[MUSIC PLAYING]
SPEAKER: Elizabeth is going to
go first, and then [? Erin. ?]
But we can adjust to that also.
OK?
So, Elizabeth.
ELIZABETH CROSSON: OK, thank you
very much for inviting me here
to speak today.
I'm going to be talking
about a theoretical analysis
that we did for a particular
class of problem instances
that allows us to
obtain rigorous results
on the asymptotic performance
of simulated quantum annealing,
and also compare them with
rigorous results for quantum
annealing and classical
simulated annealing.
So the first version of this
work we put on the archive
in January, but I just put
an updated second version
on the archive last week
that made some corrections
and also expanded the scope
of the results by a bit.
And just recently,
this work was also
accepted to Fox 2016,
which is a computer science
conference that will
be held this November.
So I'm going to be taking
the approach of comparing
quantum annealing to other
meta-heuristic optimization
methods.
And in the language we
heard earlier this morning,
these are non-tailored
sequential algorithms,
that is, the big three-- quantum
annealing, simulated annealing,
and simulated quantum annealing.
Now, how are we going to answer
the big question of how useful
is quantum annealing?
Well, we've heard some
about impaired [INAUDIBLE]
so far at the conference.
You can build or numerically
simulate quantum hardware
and benchmark it on
random instances.
But another kind of approach,
the approach I'll be taking,
are theoretical tests.
So here you may have things
like analytically bounding
the spectral gap of
a quantum annealing
system in the regime
where the noise is
low enough for the
adiabatic theorem to apply,
or you could do some
work to theoretically
bound the efficiency of
classical simulations
of quantum annealing.
And in particular, I'm
going to be focusing
on simulated quantum
annealing, which
is a classical
algorithm for sampling
the output of some quantum
annealing systems, including
the transverse field
that we care about.
But in general, the performance
of simulated quantum annealing
is still very much
an open question,
at least theoretically.
We're starting to get a pretty
clear picture empirically.
So let's go over the
special properties
that make simulated
quantum annealing possible.
Well, we know it's possible
when the Hamiltonian has no sign
problem.
Or put another way, the
quantum annealing Hamiltonian
has non-positive
off-diagonal matrix elements
in the computational basis.
This immediately implies that
the thermal density matrix
is a non-negative matrix.
It has non-negative
entries everywhere.
And you can see this just by
expanding that exponential
as a Taylor series
and noting that each
of the terms in the
sum is non-negative.
But if we take the limit as
the temperature goes to zero,
then the thermal density matrix
becomes the ground state.
So this is also an
argument that says
that the ground state of a
Hamiltonian without the sign
problem has non-negative
amplitudes everywhere.
And this property
of the Hamiltonian
is called being
stoquastic, which
is a combination of the
terms quantum with the term
stochastic from the field
of stochastic matrices.
So let's talk a little more
about the specific kind
of simulated quantum annealing
that I'll be analyzing.
So this is based on the path
integral quantum Monte Carlo
method.
One way to quickly
explain how this works
is that you want to express
the partition function
as a discrete path integral.
So you see in the
first step, I just
expanded the definition
of the trace.
And I can do this trivially
because the Hamiltonian
commutes with itself.
And then here,
I've just inserted
several copies of the identity.
L is some integer parameter
that sort of controls
the discretization that we've
done for the path integral.
And so the property we talked
about on the previous slide,
that the off-diagonal matrix
elements are non-positive,
that implies that these
weights are all positive.
The terms of this
sum are all positive
because they're E to minus
some positive real value.
So this creates a
probability distribution
on the space of paths.
I'm not going to have to get
into the Trotter-Suzuki formula
or anything because
this kind of illustrates
the fundamental
difference between more
general kinds of quantum systems
and this restricted class
of stoquastic Hamiltonians.
So Feynman told us that, when
you have a genuinely quantum
process, you have all paths
contributing with some phase.
And all those phases, of
course, have the same magnitude.
But for a stoquastic
system, some paths
are more important than others.
So it's somewhere in between
the classical regime,
where you have one path
that minimizes the action,
and the quantum regime,
where you really
need to count all of the paths.
So when you have a distribution
that places more importance
on some paths than
others, you can
start to think of exploiting
a general randomized algorithm
called Markov chain Monte Carlo.
