year with probably five to 10 people. That's a lot of money. To set up all the PR that went with it, come up with the idea, design it. You know. So that's removing the patents was probably done by a couple of PhDs? That was just done by us, basically. Four people in the lab. 10 days. Right? We didn't get any money for it, of course. We get paid. That's the money we get, basically. Okay. Thank you guys very much. I'll be around tomorrow and on Wednesday. So if you guys have more questions, feel free to approach me. And let's have some beers. Thank you guys. Thank you. That was classic VR, to be honest. I blind only 5 if people need something to do. Imagine...
 Hi everybody. My name is Andreas and I'd like to tell you a little bit about some of our  work on spins on surfaces as you probably know. I also have a talk tomorrow night where  I'm going to talk about something quite different. So you won't hear the same thing again. So  that's good I guess. And also I want to do something which I've never done before which  is to try to teach you that these spin matrices, these poly matrices that you probably hated  in quantum mechanics just as much as I hated them at the time, that they are actually pretty  useful to calculate things that you can actually measure with a scanning tonic microscope. So  that's basically the goal I have. And that might be a little bit boring sometimes because  I actually want to use them. I want to show you that you can actually do real calculations  with them. So if that's too boring, let me know. And so I don't really have a detailed  agenda. I put something together and you have a copy of that. That's not exactly what I'll  be using. So you'll see some differences. I'll be happy to provide the exact slides that  I'm using if you guys want to have those for later. That's fine. Before we get started though,  I would like to take maybe five minutes or so and have each one of you tell me in a few  sentences what your background is. Just so I have a little bit of an idea what to expect.  So maybe we start over here. Just your name and a couple sentences on what you're working  on.  Hi everybody, my name is Xenia and I'm from the University of Twenton, and I'm working  on the chess transfer through organic molecules.  Thank you.  Hi, I'm Matthias from the same university and I'm actually trying to make quantum thoughts  in silicon-driven nanowires.  Hi, I'm Daya from the same university. I am working on electronic quantum inventories.  I'm Pessa Pockemer from the University of Amsterdam and I'm working on non-dimensional  chemical reactions. Hi, I'm Eningen. I'm from Radbaal University and I'm working on the  interaction between oxygen molecules and surface molecules. I'm doing some calculations.  Hi, I'm Marc Sudej. I'm from the University of Amsterdam and I'm doing work on quantum  reaction in quantum gases. Hi, I'm Eningen Kapitza from the University of Birmingham and I'm  working on spin transport in motion metallic also.  And I work on coupling with the spin heat transport in gravity and electrical motion.  I'm Marc Sudej. I'm also from the University of Birmingham and I work on spin transport  in the field and I recently also started a little bit on magnetism.  Spin transport is not magnetism?  Well, yes.  Well, yes, you're right.  My name is Jorma, I'm from the Tractinian Earth Dragon and I work on optical gas in the  field of ultra-pestuous spin sometimes.  My name is Daphne Ruekom. I'm in PUNL. I work on my run-off females in the cookware boxes  and today you are actually in my lab.  I'm Iman, also from DL and I work on integrated quantum object torsors.  I'm Dimak from also from DL and I work on spin qubits and getting more sorry for the  subject.  Hello, I'm Nick from Leijf, University of Birmingham and I work on spin transport in molecules  and then I'm working on the molecular charge in spin transport in qubits and in qubits and  then I'm working on the spin transport in the molecules.  Hello, I'm Davide Grossi and I'm from the University of Birmingham and I'm working with  Tesla.  You don't have much to learn here anyhow.  Great.  So some of you will be really bored if you're working on spin qubits.  You probably know more about those spin operators than I do.  So you'll really be bored and the rest of you, you know, I see some of you will actually  learn something today which is what we want to achieve.  Okay, so let me get going.  I was making a little bit of fun of the magnetism versus spin, right?  But this actually is actually a very, very serious distinction I think between the two  and the distinction really lies in whether you think of them as quantum mechanical or as classical objects.  But classical here, I mean, and that's probably what you also meant is you use something like a  Landau-Liffschild-Gilbert equation or just the direction of the magnetization as the magnetic moment  and you can determine that magnetization direction to any precision, right?  That's in contrast, of course, to the quantum mechanical description where you use spin operators  and you have quantum computing and all these applications in mind.  So maybe that's the distinction that you have.  This is something that I'm actually very interested in is, you know, if you take a single atom, for example,  and you put that on a surface like I'll be using in this discussion today,  then we mostly use quantum mechanical concepts.  I won't be telling you anything about phase coherent quantum mechanics such as you would need for quantum computation,  but we're not going to be trying to get there, but we don't have that with our systems.  So it's all going to be the energy eigenstates, essentially, of such systems.  On the other hand, if you want to do data storage, then you don't really want to do quantum mechanical systems.  You want to have something like a compass needle that's points north or south.  That's your magnetic state that you want to use for data storage.  And so quantum mechanics is a bad thing for those kind of spin systems.  So I'm just in question how you go between, sort of, the million atoms that you have in these things here  to the few atoms where you use quantum mechanics and you can ask yourself how that transition region works,  which is really what I'm mostly working on.  Now, since this is a lecture and I don't really want to talk the whole time,  I really need you guys to break this up, right, and ask questions and give me your thoughts on things.  So what do we want to learn from such a quantum system?  In the end, we want to do basically quantum computation with those.  So in the really far future, I'd like to put atoms on surfaces,  and I'd like to use a scanning and tunneling microscope as the tool to interact with those guys  and do full phase coherent quantum mechanics like some of you are doing if you do qubits in semiconductors, for example.  Now, we don't have this yet. For this, you need to know a lot of things and a lot of detail about the interaction  between this atom here or the spin, and, for example, the environment that is in there.  Now, what we will talk about today to quite some extent is the energy landscape that these spin states here will have.  So here, I'm sketching a spin that is larger than the spin one-half, which you probably are using for qubits.  So here, it's a spin two, which is actually an iron atom in our case.  And this iron atom has five spin states.  Those spin states are not degenerate at zero magnetic field, and we will talk about this in quite some detail.  I want to show you how we can calculate the so-called zero-fuel splitting from some knowledge about the spin operators.  That's one of the goals that I have for this talk, and also how we can then learn something about the dependence  of these energy states as a function of, say, an external magnetic field.  If we have time, then we can also talk about how we can measure relaxation time.  So if you take one of those states and you put it into an excited state, then you might also want to know how long does this spin actually live in this excited state,  how fast does it relax back down.  That's a T1 relaxation time for those of you working in quantum computing, so not T2, not phase coherent, just the energy relaxation time.  And those we can measure with the time resolution down to approximately one nanosecond.  If we have time, we can talk about this again.  And basically, the technique that I'll be using is spin excitation spectroscopy with STM.  So using scanning tunneling microscopy to be sensitive to the magnetic states of those single atoms.  Okay, now I promised for some of you who do quantum computing, this is going to be trivial, so you have to be quiet, everybody else.  This is sort of the introduction to what is the difference between classical spins, classical magnetization description, and quantum mechanics.  So you've probably all learned that electrons have a spin, and you can measure the spin with a Stern-Gerlach experiment.  In the Stern-Gerlach experiment, you take a beam of atoms that have a simple magnetic state, just a single electron.  Essentially, you can think of this as a single electron flying through space if you want.  And you put the single electron in this apparatus, which is basically measuring its magnetization in the z direction, in one particular spatial direction.  I call that zero here.  And because this is a quantum mechanical thing and it has two states, we see two deflections, the beam deflects into two states with equal opportunity.  So you see a signal here, which we call SC plus, and a signal down here, which is SC minus.  In practice, this is done with an inhomogeneous magnetic field that deflects the spins depending on the magnetization of their spin and gives rise to this.  But don't worry about the details, because we're actually not going to do this, we're just going to do this conceptually here.  Okay, so that's not that difficult, right?  What you learn from this is that there is such a thing called the spin, and that the spin has two levels that it can obtain.  Now, this becomes more interesting if you then use a similar apparatus that you've now rotated by 90 degrees.  So now you're measuring the magnetization of this beam of atoms in the z direction.  Here I'm blocking out one half of those atoms, so I know that all of these guys are in the SC plus state here.  And now I'm sending them through the Stangale experiment here in the x direction, 90 degrees rotated, and I get two beams coming out.  I can repeat this now.  I block the x direction here, so block out the Sx minus.  And so now I have SC plus and Sx plus, and so the question then is what comes out of this apparatus?  I've already measured that it is SC plus, I've already measured that it's Sx plus, so if I do the z measurement again, what do I get out?  Are you working on quantum computing?  What do you get out?  Half and half, right?  And you probably learned why this is, but I would like to take a moment and maybe take some time with this and show you why this actually works.  So let me do this a little bit slower.  You have the slide actually in your hand out there.  So really what the, I mean the answer for why this is one half and one half is of course because the Stangale experiment in the x direction has destroyed the information from the Stangale experiment in the z direction.  These are non-commuting operators we're saying quantum mechanics.  But let's look at this a little bit more in detail.  So I'll be very sloppy today.  The Sx operator is defined as h bar over 2, 0, 1, 1, 0.  And we're just going to make this simpler by using just the Pauli matrices here.  So I'm leaving out this one half here and I'm also not worried about h bar.  Okay, this is just, just don't use this for detailed computations at home or something like that.  Now, also what you need to know, the two states that we have, the upstate here is defined as 1, 0 and the downstate is defined as 0, 1.  And so you can just, for example, you can apply sigma x, the upstate, right, the 0, 1, 1, 0, times 1, 0.  You do this, you get a 1, you get a 0, right?  Which just means it's an eigenstate.  That can't be right.  So see, too fast.  Not an eigenstate.  So what comes out?  0, 1, right?  See, you're paying attention.  That's good.  So basically applying the sigma x operator on the upstate gives us the downstate, right?  But what we're doing in this experiment here is we're then blocking out the Sx minus component here, right?  We're blocking out the Sx minus component with basically preparing this guy in the Sx eigenstate, which is, I haven't written this down here,  the Sx eigenstate is something like, I don't know, what's the proper terminology here?  The sigma x upstate or something like that is something like 1, 1, right?  And now we're asking in this last experiment what is the probability of finding this guy in the upstate and the downstate.  And this is just of the 1 over square root 2 here.  This is 1 over square root 2 times 1, 0, plus 1 over square root 2 times 0, 1.  And this is the 1 half that you get, the equal probabilities here, right?  Now, then we can also do something maybe a little bit more interesting here, which is, that's what I just told you here.  So you can ask yourself what would happen in this experiment here?  Basically I've already given you the answer here, I think, right?  So the only difference is we're not blocking out the Sx minus here.  And so the question is what comes out here, is it still 50%?  Right?  So we're preparing the system in the SC plus state.  We're operating on this with the Stegale experiment in the x direction,  but we're not blocking out one of the two beams.  And then we're asking the question, what is your state in the z direction?  50, 50?  What about the quantum computing does?  I don't think it's 50, 50, because the 50, 50 came about because we prepared here an Sx plus state by blocking out the Sx minus state.  What we're doing now is we're doing something that is more akin to actual quantum computation.  We're operating on a well-prepared state here with this operator, which gives us exactly this here.  And then we're asking this state, what is your decomposition in SC plus and SC minus, which is 100% SC minus?  Right?  So it's not magic.  That's really in a nutshell what this whole talk is about.  This is probably the only thing that you're really going to learn,  that you can use these matrices if you apply them halfway reasonably to learn something about these kind of systems.  So we can take our time here.  You see the difference between these two experiments?  It's just the fact whether you're blocking out this Sx minus here.  So this experiment, you don't have to do this at all.  This doesn't give you anything, doesn't make any sense in the end.  You could just leave this out.  That would be exactly the same outcome.  This just means that you're preparing your system in the Sx plus state.  And then you're asking if you are in the Sx plus state, how much probability do you have to be in the SC plus or in the SC minus state?  Which is what this guy asks.  