Quantum Computing for Computer Scientists

>>Welcome to Quantum Computing
for Computer Scientists. Happy Valentine’s Day. We will learn about
this most romantic of all subjects, Quantum Computing. This is a talk aimed
at Computer Scientists. We will not go over very much, if any, physics
during this talk. We won’t go like
for the Double-Slit Experiment or
Uncertainty Principle. We’re not learning
about any of that. We’re going to cover
the Computation Model which you’ll all be finding, fairly intuitive I think,
is a state machine. And we will run
a Quantum Algorithm on it, show that it outperforms
a classical Computation Model, and then we’ll end
with some demos, including a running program in Q-sharp, Microsoft’s
quantum language. This is the Gate Quantum
Computation Model. It looks a lot
like the classical Computation Model
where you have bits, and you send them
through the logic gates, and transformations
are applied to them. There’s another
Computation Model called the Quantum Annealing Model which is used by D-Wave,
if you’ve heard of that. That’s something
different from what we’ll be going over but people think they
might be equivalent. The jury is still out on that. Anyway, I’m sure you’ll
have a great time. So, I’m sure all of you have your reasons for being
here but here are three reasons I’ve come
up with if you need some additional convincing why you should learn
Quantum Computing. The first is that
quantum supremacy is expected this year,
this very year. Quantum supremacy means we have a real problem running on a real quantum computer
which in real-time, runs faster than
the same problem on a classical computer. So, Google has announced they think they’ll
do it this year. If they do, it’s
a really big thing. So, it might be in
your future. Who knows? Five, 10 years, we
could all be running some really tricky problems
on quantum computers. So, we have lots
of large companies, Microsoft, Google, Intel, IBM, they’re all investing
billions of dollars in quantum computer
developments. There are lots of
really exciting applications. This is the one that everyone’s heard about, Shor’s algorithm. When this paper came out,
it lit the world on fire. People were like, “Wow,
quantum computers could actually have real
economic impact.” We could factor RSA, and undermine our
global financial system, which is pretty cool. Here’s one that
programmers will I think appreciate which is we can search an unordered list
in square root N time. So, you think for
an unordered list, you have to check
every element. That’s ON. You can just get
a nice speed-up, just square root n queries
on a quantum computer, which is pretty, pretty
nifty, generally useful. And here’s the one
which, I think, probably is the best probability
of changing the world. Sorry, Programmers scroll and search probably won’t
[inaudible] too much. Which is, we might
be able to have an exponential speed-up in simulating quantum
mechanical systems. So, things like drug design, if you simulate like
biological molecules interacting, things like that, it could massively speed up a lot of our
biological research, and everyone talks about
nitrogen fixation for this. That’s like part of
the main Microsoft sales pitch. But this is the one that actually motivated
me, the bottom one, which it’s, I think, from talking to couple of
you before the presentation, we can also motivate you. It’s intellectually
interesting, and it’s interesting because it’s just kind of outside my intuition. Like I think all of us, at
this point in our careers, can basically look at any
digital or mechanical system, and have like a ballpark idea
of how it works, right? But if you look at
a quantum computer, like how can a quantum computer outperform classical
computation? It doesn’t make any sense. Like there’s no way I could even start to
guess how it would work. And so, I really
wanted to learn it. And I think there’s a reason
for this, and this is kind of getting
a bit philosophical. But our language, our informal
language that we use developed in
very classical world. It is simply not equipped to
deal with the quantum world, and this is why
any pop science article you’ve read on
quantum phenomena, it doesn’t really ring true. All the metaphors,
they don’t really make sense because we’re trying to express it in
this language which is developed in
the classical world. We need to learn a new language which is
the language of mathematics, and that is the only thing
which will actually let us understand quantum mechanics is to learn it mathematically. All metaphors and analogies
will lead you astray. There’s a famous quote
here by a physicist named David Mermin which
is “Shut up and calculate”, with regard to
quantum mechanics. So, some grad students, maybe they lean back
in their chair and they’re like, “Oh, what
does this all mean?” Shut up and calculate.
