>> Krysta Svore: So today we are excited to have Fernando Brandco
visiting the Clark Group. He's a candidate for a research
position in Clark. And Fernando is from actually Brazil, hails
from Brazil. He obtained his PhD. in physics from Imperial
College London. Currently he's an associate professor or lecturer
at University College London, where you've been for about a year,
I think. He's held positions at ETH Zurich, Universidade Federal
de Minas Gerais. Oh, shoot, I speak French and not Spanish or
Portuguese. And also at University College London, Imperial
College London. He's interested in quantum information theory,
quantum computation, optimization theory, many-body physics,
computational complexity and entanglement, which is what we're
going to hear about today. So he's going to talk to us about the
complexity of quantum entanglement. Let's welcome Fernando.
>> Fernando Brandco: Thank you [applause]. Thanks. So
thankfully I'm able to come. I'm happy to be here and to interact
with you during these two days. Okay, so I'm going to talk about
some work that I did in collaboration with several people. Here
are their names, in like over three, four papers. So the
beginning, I wasn't sure what the audience would be, so I put some
like very basic like motivation for doing quantum computation. I
don't see it's necessary here, but let me go really quickly over
it. So the reason why we are doing what we do is [indiscernible]
two different problems. One is this problem of simulating quantum
mechanical systems. So there is some thought that I took from
[indiscernible] that many people use it now. It just shows the
time of supercomputer power from the Department of Energy in terms
of the disciplines. Then you can see that like roughly 25 percent
of this supercomputer power is evoked to simulate quantum physics.
This is a very time-costing simulation problem, and you'd like to
get a better handle on this problem. So another challenge is that
we have cryptography. This works very well, but this is all based
on unproven hardness assumptions. On computation, the hardness of
the problems that we don't for sure if they are hard or not. Of
course, it would be much better if you can base security on really
more firm grounds. For example, the laws of physics, for example.
And what quantum information and quantum computation gives is
really a pathway for solving both problems. So one day we can
boot it on the computer or like long scale quantum cytography
systems, we can get much better answers for these two problems.
So the area of quantum information theory, really the goal is to
try to lay down a theory for this future quantum-based technology,
quantum computation, quantum cryptography and so on. There's a
lot of like subareas that we explore here. You guys work on
mainframe here, so in quantum communication, we understand that
you have to make limits of information transmission. So we have
quantum mechanic systems, and therefore we have to understand how
we can store and manipulate and send the information in these
systems. Then entanglement theory, we see these quantum
correlations entanglement that would be the focus of this talk as
some resource in quantum information science, and we'll be able to
understand how to use this resource. Then quantum error
correction and fault tolerance is really about this idea that this
quantum computers, they are actually in a sense digital computers.
We have a many [indiscernible] of correcting errors and scaling
them up. Then the quantum computation, I try to find ways of
using a quantum computer once we have one, trying to get
algorithms with exponential speed-up. And quantum complexive
theory just comes from the idea that once we know the fundamental
limits for computation due to quantum mechanics, we understand
what are the limits for quantum computation, so what we cannot do
even with a quantum computer. There is a lot of work in all these
directions, and they are all very interesting. But I guess a
bonus of all this theory is that all this theory developed, it's
not only useful for trying to understand what we can do with this
new technologies but also for studying other problems in other
areas of physics and in computer science. And one of -- this
would be one of the focus of the talk. So there is a lot of like
emerging connections of quantum information theory before the
branches of science. So just put some like random examples here.
So in condensed matter, there is a lot of stuff going on between
strongly correlated systems and entanglement theory or topologic
order and quantum error correction. As you know very well, in
high energy physics and general [indiscernible], there is some
very nice recent connections as well. What I want to focus here
is on connections of field quantum information with the field of
computer science, and in particular to optimization. There's been
some work, for example, on compressed Sensing better using quantum
information ideas. What I want to focus on is to trying to
understand this very useful algorithmic techniques known as SDP
hierarchies in terms of quantum information ideas. So we'll get
to that soon. So this talk is really just to give an example of
these connections of quantum information theory with optimization
theory in particular and some of the same programming. So the
problem that we want to address from a pure mathematical point of
view is a very basic one. It's just like a relation of quadratic
versus biquadratic optimization. So let's start with some
problem, which is a little elementary. This is just we're given a
matrix, for example, d by d, and we want to compute this. The
maximum over unit vectors of x [indiscernible] conjugate Mx. So
this would be the expression. So it's just a quadratic
optimization problem in the interest of x. And of course this is
a very easy problem. This is just the maximum value of M. So we
just use a little algebra and we can compute this very efficiently
and [indiscernible]. Now, let's try to generalize this problem a
little bit, just add another variable Y. So now we have M as the
[indiscernible] matrix, but it's a tensor product of two vector
spaces, one the dimension, the other L dimension. Another one, we
want to solve the analogous problem, but instead of optimizing
over any vector, we are going to optimize over product vectors, so
x stands for y. Both are unit vectors. So this is just like a
general biquadratic optimization problem in x and y, so over some
homogenous [indiscernible] and we want to calculate the maximum of
this. Now, it turns out that this is a much harder problem. So
we don't really understand very well the complexity of this
problem anymore, but using some ideas from quantum information
theory we can both get the best known algorithm that we know for
this problem and the best hardness result as well. This is what I
want to show you. So just interrupt me at any point if you have
questions during the talk. All right.
>>: What is the difference between the two problems?
[indiscernible] just to consider x tensor y to be a vector.
>> Fernando Brandao: No. You can. It's definitely a vector, but
here you optimize over all vectors in the space. Here you only
optimize over product vectors. We also have vectors which are not
products, so that's why.
>>: We would look at it as Hs and tensor?
>> Fernando Brandao: That's right. The way that this problem
will appear in the context of quantum information is it's trying
to solve computational problem that people study a lot in quantum
information, which is problem of [indiscernible] entanglement. So
what I guess I would just -- you all know that, so in a pure state
just a vector. I mentioned the complex vector space. We also
care about mixed states, so these are just positive semidefinite
matrix of unit traces. So positive semidefinite unit trace. We
can always write as a complex combination of pure states, if we
want. And quantum measurements is just some pure [indiscernible].
So let me just skip this. Now, this is the point. We want to
understand when a quantum state is entangled or not, right? So we
assume that we have a [indiscernible] of quantum states. Suppose
it's a pure state, for example. It's just [indiscernible] on the
tensor product of the dimensional versus the L dimensional complex
vector space. Another way that we define entanglement is first to
say what is not entangled, right? So we say the states is not
entangled if it is a tensor product of two local states,
[indiscernible]. Then we say separable, and separable means it's
not entangled. So if you do a measurement on A, it's completely
independent of the state on B or for the information that you do
on B. Now, for mixed states, we have to -- we also are interested
in understanding when a mixed state, quantum state is entangled.