In the other hand, if I do this experiment here, then this one really matters.  I'm preparing the system in the SC plus state,  and now I'm doing something that you would call a quantum computation if you are in this field.  You're operating on this state with this operator here,  and you're keeping the full result of it and then asking what is the outcome of this.  Yeah.  This actually means the exact same thing as having no box at all.  As having no box that does this experiment in the X direction at all.  It does because the outcome of this is the spin down, the flip the spin.  But like if you made up a different position of Sx plus and Sx minus, and you made them separately, you would get the result that would be exactly the same as the box.  Say that again.  If I made an experiment with Sx plus and Sx minus of equal weight,  so like in SC, if this guy here, if you express this guy as a superposition of Sx plus and Sx minus, that's going to have equal weight of those two.  Right.  Well, okay.  No, no, no.  This is what this is all about.  I mean, there's no need to not answer your question.  I mean, I get the point that I'm just trying to make a remark that it's very important to understand what the box actually means.  Okay. What I mean by those boxes is in the end some way to make these matrices, bring those matrices to life.  Right.  So this is sort of, I mean, I didn't make these up, right?  This is standard textbooks kind of, some of them.  Sakurai is my favorite one that actually starts the quantum mechanics with this experiment to show you that classically with the first one.  That classically, of course, the outcome is if this is a classical spin, not a quantum mechanical spin, right, you can measure the z component and you can measure the x component.  That's fine.  Classically, you can determine the magnetization to any precision in all three spatial directions at the same time.  That's what we mean by a classical spin.  Basically, right?  And so the outcome here is that you get SC plus classically, right?  And I've used as an example of why quantum mechanics is different from classical mechanics.  Okay.  Now, I really like your question.  I want to make sure that I don't come across as not liking your question here.  Okay.  So now I want to take maybe 10, 15 minutes to introduce the basic measurement technique that I'm using, which is an STM scanning tunneling microscopy technique  based on inelastic electron tunneling spectroscopy, which tells us something about the energy separations between the ground state and the excited states of these spin systems.  This is going to be a little bit more like a regular talk.  But again, I'd like you to interrupt me and ask questions as we go along.  So the experiment we're doing, some of you are doing transport in graphene or other plain out junctions.  The sense is related. We run electric current through a tunnel junction where we have a single magnetic atom in the beginning or a chain of atoms on a surface.  And we're taking the current from, in the first case, a non-magnetic tip to a non-magnetic sample, but running it through our magnetic atom in the middle.  One of you is working on molecular structures, what?  One of you two, right?  So with the break junction, basically?  In a network, you have a sort of gold nanobarber for network, and then I can insert speakers over molecules in the field.  And how do you measure the states?  In this network, you can stand them on another trans device and then into a complex, you can measure the resistance.  Electrical transport going through several molecules at once.  This is, in some sense, the single molecule equivalent if you want, right?  If you had one of these molecules in between, break junction experiments, for example, can do this to some degree.  Okay, so in the simplest case, if we just have a spin one-half, like these guys over here, you apply a magnetic field to a spin one-half.  At zero magnetic field, it's degenerate.  At finite magnetic field, you have an energy separation.  That's the Zeeman energy.  Zeeman energy is proportional to the external magnetic field, right?  And the property that you can learn here is the so-called G-value, which tells you something about orbital angular momentum and spin angular momentum of the structure that you have here.  Now, I'm going to be using only one material here, which is a thin layer of copper nitride, which is prepared on copper.  And this is a scanning tunnel in a microscopy image, approximately 100 nanometer by 100 nanometer large or small.  And what you see here is in green, the bare surface of copper, the particular spatial direction of copper, 1-0-0 direction.  And you can see that that's basically atomically flat and beautiful.  Then we prepare on these patches of blue here, which are a single atomic layer of an insulator, copper nitride.  That's an area of approximately 10 nanometers by 10 nanometers here, and that's one atomic layer, thick or thin.  And then we put magnetic atoms on there, and here says Manganese copper and iron.  Yes?  So, you can see this picture also telling us something about the height of the copper nitride, so you can collect it in the copper.  Right. It does.  So, the operating mode here is to keep a constant tunnel current.  And so, I'm showing you here the height of the tip as a function of lateral position under the assumption or under the condition that I want the constant current of, say, 1 nanom.  And so, if you have an insulator there, then the tip has to be brought in closer over the insulator to keep that 1 nanolamp of current requirement.  Yes, and the copper nitride is actually laying on the copper.  The copper nitride here physically, actually, this is more likely or more like this.  So, you have your copper here, and then the nitrogen is almost sitting inside that copper.  So, you can think of, and if there's an end to the copper nitride, right, so there's no nitrogen, and here's nitrogen, see it?  And that's basically a planar structure to a certain, the first approximation.  So, it's not sitting on top, it's sitting inside the top layer, and it appears lower here, right?  It appears like it's a step down because of this requirement to keep a constant current.  This we only know from calculations, density functional theory calculations to be precise, but the measurement of the height, we can do that directly here.  So, it's about 30 times less current if you keep the same height over the copper nitride than on the bare metal, isn't it?  Okay, and so then we can take our STM tip and we can park it on one of these atoms and we can run our current through that.  As a function now of the voltage that we apply between the tip and the sample, and this is the conductivity, the IDV here, as a function of this voltage, yes?  How do you make this copper nitride? Is this a matter of adding nitrogen?  Yeah, so what we have is we have a low temperature scanning tunneling microscope, and attached to that we have room temperature, essentially a simple MBE chamber, metal beam, epitaxy chamber.  In the case of nitrogen, so first step is to prepare clean copper.  So there are many recipes that other people developed a long time ago, use sputtering to blast off some of the dirt, and then use heating to bring out, you know, make it a clean, flat copper surface.  So that's relatively easy to do.  And then the nitrogen we have to crack the nitrogen bond.  But if you just let gas into a vacuum chamber, you will not make copper nitride because the nitrogen bond is very strong chemically.  So you need to break that.  We do it very simply by using that same sputter gun that accelerates the nitrogen to something like a kilovolt.  That's definitely enough energy, right?  That kilovolt of nitrogen then hits the surface and gets implanted somewhere in that material.  And if we then heat that up to proper temperatures like 300 degrees C, then this nitrogen comes out and self limits to only a single layer of this copper nitride.  So we're using recipes developed by other people mostly to do this.  But we can do other materials as well, like MGO, magnesium oxide is one that we're working on right now.  So, yeah.  So do these nitrogen atoms just arrange themselves in these 10 by 10 projects?  They do.  So you can see that this is really a fairly regular kind of looking structure, right?  They really want to be a certain size.  And they also keep, if you see here, they keep only one layer of missing nitrogen here.  There's one layer of missing nitrogen. There's one. There's one.  So you can get a fairly regular pattern, nano pattern if you want to call it that.  And so some people are interested in just using this as a template to grow something else in a regular pattern, for example.  For us, that's not that important because we can always find an area where we have copper nitride or not.  But this is limited by the strain. The copper nitride has a slightly different lattice constant than the copper.  And so it wants to relax that strain.  And so if it has space, it will keep some copper in between.  If it doesn't have the space, it will leave just one layer, one roll of missing copper.  And if you even force more nitrogen in there, you can make a continuous film.  But that will also have some defects in it to get some strain relief.  So lots of things that we can learn about the growth of this.  Yeah.  These guys? Those are single atoms that we put there.  So every one of these mountains here is a single atom.  And in this case, these are manganese copper and iron atoms.  Defects are these guys here.  So there's some defects in there.  But again, it's the beauty of a local probe, right?  If you have a defect nearby, just don't look at that guy.  Use a different one.  So in some sense, that's why I like this technique,  because it allows us to make measurements that are extremely well characterized.  I know exactly that this is, for example, a manganese atom on the surface.  So I can tell DFT people, this is the structure that I have.  Calculate for me what does this look like?  What does the atomic arrangement look like that I can't see here in detail?  So you can do these kind of experiments or these kind of comparisons relatively easily.  I mean, if you guys have a bit more technical questions about how this technique works,  we can do that too.  So I have no agenda in which direction you guys want to take this.  OK.  Now, so the basic measurement then is we have our SDM tip parked,  in this case, on a manganese atom on copper nitride, on copper.  And we measure the conductance, conductivity.  Yes?  Tell you about the no-prosthumance of the manganese atom or iron or whatever.  So here you actually see that that's not so easy to tell, right?  I mean, if you look at these four guys, they basically look the same.  So without prior knowledge, they might have a very slight difference in height,  meaning how much current flows through there.  But if you don't know a priori who is higher and not, you will not be able to tell.  So what we do is, we, of course, before we do something that is complicated like this,  we only use one metal atom.  So let's say manganese.  And we look at the surface.  And we look at the spectrum, which I'll show you in the next slide,  for all the different bumps, like 100 different ones.  And of those, maybe 95 will look the same.  And five of them are close enough to defects.  They will look different.  So we just say, all right.  Most likely, this spectrum belongs to a manganese atom.  And then you repeat the same thing with another species,  and then you can definitely tell them apart.  But this is actually not a trivial question at all.  You have to do a lot of work to do that.  Okay.  So back to our manganese atom.  We run current through this, and we measure the conductance that flows through this atom here  as a function of the voltage that we apply.  This is now the basic measurement of inelastic tunneling spectroscopy  that I want to explain to you.  What you see here is the zero is no voltage applied  and corresponds basically to the Fermi energies of tip and sample  being at the same energy level, same energy.  So that's this point here.  We have a constant conductance around the Fermi energy.  And then we get a stepwise increase in the conductance  to a higher current or higher conductance at a voltage of approximately 1 millivolt here.  Yeah.  So first question, why is it not symmetric?  I would say it's pretty symmetric.  Okay. So this is not symmetric.  Well, we can talk about that. Yes.  And then the second question is...  It's a difficult question, though.  Okay.  Is there any... Okay.  The magnetic field perpendicular, is there any particular reason for having that?  In this case here, not very much.  I'll show you an example where it matters very dramatically  which direction you apply the magnetic field.  Okay.  But to give you a quick answer to your first question is,  the difference is that there seems to be a peak here.  Right?  If you ignore the stepwise increase,  if you basically subtract this step from this negative side,  you see that there is a peak here, right?  There's also this slope here and there is this...  So actually there's a peak at the same energy.  And this is, to my knowledge, not fully understood yet  why there are sometimes peaks and sometimes only steps.  It has something to do with higher-order correlation,  something like a condo effect happening,  if you know what the condo effect is.  So higher correlations often...  So this is a single...  What I'm going to be telling you in the next slide,  how this actually works, is a single electron picture  and you need more than these single electrons  to describe this condo effect.  But I cannot tell you why it happens on this side  and not on this side.  So that's actually a pretty non-trivial question.  Not that the other questions are trivial,  but that only means that I don't know the answer, really.  Okay.  Now, how do we know that this is a magnetic excitation?  Well, that's also a non-trivial question, right?  And we really do this by changing this external magnetic field  and you can see that this energy or this voltage  where the step occurs goes to lower energies,  lower voltages, as we decrease the magnetic field.  Now, you see here we have a low magnetic field  scanning tunneling microscope.  So if you had 33 Tesla here,  you would make your step somewhere way out there.  Actually, this is well-matched, though,  to the measurement temperature of 0.5 degrees Kelvin.  So we need the combination of relatively high field  and low temperature to make this visible.  If you had 33 Tesla,  you could do this at 4 Kelvin and you would get similar information.  Yes?  So the in-voltage is this amplitude of the step  corresponding to the electron temperature, actually?  The amplitude? The height?  No, no, no, no, sorry, not the amplitude.  The in-voltage on the next axis.  Yes, yes, I'll show you this in the next slide.  So the basic information is  what is the location of this step  as a function of the external magnetic field?  