Local stabilizer codes in three dimensions without string logical operators
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Haah, Jeongwan
We report examples of local stabilizer codes in 3D that have no string logical operators. Previously known local stabilizer codes in 3D all have string-like logical operators, which make the codes non-self-correcting. We introduce a notion of "logical string segments" to avoid difficulties in defining one dimensional objects in discrete lattices. We prove that every string-like logical operator of our code can be deformed to a disjoint union of short segments, and each segment is in the stabilizer group. The code space dimension depends on the number-theoretic property of the system size.
 scientists reviewed all these submissions and picked out two for special recognition. So I'm about to announce, I hold in my hand the name of the winner of one of the best student paper awards for the conference. Before I announce it though, I would like to ask the next speaker to raise his right hand and repeat after me. Okay. I solemnly affirm. I solemnly affirm. That I will faithfully correct. That I faithfully incorrect. All quantum errors. All quantum errors. Real or imagined. Real or imagined. And will to the best of my ability. And will to the best of my knowledge. Preserve, protect and defend. Preserve, protect and defend. The quantum computer. The quantum computer. Congratulations by the authority. Bestowed upon me. I bestow upon you the best student paper award. And our next speaker. Thank you. Thank you. And one pop. Local stabilizer codes in 3D without logical string operators. Thank you very much. It's my great pleasure and honor to receive your prize. And I would like to thank my advisor, John, because it was like a year ago or half ago or two that I was working on a wrong model. I was wrong. I proposed the model and it turns out to be have a string operator like after three months of calculation. But still John expressed genuine interest in my work and that encouraged and helped to accelerate and to seek further tries. So let me begin. It's all about the quantum hard drive. We wish to find the, or noise free, subsystem or subspace from a large helper space from a physical system. But we want to have some manipulation ability on that. For example, there is a billion years of old entangled system, the ground state electrons and the helium atom, but we have no ability to access them. So if it's not interesting to us, we wish to have manipulation on that. My focus is on a many by system that have so-called local indistinguishability or called topological order, which means it has a ground state, degenerate ground state, more than two ground states, but they are exactly the same when you look at the locally. But this kind of steps, and they are for a long time known to stable at zero temperature. Taira mentioned it briefly in his paper and recently just last year, Braby Hastings and Mikalakis proved, actually, for proper Hamiltonians, such a degeneracy and the gap remains stable under any arbitrary perturbations. So it defines a quantum phase. It is, but the word topological order should not be confused with the topological quantum computing with the anions. Topological quantum computing with the anions concerns about the excited states. We encode the quantum information into the excited states, the configuration of the facts, anions. But here, we are mainly focused on the ground space of the Hamiltonian. Why is this problem hard? Well, the prototypical example of the topological order is obviously the two-dimensional types code, just two terms in the Hamiltonian. There are all commuting. Thank you. So, yeah. And it has four degenerate ground states when the model is defined on the torus. And such a forward degeneracy comes from the non-trivial topology of the space it is embedded. And because such a topology of consideration leads to the conclude that the ground states are locally indistinguishable. But at a finite temperature, there may be some defects. And there has to be some defects. And since we are able to operate on the anions, we believe to do so. So anions can move around. But if we can move the anions, then nature can move. And thermal errors are unknown to us. And they will drag the defects from arbitrary far apart. And if the defects are created and propagated for long distance and realylate in a non-trivial way, then it is as if there has been some non-trivial action on the ground space, which means the encoded quantum information at the ground space has been corrupted. It's been flipped. Further, this phenomenon has happened for all low dimensional models. By low dimensional, I mean 3 or 4, 3 or 2. But around 10 years ago, in a paper by Dennis and all, the four dimensional toricos is proposed. And three or four years ago, it has been rigorous proof that it has a self-correcting property, meaning that the realization time of an encoded state grows exponentially with a system size, provided that the temperature is sufficiently low. So there has been a lot of questioning. What would happen in three dimensions? Bacon proposed a model in the three dimensions. But it seemed to work via the mean field theory. But this Hamiltonian is non-commuting. So we have no idea what the spectrum looks like. It may not be even gapped. But here's the possibility. Each term defines a quantum code. Each term is a generator for the quantum code defined in a three dimensional torus with two qubits per site. And you can easily verify that it defines a stabilizer code. Every terms commute. So energy spectrum-wise, everything is no. It's just equally spaced. And the gap does not