So here we're given a dense matrix. This is just a set of dense
matrices on this vector space. And now again, we define what are
no entangled states first. They are just complex combinations of
product states. So just take our definition of no entangled
states in the pure state case and we extend to mix states by
[indiscernible]. Otherwise, if you cannot find the probability
distribution Pi and product states [indiscernible] such that we
can write the states in this form, then we say it's entangled. It
has like quantum correlations. Yes?
>>: The problem you showed on the previous slide, is that
considered in other fields? Is there other applications?
>> Fernando Brandao: Yes. So I'll get to that in the other talk,
yes. All right. So now Y makes sense to define entanglement in
this way. This comes from considering this paradigm that we have
in quantum communication and cryptography, which is this LOCC
paradigm. So this is the situation where we have two parties,
Alice and Bob, and they are far away, right? They have like
quantum [indiscernible]. They can do whatever operation is
allowed by quantum mechanics locally, but they can only
communicate classical bits to each other. They are very far away.
To send quantum [indiscernible] is very expensive. They can only
talk on the telephone line. So now they would like -- initially
they have no correlations. They didn't interact with each other.
Now they would like to create their quantum states using
this class of operations, local quantum operations and classic
communication. Alice makes the measurements, sends some message
to Bob, classic message. Bob makes another measurement, sends a
message back to Alice, back and forth.
>>: Is the tensor product [indiscernible]?
>> Fernando Brandao: Exactly, yeah. Right. Now the hilbert
space is given by a tensor product of tensor space of L, stands
for vector space [indiscernible]. Now several states are exactly
these states which we can create by this classic operation, right?
So they can just share like, you know, some coin given by the
probability distribution of Pi. They just sample from this coin.
If they get outcome I, Alice prepares the [indiscernible] locally
and Bob prepares the state [indiscernible] locally. So it's clear
that any states of this form can be created by just like classical
coordination and turns out that any state that cannot be written
in this form requires some kind of quantum interactions.
Therefore, there must be some quantum correlation there. So
entangled states is just states that cannot be created by LOCC.
They have this non-classical correlations. So now we get this
computational problem that I would like to start in this talk.
The computational problem is the following: We are given a dense
matrix, right, it's just this [indiscernible] matrix of unit trace
and we want to know if it's entangled or not, if it corresponds to
entangled states or to separable states. So now it's useful to
allow some error in the answer, right? Some accuracy for the
problem. So we use this weak membership formulation. So it's
easy to see that the set of separable states is a convex set, this
set over here. D is the set of all [indiscernible] matrices. And
now this weak membership problem is the following: It's a
decision problem and we are given a state [indiscernible], and if
it's separable, if it's inside here, the sets, or if it's
absolutely away from separable, if it's outside here. So it's
really -- there is no separable state close by. Of course, we
have to use some norm to quantify this [indiscernible]. We'll get
to the [indiscernible]. So it's just some version of deciding
entanglement where we don't care what happens here in this bond
area. All right. I want you all to understand how efficient we
can actually solve this problem. There's also the dual problem,
right, it's an optimization problem, a weak membership problem.
There is the optimization portion of the problem, and this is
really just when optimize over elements in this convex set, which
is a set of separable states. Here you can see the disfunction
that I mentioned before appears so this eight step is called like
the support function at this point M of the sets of separable
states. The definition is just the maximum over separable states
of the linear function trace M sigma. M is just whatever matrix.
It's a fixed matrix on this biquadratic vector space. Then of
course, these are separable states. We know they are convex
combination of product states. We take the maximum over convex
set, so the optimal solution is in the extreme [indiscernible],
which are just product states. So we have exactly this
biquadratic optimization problem. So the problem of deciding
entanglement is tightly connected to this biquadratic optimization
problem. One is like the dual of the other. If it is a good
operation, you can [indiscernible] for the other . But I will get
to that more later.
>>: [indiscernible] is somehow interpreted as the distance
between M and [indiscernible]?
>> Fernando Brandao: The trace, it's just the trace of the matrix
even by the product of the [indiscernible].
>>: Right, right. Why are you maximizing that?
>> Fernando Brandao: So this problem is just that you have this
convex set and you just want to maximize a linear function over
elements of the sets. This is the definition of this problem.
There are well-known results in like convex optimization which
says that if you can solve this problem, you can solve the weak
membership problem and vice-versa. So the ability of solving one
gives you the ability of solving the other. So when we study this
problem, we can study this in place of it and vice-versa.
>>: To what extent is that notion of being separable and not
stable under a discredization of all numbers? So instead of with
reals, I work with loads, so I work with ->> Fernando Brandao: Right. Yeah, it's a good question.
>>: Could it make or break the fact that it's entangled or not?
>> Fernando Brandao: So yeah, there might be some composition in
terms of product states, which you cannot write if you only have
something [indiscernible] in the elements, right, of your matrix.
But that's why we allow for this error. Because if this error,
then it's fine.
>>: That will take care of all of these numerical issues?
>> Fernando Brandao: Yes, typically.
>>: That's what I'm worried about. If I make Epsilon very
small, can I still decide the effect of the ->> Fernando Brandao: No. If Epsilon starts being -- like if
explanation is small in the [indiscernible] ->>: Minus five or something.
>> Fernando Brandao: So I don't know for this. You know, you
just -- it's a good question. I don't know, given the fixed
error, what the accuracy you need from the -- from the odd limits
of the matrix such that it doesn't make a difference. Actually, I
don't know. I have to think about it.
>>: [indiscernible] would not work with the reals?
>> Fernando Brandao: Yes, that's true.
>>: It might have something to do with the condition of
[indiscernible].
>> Fernando Brandao: Right, yeah. So what I have in mind here is
just that ->>: Wait. How elliptical is that set? If it's really skinny,
you know, you might ->> Fernando Brandao: So here would be most interesting in
[indiscernible] Epsilon and then we just -- the numerical error is
much smaller than this. The relevance for this one is just what I
mentioned before. Entanglement is this resource in quantum
information science like for cryptography or quantum
communications and so on. So you usually know when someone gives
up some experiment for doing something useful, the first test they
should do is determine if there is entanglement or not there.
Then they should characterize it further, but this is like in the
sense a minimum requirement. So because of this, other people
studied this problem, like in the past 15 years, since like the
beginning of quantum information, and now what I want to show you
are some results on this. So but now before I get to the results,
there is this norm here, right, so there is how you quantify the
distance between two [indiscernible] matrices. There are several
ways of doing that, and it turns out this is important for us, how
we quantify this. So let me just show you some options. This is
how we quantify the distance in the weak membership.
[indiscernible] but not the most operational one is using
[indiscernible] norm, it's just the [indiscernible] norm for if it
treats the quantum state as a vector. So this is norm. It's just
trace x dagger, x to the one-half. A more operational motivated
version is what we call the trace norm. This is the mathematical
definition. But it's a useful one, because if you look to the
trace norm of the difference of two quantum states, then it is
equal to two times the maximum over measurements, quantum
measurements, of trace M [indiscernible]. So this is just a
[indiscernible] element, and we look at trace M [indiscernible].