So this is what I show you here, the magnetic field,  and here is the splitting,  basically from 0 to this energy here.  And so what you see from this is that's a linear increase  corresponding to the Zeeman energy  between the two spin states,  the ground state and the excited spin state  with a slope that you can learn something about this G-value.  It's 1.90 in this particular case for manganese  on this copper binding site on copper nitride,  so that's very specific to the atom  and to the local binding site as well.  But it does tell you that it's less than 2.00,  which is what you would expect for a free electron.  Right? That's the Zeeman,  that's not the Zeeman, that's the G-value  for a free electron, basically.  So it's significantly less than that.  And that tells you something also in a way  that I don't fully understand  about orbital angular momentum and spin angular momentum.  If you go into atomic physics,  if you look up what is the G-value  of a certain state in an atom,  then this G-value will tell you something  about orbital moment and spin moment,  in a not very trivial way, basically.  Yeah?  This basically is lowered because of the environment here.  It is lowered because it actually sits  on this copper binding site here.  It sits like this.  I don't know if I have a DFT calculation in here.  So it replaces one of the copper atoms  that used to be there in this copper nitride.  And so it has these nitrogen atoms nearby,  and these nitrations are negatively charged.  And so it sets up something that we call a crystal field,  local electric fields.  So this environment here matters very dramatically  for these effects.  It's not an atom in free space, basically.  Okay.  But I still haven't explained to you, really,  why we got a step and what we can learn about  the width of the step. Yes?  I really like the fact that you're asking a lot of questions,  by the way. Yes?  What about the offset?  I mean, from your equation, it should go to zero.  Yes.  So it should go to zero.  If you get a splitting at zero magnetic field,  you call that zero field splitting.  That's something that's well known from electron spin resonance,  which is basically the other technique  that can measure something like a G-value, right?  This is a resonance experiment,  usually in gigahertz range microwave fields.  And it's, you know, well-established technique  over 30, 40 years or something like that.  And it can measure G-values much more precisely  than we can here, but it needs, like,  10 to the 10 spins to get enough signal to do this, right?  And so they coined the term zero field splitting  in their measurements.  And I will explain to you in quite some detail  how this comes about for iron on the same binding site.  It's due to the direction dependence of the spin.  So if the spin is pointing in the horizontal direction, like here,  why did I call this a copper?  I was going to call this a manganese.  Sorry.  This was supposed to be a manganese at me, okay?  So if this manganese spin is pointing  in the perpendicular direction,  it has a certain energy.  If it's pointing in an in-plane direction,  it might have a different energy.  And that's set up by the local symmetry of your binding site here.  And that difference in energy,  we call that magnetic anisotropy,  that is something that gives rise to these finite energy offsets.  And I want to take quite a bit of time, actually,  to explain how this comes about using something like the spin operators  in a few slides, basically.  Okay.  Sometimes what I do in talks is I say,  everybody always says I want questions, right?  And then nobody asks questions.  So what I do quite frequently is I stop talking.  I say, I'm only going to talk for five minutes,  and then you're going to have to ask questions.  Unfortunately, I don't have to do that with you guys,  so keep up the questions.  Okay.  So that offset is due to magnetic anisotropy.  Now, before we go any further,  I want to explain to you how this spectroscopy actually works.  And so here I made a slide where I have four different cartoons.  They are all cartoons, basically, not reality, right?  All cartoons by trying to motivate  why we see the symmetric stepwise increase in the conductance here.  And so here I have the field states on the tip,  the field states on the sample.  Sample in my terminology, I mean the copper.  The copper underneath the spin, basically.  And here I show you the energy of the electrons that we have here  in relation to the energy splitting between the ground state  and the first excited state.  So let me just start this.  So if we increase the voltage between tip and sample,  that shifts the Fermi energies.  That's what we mean by applying a voltage between these two electrodes.  And nothing interesting happens.  We got a constant conductance in DIDV as we increase the voltage  until this situation is reached where the maximum energy available  is equal to the energy difference between the ground state  and this first excited state.  All right?  When this case is reached, then this electron up here  sometimes will give off energy during the tunneling process  and come in as a colder electron on the other side.  And somebody has to be available to take that energy,  because obviously we have conservation of energy.  And this is actually possible by this local spin here, the manganese atom,  switching from the ground state to an excited state.  Yes?  How do you know that this is a magnet, with the spin, why couldn't you go to the nitrogen atom, or is it copper atom in this case?  In the end, that's a very good question.  I mean, of course, we do this experiment also on the bare copper nitride.  We don't see anything.  We do it with, say, a copper atom instead of in place of the manganese atom.  The copper atom happens to be non-magnetic in this case,  which is also not a trivial statement.  We don't see anything.  So we do these kind of control experiments.  And then in the end, I will not, this is again a simplified picture,  but I can't really tell you this.  All I can tell you is that this cluster, whatever it is,  that might be the manganese together with the nitricians here,  that that together behaves like the spin.  So it's not atomic physics at all.  It's really this guy together with its local environment that is doing that.  Let's see.  Okay, so when we have enough energy, we can make a change.  This guy can flip.  And now what you see is that we have two ways to make it across.  We can tunnel elastically and we can tunnel inelastically.  So without any proof, you would say that it's much, not much,  it's easier for this electron to tunnel than if it cannot make this excitation.  It changes the conductance, is all I'm trying to say.  And it increases the conductance usually, not always.  And then if I further increase the voltage,  all the additional electrons can make this excitation and this guy gets kicked  more and more often from the ground state to the first excited state.  So it's not a peak in the conductance at the voltage corresponding to the Zeeman energy,  but it's a stepwise increase.  In your picture it's coming back because it's relaxing.  Right. So this only works if there is a sufficiently fast relaxation mechanism  from the excited state back to the ground state.  So I'm basically, this only works if this relaxation mechanism is faster  than the average time between tunneling electrons.  We usually do this experiment at tunnel currents on the order of one picoamp to one nanoamp.  One nanoamp corresponds to 10 to the 10 electrons per second,  which is one electron, what is that, every 100 picoseconds.  So you need to have relaxation that is on that order to make this possible.  I don't have this in this deck of slides, but we can actually get to a regime  where the relaxation is not fast enough anymore,  and then this breaks down and becomes a bit more interesting in some sense.  Okay. So it's symmetric because you can reverse this process.  You can apply the voltage the other direction.  The same thing should happen to first approximation.  And so you should see a stepwise increase on the other side.  This, by the way, is the hallmark of inelastic excitations.  That is really the difference between zero and a finite voltage or energy that matters,  and it's not the absolute value of this energy.  That's one way you can distinguish between so-called elastic excitations  and inelastic or elastic tunneling processes,  changes in the density of states, for example.  This is not too dissimilar to superconductor tunneling spectroscopy.  Pretty similar in that respect.  Okay. Now, your question was earlier, what's the width of this guy here?  The width is set by the width of the Fermi energy in the sample  and the width of the Fermi energy on the tip.  At zero, this would be infinitely sharp.  At finite, we actually get a broadening that corresponds to 5.5 times kT  because we have two Fermi functions that get broadened here.  And that also explains to you why this experiment is experimentally difficult  because most people have magnetic fields only on the range of a few Tesla.  Very few people have 33 Tesla available.  And so you need to make sure that this energy here is bigger than your thermal broadening.  So we do this at 0.5 Kelvin and magnetic fields up to 7 Tesla.  When I started at IBM, my first task was to build one of these machines.  There was no commercial machines available at the time.  It took about 10 years or so for commercial machines to become available.  Now, they are available if you have a million dollars or something like that to spend.  Okay. So what can we do with this?  We can do tunneling magneto-resistance.  I'll tell you what this is in a moment.  We can measure magnetic anisotropy, which is the zero-fuel splitting.  I want to spend a lot of time on that.  We can measure spin-spin coupling, put two spins next to each other,  and learn something about how the combined system works.  I want to spend a lot of time on this.  And if you guys keep up the questions, which I hope,  then that's going to be the end of what we have time for.  We can learn something about strong interactions between the spin and the substrate,  giving rise to these peaks, which gives you something about the condo effect.  I definitely will not talk about this.  We can drive the system out of thermal equilibrium,  which is basically if the relaxation time is not fast enough.  We can learn something about the relaxation time.  And we can also measure this relaxation time directly.  I do want to show you this at the very end a little bit,  because it's kind of a fun technique.  Now, so we'll spend most of our time on these first three issues here.  Okay.  So, Tanneling-Magneto resistance.  So, the Nobel Prize in Physics was given to two gentlemen,  Albert Ferd and Peter Grunberg.  I don't know what's his first name.  Peter Grunberg.  A few years ago, maybe four years ago or something like that, five years ago?  No way.  Old people always tell you, you know, it's amazing how time flies, you know.  It's scary.  So, it was given for the development of transport in solids  that is dependent on the magnetization, on the spin direction of magnetic materials,  so-called magneto-resistance.  Now, this was done at first by growing magnetic layers, thin films,  and seeing that if you change the relative orientation of these films,  you can get a change in the conductance.  This is actually what's used.  I still have a computer that has a hard drive here.  That's used as the read mechanism in every hard disk drive read head.  Actually, we use Tanneling-Magneto resistance.  So, we put a insulating barrier between these magnetic materials,  and that's how you read the state of your magnetic bit in your hard disk drive.  Now, we want to do something similar here on the atomic scale,  and let me show you a few experiments how we can do this.  This also answers a little bit your question how we know who is who in this case.  So, now we have a smaller image here, 6.5 nanometers of copper nitride.  That's the blue area, bare copper.  That's the green here, and four atoms on the surface,  which I've already labeled here as copper, iron, and two manganese atoms.  So, we can take this tip, we can park it on the manganese,  we do the spectrum that I just explained to you.  We get a stepwise increase at finite magnetic field, again 7 tesla here,  at approximately 1 millivolt.  Now, what we actually learned by doing a lot of experiments actually,  is that we can bring this tip into point contact,  bring it in very close, and applying a fairly high voltage at the same time.  And if we do this right, we can transfer this atom from the copper nitride to the tip.  This is actually very simple in this cartoon,  but in reality, that's a very difficult experiment to do.  Because when we do this, if I go back,  when we do this, at this point here, we apply two volts,  and approximately 10 microamps of current running through a single atom.  So, it's not surprising that you change something.  I mean, if you do that, two volts and 10 microamps through a single atom,  you should change things.  The only thing that's surprising about this is that sometimes,  you can actually make this a reproducible process.  Now, how do we know this?  Again, not a very, very non-trivial question.  One thing we know is we can take this tip now, and we can take an image of the same area.  And of course, one thing you notice is that this atom is missing.  So, we can conclude from this.  All right, we can make atoms disappear.  All right?  How many times does it work?  That's a difficult experiment.  Anna probably does similar things, so...  Yeah, it really depends.  It really depends on the shape of the dip.  So, if you have a good shape, it can work every time you do it.  But if you have a bad shape, it gets bad like hours to get it worked.  That would be my answer, too.  So, we can easily spend two days where it doesn't work.  Ever, basically.  We can always make the atoms disappear.  They can always disappear. That's easy.  But to make it go to the end of the tip, and then in particular to get it back off,  that's a difficult experiment.  I'll show you on the next slide.  Very similar process, basically.  Okay, so first thing is this guy is missing.  Now, second thing, you also see that these atoms,  the remaining atoms, look better.  They look more round, right?  So, this is a more point-like SDM tip,  because in the end, the SDM image is always a convolution of the shape of your tip  with the shape of the surface, right?  Nature doesn't distinguish between tip and surface.  So, if you have a very sharp atom on the surface,  then really what you're seeing is the shape of your tip.  The shape of the last few atoms of your tip, right?  So, that's what this one is. Not a very nice image.  You see this atom is slightly deformed,  because our tip has two atoms that are close to the very end instead of one, right?  But here you see it's better.  So, maybe two points for this mechanism, right?  The atom is gone, and the atom image looks sharper.  