For that end of the table, here come Wolves again down the left-hand side. This time it's Jota. And today's the only... Simon France is flying in. Jota got to the ball first. Sliding tackle was a late one. Left-hand side of the box, right-footed. Curling, and into the far corner. And Wolves take the lead against the Cherries. Yalmutinho with a quite brilliant free-kick. With 22 minutes on the clock. Bournemouth-Neal, Wolves won. Into the top corner past Saren Ramsdale, what separates the two sides. And again, Tairori's played in behind from the free-kick. Raising a cross-goal, and Raoul Jiménez side-footed past Saren Ramsdale, his 14th goal of the season. And that was just far too easy for the Wolves. Cutting Bournemouth open with ease. They double their lead at the Vitality Stadium. Bournemouth-Neal, Wolves 2. And now he's calling Simon Francis on. Simon Francis gets a second yellow card, and is sent off for the Cherries on his first Premier League start since Boxing Day last year. They're away to our left, in between the main stand and the north stand. Ryan Fraser delivers right-footed to the front post. Flick on by Steve Cook, and into the far corner. Bournemouth back in the game. It may have taken a deflection on his way into the goal, but the Cherries won't care. They're down for 10 men, but they reduced the deficit to just one. Bournemouth-1, Wolves 2. Lest the 2-0 up against Brighton. James Madison points to the box. Oh, Nakke! Oh, what a save from Patricio. Back out again, this time to Steve Cook, who fires goalwards. How has he kept that out? An incredible overhead kick from Nathan Nakke out of nowhere.


Poops on the hardwood is back of the Bemidji girls team. Their season and year 200 head coach Steve Schreiber begins tonight, 14 and 13 a season ago, the loss in the section semifinals, but that's all wiped clean. The fresh start this evening, they open up at home in a section battle with Steve River Falls. Let's go first half. Jax used the long ball. Clarissa Whalen spots up left wing. That rattles home to give them early points. And later in the half, this time, had any mills her first game back in over a year. That's in from the corner to tie the game. And how about one more top of the key, Allison Beard, who says goalies can't shoot. Bemidji making it rain from downtown, but Prowlers made some shots of their own courtesy of Kylie Proska. She had herself a pretty good night. Jax fall in their season opener, Thief River Falls 67-57, the final. If you've enjoyed this segment of Lakeland News, please consider making a tax deductible contribution to Lakeland Public Television.


Rotates around the axis. And all of them just rotate around. In every basis, the modes rotate. OK. So just as a reminder, this lecture is going to be, I'm going to touch on a recent experiment we did where we sent this high dimensional entanglement that I've been talking about through a complex medium, in this case, multi-mode fiber. And then I will start to tell you a little bit about multi-photon entanglement in high dimensions. So more than two. And if time permitting, a few slides on free space quantum communication and classical communication. OK. So let's start with unscrambling entanglement through a complex medium. So with quantum comms, I like to think that we are at an early stage harking back to the days of classical communication when we struggled with things like storms and bad detectors and things. And so we have to overcome similar problems with quantum communications. For example, dealing with noisy channels is a huge issue. And a complex medium, essentially a scattering medium, acts as a noisy channel. So what do I mean by complex media? This is very trigger happy here. OK. So anything can be a complex scattering medium, from paint on a wall to your teeth to sugar cubes. Essentially, modes go in. They mix up, scatter in some ways, and come out in a different basis or with loss. A multi-mode fiber is a more controlled complex medium in that it's fairly unitary. There's no loss, little loss. But the modes go in and come out, all sorts of mixed. If you've ever seen light coupled into a multi-mode fiber, it comes out with a speckled pattern, laser light. And so there's a lot of effort being made in trying to overcome this scattering for things like classical communications. So SDM is space division multiplexing. We are at the stage where the classical data crunch is happening, and we want to push more and more information so that we can all watch YouTube videos of cats doing funny things. But one of the ways that people are going to do this is by our doing it currently is by putting lots of spatial modes into a fiber. And people have reached a terabits per second by space division multiplexing. And this was a work done by some people in Scotland, actually, at the time, where they figured out how to send images through a 30 centimeter section of a multi-mode fiber. So that's real data that you're seeing there. And you can see as the fiber is bent, the image stays the same pretty much. And so we wanted to do something similar in the quantum regime. So I want to talk a little bit about how people do this classically. So a complex medium is described by a complex transmission matrix, which is basically a set of complex coefficients that map a set of input modes to a set of output modes. So a lens has a very simple transmission matrix. A sugar cube would have millions, perhaps more complex coefficients that make up this transmission matrix. So using this device that I've been talking about, a spatial light modulator, people can map or measure the transmission matrix of a complex medium quite simply by sending in a basis of probe states. So they prepare a set of probe states, I, and measure a set of output states, J, which then allows them to retrieve the complex coefficients, Tij. Now, of course, there's some phase retrieval involved, because measuring J here, you would measure the intensity. So in order to get the phase of the Tij, you would have to do some additional work. But in the end, you get something like this. Of course, I'm showing you an absolute value here. And this is all very nicely described in this early paper by Sylvan Giga. Once you find the transmission matrix, you can construct a set of eigenmodes, which is essentially equivalent to a singular value decomposition or whatever you want to call it. But basically, you find the inverse of the transmission matrix, and you can either prepare a set of eigenmodes or you can rotate, prepare a unitary transformation after the complex medium to basically send information through this scattering medium. So let's move to quantum mechanics for a second. In quantum mechanics, we have this thing called channel state duality, which basically says that statements about a state can be mapped to statements about a channel. So if you imagine this scenario where you have a maximally entangled state, and now you should be fairly familiar with this ii1 over root d state, and if you put one particle of this maximally entangled state through a complex medium or a channel in this case, essentially the resulting