So really this tells us the thing we should [indiscernible] of
[indiscernible]. If this is small, it doesn't matter what
experiment you do on [indiscernible] or sigma, they have similar
outcomes. So for all practical purposes, if trace numbers are
multiple states, they are the same. Now thinking about this
distant [indiscernible] paradigm of Alice and Bob, and they can
only operate locally, there is another operation motivated norm
that we can think of, and this would turn out to be very useful
here, which is called the LOCC norm. So this is the norm which
tells that the thing [indiscernible] of two quantum states only
with local operations and classic communication. We have a
restriction on the kind of operation of measurements that we can
implement, and this just represents this restriction. So a
particular simple set of LOCC, which is what you can say something
about, which is called one-way LOCC. This is very simple. Alice
makes a measurement, gets an outcome, communicates the outcome to
Bob, and then Bob makes another measurement conditioned upon this
outcome. Then they can implement some measurements in this way.
So that's the definition. It's exactly like the trace norm, but M
is not any measurement. It's just a measurement that you can
implement by one-way LOCC. And because there is like only forward
communication from Alice and Bob, actually there is a nice
characterization but you won't need that, so don't worry about
this. This is just a mathematical way of writing. It's just like
a sum of positive operators here on Alice and Bob with some
[indiscernible]. This doesn't matter.
>>: What is the purpose of one?
>> Fernando Brandao: The one reflects the fact that we only allow
communication, classic communication, from Alice to Bob and not
the only way -- and not like back and forth. We could also alter
the course this norm where there is as many realms of
communication as we want, but the results would be more about this
norm.
>>: So if there would be a two, what difference ->> Fernando Brandao: Here? This would be more complicated. So
then you have to have a recursive definition. Like you define for
one and for two it would be like this one and then there is
another measurement here with [indiscernible]. This gets
extremely messy like the LOCC. For example, it has just been
proved [indiscernible] even a close of sets. So this back and
forth can complicate things a lot.
>>: Do you have an LOCC around N all the way to infinity? Would
you be able to approximate all the measurements there are?
>> Fernando Brandao: No. They are definitely measurements that
can only be done locally, even like here with many rounds of
communication. For example, measuring like in the basis of -- in
the bell basis. So you measure like in the base of maximum
entangled states. This cannot be done by by LOCC, not even
approximated. All right. So the question is whether
[indiscernible] is entangled, right? There is this weak
membership problem. There is a lot of theory behind it. There is
like concepts of PPT states, entanglement witnesses. They are all
developed to try to solve this problem. But once you turn this
into worse case analysis for algorithms, they are not very
interesting. So they only give like pretty bad algorithms. So I
will denote the local dimensions and of A by this, and of B by
this. So they give like rough explanation of [indiscernible] only
in the local dimensions.
>>: In the worse case, is that important?
>> Fernando Brandao: No. But we don't understand the average
case of [indiscernible]. We don't have much results yet. But we
do know it's very costly even on [indiscernible]. So it's not
that it's some pathological case and for most cases it's very
easy. We do know that it usually takes -- it's very hard to tell
whether I say the same thing or not. Okay. And the same for
[indiscernible]. So we don't know, we have this kind of -- we
have this kind of bound and this is no better than just doing an
exhausting search. All right. So we also have some -- we also
understand why this problem is hard. It turns out that it's a
computational hard problem for some parameters of the error. So
this starts with the work of Gurvits and then many people work on
that. You can prove that the problem is NP-hard when the error is
pretty small. It's like [indiscernible] small in the local
dimensions. And then this works for any [indiscernible], either
one norm, which is stronger, or two norm. So it's always NP-hard.
Even like if you care about -- this is for very small error,
right? Even for constant error, there was a hardness result.
It's more difficult to state, but let me try. This just says that
if you look at this time complex, it says explanation constant log
of A, log of B, absent some small corrections. This is like some
explanation, right, for [indiscernible]. So it shows that under
some assumptions that I tell you, there is no algorithms of this
[indiscernible]. Even when this error is constant, it's like
0.01. And this is for any such numbers beginning with zero. I'm
being very specific because these are a very important function in
the following. So it seems that there is limit of getting an
algorithm of like faster than exponential log dimension of A, log
dimension of B. And the hypothesis is stronger than NP-hardness
[indiscernible] is what we call this exponential time hypothesis,
which says that free SAT, free SAT, cannot be solved in like
[indiscernible] time. In time like faster than 2 to the N
roughly. So under this assumption, we have even this very strong
kind of like an approximate [indiscernible], okay.
>>: So SAT is an [indiscernible]. It's a lot of things.
>> Fernando Brandao: It is. So that's a motivation for this
work.
>>: A very strong hypothesis.
>> Fernando Brandao: It is a strong hypothesis, but as far as we
know empirically ->>: I'm not saying it's not possible, but a lot of people would
stop short of declaring that NP is a subset of X.
>> Fernando Brandao: Right, okay. So, you know, you can relax
just a little bit, but the minimum that you need to have N trigger
here is that you cannot solve SAT in time less than two to the
square root of N. So you need something stronger than NP
hardness, but it doesn't have to go all the way up to exponential.
This is just to get this nice answer here.
>>: Yeah.
>> Fernando Brandao: Okay? All right. So this is what was
known, the algorithms and the hardness result. Then let me tell
you what we got a few years ago and what was the beginning of what
I'll tell you next is that if you look in terms of algorithms,
actually you also, you can get the same scaling, right. This
exponent in. So it's not a practical algorithm but shows that
it's much easier than this NP hardness result might suggest at
first sight, right.
So there is like we call a quasipolynomial-time algorithm. So
it's polynomial both in the dimension of A and dimension of B. So
it's bed in the error so this only go to an error is like constant
or something, right. And this is both for the [indiscernible]
norm or for the 1-LOCC norm, okay. So now I said this two norm is
good from some geometric point of view, this 1-LOCC norm is more
natural from operational point of view, right. And what's
intrigues is that this bound, right, okay, so I'm say here, so one
corollary is that in 2 norm, the problem cannot be NP hard when
the error is somewhat large, right, like poly log. But if it
were, we could solve SAT in some exponential time, right? So
assuming that we cannot, this shows that there is a change in
complexity. From there is more error to kind of error which is a
little bit bigger. So you have contrast of this one. Another thing to mention is
that -- well, first, the same thing is true for the optimization
version of the problem. It's SEP. But now we cannot do for any
M. We have to do for a special M. And basically for M, which is
1-LOCC, which is kind of measurement of H. We have this kind of
algorithm. And now you can compare with this. Right, so this
result by these guys, right, that there is no such algorithm with
this running time. Now, the thing is that in their result they can only prove
hardness for a slightly bigger class that's called separable.
It's almost the same as 1-LOCC, but the normalization is
different. So we have like almost, like very suggestive
algorithms versus hardness results. The function seems to match
but the norms are wrong, okay. So and this is one of the open
questions I'd like to solve in the end. So it appears that there is some suggested
evidence that that might be the right complexity for this probability problem, but we
don't know yet. So we have evidences and it's really to put them
to get this [indiscernible] problem. All right. So what is this algorithm? Okay. So this
algorithm is not new. It's actually just based on some very useful -- when
did I start? I lost track of time.