But then we can also take this tip now and measure a spectrum on the remaining manganese atom.  And what you see then is the following.  So, here you see an inelastic tunneling spectrum measured on the same manganese atom,  but now with a modified tip.  And what you have here, again, at minus 1 millivolt, you have a step,  and at plus 1 millivolt, you have a step, but the height of these steps is different, right?  This step is about half as tall as this step over here.  So, the amplitude has changed, the energy has stayed the same.  Okay, so then how do we do the reverse?  We, again, we take this tip and we bring it back down to the surface,  point contact, several microamps of current,  and now we apply a polarity difference, opposite polarity,  and then sometimes this works.  This is a difficult process, actually, to get this back off.  And, you know, I work for a company,  so we actually do not spend a lot of time trying to understand why this works, right?  We don't have time for that in the end, because I can't, you know, I don't have a small group.  We always have to focus on the highlights basically, we can't focus on this,  because in the end, who cares, right?  If it works, it works.  My bosses don't care. I can guarantee you that.  Okay, and so then we can take this empty tip and we can image again,  and you can see we can go back to exactly the same shape of this tip without the manganese atom on it,  but this manganese atom has moved from here to here, right?  So we can move this manganese atom by a few lattice sites,  actually two or three in this case, by using this manipulation process, right?  Now, if you have a tip that works, the biggest structure we've built with total precision was 96 atoms.  With the same tip, we could do this 96 atoms, 96 times in a row, without messing up, basically.  So if it works, it works.  Okay, so go back to the same atom, we can measure its spectrum again,  of course we go back to exactly the same spectrum that we had to begin with,  because we have exactly the same tip and exactly the same thing on the surface,  and that's again, it's the beauty of this, right?  If you know it to this precision, what you have, everything has to be exactly the same.  We can pick up a copper atom, which was the guy up here, so we now have a copper at the end.  You see that the spectrum is essentially exactly the same as what I call the empty tip here.  It's not 100% the same.  If you look very carefully, you can see that there are slight differences.  I don't know if you can see the brown curve underneath there, that's the old one, right?  Because the density of states changes a little bit between having a copper atom there and not having a copper atom there, right?  But to first approximation, the height of these steps is again identical.  And go back to the empty one, back to manganese, we can do this again many, many times.  Okay, so how does this work?  One problem that we have, or one complication that I haven't mentioned to you yet,  is that the manganese atom is not a spin-1-1-2 particle.  Manganese, if you look at the periodic table, is in the middle of the D-shells, of the 3D shell.  So if you just look up at the atom periodic table, you have five electrons in the D-shell,  giving rise to a spin of 5 half and to zero orbital angular momentum.  That's what the periodic table tells you, basically.  So if you have a 5 half atom, you should have six excitations or six energy levels in the Zeeman energy diagram,  but we only see one excitation.  That's kind of weird, right?  So we have some selection rules at play here, and I want to explain that to you.  Now, the selection rule stems from the fact that we're actually using electrons for this excitation mechanism,  and it's actually not so difficult to understand.  So the electrons, of course, are spin-1-1-2 particles.  The tunneling electron is a single electron, spin-1-1-2,  and we have to conserve not only energy, but we also have to conserve spin angular momentum.  Those are two, I mean, not spin, but angular momentum has to be conserved,  unless you produce orbital angular momentum, the spin has to be conserved, basically, in this tunneling process.  So really what we can do here is we can exchange spin between the manganese atom and the tunneling electron,  and so really this is an exchange interaction between these spins,  and so that means that we can only go from the ground state to this first excited state.  The electron here, the tunneling electron, just does not have enough orbital moment  to get you from 5-1-2 to 1-1-2, basically.  Can you, if you do it fast enough, can you then excite the excited atom and excite the excited atom?  Yes, yes.  So if we don't give it enough time to relax back down from this state to the ground state  before the next electron comes about, right, then we can drive it from this state to this excited state, right?  So sequential electrons will do that, basically.  Practically also working, so...  Yes, yes, yes.  And, well, I don't think we will have time, but we did this experiment a couple of years ago,  and this tells you something about how fast this relaxation back down to the ground state is,  because we can make this same measurement as a function of the tip sample distance.  So if you say at 1 millivolt, you know, at some height I have 1 nanohamb of current,  and if you bring your tip in closer, at 1 millivolt you have more current,  or if you pull your tip back, at 1 millivolt you have less current, right?  So that's something that people that are doing transport in, say, planar tunnel junctions, they can't do that, right?  You basically have a tunnel junction that has a certain resistance,  given by the thickness of your insulator, and so that's it.  So at 1 volt, you always have the same current, basically.  We can have this freedom by changing this distance.  Okay, so why does this give rise to this asymmetric spectrum here?  It's not so difficult to understand if you allow me to make a simple simplification here.  If you think that this manganese atom, together with an external magnetic field, on this tip,  you know, it's sitting on the metal here, so let me make this simpler.  Let me ignore quantum mechanics here, and let me use quantum mechanics here, right?  They're both atoms. That's not a trivial statement, right?  This is a pretty drastic approximation.  But if you think about how, what you learn about spin polarized currents, usually in solid state physics,  then what you mean by having a spin polarized material, like cobalt, for example,  what you learn is that you have a different number of electrons of one spin flavor at the Fermi energy  than the other spin flavor.  But usually you say you have more majority spins than minority spins at the Fermi energy,  and so your transport will be spin dependent, spin polarized.  Now, if we say that this tip behaves like this, so I have fewer or minus one-half electrons  and more plus one-half electrons, then you can understand why the spectrum becomes asymmetric.  Now, we have elastic current going from tip to sample in this window here.  We have elastic current going in the opposite direction.  I'm just changing the polarity here of the bias voltage.  More interestingly, if I go in the inelastic window here, I have to start as a minus one-half electron  because I want to transfer angular momentum here.  I have to start as a minus one-half electron, oops, and I want to end up as a plus one-half electron.  So in this direction, I can do this with a certain probability.  In this direction, I can do this with fewer electrons because I have less initial states available here.  And the beauty of this is that this allows us to measure this degree of spin polarization here,  usually defined as the density at plus voltage and minus voltage, minus this,  divided by the average of the two.  So we can assign the spin polarization directly to this asymmetric spectrum.  The degree of asymmetry tells us what the spin polarization is, giving this a simple formula here.  And in this case, this comes out to be approximately 0.3.  Spin polarization 0.3 means that you have twice as many electrons of one spin flavor than the other spin flavor,  just using this formula here.  Now, this is interesting because, A, it tells us that it degrees well,  that we put a magnetic atom on this tip and we apply a magnetic field to magnetize this atom, basically.  That makes a spin polarized tunnel junction.  And it allows us to quantitatively measure what the spin polarization is.  This is very difficult in practice in other materials.  If you ask somebody, some of you are doing spin transport, right?  How do you know what the polarization of your, you use cobalt electrodes?  How do you know what the spin polarization is of your cobalt?  It's very difficult to measure, right?  It's very difficult to measure.  I mean, people have done it by measuring, using a superconductor and pushing magnetic material into the superconductor,  using those kind of devices, but it's very difficult to measure.  So here we have a way to measure this for these simple tips, basically.  Yeah?  Maybe I missed it. Why don't you have from minus or half to minus or half?  Because the selection rule is basically that we start with the atom in the ground state  and we need to transfer a certain direction of angular momentum to this, right?  If you would transfer the opposite amount, you'd go from this state down one state,  but you don't have a state below that, basically.  You were first.  My question would be, is spin polarization, is it typical for manganese or can you also, well, change it?  Should I get another value there?  It's somewhat typical, but not reproducible in the sense that every time you pick up a manganese atom,  you will get exactly that spin polarization.  Of course, it depends on magnetic field over temperature, right?  At zero magnetic field, you will not get any spin polarization to first approximation, right?  You need to polarize, you need to magnetize, quote, unquote, the single atom here,  Curie-Weiss law, basically, right?  So just B over T.  So in the saturated regime, so if you apply more than one Tesla at 0.5 Kelvin,  you are in the saturated regime, and then I would say it's typically in that range.  If you pick up more magnetic atoms, you can make it bigger or smaller than this, right?  And it does not depend very dramatically on whether it's manganese or iron.  Can you, because we are coming from one manganese to the other one, can you have them aligned in anti-barrel, for example,  because now if you have to apply a certain saturated field, all your manganese atoms are in the same direction, right?  Not in the ground state, basically.  In the ground state, they will be aligned, and you get a certain current, right?  So the answer to that is really no, in the sense of having a steady state, you know,  tunnel junction where you can do this real experiment where you change one orientation versus the other.  We can do a time dependent, right?  If we kick this guy out of the ground state, then this will be rotated, if you think about it classically,  and your current will change for a short while until it relaxes back down to the ground state.  Can you, for example, just pick up another manganese atom trying to change the, for example, field, or change the total spanning for that guy?  In practice, the answer is no.  You know, but what other people have done is built structures of, say, five or ten atoms on the surface,  a big blob of iron atoms, for example, and then those will be a steady magnetic state in a certain direction  that's paramagnetically coupled to each other, and then you could have this.  If you could transfer that to the tip, then the answer would be yes, but that doesn't actually work that way.  So the subject is also the same, and that depends on the substrate?  That depends extremely heavily on the local environment around these atoms.  Could it be different from the one in the tip?  It's very, very, very different, guaranteed to be different.  Yes.  And we don't know it, right?  I told you so, you know, big word that we know everything about the structure on the surface.  We don't really know much about the structure of the tip, right?  So the tip is always a black box for us, in some sense.  I'm interested in what your next slide is.  How do you know my next slide?  Yeah, you're a student.  Oh, okay.  So I understand the picture that the high-intensity amplitude is observed by the density of states, but the topic also seems to indicate that there's an energy difference.  So it takes more energy to reach the mindful mouthpiece.  This energy difference here?  Yeah.  It's, yeah, it's, it's, this is supposed to be a picture.  I agree with you.  This is not, not the greatest picture.  Let's do this over here.  So if you think about what happens when you apply a magnetic field to a normal metal, right?  You have your parabola of states, basically, right?  Energy versus K, for example.  Well, I don't know if K is good here.  And now, and this is spin up and this is spin down.  You take a normal metal, not a ferromagnet, but a normal metal, and you apply a magnetic field to this.  You get the Zeeman energy of the electrons in the metal, basically.  And in some sense, what you do, the first approximation is you just shift one of these parabolas down and keep the other one at the same, at the same place.  And you could do this change, you could change the Fermi energies or you can change the bottom of the space here, right?  Now, what happens, of course, is that the Fermi energy will stay constant because you can get spin, the spin is going to, I mean, the Fermi energy is going to be the same, basically.  You're not going to build up a voltage inside your material by applying a magnet to a normal metal, I mean, applying a magnetic field to a normal metal, right?  And so, I just draw this, basically, as a shift in this parabola.  So, I shift one of these parabolas here by the Zeeman energy.  So, this shift there would be the Zeeman energy, right?  So, this distance here would be the Zeeman energy in the metal, basically, for the quasi-particles.  Don't take that too seriously. I mean, this is really meant to be a total cartoon, right?  If you don't like this parabola, I'm perfectly okay with this.  But just say that you have a difference of the number of electrons at the Fermi energy of the two different spin flavors, right?  That's really all that's supposed to mean. Maybe that's taken the cartoon too far.  I see what you mean.  I had one question about, so you don't say you don't know what the surface tip structure of the tip is.  But you also said when you're probing the resonance, you're basically probing the state.  Yes, you're imaging the tip to some degree.  So, we learned something about it. And you could, in principle, do sort of a deconvolution and take the image that you get, maybe for different atoms, right?  If you have one atom that's very tall, one atom that looks very short, you know, you could learn something.  We haven't done this in practice. I don't know. Has anybody done that? Probably not.  