>> Nathan: Great. Thank you very much for coming. Today, Maria Kieferova is going to be
giving a talk on her work and Maria is doing her master’s degree currently at the Comenius
University in Slovakia, in particular, doing research at the Research Center for Quantum
Information at the Slovak Academy of Science and much of her work has been supervised by
Daniel Nagaj. Maria's work for the previous two summers at the Institute for Quantum
Computing in Waterloo and today she's going to be talking about some work that I'm really
excited about on showing quantum speedup by quantum annealing. Thank you very much.
>> Maria Kieferova: Okay. Thank you, Nathan, for introduction and also thanks for having me
here. I will tell you something about my recent work with Rolando Somma and also my
supervisor Daniel Nagaj, when we were back in Slovakia. And we asked ourselves this question.
Can we get an exponential speed up with adiabatic computation? We know that there is a key
relevance between adiabatic quantum computation and [indiscernible], but could we do it
using some very natural algorithm with a sparse simple Hamiltonian and in this talk I will tell
you that the answer is yes for the given problem called glued trees that I will talk about. This
program was solved previously with quantum walks due to Andrea Childs and others, but we
use new idea about the quantum computation approach so I will talk then about adiabatic
quantum computation and finally I will present the new algorithms which we made. Let's start
with glued trees. We have two binary trees which are glued together with a random cycle. We
could glue these together just by connecting the leaves from the trees, but when we do it this
way all vertices we will have [indiscernible] except for these two guys, the entrance and the exit
which are somehow special this way. Our goal is basically to go through this webbing starting
at entrance and go to the exit point. Since the depth of one tree is some n then we have
exponential number of vertices here and how adjacency matrix would be exponential so that's
why the problem is given us an oracle problem. So we start with a key for the entrance vertex
and the oracle when we input the label of a given vertex, it will give us labels of the neighboring
vertices free if we are somewhere in the middle of trees or two if the vertices, if the vertex was
entrance or exit. Also, we could possibly try to guess these labels, but the labels can be larger
strings so it's exponentially unlikely to try a good vertex, to try a good label and succeed. Our
goal is to find an exit key, exit label to the exit vertex. We start here and we asked oracle which
way do we want to go and we need to find this vertex on the opposite part of glued trees. So
the easiest thing we could possibly do would be to try some random walk. When we start from
entrance everything is easy until we get to the other tree. Here it doesn't matter where we go.
Every time we go to the right direction, but when we cross the middle point, we have every
time bigger chance to go back than to continue to the right way. After several steps we will
start returning and it will take us exponentially short number of steps to reach the exit vertex
just with random walk. We could use also memory and try to put a thread through our
[indiscernible] or color vertices that we already know, but Andrea Childs and other people
showed that this is not possible. Even if we allow ourselves memory, no classical algorithm
could succeed and travel the glued trees in lesser than exponential number of steps.
>>: Did you say the bipartite graph in the middle is a single cycle?
>> Maria Kieferova: It doesn't have to be. It's randomly glued somehow.
>>: It's just a random degree two bipartite graph?
>> Maria Kieferova: Yes.
>>: Regular bipartite graph?
>> Maria Kieferova: Okay. Yeah. It's randomly pulled from this part to this part so that all
degrees will be free.
>>: Degrees to [indiscernible]
>> Maria Kieferova: It doesn't have to be. It's still [indiscernible] but… Okay. So what about
quantum case? Could we go somehow efficiently from here to here? And we will see that the
answer is yes, but first we need to define the problem in quantum setting, so we start in the
entrance state and we have access to these three oracles. First one is the adjacency matrix of
the glued trees which will serve as our Hamiltonian, and also we have projectors on entrance
which is easy because we already know this place and projector on exit, which is entrance state
with degree to our entrance, and the goal is to prepare exit state eventually. So we need to
end in a state which will be overlapped with the exit state so then when we measure it we
could get exit state with high probability. In this way we could prepare it probabilistically. The
way how we can traverse glued trees is you think quantum walks, which are in some sense
quantum analog of continuous time random walks or diffusion. So continuous time quantum
walks on graphs are defined by graph. It's vertices are our state and the edges give possible
connections between states, so the adjacency matrix is our Hamiltonian and the dynamics of
quantum walks is a unitary one. So the Hamiltonian will be adjacency matrix of this graph or
any graph we want to use, and then unitary evolution is given by shredding their equation
where the exponential of the matrix is defined as Taylor serious [phonetic] but the most natural
way how to implement it would be to go to eiganbasis and then exponentiate the matrix
trivially. Continuous time quantum walks have been studied in the last 10 years, let's say 15.