collapse in the thermodynamic limits. And it has no frustration. And we can prove that it has no string-like operator, the meaning of which I will describe in detail later. The key concept to understand this model is to understand the string segments, not the strings. Well, I will prefer to define a string in terms of string segments, as you will see later. And it has a very exotic degeneracy. When the system size has a lot of powers of two, the degeneracy grows exponentially. But if the system size has in a particular shape, then it drops to constant. OK, let us begin from the very primitive setting, because this is the way I found that model. From the yesterday's TOS tutorial, we know that stabilizer code is equivalent to study the classical codes over the operator algebra. Well, to be specific, the poly operator group. And you can parameterize any poly operator by bit strings. And you can interpret the multiplication of operators in terms of the modular two addition of such a bit strings. Further, you can even encode the commutation relation into some simple tactic structure. So every language, well, statement in the civilized code can be described in terms of simple tactic binary vector space. More generally, if you consider QDs with a prime dimension, then the discretion becomes the linear algebra over that final field. So what's code in this language? The commutative set of poly operator corresponds to the null space by null mean where the space on which this simple tactic form vanishes. And we call that the stabilizer group. And the corresponding symmetric group, when you think of the codes as some representation of a Hamiltonian, the symmetric group of that Hamiltonian corresponds to the law school operators. OK, what should I do with that? Ios are there may be some several qubits per site. That's very general setting. So I don't assume anything about the number of qubits per site. And there are some finite number of generators for each site. And since I don't want to describe the model with the infinite data, so I assume the translation invariance. And I want to have no single site logical operator which corresponds to the local indistinguishability of the ground space. And since we want some Hamiltonian where the defects cannot move, where the string offers an absent, so we want some model with the no obvious string operators. But this is too general. So I have to simply find myself to the case where t equals 2. Just one generator for x type and one generator for z. And no assumption on the number of qubits per site. And I wish to have a degenerate ground space so the product of all terms in a Hamiltonian should be identity. This was the, you can interpret the degeneracy of the tory code in this way too. And the no single site logical operator condition translated to this logical operator on a single site should be identity. Well, this is too restrictive, restrictive, because all we need is if a single site logical operator is found and we require the element should be in this stabilizer group, but setting to be identity is more simple. And I assume further that the logical operator on a straight line, meaning that three coordinate axes or face diagonals or body diagonals are all identity. If they are repeated, a single site operators. And apply the formalism of stabilizer code to this conditions, then you can translate every condition in terms of linear algebra equations. And the last two conditions, the logical operator on a single site to be identity and the logical operator on a straight line be identity is translated to some rank condition on the commutation matrix. So I started calculation by my desktop computer. Since I assume the two types of generators, one for z type and one for x type, there are 16 corner operators. And it turns out that not exactly representation of the operators are important. The only important and the unique parameter is the commutation relation among those. Since there are eight z's and eight x's, there are 64. So 8 by 8 matrix determines everything. And we need some condition so that our terms define the stabilizer code. And that translates to 27 linear equations, which can be easily solved by hand even. And we solve that equation by brute force. So 64 variables and 27 equations, and you are left with 37 free variables, which is like 100 billion, a not very big number for modern computers. So check all the rank conditions that demands the single side no operator or no obvious string operator. Then you get not a large number of solutions, not no solutions, but there are handful of equations that satisfy all the conditions I described before. So here comes the first one, which has the largest symmetry group of all. And I designed it from the first place that they are commuting, so you don't have to check it manually. But if you wish, you can. And there's some property that I didn't demand. There is some duality between z operator and x operator. And this duality reduces a lot of calculation in practice. For example, if you flip the body diagonals, like z, z, and i, i, and replace the first qubit and the second qubit, and relabel z and x, then you get z from x and x to z. They are exactly dual. OK, I designed it so that if it's topology ordered and it has no string, but I didn't prove anything. It was just a convenient requirement, necessary conditions. It's not sufficient, so we have to check whether it really works. And there's a quite annoying question. What do you mean by string? I didn't define a string. Well, let us answer the first question first. The first question is, is it topology ordered? And the answer is yes, meaning that there's no local observable