So I'm going to break
down these terms,
because we often just
say them together,
and I want to break down
what they mean separately.
So Monte Carlo means that you're
estimating expectation values
for a distribution
by only sampling
a small number of points
instead of summing
over the whole domain.
Now, but this
leaves the question
how do you sample these
values from some complicated
distribution pi.
And that's where you
use a Markov chain.
So a Markov chain
is a random walk
that takes place on the
domain of your probability
distribution.
And the way you construct
this random walk,
or a sufficient way to
construct this random walk
is by designing it to have
transition probabilities that
satisfy this detailed
balance property.
And so I think, in
the physics community,
we usually characterize
the performance
of a randomized Monte Carlo
Markov chain algorithm
by something that's called
autocorrelation time.
I'm going to use a slightly
different concept, which
comes from computer science,
called the mixing time.
So this is the number of
steps of the random walk that
need to be executed to be
close in trace distance
to the intended distribution.
And as we know very well,
in quantum information,
once you're close
in trace distance,
this preserves all of
your expectation values.
OK?
So that's the formula
for the mixing time.
But if you haven't
seen that before,
just keep in mind that
the mixing time controls
your convergence to be
intended distribution
you wish to sample.
OK.
So now I'll get to the
particular class of instances
that I'll be analyzing.
And of course, they're bit
symmetric cost functions
with a tall, thin spike.
And in fact, so their energy
plot will look like this,
or this is maybe
the first example.
We'll actually generalize
this a bit on the next slide.
But what you see here is,
on the horizontal axis,
you have your domain.
It's controlled
by Hamming weight.
So it's important to keep in
mind that this is really a spin
system, but we're viewing
it as one-dimensional
because the cost function only
depends on the Hamming weight.
And so the way the cost
function works is it
just scales up with
the Hamming weight,
except for this
one special point,
where you introduce a tall,
thin spike in order to create
a false local minimum here.
And this false local minimum,
it was shown in 2002,
makes the problem
take exponential time
to solve with classical
simulated annealing,
simply because it
takes exponential time
to even propose, or
to accept a proposal
to step onto that spike.
But it was also
shown in this work
that this problem can be
solved in polynomial time
using a quantum annealer.
And so this was
kind of the origin
of all these tall,
thin spike problems
that we've been talking about.
Now, let's look at a small
generalization of this.
You know, so when
you look at this,
you might say that,
well, what if I
allow my simulated
annealing algorithm
to take steps of size
two instead of size one?
Then it will still solve
this in polynomial time.
So to defeat that, we're
going to scale the width
of the barrier polynomially.
So the height of the
barrier is n to the a.
And the width of the
barrier will be n to the b.
And so now I'm going to go
through an argument that
was given in 2004 which explains
why the quantum annealing
gap of this system is constant
when a plus b is less than 1/2.
It's a nice argument, and it'll
also create a nice parallel
with what we do in the proof
about simulated quantum
annealing convergence.
So the spikeless Hamiltonian,
H tilde, has an order one gap.
And we know this because
the spikeless Hamiltonian
means that the spike
term is not here,
and so the cost function simply
depends on the Hamming weight.
And so it is a collection
of non-interacting qubits,
where each qubit prefers to
be zero in the sigma z basis.
So it's an exactly
solvable system when
the spike term is not there.
And it has a constant
gap, and you can also
see that the ground
state probability
distribution, viewed as a
function on this Hamming weight
basis, is a binomial
distribution.
And so the amount of
probability that it
can put in any one location
has to be less than 1
over square root n.
So now the next fact
we want to point out
is that adding this
spike term, because it's
a positive
semidefinite operator,
does not decrease the
energy eigenvalues.
They all go up.
So these E tilde sub i is
less than E sub i for all i.
Now, putting this
together, the argument
that Reichert gave
in 2004 is that we
can use the ground state of
the spikeless distribution
as a variational [INAUDIBLE].
We can use the variational
method of quantum mechanics
to upper-bound the
ground state energy.
And so when you form that
inner product, that expectation
of the energy, you get that
you have the ground state
energy of the spikeless
system plus some term that
asymptotically goes to zero
when a plus b is less than 1/2.
So because the first excited
state energy does not go down,
this implies that the
gap of the spike system
is constant when a plus
b is less than 1/2.