The two things which are interesting here is how they evolve and spread. This is one picture
where we started here in the middle and we let the quantum walk evolve on a line and this is
how it looked for quite short time and it spread like this when the times was longer, so you can
see that the distribution, probability distribution which can be made from the unitary evolution
is essentially different than for random walks. We don't have binomial distributions here, but
these strong peaks at both ends and also they spread faster than random walks. This property
was used for algorithms and for going through nth trees and decision trees and also Andrea
Childs and others showed that the quantum walks are universal for computation. This is a new
result which is basically an exponential improvement over the old paper. To solve everything
through glued trees using quantum walk. We first, we don't use stage as single vertices from
this graph, but columns are important here and we define this column sub space which can be
shown that its invariant under action of Hamiltonian which is adjacency matrix of the whole
graph. When we are in one column, let's say this one. And we apply the Hamiltonian we will
end up in these two neighboring columns as expected. So we can project the whole graph on a
line like this and we don't have exponentially many stays as before, but only linear. But the
definitely we also need to find the weight of the edges because the -- and this is how the line
finally looks like. All edges have the square root of two except for the middle one. This comes
from normalization of columns and calculation which is pretty straightforward, but quite
boring. It's easy to see that all of edges in this part will be equivalent and only this edge can be
somehow special and that's exactly what we see here. This is the adjacency matrix of the line.
It's a free diagonal and this is the interesting part in the middle. So we can do unitary evolution
in this Hamiltonian and it will show that this approach requires only polynomial number of
steps to reach the exit vertex with some 1 over n probability. But then we wanted to apply this,
wanted to solve this problem with an adiabatic algorithm because as I mentioned, glued trees
are hard for a classical computation, but they were solved using quantum walks in quantum
setting, so they are an interesting example of a distinction between quantum and classical
algorithms. So firstly, time-dependent Hamiltonians. We can again solve Schrödinger equation
but unlike in the common setting, here the Hamiltonian also d depends on time, so we can just
exponentiate it directly but we would need to slice the time into many small pieces and each
step we can evolve with Hamiltonian which doesn't depend on time anymore, just for a short
time update our state and update our Hamiltonian and evolve it again. This sort of evolution is
quite complicated, but we can, again, define eigenstate but it will be different for any time, for
time but for any time t we can, there is an eiganbasis which is complete and orthogonal.
Moreover, we will use these notations when we use relative time instead of time and the s
relative time parameter goes from 0 to 1. Although time-dependent, the evolution with timedependent Hamiltonians is complicated, there are two easy regimes. The first one is called
diabatic when the total evolution time is very short. So we have initial Hamiltonian and very
fast we will change it to final Hamiltonian, so for example, we have spin in magnetic field and
we turn the magnetic field to opposite direction or somehow else. In this case since the time is
so short the function and the state doesn't change almost at all, so if we were in an eigenstate
over the initial Hamiltonian, will be in the same state after the evolution, but it probably won't
be an eigenstate of the final Hamiltonian anymore. We'll end up in some super position of
eigenstate of the final Hamiltonian. The complete opposite is adiabatic regime. When we
change the Hamiltonian very, very slowly and when we change it infinitesimally small then the
system remains in its eigenstate or when we change it just slow enough we will be close to the
eigenstate or in our setting to the ground state. If the change of the Hamiltonian is smooth
enough and it can be normalized properly, then we can show that the time needed to reach the
adiabatic regime squares as 1 over the minimal gap between the ground state and the first
excited state. So this idea can be used in quantum computation. We start with a Hamiltonian
which can be easily prepared in a lab and in the ground state of it. Then we change the
Hamiltonian slowly to go to the final Hamiltonian and if the change is slow enough compared to
the minimal gap we will end up in the ground state of the final Hamiltonian which is somehow
interesting for our computation. And this way of computing is universal as the original showed.