Just trust the math. Math is the only thing which
will not lead you astray. Okay, here’s how this
presentation will break down. First, we’re going
to learn how to represent computation using
super basic linear algebra, matrices, vectors,
matrix multiplication. Then we will generalize
that to learn about Qubits, superposition, and
quantum logic gates. And finally, we will use
all those tools we’ve developed to go over the simplest problem where quantum computer
outperforms a classical computer. This is called the
Deutsche Oracle problem. I mean, there are
a bunch of problems, but this is the simplest. And finally, it’ll be almost
negligent to let you get out of here without learning
about Quantum Entanglement and Teleportation just because
with the tools we’ll develop, they’re so easy to understand. So, I’ll have bonus topics, and then we’ll have some demos. All right, let’s get to it. You’ll become very, very
familiar with these two vectors. This is how we
represent a 0 and a 1, the two values of one bit
in classical terms. So, the top one, it’s just 1 over 0, and
the bottom one is just 0 over 1. You can also write them in that weird angle bracket thing. That’s called
Dirac Vector Notation. So, you have a zero
in there. That means this is the value
of that vector. If you have trouble
remembering this, just think of it like
an array indexed from 0, and so, there’s a 1 in the 0th index,
that means it’s a zero. There’s a 1 in the first index,
that means it’s a 1. Okay, everyone’s fine with this? You’ll be very
familiar with this by the end of the presentation. We’re going to go over matrix multiplication
really quickly. So, the way to think about
matrix multiplication, just to review for you. Most of you will have
gone over this in your Linear Algebra courses
the first or second day. So, if we have a matrix
multiplied by a vector, you kind of take
this horizontal row, and then flip it,
and multiply it point-wise with
these vector values. So, a times x, b times y, and that’s how
you get the top one. And you do the same thing
in the bottom row. C times x, d times y, and you get a two vector. The same thing happens
with three-by-three matrix. A times x, b times y, c times z, and you
get this matrix. This vector, you can also
multiply matrices together. We won’t really be
doing much of that. And there’s a very
special matrix called the Identity Matrix which
has zeros everywhere, except for ones along
the main diagonal, and anything multiplied by
the Identity Matrix is itself. It’s just like multiplying by 1. In fact, usually, you refer to the Identity Matrix
by the symbol 1. Now thankfully, for simplicity, a lot of the matrices we’ll be using look a lot like this. They’re mostly 0s
with just some 1s, and you can notice, it’s
a nice rule of thumb, that relative to
the identity matrix, this matrix has the middle
two rows flipped, okay? And that’s actually
the action it has on anything it multiplies. So, it flips
the middle two rows. It’s a nice sort
of rule of thumb to remember how
matrix multiplication works. Does anyone have
any trouble with this? Because you really need to know this in order to go ahead. You got to know it. Okay, great. Operations are actually, I’d
like to make this a quiz. Can anyone tell me what are the four operations on
one bit? There’s four of them.>>One bit.>>One bit.>>AND, OR, exclusive-or.>>That’s around two bits.>>Not identity.>>Identity not.>>Set to 0, set to 1.>>Set 0, set 1. There we go. Good. Yes. So, we have identity, just f of x equals x. Negation fx equals not x. Constant-0, it’s
always set to 0. Constant-1, it’s
always set to 1. You can see with a diagram
how it kind of looks, okay? And we can write
these as matrices. So, the identity is
just the Identity Matrix. This, remember, is
our symbol for zero. Identity multiplied by 0 is 0. Identity multiplied by 1 is 1. Here’s negation. You flip
it from 0 to 1, and 1 to 0. Same thing for
constant-0, constant-1. So, I hope I’ve
convinced you we can represent very simple
functions at least, with matrices multiplied by
our 0 and 1 vectors, okay? And remember
these four functions because I will quiz
you later on them. They’ll come back up. So, let’s talk about Reversible Computing. Reversible Computing is
a neat sort of buzzword. It if you’ve read the popular
sci-fi book Accelerando, it talks about
Reversible Computing, and basically it means
that if I tell you the operation I did and
the output of that operation, you can always tell
the input of that operation, if the operation’s reversible. So intuitively, operations
that just shuffle bits around, like permute them,
they are reversible. But operations that erase bits and then overwrite
them are not reversible. And in our previous slide, identity and negation are both reversible because I
say the output is one, and I put it through
the negation gate. That means the input was zero.