>> Krysta Svore: 1:35.
>> Fernando Brandao: Thanks. So it turns out that what we do is
just we apply a well-known algorithm technique to this problem,
okay. We just found a way to analyze it better. And this is what
is called the sum of square hierarchy, okay. SoS hierarchy also
known as Lasserre hierarchy. Okay. So this problem it's SEP,
right. Let's focus on this now. It's easy to analyze. We just
maximize over unit vectors and this [indiscernible] function. So
[indiscernible] function in the elements. So this is just a
polynomial optimization problem under the hypersphere and the
hypersphere are just polynomial constraints again, right. It's
just like some over XI square equals one. So whenever we have an
optimization of a polynomial function or polynomial constraints,
we can use this hierarchy that's called SoS hierarchy. I will
show you the next slide what it is, but basically it's a heirarchy
that's also called Lasserre, was also introduced by Parrilo
independently. What it gives is a sequence of seven de Finetti
problems, right, and these seven de Finetti problems we can solve
efficiently in the dimensions of the matrices. That approximates
increasingly well this function, this number, okay?
>>: Basically, binary search between the two subspaces? Is that
what it is?
>> Fernando Brandao: No, it's more elaborate. Yes, I will show
you more.
>>: There are log log step versions of this kind of thing for
optimization.
>> Fernando Brandao: Oh, I see. No, this is something else, yes.
But ->>: Is it more like [indiscernible]? You said a product of the
hyperspheres.
>> Fernando Brandao: You're right, sure. Yes, right. So first,
yeah, you have two hyperspheres in it, yeah.
>>: But it wasn't [indiscernible] they came up with an infinite
hierarchy?
>> Fernando Brandao: It's just this one, yes. We will get, yes.
So this has two properties that are useful. First, there is like
a K, an integer which parameter rises the level of the hierarchy
that you are. For each K, there is SDP, and the side of the SDP
is polynomial dimension of M and exponential in K on the levels,
okay. And we know that when K goes to infinity, it really
conversely right solution for every M. So we're willing to wait
long enough, we are able to get as good an approximation as we
want.
So this is an nice hierarchy, because there are all different
kinds of SDP hierarchies, like [indiscernible] hierarchies like
[indiscernible] and so on.
>>: [indiscernible].
>> Fernando Brandao: That's a question I will address, yeah.
This is like for a few problems, like for example this sphere that
I showed, we come about trying to prove this converge fast in log
M steps, log M levels, yeah. But we will get to this. This
hierarchy is very interesting because first it's the strongest SDP
hierarchy that we [indiscernible] so if we want to try solve a
problem, we should try to use it and see what it gives. Also, a
lot of connections with interesting things like there's this sum
of square proof system, you know, it's a way of like automatizing
proofs. I don't know much about it but it's tightly connected -it's [indiscernible] write geometry and to [indiscernible] to this
Hilbert 17th problem, okay. So let me just show you very briefly
what this hierarchy does in a very simple case. So the problem is
suppose you have a polynomial F of X over [indiscernible] and we
just want to decide whether this polynomial is positive, okay. So
it's bigger than zero for all points. Now there is a particular
class of polynomials for which they are always positive by
construction. These are the sum of squares, right. So we say
that a polynomial is a sum of square if we can write it as a sum
of square of other polynomials, right. Of course this is always
positive, right.
Moreover, actually if I give you F and you don't want to check if
it's positive but you just want to check if it's sum of squares,
this is SDP, okay. So how is SDP work so you are given F of X of
degree [indiscernible] and it's a sum of square if and only if
there is a PSD matrix queue such that you can write F in this
form. So we just write a vector [indiscernible] which are all the
monomials from degree of zero up to degree D over 2 and then we
just say these polynomials are the same and the same means we look
at all the coefficients of this one and equate to the coefficients
of this one and if this is a sum of squares you can prove that Q
must be [indiscernible] in this presentation. So we can just look
for a presentation of this form with a PSD queue.
>>: It's a pretty simple argument.
>> Fernando Brandao: Yeah, right. So to check if a polynomial -so S, that's actually very efficient, right. And right. Of
course it gross a degree of -- exponential degree of the
polynomial. But for fixed degree, it's very good. This is going
to be [indiscernible]. Now it should be. Okay. So what this
hierarchy. This hierarchy is to try to use this argument to solve
this original problem, okay. It turns out that there is no good
prioritization for this problem. Actually, it's a [indiscernible]
problem in general, it's NP hard but there is a way of
approximating solutions of this problem using this kind of
[indiscernible]. And this is exactly this SoS hierarchy. So the
level K of the hierarchy is the following for this particular
instance. You just take F of X, you multiply by this polynomial
which is, of course, positive. And you check if this product is
sum of square [indiscernible], okay.
This is SDP size M to the order of K, right. So for every K, you
can solve this SDP and see what you get. Now, it turns out that
this is complete when K goes [indiscernible]. So it turns out
that the polynomial is positive if and only if the [indiscernible]
K is the sum of squares. In other words, every positive
polynomial can be written as a rational function of two SoS
polynomials. This was a problem, this 17th Hilbert problem
proposed by Hilbert a long time ago and solved by Artin in '25.
So this is like goes a long way in mathematics, but the innovation
of Lasserre and Parrilo was to observe they can solve this by SDP.
So this gives the hierarchies of SDPs.
All right. Very good. So now, of course, we can apply this to
our problem. So we have an optimization problem. It is turn into
a decision problem. So the maximum of this function is able to be
minimal real number such as this is a positive polynomial, right.
[indiscernible] just put these on the other side, you divide by
this to be in its vectors and you get it.
So now we just have to check that this polynomial is positive.
And, of course, we can apply this hierarchy. This is the K level
of the hierarchy, okay, so you have something of this form. All
right. So definitely you can use that, but now what I want to do
is take a look at this quantum angle to the problem. So what I
want to do is [indiscernible] this derivation of the hierarchy,
right. But actually show you how we can think about this
hierarchy in the quantum way, okay, using like quantum states and
[indiscernible] states and so on. And this will give at least in
my view, a more natural way of thinking about this hierarchy.
And then once we have this zero state hierarchy, we can prove the
[indiscernible], which is basically saying that the hierarchy
converges faster, okay, than like in exponential time.
>>: You say the state's NP hard. Is it believed not to be NP.
>> Fernando Brandao: This state here?
>>: Yes.
>> Fernando Brandao: Well, I [indiscernible] is different from
NP.
>>: No, this particular problem is not believed to be
[indiscernible].
>> Fernando Brandao: So, in general, yes because you can prove
that to solve this problem for every polynomial, it's NP hard in
the degree of the polynomial. And [indiscernible] a number of
variables.
>>: So when you guess this K is ->> Fernando Brandao: Exactly. So if you want to solve this
problem in general, in the worst case here, indeed you have to go
to K, which is linear [indiscernible] variables which use
exponential time solutions.