state rho captures all the information about the t. So the maths looks like this. You have rho equals identity tensor t. Identity acts on one particle, and t acts on the other particle, and the resulting state rho then encodes t, essentially. This is called the Choy-Jamiolkowski isomorphism. I'm sure I'm still saying that wrong, because there's a slash across the L, and I know that's not an L. So proposed independently by these two brilliant theorists in the 70s, and so we're going to use this interesting feature to recover our entanglement. So I told you how one can use a set of probe states to measure t. What we do instead is we send one entangled state, or prepare one entangled state, and we send one particle from the state through our complex medium. And so the resulting state phi t, now I'm assuming it's a pure state here, encodes all the information about the channel. Now you can do a few things. If you want to recover your input state phi plus, you can prepare an inverse t inverse and apply it here, right? Or, which would then diagonalize your channel, and you would essentially recover your correlations. However, there's an interesting property of maximally entangled states that allows you to do something interesting, perhaps clever. If you take two operations on a maximally entangled state, you can move the operation on one particle onto the other particle or vice versa. So A tensor B can be written as B transpose. It's actually complex conjugate transpose. A tensor I, or I tensor B A transpose. What this allows us to do is instead of unscrambling Bob, you can scramble Alice and recover your state. So even though Bob's photon went through a scrambling channel t, you can put a scrambling operation on Alice and you will get back your entanglement. This is a unique feature of entanglement in that the entanglement doesn't actually go away. It's just rotated in funny ways. So if one photon is messed up, you mess up the other photon in a clever way and you get back your entanglement. Okay, so let's start with the initial state. As before, we have a crystal pumped with a carefully shaped pump laser and then we measure correlations between Alice and Bob. In this macro pixel basis, this is one basis. This is the mutual and bias basis. We measure very nice correlations and this allows us to certify a fidelity of 94% or so in this seven-dimensional mode space which says we are entangled in seven dimensions. Now, when we put one of the photons through multi-mode fiber, the following happens, right? The data looks terrible. You have no correlations. But if you think back to what I told you about the channel state duality, you remember, we realized this later actually, this data that doesn't have any entanglement actually encodes information about the fiber in two different bases. And so the correlations that are lost actually tell us what the fiber is doing to some extent, right, absolute value. So what we need to do is actually recover the entire transmission matrix from this scrambled data. And the way we do this is we prepare a reference mode that copropagates with the state going through the photon, going through the fiber. It doesn't matter on which photon you have the reference. It's a two-photon state. Whatever you do here can also be done here. So we use the space between our pixels as a reference mode and we phase-step the reference. And without going into too much detail, we recover what we call the S matrix which is the T matrix of the fiber times some diagonal matrix that we need to recover. And then we recover this diagonal matrix through another reference mode process. And with this information of the S and the E, we get our complex T. Okay, so the T looks like this, a real part, imaginary part, and if you remember, I guess I could go back a few slides. It's quite similar to this, right? Looking similar at least, that. And so what we do is we take this information of our measured T and we prepare a new basis on Alice's side. So now I'm essentially messing up Alice in a clever way such that the photon that's going through the fiber comes back to being maximally entangled. And this is what we see. We see this messed up data is now perfectly correlated again, which was really nice. And this fiber is actually just a fiber we bought from Thorlabs, right? Two meter long, just a commercial multimode fiber that supports a few hundred modes. So we are kind of only looking in this seven-mode subspace of the fiber. So this is not enough to certify entanglement. To certify entanglement, as I told you before, you need at least two bases. So we take the information about our transmission matrix and we rotate it using some mats to access the transmission matrix in all the other mutually and bias bases of our seven-dim mode space. And doing that, we find we can get correlations back in all the eight mobs of this pixel mode space. And with this, we can certify a fairly high fidelity that tells us we have six-dimensional entanglement. Oh, that was too quick. So I just wanted to point out that this work was done in collaboration with a colleague of mine in Glasgow, Hugo Defian, and then of course people from my group. And we had a summer student visiting who's actually starting his PhD with me in September. Okay, any questions about that? I know I've kind of rushed through that a little bit. Yeah, so the question was, did we have any problems of scattering or absorption in the fiber? We lose modes to other higher order modes, right? The fiber can support a few hundred modes, like I said. So light does scatter out. So we are basically only looking at the light, the photons that are left in our mode subspace. The idea eventually is to try and access the entire mode space of the fiber. In fact, it's not that big at telecom. It's about 45 modes for a standard like grin fiber. So we do lose photons, but it doesn't hurt us. As long as it's, as long as the loss is kind of flat across the spectrum. How bright is it? You mean how many counts do we have? Oh, it's two photon entanglement. Yes, yes. And what's the B3 for like countering? I think we had a few thousand counts per second in one mode. Okay, so I want to move on to three qubits. Essentially tell you about how do we entangle more than two particles. And I'm gonna start with a broad sort of introduction about multi photon entanglement and why it's important and how do we do it. So if we think back to like the first days of entanglement, you know, we've all heard about the EPR paradox and the EPR paper. But so Einstein, Podolski and Rosen proposed this idea which, funnily enough, didn't get much mileage. I think it had like four citations until the 60s. But they were not bad citations. One of them was by Bohr and the other one was by Schrodinger. So good citations to have. But then in the 60s, John Bell came about and proposed a test of entanglement, essentially, which we all know as Bell's inequalities. And this was a statistical test of quantum mechanics that allowed you to make a series of measurements on an entangled state that ruled out any local hidden variable theories. Essentially, that said that