You can always tell that. But if I say the output was 1, and I put it through
the constant-1 operation, you can’t tell
what the input was. It’s not reversible.
It erases information. And the nice thing
about Quantum Computing, which you’re going to remember, is that one of the computers only use
reversible operations. Those are the only ones
we care about. So, the only operation we actually care about on
this slide is really negation. Like also identity but it’s
the same as doing nothing. A further fun fact about
chronic computers is they actually only use operations which are
their own inverses. So, not only are they
reversible but if you apply them twice you’ll just get
back the same input value. And we’re going
to go in a little, this is kind of like going
into pop science but whatever. There’s something called
the Von Neumann Vander Limit, which is the smallest amount of energy physicists
have calculated is necessary for the simplest
possible calculation, which raises is non-reversible. And reversible computing might eventually lead us
go beyond that limit, more efficient than
that theoretical limit. Now, currently, processors needs millions of times more energy than the Von Neumann
Vander Limit but, you know, 5,000 years
from now, who knows? Pretty nifty. All right. This is something you
probably did not go over in linear algebra but don’t worry
it’s still pretty simple. It’s called the tensor product. So if you have the tensor
product of two vectors, it’s kind of like you take
the second vector and you tile it for each element
in the first vector. So, X0 gets its own copy
and X1 gets its own copy. And then, you multiply it
out so you get eventually X0, Y0, X0, Y1, et cetera. It’s maybe easier to look
at this with real numbers. Here we have one times three, one times four, two times three, and two times four to get this. Here’s how it looks
like with three vectors. You get this like giant array. Don’t worry about it too much. And I’d like to point out that if we use R0 and one values you end up with a vector which has a one in
only a single position. Okay. Does anyone have any real trouble
with this concept? If I gave you
two vectors you could probably calculate
the tensor product. It’s not entirely
difficult. All right. But, the final point
animate leads into how do we represent
multiple classical bits? And we represent them
by their tensor product. So, 00 remember we tensor,
this is a zero symbol. 0 tensor with 0 is this, product state, we call it, 01 is this product state
10 is this product state, like 0 times 1, 0, 0 time 0, 0, 1 time 1, 1. One times 0, 0. Okay. And 11 is this product state
and similar to just a single bit
you can think of this is an index in
this vector array. So, this is 0. There’s a one at the zero position.
This is one. There’s value with
the first position coming from zero and, this is two value of
the second position, three value of
the third position. This is the only time
I will tenser three bits together thankfully,
because they us humus. They grow exponentially in
size but it works for this. So, four is 100 you have
01234 is where the one bit is. Okay. Yeah. So, we call
this the product state. We can always factor it out into its individual states and the product state of n bits is a vector of size
of two to the n. Here, we see a sort of whisper of the power of
quantum computing in this exponential term but hold that thought,
we’ll get back to it. So, we’re going to learn
a very important operation, very fundamental to reversible computing
and quantum computing is called CNOT or
conditional not. It operates on a pair of bits. You designate one bit, the control, bit the other bit
is the target bit. So, the control bit if it’s
one it flips the target bit. If it’s zero it
leaves the target bit alone and the control bit
is always left alone. So, if we have
the most significant bit of a 2-bit system as control, the least significant
bit is target, then 00 since this is
the most significant bit, it’s the target bit, and just
goes to 00, 01 goes to 01. Now, 10, since
the control bit is one maps to 11 because
it flips the other bit, and 11 since it again flips, the target bit goes to 10. And this matrix, you can
actually kind of write it, you see there’s a sort
of correspondence here we flip
the bottom two rows, which is the identity matrix, and this I claim which I will show in
the next couple slides, will change our product states according to
the semantics of CNOT. So, does everyone get
the semantics here? We have a control bit
and a target bit, control bit one, we flipped
the other bit. Pretty good? Now, this is a lot of math but don’t panic we’re
gonna go over step-by-step. If we apply the CNOT gate to 10, remember that this is
the control bits one, so we’re going to
flip the other bits, so we expect it to go to 11. Now, let’s expand this
out just step-by-step. So,10 corresponds to
this tensor product. This is the matrix
from the previous slide and we’re applying it
to our product state. Remember, it’s