that this thing which ground states. So logic goes like this. If an operator is local, then when restricted to the ground space, its action is proportional to the identity element. And how do you prove that? The claim is, any poly operator support, commuting with all stabilizer generators, then the poly operator itself is a stabilizer. And this implies the first statement because any operator has a linear expression in terms of poly operators. And this equation is linear. And because of the duality, I don't need to check for all of possibilities. I just focus on the x type only for convenience or z type equivalently. OK. All you need is this tool, the eraser. If the question mark is one of these four operators, well, ii means i tensor i, actually. And xi is x tensor i for convention. And if I impose that, it has to come in with zz and what's the possibilities. And you should be able to figure out. The answer is either identity or x. What if I impose two conditions? It has to come in with iz and it has to come in with zi. Then it's nothing. It's always the identity. And by similar manner, if I pose lower two conditions, then question mark should be identity two. OK. What is that? What is it good for? So I promised to show that any x type operator that commutes with z type stabilizer generators are stabilizers themselves. So suppose a really big box, well, big bed finite, and check the commutativity with the z operator here. This corner has to commute with this corner, which means this corner is either identity or identity. No other option. But this xi is this xi too. So if you multiply this generator to here, if this side turns out to be xi, then you can make this side to be identity by acting on inside of this box. So I can keep doing this. So whenever I see a vertex from this side, I can always make them as identity by multiplying this. So I would get a poly operator of support that looks like this. But look at this side. This side has to commute with i, z, and zi. By previous exercise, you know that this is identity. So you keep doing that. And then you will erase the left-hand side vertical square. And look at this side, and you apply the same logic here using this edge. And you know that this is identity. And everything disappear. So I started from a very general x-type operator and imposed that it has to commute with this and implied that it is a product of the right-hand side. So it is a stabilizer. Now, for the no-string rule, here's a challenge. This square represents the elementary excitations when there is a single bit flip on the first qubit or second qubit. And the black dots represent the flip stabilizer generators. Can you make a pair just two defects out of this? Well, the answer is no. And I'll just give you a little bit of a summary of that. Well, the answer is no. And I will show in detail how to prove that. So here's an old question. What do you mean by string? We know there's a string operator in the directory code because we seem like an operator. We named some operator string. We haven't defined string yet. For example, this looks like a string. Combination of two strings that looks like string. OK. Is it surface or string? It's not obvious. So we need some different definition for a string. So here I propose most natural definition, I think. The problem was that in the discrete lattice, one dimensional object is not well defined. You can tell a one-dimensional object when you see it, but it's not a definition. And in general, we need to deal with a family of Hamiltonians, and I have no idea. So we need to go to some finite setting. We can't deal with some infinite things. So what's a string segment? It is a finite supported poly operator that crustation at most two locations. Here, two is important because any conceivable string has two ends. And three more terminologies. The anchors mean that the box is that envelope of excitations. And width is defined to be the size of the anchors. And length is the distance between the anchors. The point is there's no geometry constraint. The string segment can be like a sausage or some weird knots or whatever you want. The only condition is that it creates only an excitation in two locations. But it's not the end of the story because if you consider an identity operator, then it is trivially a string segment because it creates no excitation. Therefore, it creates less than two locations. So we need to get rid of such a triviality. And trivial conditions are rephrased like this. Whenever you have a string segment, it has an equivalent poly operator whose support are separated by multiplying the stabilizer generators. And if you want to be more sophisticated, then you can say that a trivial string segment only creates trivial charges that are possible to create locally. So let us see some examples. In the territorial code, when you apply zzz in this manner, then it creates a defense at the end, two locations. So it conforms with the definition of string segments. And it is. And you can show that it is non-trivial. There's no string segments that you can cut out this region. You have to connect somehow between two yellow squares. It has width 1 because width is defined to be the size of the anchor. And length is potentially infinite. It is not bounded from above because you can put as many z's as you wish. OK. What about then to the Ising model that we understand very, very well? It is an example of stabilizer code because any classical code is. So suppose a black region has flipped spins, then you will create a domain wall along its boundary. And in order for such a string segment, well, this is a string segment because it creates an expectation