And the intuition is that
quantum annealing does not even
feel a modest-sized
spike in this problem.
In other words,
the wave function
really doesn't need to
change all that much in order
to minimize the energy.
OK.
So now I'll get to
the new results,
which is simulated quantum
annealing for the spike.
So we show that simulated
quantum annealing
equilibrates in
polynomial time and finds
the minimum of the spike
cost function whenever
a plus b is less than 1/2.
And in terms of
explicit bounds, we
find that it takes orders n to
the 17th time with single site
metropolis updates, or
order n to the seventh time
with world-line
heat bath updates.
Now, these O tildes
simply mean that we're
ignoring logarithmic factors.
So here are some of the key
ingredients for the proof.
So just to say
what this means is
that simulated quantum annealing
works in polynomial time.
So does quantum annealing.
But simulated quantum annealing
takes exponential time.
So from these results, you can
conclude that simulated quantum
annealing can be
exponentially faster
than classical
simulated annealing.
Now, the proof
idea is that we're
going to compare the
simulated quantum annealing
Markov chain with and
without the spike term.
We're also going to relate
the typical proper time.
So here, proper time means
time along these world lines,
these quantum Monte
Carlo configurations.
We're going to relate
this typical proper time
to the expectation values
of the quantum system.
And you might say,
wait, we already know
that, because that's the whole
reason we use quantum Monte
Carlo is that it relates to
the expectations in the quantum
system.
But we don't just want
the mean values to agree,
we want the distributions
to closely agree.
We want a concentration of
measure within this measure
that we put on the
space of paths.
We want to say that the
proportion of the world
lines that they spend sitting
in the region of the spike
is never that different than
what the quantum system itself
does.
And then finally, on a more
technical note, what we really
prove is quasi-equilibration
of simulated quantum annealing
within a subset of the quantum
Monte Carlo state space.
And we use the adiabatic
path to guarantee warm starts
in this subset.
So of course, a key
feature of Markov chains
is that they, after you
run them for a long time,
forget where they came from.
And in fact, that's what the
mixing time notion also says.
But what we're going
to do is we're only
going to let our algorithm
live in this physically
important region
of the state space,
and show that it
quasi-equilibrates there,
and that we can keep
ourselves in this region
with high probability using warm
starts from the adiabatic path.
Warm starts, if you
don't know that term,
it means that you're
in a distribution that
is close to the
intended one already
as soon as you start running
the Markov chain Monte Carlo.
OK, so now a little bit more
background on Markov chains
and mixing times
because, you know,
I want to say this is meant to
complement the methods that we
saw earlier this morning.
You know, we heard about
escape rates and instantons
and all this beautiful physics.
And I'm a physicist, so
I like that stuff, too.
But we're going to use
completely different proof
techniques that come from
the theory of Markov chains.
So Markov chains,
essentially, you
can think of a discrete time
Markov chain as corresponding
to its transition matrix.
This is a stochastic matrix
because, well, what that means
is that the rows sum to 1.
And so if you apply this
matrix to a vector that
represents a probability
distribution,
then you get another vector
that represents a probability
distribution.
A stochastic matrix
conserves probability.
And in terms of the
random walk, this
means if you start
at the point x,
you always have a probability
of going somewhere.
Now, with this definition,
the stationary distribution
will be the eigenvector
with the largest eigenvalue.
And it has eigenvalue 1.
And you can actually
control the mixing time.
That's what this tau
epsilon is, the time
it takes to be within trace
distance epsilon of your target
distribution.
You can control it with the
spectral gap of this Markov
chain transition matrix.
So it scales
essentially inversely
with the spectral gap.
And then there's this
factor of 1 over pi min,
but it has a log that makes
it typically not as important.
But this one over pi
min really characterizes
that your random walk,
we're considering
the worst possible starting
point for that random walk.
OK, so let's talk a bit
about how-- so you know,
we know as quantum
annealing aficionados,
that analytically determining
spectral gaps is very hard.
And so we need
additional tools that
also characterize the mixing
times of these Markov chains.
And one of the most
beautiful tools
is this idea of conductance
and bottlenecks.
So we're going to visualize
our Markov chain, pi p omega,
as a weighted graph.
The vertices are at the
states x of our domain.