I mean, it's also about structure. You need sort of a good structure for your tip to make this process work.  So, hand-waving, yes. Hand-waving, you know, you learn something about, all right, this tip might work just because it looks like this.  And then it works half the time or something like that, right? I mean, so it's not, that's not really scientific.  No, it's not.  No. I'm sure that, I mean, it is. In the end, it is. But we haven't put the time and effort in that.  Okay. Now, I think we already went through this. But as I said in the beginning, I do want to spend a little bit of time on these operators  trying to explain to you now how we can understand these experiments a little bit more, you know, from a theory point.  Now, what I'm going to be telling you sort of in the next few sections is not something that a theorist would actually tell you,  because it's not really theory. This is trivial for a theorist, right?  But it's just a little bit beyond what normal experimentalists learn about spin operators.  And so I think that's why it might be worthwhile for some of you.  Okay. So, first step here is how can we model this, basically?  So, we start with our spin manganese being in the ground state, M equals plus 5 half.  And we ask if I have a tunneling electron that starts as a spin-down electron, what can it do?  It can flip and become a spin-up electron, but we have to conserve angular momentum,  and that means that the manganese has to go to an excited spin state, right?  That is the inelastic tunneling channel.  But you can also do the following process. You can keep this down electron as a down electron.  And the manganese, of course, stays as M equals plus 5 half and remains an M equals plus 5 half.  That's a perfectly allowed transport process as well.  You could take an up electron and start in the ground state and keep an up electron  and still be in the ground state of the manganese.  And the fourth channel, you start as an up electron.  I think you asked that. Somebody asked the question.  And we start in the M equals plus 5 half.  We flip this to a down electron and go to an M equals plus 7 half.  That state doesn't exist. So, this probability is zero, basically, right?  So, we have three different probabilities here.  And can we learn something about what happens, basically?  Can we do some modeling here?  And through quite a bit of work, what we actually learned is that it's actually not that difficult.  So, the tunneling electron, which is the state that is either up or down,  and I'm not using full quantum mechanics here, so there's no, you know, coherent manipulation or something like that.  So, this is either up or down the tunneling electron in the copper of the tip,  or in the tip and in the copper in the metal, right?  So, this is, this is say the tip here and this is, this is in the metal underneath.  So, that's the tunneling electron operators.  And then the operators corresponding to the local atom, the manganese with its local environment here.  So, that's, and if for those of you who are familiar with this, this is an exchange interaction.  You know, S1 dot S2 times usually a parameter like J,  that's called the Heisenberg exchange interaction,  and tells you something about exchange of electrons, interactions between electrons.  And I want to spend a little bit of time understanding what these operators actually mean.  Now, we have here something a little bit more complicated.  We have an additional potential scattering term.  And I can't explain to you why this comes about.  Theorists will explain to you why this comes about, okay?  But I know that we need this parameter to do good modeling, right?  So, I'll just use it here.  You can grill me later if you want on why this works.  Is it zero field or what is it?  Do you fit it for this to something?  Yes. So, we fit, there's, there's, this holds at any magnetic field.  And we basically fit it, you know, for example, this u parameter is a single parameter,  one number basically is a real number.  And we fit this to a certain set of measurements, for example, as a function of magnetic field.  But we want this parameter to be constant, independent.  It should be a parameter that only depends on having manganese in this particular configuration.  Okay, this means u is x basically.  And then everything else should come out of that basically.  Okay, so this term here, sigma dot s is actually a pretty difficult animal,  if you look at this a little bit more carefully.  Because this dot s, what that really means, this is just a definition of what sigma dot s means.  It means sigma x sx plus sigma y sy plus sigma z sc.  All of these are operators in this sense.  Right?  That's a pretty non-trivial, trivial term if you really want to calculate that.  Right?  And let me see, I think I have this written down here in a little bit slower,  because I want to take some time here and bring this to life a little bit.  So, where is the term?  So, okay, so what we have there is, in the end, let me just write this down.  So we have four cases here.  I don't think this is in the slides.  I'm not sure.  But I'll give those to you.  Actually, I don't think I have this in the slides.  So you can either get my handwritten notes or if you care about this,  you probably have to write this down.  So, the sigma initial of the tunneling electron and the sigma final,  so it can be from up to up, from up to down, from down to up, and from down to down.  Those are the four cases that we sort of evaluate as we do these simulations.  And again, I'm ignoring quantum coherent kind of states here.  The electron in the tip could be in a simple position of states,  but I'm not dealing with this at all here.  So just sort of classical magnetization states.  So let's just take an example here.  So one of these probabilities here is up electron as final,  sigma plus as minus, sigma minus as plus, sigma z as z plus u up as initial.  All right?  So if I want to calculate what's the probability of this process happening,  the electron starts as an up electron and ends as an up electron.  This is just one of the four cases that I have to calculate basically.  So that's this case over here.  Then you'll see that what happens with, you guys are probably familiar with sigma plus and sigma minus, right?  That's sigma x and sigma y.  Those are raising and lowering operators, just like you learned in basic quantum mechanics.  So, but you know that if I operate sigma minus on the upstate, I get the downstate.  And then I multiply or I check the downstate overlap with an upstate is zero.  If I operate sigma plus on an upstate, I get zero.  So that tells you that these two terms don't play any role in this particular case.  So I can simplify this by saying this is s sigma z sc plus u up as initial.  Absolute value squared.  So you can do sort of these simple terms.  And so you learn that if you don't change the spin of a tunneling electron,  the plus and minus terms don't play a role.  On the other hand, if I do the opposite, right, if I do change the spin to down s final,  sigma plus s minus plus sigma z sc.  This by the way is the part where you can yell at me and say this is too slow if you want.  I can do worse handwriting, but not much.  If you want to calculate this one here, now you have to find the term where the up electron  changes to a down electron by applying these sigmas, right?  And that is which one?  That one, right?  So now I have down s final sigma minus s plus up as initial.  Absolute value squared.  If you want to calculate basically what are the occupations of these six m states I have  for my manganese, right, in the finite magnetic field, you have to keep track of these processes  and do something like a bookkeeping of which state are you in.  Maybe you start in the ground state, you run an electron through,  with a certain probability it goes in the first excited state.  Maybe now you do a relaxation mechanism or not, right?  And then you keep track of this and you calculate what is the current as a function of the occupation  of these states, following these rules here basically.  On the other hand I can simplify this even further because I know that sigma minus operated on up  and down excluding an h bar or something like that.  I don't know the h bars, I don't have these written down here, right?  That is s final, oops, s plus s initial, right?  I've basically applied the sigma minus here to this upstate.  I get, you know, excluding factors of two or something like that,  I don't have all these factors written down and an h bar probably, right?  You get the downstate here and then you can evaluate this  and you get a certain value out, right?  And so you are left with this process here.  So basically what this tells you is by doing some bookkeeping,  you can keep track of the tunneling electron, what the tunneling electron is doing  and you're left with some operators that depend on the manganese atom that is sitting on the surface, right?  And that's why in this description here I call this the elastic tunneling.  That's basically the tunneling where the tunneling electron does not change.  If it starts as an up, it ends as an up.  If it starts as a down, it ends as a down.  And the inelastic one where it actually does a flip basically, right?  So the inelastic one have the x and y which is the same as plus and minus operators  and the other one has this sigma z in it.  I don't have this written down any better than that, right?  I mean I think you get the idea how you can do this kind of modeling.  This is basically an exchange direction for a certain amount of plug.  We really don't make any, yeah, we treat this in a very funny way if you think about this.  Are you a quantum dot person?  In a quantum dot you actually think about the electron hopping on the dot and hopping back off, right?  Here I'm not doing this basically.  The electron is zapping through there and it leaves behind an excitation or not.  And I'm asking, this is more like a scattering process in that sense.  I don't really tell you that the electron lives on this atom here for 10 femtoseconds or something.  I have no idea, right?  I can tell you if the electron goes through and it makes an excitation in the spin,  how long does the spin live?  I can measure that.  But I can't tell you how long this electron that actually went through there lives on this atom.  I can't even tell you if this is an allowed question to ask.  I know this is a good hand-waving.  I really don't know the answer to that question.  It bugs me.  I don't really know how this transport process really works, obviously.  But we do know that it actually looks like it does this exchange interaction, obviously.  That's really what is happening from experimental measurements.  Is this something, I mean, I guess you can sort of fit that effect with the gate  and this effect of exchange interaction?  Yeah, that's in the end somehow what this U parameter does.  It's the ratio of this exchange interaction to a potential scattering.  It's only one parameter that matters.  It's not two parameters.  On the strength of the tunnel current?  No, it does not.  Yeah, I think the answer to that is that the tunneling process is always slow  in comparison to the transition time of this electron.  A while ago, just before you guys were born almost now,  but before you guys started physics, basically,  people were wondering how fast does an electron tunnel?  What's the time corresponding to a tunneling process?  And I don't really know what came out of this discussion,  but that's at least 15, 20 years ago, that was an unsettled problem.  And I don't know if this is settled by now or not.  Can you actually assign a time to such a tunneling process?  I don't know.  Anybody know?  I read about this a while ago.  It depends on the transfer of the wave function and the overlap.  That's how we can calculate, basically, how much current will flow.  That's the overlap of the wave functions.  But it doesn't really tell you anything about the time that takes that tunneling process.  It's a really non-trivial question, I think.  So, you know, this to excluding a factor of one half or something like that,  oops, sorry, you can actually equate sigma x sx with sigma plus s minus and so on.  That's true to a factor of one half or something like that.  So that's the same thing, basically.  Okay, so I'll tell you a couple slides how we can measure this, this u, basically.  And then I want to go into this zero field splitting.  So with a non-spin polarized tip, you know, we get our manganese spectrum  that we talked about quite a bit now.  And here we have elastic tunneling and inelastic tunneling,  and the inelastic tunnelings are the same because we have a non-spin polarized tip.  If we take a spin polarized tip, we have more inelastic current in this channel  than in this channel, right, that's just the polarity of the bias voltage.  And sometimes, I think one of you asked, can we make different tips?  By picking up more atoms, yes we can.  And sometimes we can even make inverse spin polarized tips.  If we pick up a lot of atoms, maybe we'll make an anti-farrer magnet on the tip, who knows?  We don't know.  But sometimes we actually see that the opposite argument holds,  and so, you know, basically what this means is that the end atom on the tip  is anti-aligned to the external magnetic field.  And maybe that was the question, can we make this?  Sometimes we can, but we can't do this reproducibly.  I can't say, all right, if I pick up 10 manganese atoms, I will make this.  It just happens to be sometimes the case.  Okay, now this U, not from this measurement, but from something similar,  let me just skip that part, we can calculate this U parameter,  the quantity that this belongs to, okay?  And for the manganese atom, the dashed line here is the value  where you get the maximum tiling magneto-resistance.  And that I will show you in a calculation in a couple slides.  So for manganese, the manganese atom looks like it wants to be fully spin-polarized,  basically.  Whereas an iron atom, that line here is 1.0 in our parameters, right here is 1.25.  That's just because the iron is not spin-5 half, it's spin-2, right?  So we didn't use a proper, you know, that, I mean, the difference between 1 and 1.25  is just the magnitude of the spin, right?  That's just the way that, I'm not a theorist.  A theorist would do this better, and this would be, you know, always 1 or something like that.  But you can see that the iron atom does not correspond to this maximum value of U,  and so this U really has some physical meaning.  It tells you something about how this tumbling process works, I think.  But what it tells me, I don't know.  So it really we treat it as a fit parameter, basically.  A fit parameter. But only one parameter, right?  If you take manganese on this binding site on carbonitride, you have a U of 1.25.  You take iron on the same binding site, you have U of 1, basically.  Okay, so we can determine, all you can learn from this, we can determine this value.  