the correlations we are seeing cannot be nature playing tricks. It is inherently non-local or you have to give up either locality or reality. Not going to go into too much detail on that subject. The main point is that the main point is that it was a statistical test. So in order to measure Bell's parameter S, you had to make a series of measurements on both particles that allowed you to then reconstruct this parameter which then tests whether nature is acting in this manner. So in the 1980s, early 90s, Greenberger-Horn and Seilinger, seen here in order, Greenberger-Horn, Seilinger, proposed a new test of non-locality that is a non-statistical test. Essentially, it's one measurement that tells you, it's called an all or nothing test. So it rules out this idea of having to do a series of tests on entanglement that tells you whether nature behaves in this strange manner or not. And the main point is that it required three particles, three or more particles to be entangled. And so here I'm showing you a three-part type G at Z state. And then Merman proposed an equality just like Bell proposed a test of this two-photon or two-particle entanglement. Merman proposed an inequality which required you to measure joint observables on all three particles. These are essentially sigma X and sigma Y. And if you measured a Merman parameter between zero and two, then it was explained by local hidden variable theories, measuring it between two and four implied that nature behaves in a non-local manner. Okay, so what are the current records of multi-partite entanglement? This is possibly a little bit outdated because every year, Janhwe Pan's group adds a few photons and possibly a few students and postdocs to the experiments. And so 12 photons have been entangled in China. People have entangled up to 20 ions, although I say five in 20 because they only showed entanglement between any five in 20 ions in Innsbruck. And people have entangled again in China up to 20 superconducting qubits. Although Google may have done more, but we don't know. Right, so all of these describe a state such as this where you have zero tensor N, right? N times zero plus one N times one G at Z. What about high dimensional multi-partite entanglement? People haven't really explored this and I'll tell you why. So the simplest extension is of course, you take your two dimensional G at Z state and extend it to three dimensions, right? So you have a, I call this a 333 state and I'll tell you why. Or what happens is, well okay, I'm describing these states in terms of these sets of states. So here's my set of 222 states and here's my set of 333 states. When you go to higher dimensions than two and more particles than two, a funny thing happens and actually this is going back to a question somebody asked me earlier about Schmidt rank being different or Schmidt rank for multi-partite states. What is a Schmidt rank for multi-partite states? It's described by a Schmidt rank vector and this in fact is the Schmidt rank vector. So the Schmidt rank vector is the rank of every bipartition, so if I look at the bipartition of this particle A with respect to BC, the rank is three and so on and so forth. So this is an example of a state where the Schmidt rank vector is not symmetric. So it's a 332 state and quite simply put, one of the particles lives in a two-dimensional space while the other two live in a three-dimensional space. And this is in the middle of these two sets of states. That's basically explaining what I just said, the Schmidt rank vector essentially. Now there are many states possible, right? There's 332, there's 422, there's a vast family and they obey a certain relationship which again is all explained in the reference in the earlier slide. So how do you create states such as these? So this very nice paper from the mid-90s, in fact I asked Anton who came up with it, he's like, oh, who's Marek? Marek Zrakowski is a senior scientist in Poland and so this is a very nice idea of basically producing multi-partite entanglement from erasing information, right? So what you do is you have entangled photons produced in two different crystals and you combine them in such a way that you cannot in principle know which crystal they came from. I can show you how this works for polarization. So a PBS polarizing beam splitter, as we all know it transmits horizontal, it reflects vertical, right? If I come in with another photon from the other side, it will do the same, transmit horizontal, reflect vertical, right? So it mixes polarization components of two input photons. Now what happens when you send an entangled state through it? So particle two will go through and particle three will go through the PBS. There are few possible outcomes. Either horizontal from particle two is transmitted and transmitted from particle three, which then imposes the condition that particle four must be H and particle one must be H because of the entanglement, correlations currently, right? Or particle three is reflected and particle two is reflected and they're both vertical, which then imposes the condition that particle one must be vertical and particle four must be vertical. So post-selecting on these four detectors clicking together, these are the only two possible outcomes. You either have H, H, H, H or V, V, V, V, right? And this is your four photon G at z state. Now if you remember back to my last talk, I was talking about coherence, the importance of coherence in entanglement. And what I have described so far is just classical correlations, right? You either have all H or all V. This plus sign here is crucial. And in order to certify that you actually have coherence, you have to do a lot of things. You have to make sure, well, okay, I'll tell you on, but the test of that is you have to do a basis rotation and look for correlations in the other basis. Right, so this setup that I just showed you is the one that's basically been extended in these 12 massive photon experiments, 12 photon entanglement experiments in China. You can almost see one crystal cascaded onto two, three, four, five and so on and so forth. So why were people able to do this for polarization? It's because there are tools that exist for polarization like the PBS, right? And also half-wave plates, which can do basis rotations from horizontal vertical to anti-diagonal. So the challenge in doing this for other degrees of freedom or higher-dimensional degrees of freedom is to create beam splitters, essentially multi-port devices that can do the same thing for spatial modes or temporal modes for that matter. So such a device does exist. In fact, it was invented back in 2002 in Miles Paget's group and Ioannis Quartiel's group. It's a Mach-Zender interferometer and it consists of two dove prisms in each arm, one rotated with respect to the other. And a dove prism is a device that basically is an image rotator. So if you go in with an image, it rotates it by twice the angle at which the dove prism is oriented. Now, thinking back to your Quantum Mechanics 101, rotations and angular momentum are connected. So generator of rotations. So when you rotate a mode