And two vertices are
connected with an edge
if there's a non-zero transition
probability between them.
So with this picture in mind
of this weighted graph where
the vertices have weights
and the edges have weights,
then the conductance of
a set is the probability
of leaving some
subset of the graph
over the probability of being
in it in the first place.
So this phi of S
quantity, here we
have in the numerator we have
this sum over all the states
that are on the boundary of S
that also have an edge leading
outside of S.
So let's think about
this conductance.
If it's large, then you
have a high probability
of leaving the set S
compared to staying in it.
But if this quantity
is very, very small,
then that means that you
have a large denominator.
You have a high
probability of being in S,
but you have a small
numerator, a low probability
of ever leaving S.
So you can already
get the idea that this is
going to imply slow mixing,
because you're trapped in
the set S. And here, just
a technical note that pi of
S should be less than 1/2.
It shouldn't be the
full state space
because, if you're going to talk
about it getting trapped, well,
there have to be important
regions outside of the set S,
as well.
So anyway, if you
take the minimum
over all of these subsets
that are less than 1/2
of the weighted state space,
if you take this minimum,
then Cheeger's
inequality is a result
that relates this to the
spectral gap of the transition
matrix, and therefore to the
mixing time of the Markov
chain.
So essentially, what this
says is that the Markov chain
is rapidly mixing-- the mixing
time scales polynomially
with the system size-- if
the state space graph has
none of these bottlenecks.
Now, even the
conductance turns out
to be pretty hard to
work with in practice.
So we're going to go
one more step removed
and look at a technique
called canonical paths.
So what you do here is you
design paths such that,
between any two vertices,
they carry an amount
of probability traffic, pi x,
pi y, for those two vertices,
and you design these
paths in such a way
that they don't overload
any of the edge capacities.
That is, the probability of
being on that edge and actually
taking that transition.
So this leads us to a quantity
called the congestion, which
is the worst-case ratio of the
total traffic of paths passing
through an edge to the
capacity of that edge.
So in the denominator,
so we're taking
a maximum over edges
of this quantity
that we think of
as the capacity,
and then we're summing
over all the paths
that pass through that
edge of the traffic.
And we also need to account
for the length of the path.
And when you define
this quantity,
it was shown maybe
in the early '90s
that this can be used to bound
the mixing time of your Markov
chain.
So our proof is going
to be very much based
on these canonical paths.
So here's the idea.
The quantum Monte
Carlo configurations,
which do not spend too much of
the proportion of their world
lines sitting on the
region of the spike,
will have a stationary weight in
the spike distribution, the QMC
for the spike
system, that is very
close to their stationary
weight for the QMC
for the spikeless distribution.
So this leads to a
picture of the state space
where we have this large
region, omega g, omega good,
colored in yellow
of states where
the weight of the
configuration-- because OK,
think about it.
So if the spin configuration
never touches the spike,
then it's going to be
identical in the spike QMC
and the spikeless QMC.
So most of the state space,
actually these distributions
actually look quite similar.
But there are also these holes.
So omega b is really the
union of all these holes.
It's the bad region.
It's the bad region where
the QMC configuration spend
a lot of time on the spike,
and so their stationary weight
looks very different than
that of the spikeless case.
So the new proof
technique we used
is that we show that you can
use most of the canonical paths
from the spikeless
distribution to construct
a set of canonical paths
for the spiked distribution
with congestion
of the same order,
albeit it's inside this
subset omega g, which
doesn't have measure one, but
measure 1 minus 1 over poly
in n.
So in other words,
we're trying to cut off
a small part of the
state space that
isn't going to mess up
our expectation values,
but does make it very hard to
see that the simulated quantum
annealing chain would
mix in this region.
In fact, maybe it doesn't
mix in this region.
At some point, we believe
the adiabatic path
should be a necessary
inclusion to get
simulated quantum annealing is
QMC plus the adiabatic path.
And we need the adiabatic path
to stay inside the good subset,
omega g.
OK.
So now I'm going
to go through maybe
some technical calculation
stuff, or just talk it through.
So when we're talking about
heat bath world line updates,
so here, x1 bar through xn bar.
Think of these as the world
lines of the n qubits.
So of course, the quantum
Monte Carlo configurations
are these length L
sequences of n-bit strings,
but here x1 bar is just a
length L sequence of bits.