in two regions where the yellow squares cover. And you can show that two anchors must be adjacent. Otherwise, each excitation in each square can be created locally. So it has some non-trivial width, the size of the anchor. But length, the distance between two anchors should be always 0. So here's the formal no string rule for my model. Our statement is that a string segment of width w is always trivial, provided that its length, the distance between anchors are sufficiently large. And the boundary is not too severe. It's just a polynomial wall. It's just linear. And this will be important when you want to show some dynamic properties of this model. So how do we prove that? Well, we can just follow the definition because we have good definition. We started with our arbitrary string segment and showed that there is an equivalent disconnected string segment by multiplying suitable stabilizer generators. And the trick is that any string segment will be reduced to a union of three flat ones. And I will show that each flat ones are trivial. So we used the eraser again. You just test them coming to activity with z operators. And since this edge has two independent ones, and that implies you can eat up the lower front part of the box. So you can transform this box to here. And that trick is used for the reduction. For a given general string segment, you can reduce some flat ones, flat means parallel to the coordinate axis. And given us flat ones, you can argue further using the same argument that using this independent edge, you reduce to the second figure. And here's some slightly complicated calculation. If two adjacent operators are commuting with i, z, and zi, and ii, and iz, then there are only four possibilities. And since the model is translation invariant, it has to make some consistent combination of the operators. And you realize that at most after two sites, it has to be identity, eventually. And you keep inductively argue line by line by eating up more and more sites. And you end up with a small triangle on the left. And a vacant thing on the right. So we have disconnected. We are done. What does that mean? No string means you can drag the D-fat, actually. If you can drag the D-fat from here to there, then the combination of those are represented by string segment. And this is not allowed. So the dragging D-fat is impossible. It's protected by the Hamiltonian. So let me finish with this exotic feature of this model. Here's the k, the number of encoded qubits, as a function of system size, l by l by l, l cubed. And it looks like something we have never seen before. But you can prove some formula. And it turns out that the calculation reduces to GCD calculation over the field of four elements, where the t. Remember that t is not the exponential 2 pi i over 3. It leaves in an extension field of binary field. So it has no connection to the complex numbers. But anyway, you can efficiently calculate the k using this formula. Let me just finish here by summarizing the basic properties of this model. Thanks for listening. Questions for Jong-un? So what happened to the other 16 codes? Are they interesting too? The seven models that are numbered like 11 to 17, it has strings but has width 2. And three other models have no strings. But the bound on the length of the string segment is exponential. It's not as good as this linear bound. And for other six things, I don't know. In Kataya's two-dimensional surface codes, you can load it with many logical qubits by removing stabilizer generators. And each generator you remove becomes a logical qubit. If we remove a plaquette type operator there, we can identify that, let's say, with logical z. And the logical x becomes a string-like operator that connects that plaquette to the nearest other plaquette, which may be the entire surface itself if it's a bounded surface. So when we create logical qubits this way, we necessarily create string-like logical operators, even if we just ignore the entire volume itself. In your codes, if you were to remove a cube out of a volume and call that a logical operator, there must be a corresponding, meaning partner to it, that anti-commutes with that three-dimensional volume and does something else. Is that object string-like, or what is the geometry of that object which anti-commutes with the cube which you've removed? Very interesting question. I thought about that for a moment, but I don't have concrete answer for that. It is, well, it is somehow related to the question that how do we define this model with the open boundary? But there could be many boundary conditions that you can impose, and there could be many poly-operators that you can put on the boundary, and I have no idea how to classify them. And even further, because K has a complex number of theoretical dependence on the system size, I don't have a complete list of logical operators, too. So for asking the geometry, it is far beyond. Do you see a way to extend this model so that it will also support computation beyond memory? I personally didn't much think about that. Perhaps one can exploit the feature that it has extensive degeneracy. But that was not the main focus of my research recent days. OK, last question. So just following on from all those comments, wonderful talk. I think you can do computation in this. If you cut a torus through this, you can have a logical operator that's a membrane in the torus. And if you have alternate logical qubits that are spheres, you can punch holes around that and drag strings and surfaces. And I think you can do computation in that scheme. So be great to talk. OK, thank you. How do you fasten your shoes, Velcro? It's just strings. OK, let's thank Zhang Lan again. Let's bring up the next speaker.