carrying angular momentum, you also impart a phase that depends on the angular momentum. So rotating this mode here gives you a phase of pi. Rotating this mode here by 90 degrees gives you a phase of three pi, which is pi modulus two pi. So you can almost see what happens. Odd modes interfere destructively. Even modes interfere constructively so that this device sorts spatial modes into even and odd values. And you can scale this up to many modes. It's just not very practical. So this is an example of a multi-outcome mode sorter for OAM, but not very practical because you need cascaded interferometers. So this is a picture of the device made in my lab. You can see the dove prisms on rotation stages, beam splitters, and this is data going in with these two modes gives us even, odd, and going in from this way gives us odd, even. So it was proposed as a mode sorter, but the key step to using it for entanglement experiments was to use the other input port, right? If you use the other input port, it's a mode combiner and a mode mixer, right? So let's see what happens if we put this device in the middle of our high-dimensional entanglement crystals, now two crystals, each producing a three-dimensional state. This is one minus one zero zero, minus one one, and the same thing here, right? There's a few possibilities. Oh, and I'm going to use the fourth photon as a trigger to herald the presence of my three photon entangled state. So one option is that photon C is transmitted through this, let's imagine this as a big beam splitter, and photon B is transmitted in mode zero zero, so even modes are transmitted, which then would herald photon D in being in mode zero, or I have odd photons reflected, so mode one B is reflected, mode one C is reflected, which then, or mode one B is reflected, and mode minus one C is reflected. Now remember, this is the herald, so there is no mode minus one B, right? And measuring these modes here imposes a condition that you have minus one and one over there. So these are only three possibilities. Combining them, and looking a bit more closely, you find that this is exactly this 3-3-2 state that actually, when we were not planning to create this, funnily enough, we were trying to create a three-dimensional GZ state, but this is one of those good cases of happenstance. So you'll notice that you have one, one minus one, zero, zero, zero, one minus one, one, photon B is in a two-dimensional space, zero and one, photon C and D are in a three-dimensional space. Now the key question is, is this in a coherent superposition, or is it just a classical mixture? The way you test for that is through two photon interference. In fact, it's four photon interference, but again, Benny showed this in his talk just to remind you. Hongo-manual interference, if two photons arrive at a beam splitter, there are four possibilities. They are either both transmitted, both reflected, and one reflected, one transmitted, and the same for the other case. And again, these two possibilities destructively interfere, they cancel out. So the only thing left is both photons going this way, or both photons going this way, this is known as bunching, photon bunching, and it's particular to bosons. Bosons like to stick together. And if you scan one of the paths here, you'll find this dip, right? And then the dip is not always perfect, and that's the key point here. The key point is that the dip not going down to zero depends on how distinguishable the two photons are. So if this photon looks a little bit different from this one in any particular degree of freedom, spectrum, polarization, spatial mode, you will not get very good interference, and this is essentially what then limits the coherence of your state. So we have to now look for Hongo-manual interference in a spatial mode basis. So what we do is we measure in a superposition basis using spatial light modulators, and we find that we, so this is again the basis rotation I was talking about, and we scan one of these path lengths, and we measure for a few hours, right? We see nothing. And then we run the experiment a bit longer. This takes a lot of patience by the way. After two days, you see a dip appearing, and it's like, oh my God, is this working? So after two days, we saw a sign that, okay, I've marked it as 15 now, but it was some arbitrary position. So that tells us that this is the point where we have the perfect sort of condition of alignment and distinguishability. It's not perfect perfect, but it's pretty close. So once we see this, we sort of have a sense that our state is in a coherent superposition. The setup, actual setup looks much bigger. It spans a big table, and if I can show you in some more detail, it consists of a pulsed pump laser, which is then frequency doubled to blue in a BVO crystal, and then the blue laser, UV laser, pumps two PPKTP crystals. The detection is done with spatial light modulators and intensity and single mode fibers. And then we have to be very careful about imaging and mode matching. So we have imaging systems everywhere. And this Mark Zender that I showed you earlier, the OEM beam splitter, had to be designed in such a way that it could be stable for several days. So we designed it in a folded Saniac configuration. So this is a Mark Zender. It doesn't look like one, but it is. And it's sort of a double path Saniac interferometer. So any changes in path length are sort of self-compensating. Lots of lenses, so anyway, you get the picture. It took us a year or so to get this to work. So now that we have this condition of perfect coherence, hopefully, it doesn't prove that our state is entangled in 332 dimensions. It just says that we have some coherence in some subspace. So to verify that the state is actually entangled, you can reconstruct the whole state. But this would take ages, especially since our count rate is so low. So we need an entanglement witness that can basically prove that the state that we measure cannot be decomposed into entangled states of a smaller dimensionality structure, essentially ruling out that our correlations could have come from 32222 or biseparable, bipartite states. So this is similar to the witness I told you before, but it's a little bit simpler and not simpler in the sense, but it's measured differently. It's a fidelity witness, excuse me. It's a fidelity witness, but it measures the fidelity not with mobs, but with two dim subspace measurements. So the way it works is you measure a state and you calculate, well, before you measure the state, you take an ideal state from the space of 322 states and you calculate the fidelity of the ideal state, 322 with the ideal state 332, and you find this to be maximal value of two thirds. And then you take your experimentally created state, which hopefully is here, and you calculate the fidelity of your experimentally created state with an ideal state. To do this, you need to measure certain number of density matrix elements. Ooh, right, 18, well, okay, I kind of went too quickly. Right, so you need to measure 18 diagonal elements and three off diagonal elements. The density matrix of this state has 