The world line of
that qubit x sub i.
OK, so how does a heat bath
world line update work?
I know a lot of you know this
because you've implemented it.
So you're going to select
one of the world lines
uniformly at random.
You're going to leave all of
the other world lines untouched.
And then you're
going to resample
the world line you selected.
You're going to delete
it and resample it
from the conditional
distribution that
is conditioned on all
the other spin values.
It's completely standard.
Now, the congestion
for this Markov chain
for the spikeless distribution--
all my tilde quantities are
the spikeless system--
the congestion turns out
to be order n squared.
And it's essentially because
the paths have n steps.
And in this step, where
you select a single i,
that means you have a 1 over
n probability of selecting
any particular world line.
So this bound, you can
already see, is not tight.
Because this is a
non-interacting spikeless
system, we know that
the heat bath world line
update should mix in time n.
I mean, the world
lines don't interact,
so as soon as you
touch a world line,
you never have to
come back to it.
You're sampling it straight from
the stationary distribution.
So this bound is not tight.
But the reason we
need to use it is
because it's the
tightest bound you
can get with the
canonical paths technique.
And our comparison method
doesn't port the mixing time.
It doesn't say the mixing
times are the same.
It says if you have a set
of paths for one of them,
then you can construct a
set of paths for the other.
So that's why we lose a little
bit of tightness on this bound.
Now, I also want to
discuss that, if we're
treating this as a non-tailored
meta-heuristic method,
then these heat bath world
line updates actually
take some non-trivial
time to implement.
In other words, without using
the symmetry of the problem.
So the way this
works-- well, first let
me say in words how we implement
heat bath world line updates.
So you've now deleted this
world line after selecting it.
You now have to look at
all the other transitions
in the rest of the system,
and break your world
line that you're updating
into these time segments.
And you also need
to introduce-- so
breaking it into all
of these time segments,
and then you compute
the effective potential
due to all the other spins.
That's what the conditional
distribution means.
Now, you also need to generate
some new jumps in the world
line that you're updating.
So let's take what
I said in words
and count up the factors
to get to this bound.
So we know with high
probability there will only
be order beta log n jumps
per any given world line.
And this is
essentially because it
follows a binomial
distribution with mean beta.
So my adiabatic transverse
field is just order one.
So that's why that
doesn't appear.
Otherwise, it would be
gamma times beta, where
that's the transverse field.
But you also have to
account for these deviations
away from the mean.
So that's why this
log n has to be there.
If it weren't there,
then we wouldn't get
the right expectation values.
Now, I said you count
up all the other jumps
in the rest of the
system to create
your effective
potential, whether it's
going to be n beta
log n segments there.
And then finally, as you're
going through and flipping
a weighted coin to decide
what these new blocks are,
you have to check all
the spin values in all
of the other world lines.
So again, you might
say, like, what
is the point of doing
things this way?
Why don't you take advantage
of some obvious speedups
you could get?
But we're really trying
to make a fair comparison.
And if you do that, then the
best you can get is order n
squared beta squared.
So now we're ready to
kind of put everything
together and sketch what
the whole proof looks like.
So the congestion of
the spikeless chain
is order n squared.
And the congestion of the spiked
chain is of the same order.
In this set, omega g.
So this implies that the mixing
time within the set omega g
is order n squared log pi min.
Now, pi min is just our
worst-case starting point
at each step of
the adiabatic path.
And in fact, here's how
we can analyze pi min.
So you know, if you're
familiar with QMC,
you know that the thing
you pay for most, the thing
that most lowers your
weight of your configuration
are having lots of
imaginary time jumps.
But because there can
only be order beta
log n jumps in any
particular world line, then
this sort of pi min scales this
way once you've taken the log.
Now, the other thing we
need is that it suffices
to take essentially this linear
schedule in order to fulfill
the warm start condition.
Let me just briefly
say what that is.
So you're running the
chain inside of omega g.
And if you start
very close to pi,
then you only have to run the
chain for a very short time
inside of omega g.
If you had to run the
chain for too long,
you would have too
high of a chance
of going into the bad subset.
So that's why you have
this sort of window--
not too short, not too long--
for which you can run the chain
and still stay in the
good subset, omega b.