>>: But for each F, there will be such a K?
>> Fernando Brandao: For each F, there is such a K, but the K
value might be very big [indiscernible]. But here we are
interested in a special polynomial, right. So there is hope we
can do something better, which is -- yes.
>>: So we started looking at these polynomials as a way of
finding the [indiscernible] functions.
>> Fernando Brandao: Yes, so [indiscernible] problem, yeah.
>>: This is sort of the converse problem, in a way, which is,
okay, so what functions can you reach with those polynomials.
>> Fernando Brandao: Right.
>>: The objective is to come up with something that actually
forces the state down [indiscernible].
>> Fernando Brandao: Yeah.
>>: And so the question is how general is that? How big -- the
history is interesting.
>> Fernando Brandao: Yeah.
>>: [indiscernible] ask Parrilo how this came about.
>> Fernando Brandao: It's true. But actually it's much before
Parrilo as well so like a [indiscernible] optimization, he also
was [indiscernible] this hierarchy a long time ago and this came
from several different communities. They all came to the same,
right, to the same algorithm.
All right. So now I want to find this quantum way of looking at
this hierarchy, okay. And this quantum way will come about
considering a very interesting feature of quantum correlations of
[indiscernible], which is the fact that this -- they are not
shareable, okay. So let me explain what shareable is.
So this is the classical correlations are shareable. What I mean,
well, suppose I have a separable state of this form. This is a
state between Alice and Bob now, right. Now it turns out that we
can find a bigger state between Alice and many Bobs such as that
all the correlations between Alice and each one of the Bobs are
the same, okay. So how do we construct this state? Well, that's
a state, right. So we just take the same probability
distribution, the same state of Alice and with the K copies of the
[indiscernible].
So this is a state for which all the marginals of A and each one
of the Bs is the same and is the original one. So in a sense,
this correlation between Alice and Bob, they can be freely shared,
right. An arbitrary number of Bobs. So this is motivation for
making this definition, which is that [indiscernible] quantum
state is K extendible if you can find a bigger state of Alice and
K Bobs such that for all J from 1 to K, if you take the state
within the partial trace of all the Bobs, except the J one, we
recover the original state, right.
So I say it's K-extendible if there is state of this form for
which all this reductions, A and B3 or A and B4, they're all equal
to the original states. So the correlation between Alice and Bob
can be shared to K Bobs in this case.
Now, the interesting thing is that the entanglement is monogamous
in a sense, right. So this is called [indiscernible]. We talk
about in quantum cryptography, for example. But here really just
gives another characterization of entangled states, right. So it
turns out that the state is separable if and only if this state is
K-extendible for all K. Okay. So if it's separable, we know it's
extendible for all K. I just showed you how to construct the
extension. But turns out that this is necessarily sufficient for
being entangled.
If it's entangled for some K has to fail based on you cannot,
like, share these correlations arbitrarily. So this gives a way
of setting up an algorithm for the problem, right. So we are
interested in characterizing this set, this set of infinitely
extendible states. And what we do, we can just come up with a
hierarchy, right. We'll say, okay, so let's test if this state's
2-extendible. If it's not, we know it's entangled. If it is,
then we test if it's 3-extendible. If it's not, we know it's
entangled. If it is 3-extendible, then we have to go to
4-extendible and so on, right?
So we have this sequence of tests and to see how well this
performs as like computationally, we have to ask these two
questions, right. So first, how close to a separable state will
it be if there is a k-extension, right. We know if or every K
there is a k-extension separable, but is there some approximate
version is one problem.
The second problem is how long does it take to check in a
k-extension exists or not.
>>: Sounds like [indiscernible] size of the [indiscernible] is
bound.
>> Fernando Brandao: That's right, yeah. That's true. And yeah.
And if like, you know, maximum entangled, then you're
[indiscernible]. Only then.
All right. So now what's this that [indiscernible]. This was,
this was discovered 13 years ago by Doherty, Parrilo was one of
the guys introduced the hierarchy and Spedalieri. They just
saw -- well, this problem, this is for the hSEP version, right.
So you have the weak membership to solve this problem
[indiscernible] is equivalent to -- of course, you can write the
sum of square hierarchy, but this is also equivalent just to
optimization over k-extendible states, okay.
So like when you go to high extensions, what really you're doing
is to go to higher levels of Lasserre hierarchy. Plus some other
tests, these computation tests, but these won't play a role in
what we follow. So what's the way of writing this SDP? Well, we
just maximize trace, M is the input of the problem over matrices
pi such that there exist as big PSD matrix of this trace for which
all of the reductions between A and BJ equals to pi. Okay. So
basically, you just optimize the linear functional over the set of
K-extendible states.
Now we add something more that's not only the K-extendibles but if
you take the partial transpose of some set of these, this is
[indiscernible] Bayesian matrix, okay. Don't worry about this
constraint. This is just [indiscernible] Lasserre, but we don't
know how to use this. This is an open question.
So basically, think that Lasserre is really just, instead of
optimizing over several states over this [indiscernible]
optimization problem, you relax over some set of bigger quantum
states which gives a [indiscernible] to probabilistics. So this
is what's happening here.
>>: So is it like -- many people try to show that this is
actually finite in the sense that given the state, there will
always be a K.
>> Fernando Brandao: That's true. It is true.
>>: It's unknown back there. Is it now solved? In other words,
is that ->> Fernando Brandao: Well, these are solved even before they did
this because there was some works that were mentioned
[indiscernible] given the answer.
>>: I see.
>> Fernando Brandao: But right. When they were working on it, I
don't know what they do.
>>: For each ->> Fernando Brandao: For each entangled state, you can find a K
for which you solve, but this K might be, of course, very big.
All right. So this solves this question, right. So how long does
it take to check if a k-extension exists. Well, it takes time,
local dimensions to the power of K roughly because you have a SDP
for it.
>>: No, I didn't mean extension. I mean testing whether
entanglement weaknesses is a different question, right?
>> Fernando Brandao: Yeah, it's very close. Like, you know,
because ->>: I mean, it's separating hyperplane.
>> Fernando Brandao: That's right. So basically, this is
optimization over the separating hyperplane, right. So if you
solve the weak membership version of this problem if you get an
invisible solution, meaning that it's entangled, right, but if you
go to the [indiscernible] this will give the hyperplane, which is
the entangled weakness.
>>: Okay.
>> Fernando Brandao: All right. So now the question really is
how close we are from separable, right, if this is K-extendible.
So how fast does this hierarchy work. And there's a lot of work,
like starting -- I didn't put the reference here but starting from
paper by, like, Werner in '88, I think, studies these bounds
called [indiscernible] theorems, okay, that [indiscernible]
things. But now we are just -- I don't want to explain what they
are. I just want to show this kind of bound that was known
before.