171 elements. The three diagonals and the three off diagonals that we expect to be non-zero are the following, right? And in the lab, as I've told you before, you can only do projective measurements. So the diagonal elements are measured straightforwardly. You just put 0, 0, 0 on all three SLMs and you get this diagonal element. For the off diagonals, as I explained before, you have to break them down into measurable quantities. So I will not go into too much detail, but just an example of the real part of one of these off diagonals requires you to measure this monster, which can then be decomposed into projective measurements in two-dimensional subspaces and so on and so forth. So just the real part requires 32 measurements and the imaginary part requires another 32 measurements. So each off diagonal requires 64 projective measurements. If you multiply that by 171, you can see where this is going. So that's why tomography is prohibitive, but the witness only requires us to measure three, right? So let's go back to this witness. We took data for four days and we find that our expected density matrix elements look like this. The off diagonals are fairly sizable, which means that our state is coherent in these probability amplitudes and we get a fidelity of about 80%, right? Which means that our state is above this bound of two-thirds and hence it can only be explained by entanglement in three times three times two dimensions. Okay, so what do you do with a state like this, right? It's beautiful in my opinion. I don't think it needs to be used for anything necessarily, but we like to get money from the government. So one possibility is to do layered quantum cryptography. Essentially, the idea here is as follows, you can have secure communication where three parties share multiple layers of information. So Alice and Bob can share, have an extra key that is not known to Carol. So Alice and Bob could share a message, sorry, Alice and Bob and Carol could share a message and then Alice and Bob could share a second layer of message or key that is not known to Carol. That's just one sort of cute idea. So what about three-dimensional G at Z, right? That's still not, we're still not there yet. We searched for a while to get to this state. We tried on paper to figure out different setups and how do you actually produce a state that looks like this to no avail. Until we decided to try a different approach. This was a while ago, it was three, four years ago now, where we said, why don't we get a computer to do it, right? Clearly, our brains are not up to the task of combining lots of optical elements and seeing how modes mix. So we took a toolbox of elements that we programmed. This is a beam splitter, this is a mirror, this is a spiral phase plate, this is an OEM beam splitter and we told the computer to create random setups. Calculate the resulting quantum state, analyze its properties including things like its dimensionality of entanglement and we also use it to come up with ways to do unitary transformations. If it satisfied the criteria, if it didn't satisfy the criteria, start again with a random setup. If it did, try and simplify the setup and then finally learn the setup and go back to the toolbox. So basically save that setup in the toolbox and use it to create more random setups. So while looking for the three, three, three state, we found that not only did we find the three, three, three state, we found about 51 other high dimensional entangled states or ways to create them. So that's what the three, three, three setup looks like. It's really simple in the end. We're like, why couldn't we figure this out? And that's what a 10, 6, 5 setup looks like. So that's a spiral phase plate that shifts the mode by minus five and this is minus two. This is an OEM beam splitter, this is a regular beam splitter. These ones were states that were not found and these are states that are not allowed to exist. So this is what the setup looks like. We got it to work after another huge effort in the lab because now it involves hongomental interference not only here, but also here. So it's quite a challenge to actually make it work. And this is where actually going back to Benny's talk, you realize that you don't necessarily want to keep doing this with bulk optics, which is why we are trying to do other things. So again, I'm not going to explain every detail about this setup. If you're interested, you can look at the paper but just to give you the gist of it, we start with two entangled states and here I'm showing you kind of a graph that shows you that you start with a tensor product of these two entangled states where you have a connection between every possible probability amplitude and then this part of the setup removes certain connections and then finally this part of the setup results in this state which looks like a three dimensional G8Z state and actually this will tell you why we couldn't find it because we weren't looking for a state like this. We were looking for a symmetric state and the computer, the condition we gave the computer was find us a state that's three dimensional entangled but we didn't say it must be symmetric, right? So the computer found a state that was highly asymmetric but each particle lived in a three dimensional space. So minus three, minus two, one, minus one, zero, one and the same for the third. This experiment involved multi mode in harm interference between several different modes. I'm not going to go into detail here and the actual setup looked like this, right? It was two main components. We had to develop a much brighter source of four photon entanglement, four photon correlations and then this massive 27 dimensional multiport essentially. 27 because it was three OEM modes raised to the power of three paths. So we verified three, three, three entanglement in the same way as before except this time it took 21 days to measure and we also did these Merman non-locality tests between in each two dim subspace and we found that there was entanglement across in every two dimensional subspace. The people who did this was Manuel who recently finished his PhD or is almost finished with his PhD with Anton and Mario Cren who is in Canada now. Okay, so that brings me to the last few slides of my talk which is we're doing all this stuff with spatial modes. What about actually doing some real experiments outside the lab where we try to send spatial modes of light across real distance and see whether you can actually use them not only in fibers but actually to communicate with satellites let's say or earth links. So this is a photo of myself and a colleague. We're catching an OEM mode on the rooftop of our building in Vienna after it propagated for three kilometers over the city and we were trying to guide, at the time we were trying to guide it into the telescope and I'm on the phone actually telling my friends to move the mode a little bit to the left, a little bit to the right and this was the experiment you can see the beam going from this one old radio tower that was not being used where we set up the sender and we sent it across to the