And the histogram here is taking different atoms, measuring this, you know, about 100 times  for different atoms, doing all this fit procedure, and you can see that it does vary a little bit.  Who knows why? But on average, this is what comes out.  I want to spend a moment on these two, both of them.  So I'll have to figure out how to do this.  But, okay, so magnetic anisotropy.  We talked a little bit about that zero field splitting, right, for the iron atom.  At zero magnetic field, there seemed to be some energy required to make an excitation.  And I want to tell you in a little bit more detail how this works.  And I'm going to use a different atom where this is much more dramatic than the iron.  Now, what this all comes down to is, of course, if you have only the Zeeman Hamiltonian,  at zero magnetic field, all the states have to be degenerate.  This is proportional to B, so at zero field, B is zero.  If you take the spin and you break the rotational symmetry of space,  and you can do this either by putting this spin on a surface like I'm doing here,  or by putting the spin in a molecule, or doing anything else that breaks the rotational symmetry of space,  then in general, the energy of the eigenstates will depend on the direction that the spin is pointing in.  It is allowed to depend on this direction, basically, right,  because the local environment is making the directions different.  And this gives rise to two effects.  One, it gives rise to the zero field splitting.  And I will use this kind of picture to describe the zero field splitting.  This is based, again, on these quantum models, and it's called a spin Hamiltonian,  which is used all the time in electron spin resonance.  This is how people describe electron spin resonance.  But it's not so well used in normal solid state physics, I would say.  So molecular magnets people, chemistry people, they use this all the time, basically, as their picture.  Now, here's the first term that plays any role in this.  dSc2.  So now the z is not the direction of the magnetic field.  It's the direction given by, for example, the surface, perpendicular to the surface.  In the manganese case, I applied the magnetic field perpendicular to the surface,  because that's the easiest case, because the spin wants to point perpendicular to the surface.  Now, what happens here, it's an sc2 term.  So you see that if I apply my magnetic field along the same spatial direction,  then this is a term that is b times sz.  And it's sc and sc2, they are commuting operators.  So that means that the energies just add up, basically.  That's this picture over here.  So you have a certain energy at zero field, and on top of that, you add the linear term of the Zeeman splitting.  If I apply this magnetic field in a different direction, let's say x, right?  Now I have sx here and sc2 here.  Now, sx and sc2 do not commute, and then the world is more complicated.  So as I increase this term here, the Zeeman term, I'll get some complicated dependence  of the energy levels on the magnitude of this magnetic field in this direction.  So having magnetic anisotropy in my pictures gives rise to zero field splitting,  and it gives rise to a direction dependence of my magnetic field.  And they are very closely related to each other.  Now, let's do this experiment. We take an iron atom on the same binding site on copper nitride.  So all I've really done is I replaced the manganese atom with an iron atom.  And I take the spectrum here at zero magnetic field and at low temperature,  and you see something quite dramatically different.  Remember, the manganese atom had just a tiny little feature at zero magnetic field,  a tiny little dip, basically. This guy, and this is plus minus 10 millivolts.  So this is 10 times bigger energy scale, basically, and it has three excitations at zero magnetic field.  One at 0.2 millivolts, one at 4 millivolts, and one at 6 millivolts.  So in some sense, the zero field splitting here is dramatically bigger than in manganese,  about 500 times bigger to be precise.  And so this is just because we're using a different atom here.  So definitely this guy has zero field splitting.  To be more basic question, how do you, it's probably a story,  where does the choice come from for iron and manganese?  Why those guys? Or why do I know?  No, no, why those guys?  Why those guys?  Well, they are the 3D transition metals.  Those are used in magnetism all the time.  That's not really a good answer, but that's why we did it, basically.  And so we've looked at most of the 3D transition metals,  and I'm just showing you those two cases here today.  If you use F electrons, you should be able to see something similar,  but the F electrons are much closer confined to the core of the atom.  And so it's not clear to me if you can actually see this current or not.  So there's a lot of experiments basically to do.  But for us, it's just that's the way we chose it to be for no particular reason.  Okay, so we definitely have zero field splitting.  What about direction dependence?  So let's apply the magnetic field in plane here, along this particular direction in plane.  And what you see here is the zero Tesla field, the spectrum.  I just showed you that before.  And now one Tesla, three Tesla, five Tesla, and seven Tesla.  And what you see is that the 4 millivolt zero field mode goes up in energy  as an increased magnetic field, and the 6 millivolt also goes up in energy.  Right?  And also the middle guy here disappears pretty rapidly as we increase the magnetic field  along this spatial direction.  If I apply the magnetic field in a perpendicular direction,  also an in-plane magnetic field, but 90 degrees rotated,  with respect to this iron atom and this particular spatial direction here,  you see, of course, at zero Tesla, they're the same, have to be the same.  And then we increase the field.  Well, the 4 millivolt mode stays roughly constant.  The 6 millivolt mode goes down in energy.  And the central mode seems to be sticking around much longer.  Right?  So quite different dependence here.  Yes?  So you understand what the two forward arrows, what are the top two are creating?  This one, with the model I'm showing you in the next couple slides,  this would be the location of this excitation at seven Tesla.  So you should be seeing something there if it was visible.  But it's not the transition, or you just cannot measure it?  The transition amplitude is zero, basically.  It's the original.  We just don't see it.  I'll show you this in some quantitative detail, actually,  which is one of the goals I have.  Okay.  And the third direction out of plane magnetic field is yet another dependence.  Okay.  So by now you know what I do.  I plot the center of the step, right?  That's here as a function of the external magnetic field.  That's basically extracting the information from this IETS measurement,  inelastic tunneling spectroscopy measurement.  And then I can get a very good fit to this.  With a simple Hamiltonian that has only a few parameters.  Okay.  So that's the fit that I get when I use this Hamiltonian.  So I have a Zeeman energy, not shown here.  Then I have this dS e squared term.  So that's a certain spatial direction that I have a so-called  easy axis magnetic anisotropy here.  That's this term here.  And then I have a transverse magnetic anisotropy.  Now, if you look at these terms, again you have S c and S x and S y.  So again we have non-commuting operators.  So you have to do the calculation.  You can't just simply see what this does basically.  Right?  That's kind of the outline of my talk in the end.  You can actually do this by using these spin operators in a fairly simple way  that you can all do without, you know, go into quantum mechanics two or three  courses or something like that.  Right?  So, now the parameters that we measure from this, this fit we get by getting d  equals minus 1.5 millivolts and e of 0.3 millivolts.  Right?  Just take those as fit parameters.  This, yeah.  So do they correspond to something physical?  So can I?  Yes.  So this minus 1.5, will you tell me?  Yes.  So this d equals minus 1.5 together with the fact that this is the simple direction,  meaning the direction where all the energies are just adding up together.  Right?  This is much more complicated than this in some sense.  It tells you that this is the preferred direction of the spin.  So this spin wants to point in plane, whereas the manganese spin that we talked about before  wanted to point out of plane, but very weakly.  It has very weak preference to point out of plane, where this parameter for manganese  is 50 times smaller.  Right?  So the magnitude of this parameter tells you how much does it want to point along a  certain spatial direction.  And 1.5 is very, very big.  The cobalt atom in the bulk of cobalt, for example, would have a, you know, d equivalent  that is about 500 times smaller than this.  So because it has this, this has these local nitrogen atoms nearby, these guys here, they  set all the energy scales for this magnetic interaction here.  They play a very important role.  Okay.  So for the next 20 minutes or so, I want to take time to show you how this kind of model  actually works.  Yes.  Yes.  Yes.  Yes.  It's completely set by the local environment, basically.  And also depends on the atom itself.  Yes.  Manganese is very different from iron, very different from cobalt, basically.  So is there a way that you can measure like a normal spring atom by...  That's why I chose the manganese to begin with.  So if you exclude that tiny little wibble at zero magnetic field, right, that was a small  fraction of the Zeeman energy.  Right?  And this here, here the zero field splitting is a large fraction of the Zeeman energy,  right?  So this manganese is the closest I have to a free spin.  All right.  Basically.  You know, it's... why is this?  Because manganese is an S equals 5 half system and that's a half filled D shell.  And a half filled D shell to a very good approximation is spherical symmetric, just like a fully filled  shell.  And so that's why half filled shells are very, you know, less susceptible to those kind  of influences.  Okay.  So how can we understand how this stuff works?  So, of course, you know that if I just have the Zeeman energy and I have zero magnetic  field applied, I don't get any excitation, right?  I have only a zero energy excitation.  Nothing to learn from that for this.  So, and I have five energy states, right?  All degenerate here.  And so if they're all degenerate, I can label them any way you want.  Ignore this for right now.  It doesn't make much sense.  If I look at finite magnetic field, this is something you learned in quantum mechanics.  What you learned in quantum mechanics is that the magnetic field sets the quantization axis.  Right?  That's what's typically taught.  And so that's what you see here.  So if you apply a magnetic field along a certain direction, let's say, SZ, but doesn't really  matter which direction that is, then this direction is a good basis to use.  And so what we do is we use the M eigenvalue of the SZ operator along that particular spatial  direction, which is set by the operator choosing a certain direction of the magnetic field.  And so that's what these are here.  So M equals plus two, plus one, zero, minus one, and minus two.  And those are very good.  Those are good eigenstates here, right?  So the eigenstate corresponding to the ground state is 100 percent M equals plus two.  The ground state corresponding to the first excited state of 0.9 millivolts is 100 percent  M equals plus one.  This is just repeating your basic quantum mechanics just written down maybe a little bit differently.  Right?  You all know this, basically.  OK.  But now let's take a moment and look at this term, dSZ square.  OK?  And so I want to write this down a little bit for you to not go too fast here.  So first off, we have an S equals two system, which I haven't proven to you, but let's assume  that that's the case here.  So how do you write the SZ operator?  Right?  Remember for an S equals one-half, it was just minus one and one, basically, or minus  one-half and one-one, plus one-half, if you want to be precise.  So here, the SZ operator is two, one, zero, minus one, minus two, with everything else  being zero.  Right?  And if you want to know what SZ square is, you can do the matrix product if you want.  Put another matrix next to it, do the matrix multiplication, and what you'll see is that  this is four, one, zero, one, four.  And everything else is zero.  Right?  This happens to be the case because it's a diagonal matrix.  Easy, basically.  Just for shits and giggles, what if we want to use this and we want to know what's the  energy SZ square, DSE square, if I'm in the plus two state?  Right?  I have to convert this to a vector because I'm using a matrix here.  A vector that corresponds to the plus two that's written up here, that's one, zero,  that's the first line up here.  Right?  That's one, zero, zero, zero, zero.  Okay?  So I multiply, let me make that simpler here, and let's say DSE square of plus two.  All right?  So you get my D here, and SE square is this matrix, and I have, what did I say, plus two,  one, zero, zero.  Right?  Do the matrix multiplication here, and you get four times one, zero, zero, zero, zero.  So this is an eigenvector of this operator with eigenvalue four.  Right?  Do the same for the minus, M equals minus two state.  You also get four.  Right?  So you have a two-folded generacy of these states.  Nothing magic about this.  Simple matrix multiplication, basically.  Yes.  Yes.  Now, let me take this one step further.  What do I do with the, where is it?  There.  Okay?  So this is really simple.  Now, let's ask ourselves, for example, what is the current that I get for an up electron  if I don't change the spin in comparison to the current that I get for a down electron  when I do the same thing?  What?  This is something like, I'm not looking at the inelastic components, I'm just looking  at the elastic components.  Right?  So those are four of the tunneling processes that we have to worry about if we want to  calculate the current here.  Now, we can do this, right?  SZ operated on plus two gives us two times plus two.  Right?  Let me make this slowly.  Plus two gives us two sigma Z.  A little bit tricky because I have the U in there.  I can't quite do that so slowly.  Let me operate both of them.  So if I operate the SC on the upstate, I get two times the upstate.  If I operate the sigma Z on the upstate, I get one half times the upstate.  So I get one half times two.  Right?  One half times two plus U times up plus two absolute value square.  Right?  Nothing magic, this is just matrix multiplications or basic quantum mechanics.  Right?  Now, so this positive gives you, if you do one step further here, gives you something  like, right, this is now just numbers.  So I can take those numbers out and this operated on this is one.  Right?  So I get out of this one plus U in brackets square.  Do the same over here.  Right?  Now I have SC on plus two, this plus two.  