Okay. How can we use adiabatic time evolution to solve our glued trees? The first thing we can
think about would be to start in enter, in the entrance and change the Hamiltonian from
projector to the entrance to the projector of the exit. First we project only on the enter and
then we project on the exit and how evolution is linear, let's say. This simple approach doesn't
work because in the middle we encounter an exponentially small gap. Another idea which is
pretty straightforward is to start again in the entrance and evolve to the ground state of edges
and symmetrics over the trees, but here we won't have any problems with small gaps, but the
ground state of adjacency matrix isn't the ground state of, isn't the exit state. Even more, it has
only exponentially small overlap with it. But we can combine these two approaches. So we
take the projectors of enter and exit and change them linearly as in the first shot attempt, but
we add the edges and symmetrics in the middle to help us overcome the problematic region.
>>: Alpha plus beta equals 1?
>> Maria Kieferova: No. We have some weight on this part and this part and the ratio here is
important, so if, for example, we set alpha to 0, then we have the first attempt which doesn't
work, which we have only alpha, then it doesn't work again as we saw previously. Maybe if we
[indiscernible] here with some weights then we can make it work. And now we see how the
spectra differ for different values of alpha and beta. This part will be always 1 because we are
interested only in ratio and we changed wedges [phonetic] on projectors. So we had only the
adjacency matrix then we have spot like this which looks nice. There are no exponentially small
gaps but it doesn't have the desired ground state or practically it doesn't do anything
interesting at all. When we start adding any projectors then we can see the distinction
between the excited state and the ground state, where here the ground state will be the
projector on the entrance and here on the exit, but there are some complicated parts here and
here and also when we make the projectors on entrance and edges even stronger we get a
complicated region also here. If we turn out the edges and symmetrics in the middle
completely they would get almost crossing exponentially small gap as I told you before. It's still
not clear from plots of spectra, so here are the corresponding gaps. The logarithmical scan of
the y-axis so this is the case only with adjacency matrix, all night but doesn't work. And when
we add projectors we see two exponentially small gaps at the edges and these strong gaps at
the edges and something which looks smaller optimistic in the middle. And when we make
weights on edges even stronger, then we have more gaps in the middle. So from these pictures
it might seem that every time we do something interesting we will have nasty gaps here but for
this picture it still looks quite optimistic because these two gaps are symmetric. That's what we
are going to use. We allowed us to differ from the standard adiabatic scheme and jump here
when we can go slow enough through this gap we instead of, we decide to jump and go on the
first excited state in this region and since it's all symmetric in this part we are supposed to jump
back. This was our idea. Again, we start in the ground state and we want to go beyond the
ground state up to this point with the exponential gap, then jump to the first excited state.
Here we need to go slow enough because we don't want to propagate more glue [indiscernible]
our excited states and if all are symmetric at the second gap we should jump again back to the
ground state and end in a ground state.
>>: What's the probability of not jumping the second gap?
>> Maria Kieferova: This one?
>>: No okay. The second one.
>> Maria Kieferova: This one. These two gaps are the same.
>>: The first one you don't care about. If you don't take the jump it's fine. You're still in a
ground state. Everything will still come up.
>> Maria Kieferova: Oh, here?
>>: Yeah. It doesn't matter whether you take that gap. Taking any of those lines will still lead
you to the same place.
>> Maria Kieferova: No. Here we need to jump.
>>: Why?
>> Maria Kieferova: Because the gap is very small and we know that if we go faster than one
over the Gibbs squared we will jump.
>>: So what?