institute building, very simple setup and with this we were able to send basically information encoded in classical modes and then we took this experiment to the Canary Islands and to see how long could spatial modes of light survive over the ocean. So 143 kilometers, this is a long exposure. I took 25 seconds of Hermit Gauss mode. So it's a superposition of plus and minus one OEM or Hermit Gauss mode 01, propagating 143 kilometers to the island of Tenerife and actually you can even see the Milky Way. It was fun, I would say, but then a storm came and then everything was gone. This is what the mode looks like after propagating 143 kilometers, it's huge. So again, not necessarily practical, but this is the mode on the wall of the telescope, right? So we're beyond the point of trying to get it into the telescope but we have it on the telescope and we encoded a simple message and it was received with a slight grammatical error. So what are the open problems in this field of high dimensional quantum photonics? One is manipulation, we need to figure out how to do simple things like half wave plates, quarter wave plates in high dimensions, rotations. We've made some strides towards doing this but there's still a lot of room for improvement. How to generate them and measure them, both the theory and the experiment, transporting them. I showed you some things about transporting spatial entanglement through a fiber to couple them with quantum states of matter and finally the theory of how to measure entanglement, how to certify high dimensional entanglement both for non-locality as well as things like steering. So these are various things that my group works on right now. Okay, any questions that bore everybody in the end?


Welcome to point set topology part 1, module 38 today. So we are going to study compactness and linear look properties in the larger topic of smallness properties of topological spaces. You have to start with the definition, take a topological space x and a subset a. Recall that a family UI of subsets of x is called a cover for a, if a is contained the union of UIs. So this family will be called an open cover if in addition each member UI is an open subset of x. By a sub cover of a given family, the family must be covered, sub cover means what? For a given set a, we mean a sub family of UI which is a cover for a, it is a sub family and it is a cover so that will be a sub cover. Now we make a proper definition, subset a is called compact if for every open cover of a, we have a finite sub cover for a. Likewise we say a is Lindelof if every open cover for a has a countable sub cover. Remember countable includes finite therefore by this definition compact will be automatically Lindelof but Lindelof may not be compact because countable includes counter infinite also. So compactness Lindelof-ness but Lindelof-ness may not imply compact. However because of the similarity these two concepts can be studied together to some extent. Lots of more properties will be there for compact spaces which are not true for Lindelof spaces in general. Most of the properties for Lindelof space will be automatically true for compact space, compact space or Lindelof space. Ectopological space is called compact or Lindelof if it is compact or Lindelof as a subset of a given space. So if the whole space if x is the whole space a itself is equal to x then what is the meaning of that? a being a compact subset is simply a compact space that is what it is. To immediate that a subset is compact if you know only as a subspace as a space as a take subspace topology etc. if a space is compact or Lindelof. These two concepts are the same because the subspace topology if UIs are open subsets in a given a UI intersection a those will be open subsets inside a that will form a cover whenever that sub cover is there from a intersection UIs the corresponding UIs if you take that will be a sub cover for the original thing and so on. So going to the subspace and back to the space is immediate so compact space is simply as being a compact subspace of a given space that is no problem. Also it is immediate from the definition that a closed subset of a compact space see every time I am saying something of a compact space it is true for Lindelof I am writing it in the bracket here. So here a closed subset of a Lindelof space is limited low a closed subset of a compact space is compact. Similarly it follows that a finite union of compact spaces is compact when you take Lindelof of course finite you know Lindelof space compact you can say more namely any countable union of countable union of Lindelof space is also Lindelof. Now let us do this little more carefully why take all open subsets let us see whether we can do it some economically start the topology space suppose b is a base for the topology then x is compact or Lindelof respected if we do not live every cover of x by members of b admits a finite sub cover admits a countable sub cover respected instead of every open cover we are having now covers which are coming from members are coming from a given base so family UI is a sub family of b that much I can get you if that is true for all sub families of b itself then it will be true for all open covers so this is the lemma here let us see how this is true clearly if you have members of b they are also open they must admit finite that is all right but why that is sufficient is what we have to see to start with the condition that uj be an open cover for x I have to produce a finite sub cover under the assumption that for a basic open subset this set this is true okay for each point x in x this x will be in one of the uj's that j will depend upon x so I write the uj x select one x is inside ujx for ujx being jx being 1000 these j's okay but by the definition of a base what happens there will be a member of this curly b namely bx such that x is in bx and bx is contained u of bx then you vary x all these bx's form an open cover for x but that are members of this curly b only the basis okay that will admit a finite sub cover this bx1 bx2 bxn's so these will cover the whole of x but now each bxi I can replace by uxj's ujx's so they will cover the entire of x so I have got a finite sub cover exactly if you if you allow this one to be countable here instead of ironed 1 to n I want to infinity 1 to infinity a countable cover then you get little of property okay now another important thing is to investigate what happens under continuous maps okay the image of a compact receptacle in the low space under a continuous map is compact respectively in the low so I will now on won't say this in the low part at all this will be automatic and it will be written down here okay okay so the proof will be also exactly similar and so on so let us look at how why this is compact and so on you start with a contained zx, p contained inside y, x and y are topological spaces and f from a to b is a continuous function such that f of h b that is subjective continuous function okay I have to show that if a is compact then b is compact okay assuming that a is compact I am sure b is compact okay b is a compact subject