Local stabilizer codes in three dimensions without string logical operators
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Transkript: Englisch(automatisch erzeugt)
00:00
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
00:21
his right hand and repeat after me. I solemnly affirm that I will faithfully correct all quantum errors, real or imagined, and will to the best of my ability,
00:41
preserve, protect, and defend the quantum computer. Congratulations. By the authority bestowed upon me, I bestow upon you the best student paper award. And our next speaker, the local stabilizer codes in 3D
01:06
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, too,
01:25
that I was working on a wrong model. I was wrong. I proposed the model and it turns out to have a string operator after three months of calculation. But still, John expressed genuine interest to my work
01:41
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 noise-free subsystem or subspace
02:03
from a large Hilbert 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 in the helium atom.
02:21
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
02:40
means it has a ground state, degenerate ground state, more than two ground states. But they look exactly the same when you look at it locally. But these kind of steps, and they are, for a long time, known to stable at zero temperature.
03:00
Kitaev mentioned briefly in his paper. And recently, just last year, Bravais, Hastings, and Michalakis proved, actually, for proper Hamiltonians, such a degeneracy and a gap remain stable under any arbitrary perturbations. So it defines a quantum base.
03:24
But the word topological order should not be confused with a topological quantum computing with anions. Topological quantum computing with anions concerns about the excited states. We encode the quantum information into the excited states, the configuration of defects, anions. But here, we are mainly focused
03:43
on the ground space of the Hamiltonian. Why is this problem hard? Well, the prototype typical example of the topological order is, obviously, the two-dimensional types code, which has just two terms in the Hamiltonian.
04:04
They are all commuting. So yeah. And it has four degenerate ground states when the model is defined on the torus.
04:21
And such a fourth degeneracy comes from the non-trivial topology of the space. It is embedded. And because such a topological consideration leads to conclude that the ground states are locally indistinguishable. But at finite temperature, there may be some defects.
04:43
And there has to be some defects. And since we are able to operate on the anions, we believe to do so. Anions can move around. But if we can move the anions, then nature can move. And thermal errors are unknown to us.
05:02
And they will drag the defects from arbitrary far apart. And if the defects are created and propagated for long distance and re-annihilate in a non-trivial way, then it is as if there has been some non-trivial action on the ground space, which means
05:22
the encoded quantum information at the ground space has been corrupted. It's been split. Further, this phenomena has happened for all low dimensional models. By low dimensional, I mean three or two.
05:43
But around 10 years ago, in a paper by Dennis et al., the four dimensional toric codes is proposed. And three or four years ago, it has been rigorous proof that it has a self-prolting property, meaning that the relaxation time of an encoded state
06:01
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. And it seemed to work via the mean field theory.
06:22
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.
06:41
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 term is commute. So energy spectrum wise, everything is known. It's just equally spaced.
07:02
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.
07:24
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.
07:40
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 a particular shape, then it drops to constant.
08:04
Let us begin from the very primitive setting, because this is the way I found that model. From yesterday's TOS tutorial, we know that studying stabilizer code
08:23
is equivalent to study the classical codes over the operator algebra. Well, to be specific, the Pauli operator group. And you can parameterize any Pauli operator by bit strings. And you can interpret the multiplication of operators
08:41
in terms of the modulo two addition of such a bit strings. Further, you can even encode the commutation relation into some symplectic structure. So every language statement in the stabilized code can be described in terms of symplectic binary vector
09:01
space. More generally, if you consider a QD with a prime dimension, then the description becomes the linear algebra over that final field. So what's code in this language?
09:22
The committing set of Pauli operator corresponds to the null space by null mean where the space on which this syntactic form vanishes. And we call that the stabilizer group. And the corresponding symmetric group, when you think of the codes as some representation
09:43
of a Hamiltonian, the symmetric group of that Hamiltonian corresponds to so-called the logical operators. OK, what should I do with that? There may be some several qubits per site. That's very generous setting.
10:00
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 infinite data, so I assume to the translation invariance. And I want to have no single site logical operator, which
10:22
corresponds to the local indistinguishability of the ground space. And since we want some Hamiltonian where the defects cannot move, where the string operators are absent, so we want some model with no obvious string operators.
10:43
But this is too general. So I have to simplify 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,
11:03
so the product of all terms in a Hamiltonian should be identity. You can interpret the degeneracy of the Tory code in this way too. And no single site logical operator condition translated to this logical operator on a single site
11:23
should be identity. Well, this is too restrictive, because all we need is if a single site logical operator is found, and we require that the element should be in this stabilizer group. But setting to be identity is more simple.
11:41
And I assume further that the logical operator on a straight line, meaning that three-coordinate axis or face diagonals or body diagonals, are all identity if they are repeated single site operators. And apply the formalism of stabilizer code
12:02
to these conditions. And you can translate every condition in terms of linear algebra equations. And the last two conditions, the logical operator on the single site be identity, and the logical operator on the straight line be identity, is translated to some rank condition
12:21
on the commutation matrix. So I started calculation by my desk at the computer. Since I assumed two types of generators, one for z type and one for x type, there are 16 corner operators.
12:43
And it turns out that not exact 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.
13:03
So eight by eight 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.
13:21
And we solve that equation by brute force. So 64 variables in 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 site null operator or no
13:44
obvious string operator. Then you get not a large number of solutions, not a null solutions, but there are a handful of machines that satisfy all the conditions I described before. So here comes the first one, which
14:01
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.