Sigma Z on the down is minus one half.  Right?  So I get here minus one plus U square.  Right?  So and remember that I told you for iron, the U was 0.75 in these arbitrary units which  I'm using here consistently for this U.  And I said that the maximum value for spin polarization for iron is one.  And what this means is that if you had a U of one, that means that if you have an up electron  that's tunneling, you get a certain current.  If you have a down electron that's tunneling, minus one plus one is zero.  You have 100% spin polarized current.  Right?  This channel is exactly zero if you have the maximum value of spin polarization.  Now it's not exactly because it's 0.75, but you can see that there's quite a dramatic  difference between an up electron tunneling through this iron atom and a down electron  tunneling through this.  Right?  And I'm not, and again, I'm not using any magic here.  This is just simple matrix physics basically.  Okay.  So that's what we learned from this, but going back to our measurement, we only get one  excitation at zero magnetic field and not three.  So this is not the full story.  Right?  We had one excitation at one millivolts, at 0.2 millivolts, one at four millivolts and  one at six millivolts.  And so that part we just did.  What transitions are allowed, that's that.  Okay.  Now, if we include this transverse magnetic anisotropy term, right, then what you see  is that you get a much more rich picture of what the quantum states are here.  The ground state is now a mixture of m equals plus two and m equals minus two.  That means that this iron atom in the ground state is pointing its spin along one particular  spatial direction equally up and down.  Quantum coherence of a position of these two states, but if you're pointing at the  same time up and down, you have no net magnetic moment left.  Right?  They cancel out.  So in other words, this iron atom at zero magnetic field is non-magnetic.  It's canceling its own spin.  It's only possible because you're doing it quantum mechanically, right?  Classically it's either going to point this way or that way.  This comes out if we believe this Hamiltonian just from diagonalizing this matrix.  If you look at the first excited state here at 0.17 millielectron volts, that's also  a superposition of plus two and minus two, but with a different phase here, with a minus  one.  Right?  Just comes out of the matrix.  That's just what comes out if you diagonalize this matrix basically.  And so on.  Right?  So this is, again, not too difficult if you believe in this kind of matrix stuff or in  this kind of quantum mechanics.  Now let me see what I was going to show you here.  Okay.  Let me not write this down because I do want to show you the second part too here.  Now I showed you the SC operator, right?  That's the two, one, zero, minus one, minus two.  You can also look up what the SX and the SY operators are.  Right?  So let's have zero on the diagonal and have some finite values on the one off diagonal.  This is just looking up in the textbook basically what is SX and SY for these guys.  Now you can matrix multiply, which I actually did, you know, paper and pencil kind of guide  that I am.  You can just multiply it by itself and you get this matrix out for SX square and this  matrix out for SY square.  They're not diagonal, so it's not trivial.  You can just square these components here and get your matrix out, right?  That doesn't work.  We have to actually do the matrix multiplication here.  Then we can look at this term as actually E times SX square minus SY square.  So we look SX square minus SY square has only six components in that matrix here.  And so then our full Hamiltonian excluding magnetic field is DSC square plus E SX square  minus SY square.  This term here gives you the diagonal elements and this guy here gives you these off diagonal  elements.  Right?  You can just do that.  That's not difficult.  You can just calculate this and then you can diagonalize this matrix that I did not do  myself, right?  That's beyond my powers of matrix algebra or something like that.  But you get out that this is the ground state.  If you take this and put in the D and the E value that we measured basically and this  is the first excited state of this.  All right?  So this is again not particularly difficult.  But what I want to do now for a second here is I want to show you what happens when I  operate the S plus operator here on this guy.  Remember that one of the terms that we had is S plus operated on the ground state and  then we can ask what that gets.  So S plus operated on the ground state.  Whoops.  Ground state.  Now the S plus operator is related to the SX, right?  And the SY.  It's a simple operator really.  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,  0.  All right?  This is just again looking up what does the operator for S equals 2 look like.  And now we want to operate this on the ground state.  Now the ground state has some components of plus 2, some are 0, and some are minus 2.  So that's the plus 2 component here.  So that's the 0.69 something rather.  0.70, 0, minus 0.17, 0, 0.70.  Right?  This is just another way for writing this.  Really the same thing.  This is just how you translate this thing to a matrix.  And then, right, I mean you know how to do this, right?  You go this one times this one times this one times this one times this one gives you  0 in the first one.  Right?  This one times this one times this one gives you a finite value.  This one, this one, gives you a finite value of minus 0.42.  The next one, ding, ding, ding, you do that and so on.  Just basic matrix multiplication and you get 0, 1.400.  Right?  Nothing magic about it.  And what this, another way to write that is, that is basically times minus 0.42 times plus  1 plus 1.40 times minus 1.  Just the same thing basically.  Now the next state that I have here, one higher than this is, I go back to this in a second.  So the next higher state, I want to skip that, is, I don't have this written down here.  If I go back, if you go back to this one here, the, so this is the ground state, first excited  state, the second excited state.  And this is a state that has plus 1 and minus 1 components in it.  Right?  And remember for our overlap, we always have final and then the s plus here, initial absolute  value square.  And so you can project, this one is, we just calculated this.  You can project this on the final state and you see that that has a lot of overlap with  the final state that has plus 1 and minus 1 components.  Yeah?  What's that again?  Why is it not normalized?  Why is it not normalized?  Because I'm very sloppy.  Very sloppy when it comes to H bars and normalization and so on.  Right.  This one probably has a factor of square root something missing in normalization.  Right?  I just sort of try and do, highlight kind of the ways that this works, basically.  Not to be used for actual computations.  Right?  Now, this is what we get out of this and if you want to do this overlap here, you just  get the plus 1 component overlaps with this plus 1 here, 0.71 of this and this one over  here and you get a finite value coming out.  Right?  So, that means with the operator S plus, we can go from the ground state to the second  excited state.  We can also go to this excited state with these operators.  Also a mixture of plus 1 and minus 1 and the last one has only a component of m equals  0 and we can't reach that at all with our interaction Hamiltonian.  Right?  We don't have a way to make it to m equals plus 2 with only 1 S plus from the ground  state.  Right?  So, in a nutshell, we can calculate what these transition intensities are and so you  can see that forward.  So, from the ground state to this one, we have a large overlap if we operate with the  SC operator, the elastic part which I called elastic.  It's not elastic because it's changing the energy but it is non-spin flip basically and  we can also get to this first excited, second excited state here because here we have a delta  m of plus 1 and minus 1 or the other way around, sorry for that and we can get equally  to this guy here.  Now, they have a different amplitude.  Why is the one in blue smaller than the one in green?  Right?  If we had a little bit more time, I could just write this down.  It's just the fact that here we have a certain combination of these numbers and you get this  together with this phase here, multiply this and you'll see that one is smaller than the  other one by a factor of 3.  Right?  So, all of this detail, all of this, you can get from this, from this spin Hamiltonian  which has a few parameters, the D and the E and the direction of the easy axis and the  sigma dot s plus u, right, which has one parameter.  All of this, the height of these things and the energy falls out of this space here.  What does not fall out is the fact that there's a peak here somewhere, right?  It's not just a step, it's a peak also and that goes back to this maybe condo effect  kind of thing which I don't fully understand.  What?  I don't understand, not fully, I mean, I don't understand it, not even halfway, basically.  Okay?  So, in a nutshell, using relatively simple matrix stuff you can look up in a simple textbook,  basically, you can get this full understanding relatively easily.  Okay?  Let me skip this part here.  Because one of you guys asked earlier, how do we know the differences?  Why?  I showed you the manganese and the iron, I mean, they're very different.  You can definitely tell them apart.  A cobalt atom also very different, right?  So once you know who is who, you can actually tell them apart relatively easily and the  copper atom is just flat because it's a spin zero in this case.  Okay?  So, I'm sorry I'm going a little bit faster here because it's 20 minutes to beer time.  And I don't want to, but I do want to tell you a little bit about what you do when you  have more than one spin.  And then we'll call it quits.  Okay, so single manganese atom, you know how this works by now.  Two manganese atoms completely different.  Like this guy, this spectrum here really has nothing to do with this spectrum.  Right?  Yes?  Doesn't matter, actually.  You can put the tip here, here or in the middle, doesn't matter.  It's delocalized because the atoms are fairly close together and we really can't individually  tunnel in only one.  They really look like a cigar.  Maybe I have an image, I don't know.  I'm sure I have an image later.  You'll see it's a cigar, so we don't really, if we build them with a little bit further  spacing then we can actually tell which atom we're tunneling into and then it can depend  on which atom it is, which is what Anna is working on, actually.  And delft, not in my lab.  Okay, three atoms, very different again.  Right?  Sort of has a little feature in the middle, similar to the single one, and has some big  energy stuff, maybe similar to the dimer.  And four atoms, different again.  Right?  So, spectroscopically, if we know what we're talking about, they're very different.  If we do our experiment carefully, we know exactly that there's four atoms and not three.  Right?  And that's an experiment.  If you try to do this with other techniques, other than STM, it's very difficult.  Right?  If you want to make a spin chain, and you want to, you know, there's some materials that  behave like spin chains, but that chain has a certain length.  And if you want to make that length one less or one more, you know, you have to talk very  hard to your chemistry friends or materials friends to make that material.  It's very difficult to do.  Right?  So, we can do this here because of this combination of atom manipulation and local spectroscopy,  basically.  Okay?  So, what can we understand about this?  Going back to the dimer here, at first observation is this is zero field, four Tesla and seven  Tesla.  First observation, around the energy where the single manganese atom does all of its  physics, all of its spin physics, namely, you know, plus minus one millivolts, nothing  happens in the dimer.  Right?  Looks like it's non-magnetic and like the iron atom, basically.  It's at zero Tesla.  Now if we look at the higher energy, higher magnetic field, sorry, at seven Tesla, it  looks like we have actually three steps.  You know, this one here at five millivolts, this one at six, and this one at 6.5 maybe.  Right?  So, observation, I mean, no interpretation of this at this point.  And so, we can plot what is the center of this step and this step and this step as a  function of magnetic field.  And we can do this maybe down to three Tesla and you can see that we have one step that  stays constant, one step that goes up with magnetic field and one that goes down.  Right?  And that, again, should look familiar to you.  That looks like an S equals one system.  An S equals one has quantum numbers M equals plus one, zero, and minus one and they would  behave exactly like that.  So what's happening here?  Well, not too surprising in the end.  We have two very strongly coupled magnetic atoms here.  They happen to couple anti-ferromagnetically such that you make a spin singlet.  A spin singlet is really a non-magnetic state.  A spin singlet is a state where you have a total spin of zero.  That doesn't mean that the manganese itself is spin zero.  You can take two electrons, right?  First electron is up, second one is down, minus, second electron is down, first one  is up.  That's a singlet state for two electrons.  And that has no net spin.  Total spin of the state is zero.  Right?  So that's what the ground state is like.  It's a spin zero state.  And then the first excited state is an S total equals one made out of two S equals five half  manganese atoms by making magic combinations of spin states basically.  And then you can get three quantum numbers and they all have allowed transitions according  to our spin interaction Hamiltonian.  Pretty simple, right?  So we use the Heisenberg interaction, J S1 dot S2, as a way to model this kind of  system.  This is simple.  Heisenberg interaction is S total is S1 plus S2.  Right?  That's just definition of S total operator.  S total square is S1 square plus S2 square plus two times S1 dot S2 because the two are  commute with each other.  And that's your Heisenberg.  These guys are constant and so you really can say that the S total square operator is  the same as the Heisenberg interaction for two spins excluding the factor of two.  So this S1 dot S2 is the same as S total basically here for two spins.  And so we have anti-firm magnetic coupling.  Ground state is S total equals zero.  First excited state is S total equals one with an energy difference of J. And so this  is a way to measure this J, 6 millivolts.  So we can quantitatively determine the strength of this interaction again by just making this  inelastic tunneling spectroscopy.  And so with that, I'd like to very briefly show you how we can actually calculate things  when you have more than one spin because that's probably something that everybody in this  room who doesn't do quantum computation has never seen.  