of y so take an open cover of y take an open cover of b open subsets of y f inverse of those will open subsets of x and they will cover a because f of a is inside b f of a is actually equal to b right so start with the uj ui i inside j open cover of b then each f inverse of ui intersection b is open in a and a equal to i i range on f inverse of ui intersection b okay i f inverse of ui itself I will take then intersection with a is the same thing as take a f inverse of ui intersection b because f is not defined on the whole of x I do not say f inverse of bi I have to take a intersection here f may not be defined on the whole of x that is not necessary f is defined only on a that is why I am writing this otherwise it is just f inverse of ui okay so this is an open cover now f in a is compact so I will get a finite sub cover okay therefore the finite sub cover so i is a sub finite subset of this is that a is inside union of bi's i belong to i what are these bi's bi's or bi these f inverse of ui's okay I have put f inverse of ui intersection b as bi intersection a bi's because these are open subsets of a means what they are bi intersection where bi's are open zx okay if f for defined on the whole of x I could have taken f inverse of ui as bi here I have to separately say what these bi's are okay so this will admit a finite sub cover here a intersection bi it follows that b will be inside the union of now corresponding ui's here where this i will run only over i okay after this bi's are just damaged here to show that this is an open subset in here so an immediate consequence is that its compactness and end end over topological properties what does that mean that means if you have homeomorphism from x to y if x is compact then y is compact and conversely so that is the meaning of topological property you don't need homeomorphism you just need continuous surjective man a is compact f of a is compact a is length of f of a is length of so that will show that compactness and end end ofness are topological properties okay so notice that it just means that the compactness of subset is a topological space does not depend where a is taken as a subset okay a may be a topological space it may subspace of some y it may subspace of some x it may subspace of some the moment a is compact in one of them it will be always compact as a space itself and then it will be compact in every other every other say so it is independent of being a subset of something okay the same remark holds for little of property connectivity path connectivity etc if you think they are topola once the something of topological properties the topological space where it is embedded is immaterial so now we have proved a theorem about quotient maps with strong property then if each fiber is connected and the base space is connected then the total space is connected x to y is a surjective map all right y is compact y is connected when this x is connected so there was a converse theorem this 3.13 similar to that we may expect if y is compact and all the fibers here f in most of y's are compact will x be compact okay the answer is no even under a stronger assumption that f is an open mapping open surjective is quite strong you can assume it is closed mapping also closed surject even then this will not be true namely a very simple example is here take the open interval minus 1 to closed one it is half closed minus 1 open plus 1 closed take this open interval to 0 1 closed interval what is the map eta x equal to x square x going to x square that's surjective okay is it an open mapping if you have an open interval here what will be the image of this open interval open interval open interval or half open on this side one closed what will be the image of those will that be open subsets of of 0 1 you see you can't take uh some epsilon some r here less than 1 then the open interval you have to be open subset after you open interval itself okay but if you take one close one also then you can take the closed interval also that will be still an open subset of the half closed interval the but image will be a half closed interval here half open interval here so that will be open subset of of 0 1 all right so eta x is both an open mapping as well as a closed mapping it is surjective inverse image of every point here is just one point or two points depending upon if it's zero inverse image is zero any other point there will be two points here okay there are two square outs so fibers are all compact the image is a closed interval that's compact but we know that half closed intervals like this are non-compact so very easy example okay in fact you can produce build on this one you can produce many such examples indeed in application this is what causes a problem so you have to be careful of exactly precise in this one what you have actually done is you look at the full thing here you take minus 1 to plus 1 close interval that will be compact that will include all the solutions what you have done is certain solutions are missing here right so that is the one which causes problems okay this precisely the way how the the expected answer here y is compact fibers are compact y x may not be compact okay so keep this example in mind so one of the consequence of this example is that we cannot do compactness of product of two compact spaces in a similar way as we have done it connectedness remember connectedness x cross y x and y connected x cross y connected which can be proved in different ways simpler way is to take the quotient map from x cross y to y apply that theorem y is connected the fibers are x cross singleton wise right for each y so they are all connected because they are a homomorphic place therefore the the entire thing with the total space with x cross y that will be connected that kind of proof is not possible here is what this example says okay so we have to do a little more work here and that gives us a beautiful theorem here called valance theorem which will be used elsewhere also okay so what does go through this valance theorem x be a compact space then for every topological space y and a point y inside y and an open subset v of x cross y such that this v is a neighborhood of is capital x cross singleton y suppose if you have this situation so x is a sub x cross singleton y is a subspace right of x cross capital y the v is a subspace which is a neighborhood of is open open subset neighborhood x cross then you have some kind of a uniform neighborhood is what it is provided this x is compact compact so what is that uniformity there exists open neighborhood n which is I am writing n y depends upon y inside y such that this entire x cross n itself will be contained inside every neighborhood has a unique single neighborhood of this point because that whole x cross n will be contained inside v this is the kind of uniformity uniform neighborhood comes it is like this n little y is uniform epsilon if you write epsilon y here that will depend upon the y point y here now this capital n does not depend upon any of them for all x cross y this is this is the same thing that is what it says okay so the proof will