And to do that, you
need steps at least
this small along
your adiabatic path.
And as we saw in
the previous slide,
one step of the world
line update chain
takes time order n
squared beta squared.
So finally, if you count all
these factors together and use
the fact that we actually have
a beta that scales with n--
and this was necessary in
order to sort of defeat
the density of states and be
close enough to the ground
state to make those
properties I said
about the distribution
being very close hold.
And so if you put all
these things together,
then you can get this
upper bound of order n
to the seventh run time.
And I think that's
essentially the best you
can do with a fair comparison.
You know, we haven't optimized
the adiabatic path or anything
like that.
We're just trying to
run it as a black box.
And I think that's
about as tight
as it's going to get with
this method of analysis.
OK.
So we've shown that
simulated quantum annealing
can inherit some of the
advantages of true quantum
annealing.
And this allows for a
provably exponential speedup
in the asymptotic
performance of-- oh yeah,
prove the exponential separation
in the asymptotic performance
of simulated annealing and
simulated quantum annealing.
Can we leverage the
ideas in the proof
to better understand the
stationary distribution
and convergence properties of
SQA and more general systems.
You notice a lot
of the techniques
that I sketch don't
actually use the bit
symmetry of the problem.
Really the best
thing that gave us
was that we were able to
solve the spikeless case
and we knew so much
about the wave functions.
Lastly, the future is bright.
There is a lot to do.
Thank you.
[CLAPPING]
SPEAKER: OK.
We have tough questions.
OK, Daniel, I think
I'll give you this.
AUDIENCE: So are
you able to relate
this speed-up to tunneling
or some other nice quantum
feature?
ELIZABETH CROSSON:
Interestingly, we did not
necessarily take that approach.
So the previous talk
did a very good job
of showing us what
the tunneling looked
like in terms of the
generation of this instanton.
But here, we had almost a
completely different set
of proof techniques.
So admittedly, we don't actually
see the tunneling necessarily.
But I mean, I think what
generalizes about this
is the fact that we had
a case we could solve,
the spikeless case,
and we had a case where
we introduced the spike, and it
didn't change the wave function
too much.
And so now that we've shown that
the simulated quantum annealing
distribution picks up not only
its mean values but its higher
moments from the quantum wave
function, that this situation,
if you have rapid mixing
in the spikeless case,
your base case, then in the case
that's not too different, now
you can have rapid
mixing there, as well.
So tunneling is
great, but that's not
the proof technique we used.
SPEAKER: OK.
Go ahead.
Go ahead.
AUDIENCE: Yes.
So, hello.
ELIZABETH CROSSON: Hi.
AUDIENCE: I admire
very much the ability
to do analytic bounds like this.
ELIZABETH CROSSON: Sure.
AUDIENCE: If I tried
to do that, I would not
be able to compete with you.
The best I can do is numerically
simulate things brute force.
And I've done that for
some of these problems
that you discussed,
and I find that when
you include the interaction
of the system with the bath--
ELIZABETH CROSSON: Yes.
AUDIENCE: --sometimes
we can get things
that are slower,
sometimes faster
than the adiabatic
theorem would tell us.
ELIZABETH CROSSON: Yes.
AUDIENCE: So in one
of your last slides,
you said that this is the
tightest that this bound is
going to get.
And I wonder if
you've also explored
what these bounds start to
look like when you-- I haven't
seen analytic bounds when
you have a system coupled
to a bath.
ELIZABETH CROSSON: Absolutely.
That's a good question.
Let me say a few things.
So when I say this
is the best I can do,
please let me qualify
that by saying I'm
very much following
this black box approach.
You must run it adiabatically.
You must converge
to equilibrium.
Once you start relaxing
those conditions,
you can solve these
problems in constant time.
You can just rush to the
end in a fully diabatic way.
Daniel's group has
this paper where
there's a diabatic
cascade that allows
you to solve these types of
problems in constant time.
So it's meant to
be a toy problem.
I mean, the value
of the toy problem
is to run our methods the same
way that we run them on more
complex problems, as well.
Now, as for finding a
theoretical analysis
of these diabatic transitions,
essentially the specialness
of stoquastic Hamiltonians
pertains to the ground state,
or the thermal state.
And once you leave these
states, well, you're
running into sign problems
all over the place,
and we might have to download
that open source package.