 scientists reviewed all these submissions  and picked out two for special recognition.  So I'm about to announce, I hold in my hand  the name of the winner of one of the best student paper  awards for the conference.  Before I announce it though, I would like to ask  the next speaker to raise his right hand  and repeat after me.  Okay.  I solemnly affirm.  I solemnly affirm.  That I will faithfully correct.  That I faithfully incorrect.  All quantum errors.  All quantum errors.  Real or imagined.  Real or imagined.  And will to the best of my ability.  And will to the best of my knowledge.  Preserve, protect and defend.  Preserve, protect and defend.  The quantum computer.  The quantum computer.  Congratulations by the authority.  Bestowed upon me.  I bestow upon you the best student paper award.  And our next speaker.  Thank you.  Thank you.  And one pop.  Local stabilizer codes in 3D  without logical string operators.  Thank you very much.  It's my great pleasure and honor to receive your prize.  And I would like to thank my advisor, John,  because it was like a year ago or  half ago or two that I was working on a wrong model.  I was wrong.  I proposed the model and it turns out to be  have a string operator like after three months  of calculation.  But still John expressed genuine interest in my work  and that encouraged and helped to accelerate  and to seek further tries.  So let me begin.  It's all about the quantum hard drive.  We wish to find the, or noise free,  subsystem or subspace from a large helper space  from a physical system.  But we want to have some manipulation ability on that.  For example, there is a billion years of old  entangled system, the ground state electrons  and the helium atom, but we have no ability to access them.  So if it's not interesting to us,  we wish to have manipulation on that.  My focus is on a many by system  that have so-called local indistinguishability  or called topological order,  which means it has a ground state,  degenerate ground state, more than two ground states,  but they are exactly the same when you look at the locally.  But this kind of steps,  and they are for a long time known to stable  at zero temperature.  Taira mentioned it briefly in his paper  and recently just last year,  Braby Hastings and Mikalakis proved, actually,  for proper Hamiltonians, such a degeneracy  and the gap remains stable  under any arbitrary perturbations.  So it defines a quantum phase.  It is, but the word topological order  should not be confused with the topological  quantum computing with the anions.  Topological quantum computing with the anions  concerns about the excited states.  We encode the quantum information into the excited states,  the configuration of the facts, anions.  But here, we are mainly focused  on the ground space of the Hamiltonian.  Why is this problem hard?  Well, the prototypical example of the topological order  is obviously the two-dimensional types code,  just two terms in the Hamiltonian.  There are all commuting.  Thank you.  So, yeah.  And it has four degenerate ground states  when the model is defined on the torus.  And such a forward degeneracy comes  from the non-trivial topology of the space it is embedded.  And because such a topology of consideration leads to  the conclude that the ground states are locally  indistinguishable.  But at a finite temperature, there may be some defects.  And there has to be some defects.  And since we are able to operate on the anions,  we believe to do so.  So anions can move around.  But if we can move the anions, then nature can move.  And thermal errors are unknown to us.  And they will drag the defects from arbitrary far apart.  And if the defects are created and propagated for long distance  and realylate in a non-trivial way,  then it is as if there has been some non-trivial action  on the ground space, which means the encoded quantum  information at the ground space has been corrupted.  It's been flipped.  Further, this phenomenon has happened for all low dimensional  models.  By low dimensional, I mean 3 or 4, 3 or 2.  But around 10 years ago, in a paper by Dennis and all,  the four dimensional toricos is proposed.  And three or four years ago, it has been rigorous proof  that it has a self-correcting property,  meaning that the realization time of an encoded state  grows exponentially with a system size,  provided that the temperature is sufficiently low.  So there has been a lot of questioning.  What would happen in three dimensions?  Bacon proposed a model in the three dimensions.  But it seemed to work via the mean field theory.  But this Hamiltonian is non-commuting.  So we have no idea what the spectrum looks like.  It may not be even gapped.  But here's the possibility.  Each term defines a quantum code.  Each term is a generator for the quantum code  defined in a three dimensional torus with two qubits per site.  And you can easily verify that it defines a stabilizer code.  Every terms commute.  So energy spectrum-wise, everything is no.  It's just equally spaced.  And the gap does not collapse in the thermodynamic limits.  And it has no frustration.  And we can prove that it has no string-like operator, the  meaning of which I will describe in detail later.  The key concept to understand this model  is to understand the string segments, not the strings.  Well, I will prefer to define a string in terms of string  segments, as you will see later.  And it has a very exotic degeneracy.  When the system size has a lot of powers of two, the  degeneracy grows exponentially.  But if the system size has in a particular shape, then it  drops to constant.  