It's known that when K grows, the set of K-extendible states,
right, if grow is a K-extendible state, it gets closer and closer
to separable. But the bound wasn't so useful for algorithmics,
algorithms because it was very -- as a quadratic -- as a square of
the local dimension. So if you look at the K that you need, K is
the level of hierarchy, right, to get a good approximation this K
has to be proportionally dimensioned. And because the SDPs
exponential in K, this only give exponential time algorithm,
right.
So this shows that the hierarchy is complete but despite give
something in terms of algorithms, and actually, this, you know,
turns out they cannot improve on that. There are examples which
saturate, almost saturate [indiscernible]. So this dimension, if
you want one norm, which is a [indiscernible] dimension, if you
really want in one norm, there is not much you can do here.
So now, this is the theorem that I mentioned before comes from
information theory result from like resulting quantum information
theory, which is the following. That if a state is K-extendible,
but now we don't care about trace norm but we allow 1-LOCC norm or
2 norm, then we have an exponential improvement in the performance
of the de Finetti, right. Doesn't go with dimension, local
dimension, but goes with log of local dimension, okay.
And once we put this into the problem, we get this exact running
time dimension, okay. So this is the main result behind this
initial theory them I show you is really that how well
K-extendible states approximate separable states goes with the
number of qubits with the states and not with the dimension.
If you use these two norms. If you use one norm, it is much
worse, right. So there is this difference between the norms they
use. And if you want something more efficient, you have to use
these weaker norms.
All right. So this is just, I guess, right, so if you use this
number of rounds of SoS, right, we know that we can solve putting
here, we know we can solve the problem of error epsilon and the
size of the SDP is just the size of the extension and the size of
the extension, the dimension of the extension is dimension of A
times dimension of P to the power of K because the state has one A
part and K B parts. So put in the number, we get exactly this
running time.
All right. So I guess let me see. I'm running out of time and I
want to mention some applications of this result to purely
classical problems, which was the question [indiscernible] asked.
So I have a sketch of the proof. So the proof basically is use
quantum information theory ideas. So like, you know, use
[indiscernible] information, right so you use movement information
and you use chain rules, and so on. But I will skip that because
I don't have much time.
So this was just to show how we can prove it using some quantum
information ideas and let me get to these classical problems, then
to finish the talk.
So now it turns out this hSEP is useful for [indiscernible] this
is the problem we care about in quantum. But it's also equivalent
to a lot of other problems that people consider elsewhere and I
just want to mention these connections and what you can learn
about these problems, right.
So what I just show you is that this Lasserre hierarchy is sum of
square hierarchy. There's a way to think about it in quantum
terms, right. You can think in terms of quantum states and
extensions, right. And I didn't show you, but once we have this
way, we can use quantum information tools to bound how fast it
works, right.
So it's just natural to apply these bounds that we have to other
problems, right, to classical optimization problems, right, the
[indiscernible]. So this is what I want to show you.
So I don't like a clear result yet, but we have actually some
suggestive evidence that something useful might come out of it.
And I want to show you now.
So there are several, like, equivalents which are pretty easy to
show and won't be the focus, like one thing is equivalent to
compute the injective norm of tensors. The injective norm is just
this generalization of the maximum eigenvalue, of single value.
For other problems in quantum information, we can connect them to
like compute the minimal output entropy of a quantum channel, and
[indiscernible] is equivalent to hSEP or computing the optimal
acceptance probability of quantum [indiscernible] proves with two
independent provers. It's the same as hSEP. So it's a lot of
other problems in quantum information that are equivalent.
Now a problem that I want to focus is this problem of computing
the two to four norm of a projector or of a matrix in general. So
this is just some hyper contractive norm. So the definition is
that we maximize of a unit vectors the four norm of P times X,
okay over X of [indiscernible].
And it turns out that you can relate these two things so you can
prove that if P is in a matrix, actually doesn't have to be a
projector. The two to four norm of this matrix equals to hSEP for
M of this form, okay. So K is just a computational basis and this
what is you get. So if you can compute hSEP in general, we can
compute this for all this.
Okay. So why is that? It's very simple so let me just go really
quick. If you have a matrix, A, right, so you want to go two to
the four norm of A [indiscernible] before. So this is the
computational basis, whatever you want and here is vectors, right.
So you can always write A in this form. You can always find
vectors A of I for which capital A can be written in this form.
Now, the easy direction write, okay, so now, you know, of course
by this [indiscernible] which is very easy to verify, you see that
if you can compute hSEP, you can compute two to four norm as well,
right, just because this two to four norm is a particular case of
hSEP, right. And so this is just a [indiscernible].
Now, actually other way is also true, which is the more
interesting direction. If you can compute or list, like if you
can approximate two to four norm, then you can approximate hSEP as
well. Okay. So turns out that they have similar complexity,
these two problems. And this use some quantum algorithms as well
that I will just keep. If you want you, can ask me later. We use
some idea from quantum information to prove that.
So now why this two to four norm is interesting. Well, for one
thing, because this is a hyper contractive norm, right. So it
appears in Markov chain if you want to prove [indiscernible] of
the Markov chain, you want to bound this two to four norm. But
more really because it's connected to two other problems that I
just state what they are, which are like, you know, very
[indiscernible] problems in current research and [indiscernible]
optimization, which are problems that we don't understand the
complexity and we don't understand what we are can do in terms of
algorithms, but they will light on the [indiscernible] and this
tells us something. So let me just tell you what the problems are
if you don't know.
One is related to the unique games conjecture, okay. So before I
tell you the conjecture, let me tell you what a unique game
problem is. This is the problem. This is the following. Oh, no,
this is the conjecture, right. So this tells you that suppose you
have a system of equations for this form. Just two variables per
equation, XI plus XJ equals C mod K. These are like two integers.
So it's solving a linear system of equations of two variables over
some finite field. Now, this conjecture tells us that this is a
very hard problem, actually. Even to estimate. It tells us that
it's NP hard to tell whether more than one minus epsilon fraction
of the constraints is satisfiable, meaning that there is a
assignment for the X which satisfies 1 minus epsilon fraction of
these equations. Or if you cannot satisfy even an epsilon
fraction of them, okay.
So your promise that either of them is satisfiable or almost none
of them are satisfiable and you want to decide which is the case.
And the conjecture is that this is an NP hard problem. So this is
just a problem. Of course, you can make this conjecture. Yes?
>>: Unique game?
>> Fernando Brandao: Unique because the constraints, these are
constraints [indiscernible] problem and it's unique in the sense
that given one assignment for this variable, there's a unique
assigner for the other variable which satisfies a constraint, and
vice versa, right.
So if you choose XI equals one, there is only one choice for XJ
which would satisfy this constraint. And vice versa.
>>: Unique solutions.
>> Fernando Brandao: It's not a unique solution. It's like -it's a unique choice for one of the variables in a constraint
given the choice of the other one. So it's ->>: Right. Unique strategy.
>> Fernando Brandao: Right, yeah.
>>: Yeah, given something like K ->> Fernando Brandao: C and K are just parameters of the problem.
(Multiple people speaking.)