actually give you better view of why I call that uniformity and so on for each x comma y as x varies x cross y inside contained inside v by definition of product apology you will get open subsets ux and vx if you have matrix basis you could have get epsilon same epsilon x epsilon y and so on so now I have to write no metric here some open subsets ux vx and x cross y x as well as y sorry not x cross y respectively such that if you take the product x y will be x y will be neighborhood of ux cross vx will be neighborhood of x comma y and that neighborhood is contained inside the open subset v that is the meaning of an open subset in the product space right so these are basic open subsets here by compactness of x okay what happens we will get finitely many x i's such that x is contained inside i read 1 to n ux i's because as x varies over x ux will cover the whole of x so they will get a finite now put n equal to this will depend upon the second coordinate y for each of them the y is fixed here that is why I am not writing uxy vxy and so on I am just writing ux vx n equal to n y which is intersection of the corresponding vx1 vx2 vxn you depending upon ux1 ux2 vxn here right so take the intersection finite intersection of open subsets will be an open subset all of them see vx is for all vx they are neighborhoods of y okay so intersection of all of them finite many of them that n is an neighborhood of y so we claim that x cross n is inside v now so this is a typical way you take the union here you take the corresponding intersection there okay so why this is contained inside v look at any x comma z inside x cross n z is in the intersection means z is in which vx i's okay so choose an i such that x is in one of the ux i's then ux i cross vx i is inside v so x comma z is inside v that is so okay so vales theorem says that if you have a slice it is a compact space inside a x cross y take a slice and x cross little somewhere a neighborhood of that you can take uniformly okay one single neighborhood of y cross the whole of x will be contained in that neighborhood so this will be very useful in proving many other things also so we are going to use that one over here in the proof of product of two compact spaces compact now you notice that I have stopped telling about lindeloff if you take lindeloff what you get you will get some somewhere countable infinite things may be get intersection of infinitely many opens of such may not be open so that is where these results will not be true for lindeloff spaces okay so we are we are not having those results for lindeloff spaces okay so x and y are compact then x cross y is compact so how does it prove that start with an open cover of x cross y for each y inside y singleton x cross y is compact right and this ui's will cover that also therefore you will get a finite subset j y that depend upon little y contained inside j subset x cross y is contained in the open cover which is sub sub family or depending on j y only y side is at j y union of all those ui's so let us call that open set v y it depends upon y now this is true for every y x cross y since that vi if i range y that will cover the whole of x so before that before that since x cross y is contained in that vi i am going to use valet's theorem here now by valet's theorem there exists an open subset n y of y such that x cross y is contained inside x cross n y contained inside v y i don't have to take this funny v y but i can take now n y and forget about x so these n y's as y varies over capital y they will cover capital y that is compact so i will get a finite even y 1 y 2 y k such that y is contained inside union of n y i i 1 2 k okay now take i equal to the corresponding j y 1 j y 2 j y k take the union of those so that is a subset of capital j okay each j y i is finite set and this is the finite union so the actual i is a finite subset of capital j the claim is that if you take now only u i's inside i this finite set that will cover the whole of x cross y and that is very straight forward start with any x comma y in x cross y this y will be in one of the n y i's because y is contained in n y i's here then x cross y will be contained in x cross n y y but that is contained inside v y j corresponding y i j y i whatever y n y j i okay v y j okay but v y j see v y j's union now is u i right where i is in j y j here okay one single set here j y j and take those unions so that is just a union of under j y i's okay so that is definitely contained in all the u i's where i rings over capital i where capital i is actually union of all j y i's okay so i have shown that x cross y is contained inside this union y is arbitrary so the whole x cross y is contained inside that that's all so product of two compacts space is compact is what we have proved immediately it follows that finite product of compact space is compact so if you have finite product like this then it's compact okay and if x is compact each xi is compact because they are we can take the projection maps so you have a product space like this is compact you spend on this each factor is compact okay so for it immediately you want to prove this for arbitrary products that is for techno's theorem okay so there is two for infinite product as well goes under the name techno's theorem but for that you will have to wait okay but let us do a little more of compact space here techno's theorem is a bit special thing we will do it's on its own at its own time after one or two more modules it x be a compact product space f1 contain is f2 contain is monotonically decreasing it's a nest of a of sequence of non-empty closed subsets okay remember such a thing we had in cantor's intersection theorem wherein we had a complete metric space then the diameters were going to 0 etc right so here there is no metric no diameter and so on all that I assume is that instead of metric space complete matrices a compact logical space then the intersection of all these f is is non-empty so here the proof is surprisingly much much simpler than cantor's introduction theorem even that was not very difficult anyway all that I have to do is apply de Morgan's law f i's are closed compliment of f i's are open f i's are decreasing compliments of f i's are increasing union okay intersection is non-empty is what we want to show is the same thing as that union is not the whole of x is the same thing intersection is non-empty by de Morgan's law if the union is the whole space what you have got is an increasing sequence of open subsets such that the union is the whole space x is compact therefore it has a finite sub cover an increasing union of open subsets the finite sub cover means at one of the the maximum one of maximum one you take that itself will be x what does that mean one of the ui's itself is x which means when you go to compliment those one of these f i's must be empty but I have taken an unempty close subsets okay so that is the full proof I have just given you a prior de Morgan's law rest of them you must be able to do anyway now I have told you all right so next time we will do some more things about compact metric spaces okay thank you