OK, let us begin from the very primitive setting, because  this is the way I found that model.  From the yesterday's TOS tutorial, we know that  stabilizer code is equivalent to study the classical codes  over the operator algebra.  Well, to be specific, the poly operator group.  And you can parameterize any poly operator by bit strings.  And you can interpret the multiplication of operators in  terms of the modular two addition of such a bit  strings.  Further, you can even encode the commutation relation  into some simple tactic structure.  So every language, well, statement in the civilized  code can be described in terms of simple tactic binary  vector space.  More generally, if you consider QDs with a prime  dimension, then the discretion becomes the linear  algebra over that final field.  So what's code in this language?  The commutative set of poly operator corresponds to the  null space by null mean where the space on which this  simple tactic form vanishes.  And we call that the stabilizer group.  And the corresponding symmetric group, when you  think of the codes as some representation of a  Hamiltonian, the symmetric group of that Hamiltonian  corresponds to the law school operators.  OK, what should I do with that?  Ios are there may be some several qubits per site.  That's very general setting.  So I don't assume anything about the number of qubits per  site.  And there are some finite number of generators for each  site.  And since I don't want to describe the model with the  infinite data, so I assume the translation invariance.  And I want to have no single site logical operator which  corresponds to the local indistinguishability of the  ground space.  And since we want some Hamiltonian where the defects  cannot move, where the string offers an absent, so we  want some model with the no obvious string operators.  But this is too general.  So I have to simply find myself to the case where t  equals 2.  Just one generator for x type and one generator for z.  And no assumption on the number of qubits per site.  And I wish to have a degenerate ground space so the  product of all terms in a Hamiltonian should be  identity.  This was the, you can interpret the degeneracy of the  tory code in this way too.  And the no single site logical operator condition  translated to this logical operator on a single site  should be identity.  Well, this is too restrictive, restrictive, because all we  need is if a single site logical operator is found and we  require the element should be in this stabilizer group, but  setting to be identity is more simple.  And I assume further that the logical operator on a  straight line, meaning that three coordinate axes or  face diagonals or body diagonals are all identity.  If they are repeated, a single site operators.  And apply the formalism of stabilizer code to this  conditions, then you can translate every condition in  terms of linear algebra equations.  And the last two conditions, the logical operator on a  single site to be identity and the logical operator on a  straight line be identity is translated to some rank  condition on the commutation matrix.  So I started calculation by my desktop computer.  Since I assume the two types of generators, one for z type  and one for x type, there are 16 corner operators.  And it turns out that not exactly representation of the  operators are important.  The only important and the unique parameter is the  commutation relation among those.  Since there are eight z's and eight x's, there are 64.  So 8 by 8 matrix determines everything.  And we need some condition so that our terms define the  stabilizer code.  And that translates to 27 linear equations, which can be  easily solved by hand even.  And we solve that equation by brute force.  So 64 variables and 27 equations, and you are left  with 37 free variables, which is like 100 billion, a not  very big number for modern computers.  So check all the rank conditions that demands the  single side no operator or no obvious string operator.  Then you get not a large number of solutions, not no  solutions, but there are handful of equations that  satisfy all the conditions I described before.  So here comes the first one, which has the largest  symmetry group of all.  And I designed it from the first place that they are  commuting, so you don't have to check it manually.  But if you wish, you can.  And there's some property that I didn't demand.  There is some duality between z operator and x operator.  And this duality reduces a lot of  calculation in practice.  For example, if you flip the body diagonals, like z, z,  and i, i, and replace the first qubit and the second  qubit, and relabel z and x, then you get z from x and x to z.  They are exactly dual.  OK, I designed it so that if it's topology ordered and it  has no string, but I didn't prove anything.  It was just a convenient requirement, necessary  conditions.  It's not sufficient, so we have to check whether it  really works.  And there's a quite annoying question.  What do you mean by string?  I didn't define a string.  Well, let us answer the first question first.  The first question is, is it topology ordered?  And the answer is yes, meaning that there's no local  observable that this thing which ground states.  So logic goes like this.  If an operator is local, then when restricted to the ground  space, its action is proportional to the identity  element.  And how do you prove that?  The claim is, any poly operator support, commuting with  all stabilizer generators, then the poly operator itself is  a stabilizer.  And this implies the first statement because any operator  has a linear expression in terms of poly operators.  