>> Maria Kieferova: And we can't go slow enough to stay here, so instead we say okay. We
don't care, and we jump here. And the same right here, we will jump again to the ground state.
>>: That's not obvious that it would jump to the ground state. I'm asking the probability that it
will jump to the ground state.
>>: [indiscernible] it's very likely that she jump and now she wants it to jump because it's likely
that she'll jump on the second time and if you don't jump on the first time then you go the
wrong way the second time. It's like two wrongs make a right.
>>: And another thing to note, Dave, is if you take a look at the blue curve down at the bottom,
you'll notice of the brown curve actually is very smooth, approaching the blue curve over there.
So what's actually happens at that crossing is that that leads to an extremely high probability of
those transitions happening.
>>: I understand that. I'm saying if you didn't take it, it wouldn't matter.
>>: That's what I'm saying, it would matter, because…
>>: Because it affects the one later?
>>: Yeah. Because you are probably then going to go the wrong way.
>>: You need to jump -- you need to…
>>: Okay. I'll talk to you later about it. Okay.
>>: You need to pop up and then you need to jump down.
>>: Understand the desirability and I understand why the first one matters and the second one,
I get why it matters. I don't understand the first one, so someone can explain it to me later.
Thanks. [laughter]
>> Maria Kieferova: Okay. First we needed to analyze the whole spectrum analytically because
I said okay. This works exponential and this looks not so bad so it will probably [indiscernible]
but we need to know for sure. That's why we try to compute spectrum of the adjacency matrix.
It's quite similar to the one from glued trees, but we got these two self loops at the entrance
and the exit vertex which some way, which comes from the weights between projectors and
the edges and the symmetrics of the glued trees and also they will be different in different
times. Here we use the trick with plane wave onset because in this region it looks mostly in
line. It doesn't matter if we are in this vertex or the next vertex. Our neighbors are still the
same and we know that in case why this plane wave inside of this form should work and we get
an eigenvalue of this form. So we applied answers like these in these two parts and we allowed
linear combinations of it to satisfy conditions also on the ends and in the middle. And so we
with a couple of equations we end with an equation for p which was translational but we were
still able to solve it analytically when n went to, in the limit of large N and from this we
computed the gaps exactly. This is how the spectrum looks like when we take everything into
account. So firstly here is the first gap, exponential but here the depth of trees isn't as large so
we can see that we don't have crossings here, but the gap in this region turns out to be
exponential and scales like this. The second gap is the same as we saw from [indiscernible] and
also from symmetry of the edges and a symmetrics and get in the middle between the first
excited state. And the second excited state squares are only as one over n cubed so it's
polynomial. Okay? therefore, if we're right than it will be somewhere between these two
limitations. When we go this way we will jump to the first excited state and because an item
set it approach here the connection is small. As this increases here we don't want to jump and
in this part we will return again to the ground state. We have analytical result that the behavior
is as I mentioned, but I will show only the one from our numerics. So we started in the ground
state and until we reached this point we were still in the ground state. Here we went fast
enough so we jumped almost 100 percent here. And in the middle region we lost some
propulsion on the first excited state, but we had still enough to reach the second jump place
and return to the ground state still with very high probability. This is how it all worked out. We
solve this problem for glued trees which is known to be exponentially hard for any classical
computers. Then we applied ideas from quantum walks from older papers and the adiabatic
quantum computation, but we didn't go through the classical, through the common adiabatic
scheme and we didn't take the ground state all the time, but we started in the ground state.
Then jumped to the first excited state and then jumped back, so this way we came up with a
new algorithm in which we had exponential gaps but we didn't mind because of the
asymmetry. With this approach we gain exponential speedup against classical algorithms and
we did this with a sparse Hamiltonian which is even more stoquastic, although we didn't follow
the [indiscernible] regime so we didn't learn more about the possibilities of the stoquastic
Hamiltonians in quantum [indiscernible]. And also there is the question if we could use possibly
jumps for some other problems and to gain an advantage over the way that people thought
before about adiabatic computation. So yeah, that's all and thank you for your attention.