>> Fernando Brandao: The conjecture is there for any constant C
and K, this is the ->>: Yeah. Doesn't it need to do with the number of unknowns?
>> Fernando Brandao: So you don't know any of the Xs, right. You
have N Xs, N variables, and you don't know what they are.
>>: Okay.
>> Fernando Brandao: Then you have to ->>: [indiscernible], right.
>> Fernando Brandao: Yeah, the problem scales with M, the number
of variables.
>>: So if K is very small, compared to N, then I would -- I think
behave like SAT, like kSAT, in that if you have too many
constraints, then there's no solution. And if you don't have very
many ->> Fernando Brandao: Oh, sure, sure. Yeah, right.
>>: Then it's easy and it's only for critical K, K is some
function of N to make this hard, I think, is the point.
>> Fernando Brandao: So what he says is true, but it's not in
terms of K. I think maybe also in terms of K, but in terms of how
many equalities you have, right.
>>: Which is N, right?
>> Fernando Brandao: You have N variables and you have other N,
like, equalities, right. Constraints.
>>: Right.
>> Fernando Brandao: And then, of course, there is some ratio for
which this -- yeah. Yeah. All right. So now why people care
about this problem, why this is an important problem. Well, for
many reasons, but one of the reasons is that Raghavedra proved
that if the unique games conjecture is true, then if you look at
the second level of this SoS hierarchy, okay, this is the best
possible algorithm, give the best possible approximation for any
constraint satisfaction problem. So this constraint satisfaction
problem is a huge class of combinatorial optimization problems.
If you can prove that, we really understand what you can do in
terms of algorithms here, right. We just use Lasserre second
level. We get the best approximation possible. And to try to get
an approximation better than that is NP hard, okay. So it's
really [indiscernible] the problem so if you want to
[indiscernible] we can understand a lot of other problems.
And you can see this is a major barrier in our current knowledge
of algorithms, right, for these problems.
Now, this is not really a conjecture that everyone knows is true.
In fact, like this really could go either way, right. It's really
like where our knowledge goes. And one of the ways to see that is
that actually some years ago, Arora, Barak and Steurer found a sub
expansion time algorithm for the problem. So given any epsilon,
you can solve this in time exponential and to the order epsilon,
right. So of course, it's not really efficient, but it's not like
SAT, right, which appears to be these exponential lower bounds.
All right. So this is one problem. The other problem, okay, just
mention that. Sorry for introducing so many problems at the end.
I just want to show what is this implication. So this is a
problem in graph theory, which is very tightly connected to unique
games. So we believe they are equivalent, but there are some few
caveats.
So this is the problem of telling given a graph, right so we have
a group graph and we have two parameters, we don't understand
whether for all these small subsets of [indiscernible] of the
graph, they expand really well. All the verts leave them or all
the verts stay inside them they stay [indiscernible].
So the conjecture say it's NP hard to tell whether for a graph
with edge set V and edge set E expansion, this is the definition
of expansion, this is the probability that our edge from M leaves
M. Whether the expansion of all the small sets is more than
epsilon. Small set means just a region of size like sub volume,
okay, so it's more [indiscernible] times the total vol you'll. So
[indiscernible] more regions.
And we want to know whether there is just one region which is
really not expanding, right. Everything stays inside or if O is
more regions, they're almost completely expanding, everyone
leaves, like all the edges leaves them. Like almost all of them.
And this is small set expansion conjecture. Again, it's
conjecture to be NP hard. Now why I tell you all these problems?
Because you have these equivalents or, like, approximate
equivalences in some cases. All these problems, they appear to be
like the same complexity, okay.
So there are a lot of, like, [indiscernible] that I'm glossing
over so this one direction is known to be true and the other
direction is conjectured to be true. This is just [indiscernible]
at the end. Again, there are some caveats in the conversion of
errors but that's roughly true. And this what is we prove in this
paper that connecting hSEP to two to four that I just showed you
and two to four is small set expansion, okay.
So now given these connections and given that we know we have some
new results about like algorithms for this guy and did I mention
also hardness for this guy, we can try to convert them, right.
And this what is we did. This is the last slide just mention the
results. This is what we could prove. Using this quantum ideas
for Lasserre hierarchy, this is what it gives for two to four
norm, for example. Even a matrix M by M, we can compute in this
number of rounds of sum of squares so logarithmic in the size of
the matrix. Additive approximation to the two to four norms, what
the error is this one. So this is a natural error because you can
also prove that the two to four norm is smaller than this error,
okay. So we just showed that there is this natural upper bound
and it showed that you can have this kind of [indiscernible]
approximation. So it's not clear if this is good for anything,
right. So but actually ->>: [indiscernible] epsilon minus three up out in front is large.
>> Fernando Brandao: Oh, it's epsilon minus two. But
[indiscernible], yeah. Sorry, it's a mistake. It's a typo. My
typo, yeah. But you're right.
>>: It's better.
>> Fernando Brandao: Well, yeah, but [indiscernible] because the
running time would be N to the order N over epsilon square, right,
which is exactly what we saw before, right. So two to the log
square N. So just right, so you see why -- this is more the kind
of bounds we had for the quantum problem, right. So if you
convert to this problem, we get this. It just comes from the
normalization of, like, one [indiscernible] measurements, for
example. I'm skipping over the details.
Now if you apply this [indiscernible] bound for two to P norms you
can record this sub exponential time algorithm that I mentioned
before, okay, which is the state of the art for small set
expansion. So we don't have new algorithm results using the
quantum ideas, but at least we can recover the best that we know
classically. Which is a sub exponential time algorithm for a
small set expansion. It just gives an alternative analysis for
the algorithm.
Now, it also list as new open problems, okay, which I find
interesting. So if we can improve this quantum analysis several
ways, this would lead to new classical algorithms and I think
that's something interesting.
So, for example, if I improve in the bound, right, like I show
this different [indiscernible] bound which was obviously an error.
If you could have a multiplicative error, then it could solve
small set expansion quasi-polynomial time, okay, so this would be
strong evidence against unique games, right. So they could not be
NP hard with this running time. Or else something very weird
happens.
We can also ->>: Unless you get famous, right?
>> Fernando Brandao: That's right, yeah.
>>: We can also get better approximation algorithms for
[indiscernible].
>> Fernando Brandao: No, because that's the sad thing about
unique games conjecture. If it's true, we are missing everything.
If it's not true, then all they've equivalents are lost so you
have to work case by case.
>>: [indiscernible].
>> Fernando Brandao: That's right, yeah. So there are also
quantum algorithms for hardness results. If you could improve on
those, we would also get this kind of lower bound for unique
games, which seems to match. So improvement -- okay. It's
something that I didn't mention. So the point is that this
improvement can be cast in these open problems in quantum
information theory. We don't understand yet and we would like to
understand. So one possible strategy for trying to solve unique
games conjecture is to learn more about quantum information
theory, right. So there is this new breach and I think it's worth
exploring more and see what we can get in terms for these
classical algorithms, considering this new quantum view that we
have.