[LAUGHTER]
SPEAKER: OK.
Would you like to
ask a question?
Here you go.
AUDIENCE: Can you
generalize your technique
to treat other kind
of world line moves?
For example, world
line moves where
you open world lines,
which typically are more
efficient than the closed one.
ELIZABETH CROSSON: Absolutely.
So first of all, the
proof as written--
so when I started this work, I
didn't know that you were going
to discover how good open
boundary condition quantum
Monte Carlo are, or
some people in this room
have now found that.
That's a relatively
recent development.
But the same proof applies.
And I've also been
looking at things like you
have worm algorithms, right.
These are more for
condensed matter systems.
Yes, some of the same
proof techniques do apply.
I just want to say that when
you have heat bath world line
updates, these didn't appear in
the first version of the paper
because these
canonical paths, you
think of them flipping
one bit at a time.
Once you have these
hugely non-local updates,
it does kind of make the
analysis more difficult.
But yeah, I have some
projects in the works that
are looking at things
like the worm algorithm.
Yes.
SPEAKER: We'll have
just one more question.
AUDIENCE: I've got a mic.
Thanks anyway.
So I've got sort
of two questions.
One is just clarification.
So in your final
algorithm, you want
to quasi-equilibrate, not fully.
ELIZABETH CROSSON: Yes.
AUDIENCE: So that means you
don't want to run it too long.
ELIZABETH CROSSON: Yes.
AUDIENCE: So you've
put a lot of effort
into making sure
that you're just
treating this as a black box.
But how do you know
when to stop running it?
Does that depend on
the specific problem
and specific Hamiltonian?
ELIZABETH CROSSON: OK.
I mean, I think that's
a fair question.
And the most I can
answer there is
that when you've gotten into
the bad region, you know it.
You can see it immediately.
It's just defined as the time
you spend in the spike region.
So then you just
abort the algorithm.
AUDIENCE: With a black box, you
don't know that it has a spike.
I mean, you're sort of treating
it as a toy problem, of course.
You could just now
put the answer.
But how are you going to
know that you're going to be
in the bad region if you
don't use the specific form
of the Hamiltonian that you're--
ELIZABETH CROSSON: OK.
That is also a good question.
So I mean, I think that
similar to the number of steps
we need in the adiabatic
path, this is something
that if you have a
true, true black box,
you're going to need
to do multiple runs.
I mean, that's the
only way to see it.
So yeah, we do have to
be careful qualifying
the order into the seventh.
But I think it's a reasonably
black box approach.
AUDIENCE: And my slightly less
mean question was, so given
that you have put all
this effort into making
it relatively
black box-- I mean,
trying carefully not
to exploit that--
so can you then go back
the other way, or have you
thought about, say, OK,
take your proof technique,
and then just see
how far does it
generalize to
other distributions
where it will just
go through again?
ELIZABETH CROSSON:
Yeah, I mean I
think that's very interesting.
And in theory, the
machinery could work.
It doesn't depend
on bit symmetry.
But the question is, can
we get these examples.
So for example, these
weak cluster instances,
I thought a bit about that.
But it's very hard to
see how we can fully
understand the quantum
wave functions well enough
to apply these techniques.
So what I would
like to work towards
is making these techniques
a little less sensitive
to needing to know all these
details of the quantum system
so that we can move to less
exactly solvable systems.
AUDIENCE: But I guess you could
also start with any exactly
solvable system and
just-- essentially you're
sort of adding a perturbation
to something exactly solvable,
and showing that things
still go through.
ELIZABETH CROSSON: Essentially.
AUDIENCE: So you could start
with any exactly solvable,
you know.
ELIZABETH CROSSON: Yeah.
I mean, so we should be
careful about calling it
a perturbation, because the
spike term is tremendously
large when you're sitting on it.
But yes, one could
think about this.
And I think the other
important question
is would these exactly
solvable instances
be sort of prototypical of
interesting quantum annealing
problems.
If you know of any, then
sure, talk about it,
and we can maybe
apply some of this.
SPEAKER: OK, thank
you, Elizabeth.
ELIZABETH CROSSON: Thank you.
SPEAKER: Thank you so much.
[CLAPPING]
[MUSIC PLAYING]