How do you write a spin matrix for a spin that is composed of two spins?  The quantum people here I'm sure have seen this, but the other guys probably haven't  seen this.  So let me go back to the x operator, sigma x for a single spin one half particle is zero,  zero, one, one.  Now how do you do this if you have two spins?  How do you write this down?  How do you even do the bookkeeping basically?  I'm not talking science here.  The science, that's all the Heisenberg interaction has to offer.  But how do you write this down?  How do you diagonalize a matrix that has more than one spin in it?  How do you do the bookkeeping?  How do you set up this matrix basically?  So if we take two spins, well one spin sorry, we have two states that we keep track of,  up and down.  We call the up one one zero, we call the down one zero one in this picture.  If we take two, two spins this is one spin, two spins, how do we do this?  Two spins, our states are basically first one is up, second one is up, that's now one  zero zero zero.  How do I write this best?  Up, down, zero one zero zero.  Up, zero zero one zero.  Down down, zero zero zero one.  So with two spins you have four basis states that you have to keep track of.  In general the number of these states is two to the power of n, which is why quantum computation  works.  It's exponential in the number of spins so your Hilbert space becomes big very fast.  But two we can still write the matrices down, that's simple.  So how do you write an SX operator in this?  First off what we know, what we want and the moment the, that was my time to end, so give  me a couple minutes.  So if I want to flip the first one and I don't want to flip the second one, I can write this  as a sigma operator on the first guy and identity on the second guy and I know that my outcome  of this should be first one flipped, second one not flipped because the sigma X is a flip  operator.  But how do I make this matrix here?  And that's actually not that difficult.  If you keep this, this definition basically, this is not science, this is just definition  and bookkeeping.  So how do you make this operator here?  So basically what you want to do, the sigma X is this guy here.  So you set up your matrix zero, identity, identity zero.  And by this I mean that you get your four dimensional matrix here by saying this is zero, zero,  zero, zero.  This guy is zero, zero, zero, zero.  And then I have an identity matrix here, identity matrix there.  Just bookkeeping, no science in this.  Just the way that quantum computing people basically do the bookkeeping and that's how  you calculate a spin chain if you want to set up the SX plus SX square plus SY square  plus SC square for chain of two spins of spin one half basically.  So let's just take a quick look what this does if I operate on one, one.  One, one is one, zero, zero.  So one, zero, zero.  First line is zero, second line is zero, third line I get this one times this one is one,  fourth line is zero.  Look at this again, that's just down, up.  Final thing, what if we want to operate on the other guy?  We want to flip the second one.  Then I do identity times, oops, times sigma X and my matrix looks first I do the identity.  So I get sigma X, sigma X, zero, zero.  And then I fill this into my full dimensions, sigma X is zero, one, one, zero.  This is zero, zero, zero, this is zero.  And then I get another sigma X here, zero, one, one, zero.  Right?  That's different than this matrix here because somehow the matrix has to do the bookkeeping  for you, which operator you're actually operating on, which space you're operating on.  Does that work?  Well, let's do the same thing.  We operate on one, one, one, on up, up, which is one, zero, zero, zero.  And we get this guy is zero, this one is one, this one is zero, this one is zero.  If you look that guy up, that is the second one there which is up, down.  Right?  Okay, that's not really telling you any science, this is just telling you how the bookkeeping  works and with that I'm almost finished.  So I brought this down in your slides here so you can look at this a little bit more.  I was going to have you do a few exercises too but that's not going to work.  We do the Heisenberg interaction, okay?  So, finally, we do this, of course, with Mathematica or Matlab or something like that, set up these  matrices for a spin chain up to length six of magnitude, you know, s equals five half.  You get a very big matrix very quickly, like a million by a million or so matrix, right?  Something like that, I don't know.  Very big matrices very quickly.  We know the j from the dimer, from the trimer, we can actually determine that the magnitude  of the spin of the individual is five half, right?  Using the j from the dimer, we can look at, all right, the best agreement is with s equals  five half and that's it, that's all the parameters that there are, right?  Only the j and the s for the individual one.  We're using the simplest, not only nearest neighbor Heisenberg interaction here basically.  And then for the even chain, you can see that this fits very well and for the odd chain,  also fits very well, right?  So if you want to do this, you know, you have to know how you translate this into matrices  and how you diagonalize these matrices, right?  This is sort of the beginning of how you set up these matrices, okay?  And with that, I think I would like to stop and just in a summary, I guess I tried to  teach you two things.  One is that you can learn something about these quantum mechanical systems of spins  on surfaces using scanning tunneling microscopy, right?  And we in particular use inelastic tunneling spectroscopy to learn something here.  And second, that you can actually model these things with fairly simple basic physical models  if you are not afraid to use these spin operators and spin matrices, right?  So those are kind of the two messages.  So thank you guys very much for coming.  And I promise tomorrow will be completely different.  So I think we have about 10 minutes or so, actually five minutes.  So, you know, please shoot.  So if you want to do phase coherent measurements, that's one of the many parts that I didn't  have time.  There's a lot more stuff in the slides that you have.  So what do you need to do phase coherence?  You need something that keeps the clock for you, right?  You need usually you use a time-oscillating magnetic field or time-oscillating electric  field, something that oscillates in sync with the energy spacing of the energy levels of  the system here.  And that can give you coherent oscillations between the ground state and that first excited  state, right?  So what we need is two things.  One, we need to make a system where the relaxation times are long enough that you could have any  chance of doing that.  If you do this on copper nitride, the relaxation times, T1 relaxation times are on the order  of a nanosecond.  Way too fast for anything phase coherent to work, right?  So that's one.  Make them longer.  Maybe make them a microsecond or something like that.  And the second one is include something like a time-dependent field that sort of sets the  clock basically.  So we are trying to do both of these processes, both of these steps.  How do we want to extend T1?  The T1 is currently limited by the fact that this copper nitride is so thin.  So we have, for example, the iron atom here and the relaxation time.  So if I put the iron atom into an excited state, this relaxation process is completely  dominated by an electron from the copper coming in and going back out.  So we need to make this insulator thicker.  So that's how we think we can get rid of this relaxation mechanism.  Then you will have other relaxation mechanisms, but you have an atomically small system.  So nuclear spins will not be as bad as they are, for example, in gallium arsenide quantum  dots or something like that.  But we don't know if this works.  We have never seen anything that is T2 phase coherent.  Is there any prediction for the T2?  Well, you know what kind of good theorist I am now, right?  So you're not going to get a good answer from me for that.  I don't know.  Nobody knows.  I mean, there's many examples where point defects in a material can have very long relaxation  times.  Like the fashionable one today is the NV center in diamond that has relaxation times that  can be up to milliseconds or longer, T2 and T1.  Phosphorus in silicon is another one.  Relaxation times can be very long.  So I don't think there is a very general reason to expect them to be long or not.  You have to really work hard on your local symmetry and so on.  The beauty that we have is you sort of got the feeling for we have quite a bit of flexibility,  right?  So we can build different atoms, different surfaces, different structures.  So the phase space of possible experiments is huge.  But we have a problem with these conduction electrons, which we have to solve for sure.  So one step at a time, I would say.  What's the big downside of increasing these insulators?  Things become more difficult.  Yeah, I mean, part of your SEM, the general curve.  That's right.  So you need to find a balance.  Of course, we want to maybe able to tunnel through this.  So we need to find a balance where we can still get enough current through this to do  our measurement, but where this process here becomes long enough that it doesn't really  matter.  And so what we're doing is, on copper nitrate, we can't make this any thicker.  It's just self-limiting to one layer.  Not in the neck, but that's the way nature works in this case.  And so for other materials like magnesium oxide, we can make them one, two, three, four,  five layers thick.  And then we're in the process, basically, of measuring these things that I told you  here as a function of the thickness of these materials.  Sort of the space of experiments that I showed you is maybe five years of work.  So this is not fast, basically.  But if you change instead of copper, you use HOPG for a five, for example.  Would you expect changing things?  Oh, it certainly will change everything.  I mean, I think you got a feeling for that the local environment plays a very important  role.  HOPG is probably hexagonal, if I'm not mistaken.  So that's very interesting because you can set up frustrated spin systems and whatnot.  I haven't done it, so I don't know if that would actually work.  The magnetic anisotropy will also be different because the symmetries are different.  You probably don't get charges nearby, like the nitrogen atoms.  So magnetic anisotropy will probably be much less.  So you can see I can make speculations, but one has to try it.  You can, I mean, HOPG is definitely conductive enough that one can work with this, with STM.  I have another question.  Sure.  I don't know if there is any experiment with the spin-polarized STM working on the spin  valve, a planar planar, but to see all the spin accumulation in the channel and how it  behaves and a probe spinalization of systems like this.  There is a vast amount of stuff known about spin-polarized STM.  I mean, the most prominent group in that respect is in Germany and Hamburg.  We've done quite a bit of work with spin-polarized as well.  So add another layer of spin-polarized stuff on top of this.  Now to actually see a spin valve.  So basically you want to make a planar junction.  So the problem will be that you don't have two tips.  So if you use your tip for the injection, you can't measure away from the tip what is  the contact.  Yeah, so you'd have to do that.  So you'd have to make a lateral structure and that has not been done as far as I know.  Anytime you go for this level of more complexity, it becomes very difficult.  This experiments work beautifully in UHV clean samples, right?  No dirt at all.  That's what we can do with STM.  As soon as you do lithography or any stuff like that, the world becomes much more difficult  for us.  That's probably the best answer to that.  One slide further.  Let's see.  Oh, there you go.  Did you program it or did you help to program this?  This one here?  This was, this is the original IBM that was built by Don Eigler, the guy I inherited the  lab from.  That was the first atom manipulation, 1989, 1990.  Tomorrow I'll talk exclusively about atom manipulation and how we can build structures  and have fun with that.  No spectra.  No inelastic tunneling spectroscopy at all.  Just pictures, that's right.  Just pretty pictures.  What does IBM expect from you?  Science papers.  And even more important than that, New York Times articles or mentioning of IBM as a leader  in nanoscience.  If we can portray to the academic community and even more importantly to the general,  to the shareholders basically, that IBM is a leader in nanoscience and that this stuff  might have applications to the future of IBM's business, then we win.  So tomorrow I'll show you two examples where we use atom manipulation to do computation  and where we use atom manipulation to do data storage.  And that, I think, you know, that's why IBM funds this kind of work.  That's also why we can't focus on figuring out exactly what happens in this manipulation  process.  IBM doesn't care about that.  And I have to be careful.  I have to do stuff that is interesting to the company.  So even though I don't work on anything that is product related, nobody at IBM expect  any of the stuff I do to make an actual product, I still have to be able to demonstrate faithfully  that the work I do has net benefit to the company.  So in principle, you work in the marketing department?  Yes.  Yes.  I often say that I actually work for the marketing department.  Yes.  Of course, the marketing department doesn't pay for it, but they should.  I mean, maybe some of you saw the movie that we made, The Boy and his Adam.  That was actually paid by the marketing department.  That was approximately a million-dollar kind of budget to promote this and produce this  and all that stuff.  Because it's a marketing firm that's involved in it.  They have a lot of...  It's not IBMs.  I mean, this is...  You know, we use a commercial marketing firm in Manhattan.  And you know, I don't know how much...  I don't know the number, basically, right?  But it's definitely more than $100,000.  They were involved in this for about a half year with probably five to 10 people.  That's a lot of money.  To set up all the PR that went with it, come up with the idea, design it.  You know.  So that's removing the patents was probably done by a couple of PhDs?  That was just done by us, basically.  Four people in the lab.  10 days.  Right?  We didn't get any money for it, of course.  We get paid.  That's the money we get, basically.  Okay.  Thank you guys very much.  I'll be around tomorrow and on Wednesday.  So if you guys have more questions, feel free to approach me.  And let's have some beers.  Thank you guys.  Thank you.  That was classic VR, to be honest.  I blind only 5 if people need something to do.  Imagine...
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