So that's all. So what I told you is just that I hope the message
was SoS is a very useful algorithm tool for problems in quantum
information. I give you just one application. There are many
more in quantum information itself. And I show in [indiscernible]
how we can get this quasi-polynomial algorithm for deciding
entanglement from this hierarchy.
What was unexpected to us, and I think it's a nice bonus is that
conversely, quantum information theory is also useful theoretical
tool to understand the performance this hierarchy, right. So
[indiscernible] to this hierarchy, and I show you like some
connections like two to four norm is small set expansion with
[indiscernible] probability, and there's this new connection that
gives potentially a new approach to resolve this unique games
conjecture, right.
So, of course, this is a very hard conjecture. I don't claim we
are close to solving it. There is a lot of barriers that might be
very hard, but I think it's something new to think about, which is
interesting. So that's all. Thanks.
>> Krysta Svore: Questions?
>>: So in some sense, [indiscernible] algorithm entangles
everything.
>> Fernando Brandao: Right.
>>: And achieves a very nice speedup as a result of that. This,
it seems like this suggests that if you have more fancier ways of
generating entangled states, then you might be able to achieve
exponential speedups if this -- if this [indiscernible].
>> Fernando Brandao: At the decomposition of the separable state?
>>: Well, the approximation of the computation of the separable
state.
>> Fernando Brandao: Yeah. I guess a different [indiscernible]
here so here [indiscernible] in terms of dimension of the states.
>>: Sure.
>> Fernando Brandao: To have an N qubit system that
[indiscernible] particles and this complexity is nothing we can do
about. So if it is one level below it, right. So even for
various small quantum states, it is very hard to decide
separability for them. But there is exponential growth of the
dimension of the number of particles. But even if the number of
particles is quite small, it is already becomes a [indiscernible].
>>: I'm just thinking that, for example, quantum [indiscernible]
suppose I have -- I want to speed up the rate at which I can
compute, let's say the occupancy of the state after a
[indiscernible]. If I can use entanglement to make up for
randomization in the [indiscernible] process, I entangle things in
such a way that I superimpose the things that need to be
superimposed, the things that are really additive, not the things
that aren't, then I might be able to design an algorithm which
would use entanglement more surgically over those -- all those
paths if you know what I'm saying.
>> Fernando Brandao: Yeah, I guess, okay.
>>: So, I mean, yeah. I'm just -- the problem is how you do that
or how do you describe the entanglement you want.
>> Fernando Brandao: Right.
>>: That will result in the right answer. So this is interesting
from that point of view.
>> Fernando Brandao: I see.
>>: I mean [indiscernible] you think it's exponentially good
unless you entangle, I think. I don't know how in the world
you're going to do it if you just, you know ->> Fernando Brandao: Yeah, for some problems, yes.
>>: UGC one provides a natural header example to where you have
very small amounts of entanglement and exponential speedups.
>> Fernando Brandao: Right that's true. [indiscernible] quantify
it.
>>: What I'm saying is something allowing your [indiscernible] if
there's no entanglement, then it's hard to get an exponential
speedup, that's all. Not small amount, but I mean, it might not
be much ->>: [indiscernible].
>> Fernando Brandao: Sorry?
>>: If there's no entanglement, aren't you classical? By
definition?
>> Fernando Brandao: Yeah, it's true, but it's actually an open
problem. So suppose you have a quantum computation for which your
promise that -- with mixed states, right, with few states.
>>: Few states.
(Multiple people speaking.)
>> Fernando Brandao: You always have separable states
[indiscernible].
>>: Okay, yeah. You have [indiscernible] between five problems.
So looks like there is a -- I don't know how natural a complexity
[indiscernible] try to develop [indiscernible] programs.
>> Fernando Brandao: Right, so, you know, because this unique
games is an important problem, people fight about that. But ->>: [indiscernible].
>> Fernando Brandao: Yeah, you know, so that was the reason for
introducing this [indiscernible] is small set expansion is a
problem that [indiscernible]. One way is prove it. The other way
is conjecture. There is strong evidence, but there is no proof.
Now, unique games, we know that if unique games are hard, a lot of
other problems are hard as well. That's why it's a very useful
one. So it's like to prove unique games hardness ->>: Is space, yes.
>> Fernando Brandao: Yeah, unique games is, yeah.
>>: But there is [indiscernible].
>> Fernando Brandao: That's right, but nondirectional, that's the
problem. Like people have tried to work out, for example, max
cuts. They really true to prove that max cut was equivalent to
unique games but they couldn't do it so it's more like, yeah,
there are a lot of problems which are unique games hard which are
not known to be NP hard, but we don't know if they're equivalent
to unique games.
>>: [indiscernible].
>> Fernando Brandao: Yeah if you could get [indiscernible] this
prove it's NP hard.
>>: It's not noted at NP.
>> Fernando Brandao: That's true, yeah. And it would be -- yeah.
>>: So second question, if all these optimization problems are
equivalent to some question about whether a quantum state is
separable or not, is there any hope for maybe some quantum
algorithm that at least does some, like, low levels of hierarchy
faster or something like that than the classical algorithm?
>> Fernando Brandao: Yeah. I have never thought about it. It
would be nice, yeah, but -- yeah. Maybe, but I don't -- I don't
know. I could search for this [indiscernible], right. But
it's -- yeah, I guess because it seems to be intermediate
complexity, there is some hope that it can do something there.
It's a good question, but I've never thought about it.
>>: You mentioned a quantity called the injective norm of the
tensor.
>> Fernando Brandao: Yes.
>>: Can you explain what that is? Is it related to the rank of a
tensor?
>> Fernando Brandao: It's different from the rank.
>>: [indiscernible].
>> Fernando Brandao: Oh, well, it's really just -- it's just a
naturalization of [indiscernible].
>>: It's kind of like the value of the tensor.
>> Fernando Brandao: Right, exactly. Because if you just have
two, it's a single value so if you have more instance so it just
[indiscernible].
>>: If there has any implication of the most famous
[indiscernible] tensor matrix modification free tensor. Does it
say anything about the [indiscernible].
>> Fernando Brandao: This is connected to the tensor rank, right.
>>: Which is unknown, as far as I know.
>> Fernando Brandao: That's right. So I -- there might be, and
I'm not aware of, but this is not the same as tensor rank, but I
don't know if there's relations. There could be. Yeah.
>>: [indiscernible] improvement on that rank, like over
[indiscernible].
>> Fernando Brandao: Right.
>>: [indiscernible] follow optimization approaches at all
[indiscernible].
>> Fernando Brandao: [indiscernible], yeah.
>>: Brought down the exponent a little bit.
>> Fernando Brandao: That's right. Very recently, right, like ->>: [indiscernible].
>> Fernando Brandao: Right. But this was by this Virginia
Williams somebody, yeah. I don't know what they do.
>>: Okay.
>> Fernando Brandao: I don't.
>> Krysta Svore: Any other questions? Let's thanks Fernando.
>> Fernando Brandao: Thank you.