And this equation is linear.  And because of the duality, I don't need to check for all  of possibilities.  I just focus on the x type only for convenience or z type  equivalently.  OK.  All you need is this tool, the eraser.  If the question mark is one of these four operators, well,  ii means i tensor i, actually.  And xi is x tensor i for convention.  And if I impose that, it has to come in with zz and what's  the possibilities.  And you should be able to figure out.  The answer is either identity or x.  What if I impose two conditions?  It has to come in with iz and it has to come in with zi.  Then it's nothing.  It's always the identity.  And by similar manner, if I pose lower two conditions, then  question mark should be identity two.  OK.  What is that?  What is it good for?  So I promised to show that any x type operator that commutes  with z type stabilizer generators are stabilizers  themselves.  So suppose a really big box, well, big bed finite, and check  the commutativity with the z operator here.  This corner has to commute with this corner, which means  this corner is either identity or identity.  No other option.  But this xi is this xi too.  So if you multiply this generator to here, if this  side turns out to be xi, then you can make this side to be  identity by acting on inside of this box.  So I can keep doing this.  So whenever I see a vertex from this side, I can  always make them as identity by multiplying this.  So I would get a poly operator of support that looks like  this.  But look at this side.  This side has to commute with i, z, and zi.  By previous exercise, you know that this is identity.  So you keep doing that.  And then you will erase the left-hand side vertical  square.  And look at this side, and you apply the same logic here  using this edge.  And you know that this is identity.  And everything disappear.  So I started from a very general x-type operator and  imposed that it has to commute with this and implied that  it is a product of the right-hand side.  So it is a stabilizer.  Now, for the no-string rule, here's a challenge.  This square represents the elementary excitations when  there is a single bit flip on the first qubit or second  qubit.  And the black dots represent the  flip stabilizer generators.  Can you make a pair just two defects out of this?  Well, the answer is no.  And I'll just give you a little bit of a summary of that.  Well, the answer is no.  And I will show in detail how to prove that.  So here's an old question.  What do you mean by string?  We know there's a string operator in the  directory code because we seem like an operator.  We named some operator string.  We haven't defined string yet.  For example, this looks like a string.  Combination of two strings that looks like string.  OK.  Is it surface or string?  It's not obvious.  So we need some different definition for a string.  So here I propose most natural definition, I think.  The problem was that in the discrete lattice, one  dimensional object is not well defined.  You can tell a one-dimensional object when you see it, but  it's not a definition.  And in general, we need to deal with a family of  Hamiltonians, and I have no idea.  So we need to go to some finite setting.  We can't deal with some infinite things.  So what's a string segment?  It is a finite supported poly operator that  crustation at most two locations.  Here, two is important because any conceivable string has  two ends.  And three more terminologies.  The anchors mean that the box is that  envelope of excitations.  And width is defined to be the size of the anchors.  And length is the distance between the anchors.  The point is there's no geometry constraint.  The string segment can be like a sausage or some weird  knots or whatever you want.  The only condition is that it creates only an  excitation in two locations.  But it's not the end of the story because if you consider  an identity operator, then it is trivially a string  segment because it creates no excitation.  Therefore, it creates less than two locations.  So we need to get rid of such a triviality.  And trivial conditions are rephrased like this.  Whenever you have a string segment, it has an  equivalent poly operator whose support are separated by  multiplying the stabilizer generators.  And if you want to be more sophisticated, then you can  say that a trivial string segment only creates trivial  charges that are possible to create locally.  So let us see some examples.  In the territorial code, when you apply zzz in this  manner, then it creates a defense at the end, two  locations.  So it conforms with the definition of string  segments.  And it is.  And you can show that it is non-trivial.  There's no string segments that you can cut out this region.  You have to connect somehow between two yellow squares.  It has width 1 because width is defined to be the size of  the anchor.  And length is potentially infinite.  It is not bounded from above because you can put as many  z's as you wish.  OK.  What about then to the Ising model that we understand  very, very well?  It is an example of stabilizer code because any  classical code is.  So suppose a black region has flipped spins, then you will  create a domain wall along its boundary.  And in order for such a string segment, well, this is a  string segment because it creates an expectation in two  regions where the yellow squares cover.  And you can show that two anchors must be adjacent.  Otherwise, each excitation in each square can be created  locally.  So it has some non-trivial width, the size of the anchor.  