[applause]
>>: The only comment I was going to make is if you went slower, but you don't want to, but if
you went slower you would still follow the path to the bottom to get there.
>>: Yes.
>>: You could also go ridiculously fast before the gap closed because you got a long break
before the first excited state, then go the optimal speed since you know where that happens.
At that point, then speed up again as long as you don't go past that polynomial gap in the
middle and then slow down to the right speed at the end and then go as fast as you want to
and collapse out of the [indiscernible]. I get it. Okay.
>>: Yeah. You could optimize this by using a local idiomatic schedule.
>>: Yes.
>>: Anywhere but near that exponentially small gap.
>>: Right. Which is exactly what we see in the stuff we've done. The neat thing here was
taking advantage of the exponential closing and using it to your advantage. That's very cool.
>>: And one of the things I would again like to emphasize is just how hard it was to find a good
example of this. I remember I spent quite some time struggling trying to make quantum
Fourier transform or, you know, like sure like algorithms, try to work naturally in an adiabatic
setting and made absolutely no progress. This trick was I think a really big event.
>> Maria Kieferova: Okay. But we needed asymmetry to break, to overcome the problems
with exponential gaps which we don't have in many common problems.
>>: But you did it nicely because it makes those two into three schedules, so you could make it
symmetric and do the same thing.
>>: So is the symmetry absolutely necessary?
>> Maria Kieferova: In this case, yes.
>>: It is?
>> Maria Kieferova: It would need to go exponentially slow to form the ground state all the
time because here you are promised if you jump then you will jump back in later time.
>>: And that's only true in the case…
>>: Or is that the case of one schedule for the whole thing? One speed, one schedule. It
doesn't have to be symmetric, but you have to know the speed and the size of the two
exponential gaps.
>>: Suppose you have a ring that was computed, based on the gap, actually built a function or
step numbers that said how fast that step wants to go based on a certain named energy, right,
which makes sense then if you think about it. I mean it's a Heisenberg [indiscernible] so
wouldn't that be enough? The problem it's generating is figuring out what that function is. But
if you had a function you could replace alpha and beta and all that other stuff with this
continuous [indiscernible] function tells you how fast to go.
>>: But if you think about it, the best that you could end up getting at those sorts of strategies
is say you [indiscernible] search will get you a polynomial [indiscernible] speedup at best and
within this case the algorithm isn't particularly a practical one, so the key point is just showing
the exponential speedup. I think these strategies could work, but it's not clear that for the
purposes here that it's necessary.
>>: This proves a point.
>>: And that's the whole point of this algorithm.
>>: It is interesting but do you really need a perfect symmetry or do you just need two level
crossings?
>>: You need two level crossings.
>>: And can you steer yourself through the state [indiscernible] effectively in the appropriate
way? How much control can you really have at these crossings?
>>: I suspected that this question will come up earlier and I think that is one of the key issues
with this work that hasn't been addressed. I mean, correct me if I'm wrong about that, the
issues about the sensitivity to small mediations in the derivatives near the crossing or avoiding
crossing.
>> Maria Kieferova: Yeah. I don't know, but I would suspect that even small deviations could
break the symmetries here.
>>: Again, if you could control it maybe by coupling to a heat bath or something, if you can
control where you come out when you go past the crossing, you don't need symmetry at that
point, if you can effectively steer your approximation.
>>: Controlled adiabatic evolution.
>>: Yeah, exactly.
>>: Exactly. Right.
>> Maria Kieferova: But then we still need at least two gaps. We get one in the middle and we
can use…
>>: Even numbers. [laughter]
>>: Gaps. Like.
[laughter] [multiple speakers]
>>: We started in an excited state.
>>: Right. This is where you start. [laughter]
>>: But how do you use to prepare? I mean I would say necessarily where you would start and
circle the answer. Don't start at the beginning. Start somewhere else. [laughter]
>>: [indiscernible] start [indiscernible] [applause].