14:22
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 zz and ii, and replace the first qubit and the second qubit,
14:41
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 it is topologically ordered, and it has no string. But I didn't prove anything.
15:02
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 string.
15:21
So let us answer the first question first. The first question is, is it topologically ordered? And the answer is yes, meaning that there's no local observable that distinguishes ground states. So logic goes like this. If an operator is local, then when
15:41
restricted to the ground space, its action is proportional to the identity element. And how do you prove that? The claim is any Pauli operator support commuting with all stabilizer generators, then the Pauli operator itself is a stabilizer.
16:02
But yeah, and this implies the first statement, because any operator has a linear expression in terms of Pauli operators. And this equation is linear. And because of the duality, I don't need to check for all the possibilities.
16:21
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, i, i means i tensor i, actually. And xi is x tensor i for the convention.
16:42
And if I impose that, it has to commute with zz and what's the possibilities. And you should be able to figure out the answer is either identity x. What if I impose two conditions? It has to commute with iz and it has to commute with zi. Then it's nothing.
17:02
It's always the identity. And in a similar manner, if I pose lower two conditions, then question mark should be identity two. OK, what is it good for? Well, so I promised to show that any x type
17:23
operator that commutes with z type stabilizer generators are stabilizer themselves. So suppose a really big box, well, big but finite, and check the commutativity with the z operator here. Since this corner has to commute
17:42
with this corner, which means this corner is either identity or xi, 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
18:02
by acting on inside of this box. So I can keep doing this. So whenever I see a vertex from this side, then I can always make them as identity by multiplying this.
18:20
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.
18:44
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
19:04
and implied that it is a product of the right-hand side. So it is a stabilizer. All right. Now, for the no string rule, here's a challenge. The squares represent the elementary excitations when there is a single bit flip on the first qubit
19:23
or a 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 will show in detail how to prove that.
19:42
So here's an old question. What do you mean by string? We know there's a string operator in the declaratory 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.
20:00
Combination of two strings that looks like string. Three? OK. Is it surface or string? It's not obvious. So we need some different definition for a string. OK. So here I propose most natural definition, I think.
20:21
So the problem was that in a 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.
20:42
We can't deal with some infinite things. So what's a string segment? It is a finite supported poly operator that creates at most two locations. Here, two is important because any conceivable string has two ends and three more terminologies.
21:05
The anchors mean that the box is that envelope 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
21:25
or whatever you want. The only condition is that it creates only 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
21:43
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,
22:01
it has an equivalent poly operator whose supports 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
22:21
creates trivial charges that are possible to create locally. So let us see some examples. In a two-deterior code, when you apply z, z, z in this manner, then it creates a defect at the end, two locations. So it conforms with the definition of string segment. And it is.
22:43
Yeah, and you can show that it is non-trivial. There's no string segment that you can cut out this region. You have to connect somehow between two yellow squares.
23:00
It has width one 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't put as many z's as you wish. OK, what about then to the Ising model that we understand very, very well?
23:23
It is an example of stabilizer code because any classical code is. So suppose a black region has flipped spins. Then it will create a domain wall along its boundary. And in order for such a string segment,
23:42
well, this is string segment because it creates excitation in two regions where the yellow squares cover. And you can show that two anchors must be adjacent. Otherwise, excitation in each square can be created locally. So it has some non-trivial width, the size of the anchor.
24:05
But length, the distance between two anchors, should be always zero. So here's the formal no string rule for my model. Statement is that a string segment of width w
24:21
is always trivial provided that the distance between anchors are sufficiently large. And the boundary is not too severe. It's just a polynomial. Well, it's just linear. And this will be important when you want to show some dynamic properties of this model.
24:42
So how do we prove that? Well, we can't just follow the definition because we have good definition. We started with 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
25:03
will be reduced to union of three flat ones. And I will show that each flat ones are trivial. So we use the eraser again. You just test the connectivity with z operators.
25:21
And since this edge has two independent ones, and that implies you can either 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,
25:42
you can reduce some flat ones, flat means parallel to the coordinate axis. And given flat ones, you can argue further using the same argument that using this independent edge, you reduce to the second figure.
26:01
And here's some slightly complicated calculation. If two adjacent operators are committing with iz 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.
26:23
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.
26:45
So we have disconnected. We are done. So what does that mean? No string means you can't drag the defect, actually. If you can drag the defect from here to there, then the combination of those are represented by string segment. And this is not allowed.
27:01
So the dragging defect 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
27:21
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.
27:48
Remember that t is not the exponential 2 pi i over 3. It lives in an extension field of binary field. So it has no connection to the complex numbers.
28:01
But anyway, you can efficiently calculate the k using this formula. Let me just finish here by summarizing the basic properties of this model. Thank you for listening.
28:29
Questions for Jangwon? So what happened to the other 16 codes? Are they interesting too?
28:42
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.
29:08
In Katya'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,
29:20
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
29:43
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. And I thought about that for a moment.
30:01
But I don't have a concrete answer for that. 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 polyoperates
30:23
that you can put on the boundary. And I have no idea how to classify them. And even further, because k has really complex number of theoretic 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.
30:49
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.
31:04
Perhaps one can exploit the feature that it has extensive degeneracy. But that was not the main focus of my research recently. OK, last question.
31:25
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
31:40
are spheres, you can punch holes around that and drag strings and surfaces. And I think you can do computation in that scheme. So it would be great to talk. OK, thank you. How do you fasten your shoes, Velcro? It's just strings.
32:00
OK, let's thank Zhang Wang again. Let's bring up the next speaker.
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