But length, the distance between two anchors should be  always 0.  So here's the formal no string rule for my model.  Our statement is that a string segment of width w is  always trivial, provided that its length, the distance  between anchors are sufficiently large.  And the boundary is not too severe.  It's just a polynomial wall.  It's just linear.  And this will be important when you want to show some  dynamic properties of this model.  So how do we prove that?  Well, we can just follow the definition because we have  good definition.  We started with our arbitrary string segment and showed  that there is an equivalent disconnected string segment by  multiplying suitable stabilizer generators.  And the trick is that any string segment will be reduced  to a union of three flat ones.  And I will show that each flat ones are trivial.  So we used the eraser again.  You just test them coming to activity with z operators.  And since this edge has two independent ones, and that  implies you can eat up the lower front part of the box.  So you can transform this box to here.  And that trick is used for the reduction.  For a given general string segment, you can reduce some  flat ones, flat means parallel to the coordinate axis.  And given us flat ones, you can argue further using the same  argument that using this independent edge, you reduce  to the second figure.  And here's some slightly complicated calculation.  If two adjacent operators are commuting with i, z, and zi,  and ii, and iz, then there are only four possibilities.  And since the model is translation invariant, it has  to make some consistent combination of the operators.  And you realize that at most after two sites, it has to be  identity, eventually.  And you keep inductively argue line by line by eating up  more and more sites.  And you end up with a small triangle on the left.  And a vacant thing on the right.  So we have disconnected.  We are done.  What does that mean?  No string means you can drag the D-fat, actually.  If you can drag the D-fat from here to there, then the  combination of those are represented by string segment.  And this is not allowed.  So the dragging D-fat is impossible.  It's protected by the Hamiltonian.  So let me finish with this exotic feature of this model.  Here's the k, the number of encoded qubits, as a function  of system size, l by l by l, l cubed.  And it looks like something we have never seen before.  But you can prove some formula.  And it turns out that the calculation reduces to GCD  calculation over the field of four elements, where the t.  Remember that t is not the exponential 2 pi i over 3.  It leaves in an extension field of binary field.  So it has no connection to the complex numbers.  But anyway, you can efficiently calculate the k using  this formula.  Let me just finish here by summarizing the basic properties  of this model.  Thanks for listening.  Questions for Jong-un?  So what happened to the other 16 codes?  Are they interesting too?  The seven models that are numbered like 11 to 17, it  has strings but has width 2.  And three other models have no strings.  But the bound on the length of the string segment  is exponential.  It's not as good as this linear bound.  And for other six things, I don't know.  In Kataya's two-dimensional surface codes,  you can load it with many logical qubits  by removing stabilizer generators.  And each generator you remove becomes a logical qubit.  If we remove a plaquette type operator there,  we can identify that, let's say, with logical z.  And the logical x becomes a string-like operator  that connects that plaquette to the nearest other plaquette,  which may be the entire surface itself  if it's a bounded surface.  So when we create logical qubits this way,  we necessarily create string-like logical operators,  even if we just ignore the entire volume itself.  In your codes, if you were to remove a cube out of a volume  and call that a logical operator,  there must be a corresponding, meaning partner to it,  that anti-commutes with that three-dimensional volume  and does something else.  Is that object string-like, or what  is the geometry of that object which  anti-commutes with the cube which you've removed?  Very interesting question.  I thought about that for a moment,  but I don't have concrete answer for that.  It is, well, it is somehow related to the question  that how do we define this model with the open boundary?  But there could be many boundary conditions  that you can impose, and there could  be many poly-operators that you can put on the boundary,  and I have no idea how to classify them.  And even further, because K has a complex number  of theoretical dependence on the system size,  I don't have a complete list of logical operators, too.  So for asking the geometry, it is far beyond.  Do you see a way to extend this model  so that it will also support computation beyond memory?  I personally didn't much think about that.  Perhaps one can exploit the feature  that it has extensive degeneracy.  But that was not the main focus of my research recent days.  OK, last question.  So just following on from all those comments,  wonderful talk.  I think you can do computation in this.  If you cut a torus through this, you  can have a logical operator that's a membrane in the torus.  And if you have alternate logical qubits that are spheres,  you can punch holes around that and drag strings and surfaces.  And I think you can do computation in that scheme.  So be great to talk.  OK, thank you.  How do you fasten your shoes, Velcro?  It's just strings.  OK, let's thank Zhang Lan again.  Let's bring up the next speaker.
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