just going to flip the bottom two rows so we
get to this product state, which when we factor it out, indeed gets us 11. Now, let’s do it
with 11, same thing, we have our familiar matrix, multiply by this product state, we go to here, which
we factor into 10. So, I hope I’ve convinced
you at this point that we can use matrices to represent more interesting logical
operations than just a bit flip. Has anyone eventually, yes?>>Is that factoring always possible or practical
for larger vectors?>>It is practical, yes. Not always possible, which we’ll get into that when we get into
quantum entanglement, basically, but good question. So, this is fine for everyone? Nobody’s lost in
the gigantic equations here? Okay. We’re gonna go over
10 then, or sorry, 00. So, the control bit 0 that means we’re going to leave the other bit alone, right? And we see that indeed this is what we do
and same with for 01. Okay. So, everyone, well, maybe familiar with
how classical computers are like real CPU’s there they’re all
built on the, sort of, NAND gate, like everything is
fundamentally built on NAND and CNOT is kind of the analogous NAND for
reversible computing. It’s like a really basic thing
that is used to build up larger and more complicated
logical statements. So, I’m sure if I
gave you a task, like build some more complicated logical formula with CNOT, you’d probably be able
to do it at this point, you just like stick
them all together and eventually comes out. I will note that it’s
not actually universal. You can’t make every logical
function with the CNOT gate. You need something
called a Toffoli Gate, but we won’t see that
in this presentation. And we’ve learned everything
we need to know about how to use linear algebra to represent
classical computation, and we can now go on
to quantum computing. So, to recap, we learned that we can
represent classical bits in vector form is 10
for 0 and 01 for 1. We can represent
operations on bits by multiplications
on their bit vectors. We know that quantum
computers only use reversible operations and in fact the operations
are their own inverses and multibit states, we write is the tensor product
of our single bit states. Finally, learned
about CNOT which is a very fundamental gate in reversible computing
and quantum computing. So, everyone’s
following along so far? This is pretty simple,
right? And it’s not that bad. Okay. Let’s make
the jump, qbits. And I will surprise
you, we’ve been learned using qbits all along actually. The cbits are just
special cases of qbits. And we represent
the qbit in general by a vector of two elements a, b where a and b are
complex numbers. Don’t worry. We’ll only use real numbers for
this presentation. We’re keeping it simple. Only using real
numbers. And we have the constraint that
a squared plus b squared is one. So, (1,0) and (0,1) fit
this definition because one squared plus zero squared is one and zero squared plus
one squared is one, right? Now, here are
some example qbit values. This is one with which
you’ll become very familiar. It is one over root
two, one over root two. Can anyone tell me what one
over root two squared is?>>One half.>>One half exactly. one half plus one half is
one. So, this is a qbit. What about one half and
root three over two? What’s one half
squared? One quarter. So, one quarter plus
three quarters is one. Here’s an interesting thing. We can actually use
negative numbers now because the negative number squared is a positive number. So, negative one squared is one. One plus zero is one.
And the same thing here, it’s just the same
as this first matrix, but one of the values is
negative. Pretty neat. Not bad so far, right? Okay, but you might say
this is kind of like saying that a qbit has a value
which is not zero or one, but actually both zero
and one at the same time. This is called superposition. Everyone’s probably heard of the Schrodinger’s cat
being both alive and dead. That’s just like a pop science the way of articulating
this concept sort of. And the interesting thing about a qbit being
in superposition, it means we can kind of, I want to really qualify
this kind of compute with the values of both zero
and one at the same time, which again is the sort of whisper of some
quantum speedup power. Okay? Now, how this works is, when we measure this qbit, it collapses to
our familiar values of zero and one with
some probability. And this is what we
usually do at the end of a quantum computation
to get the final result. We send it through
the measurement gate, we see what the result is. So, how the
probability works is, if you have a value (a, b), it collapses to zero
with probability a squared mand one with
probability b squared. So, if we look at our familiar qbit from previous
slide, one over root two, one over root two, it has a one half chance of
collapsing to zero, and one half chance of
collapsing into one. It’s like a coin-flip. Now, if we just measure
our classical qbit, our classical cbits the (1,0) has a 100 percent chance
of collapsing to zero, and (0, 1) has a 100 percent
chance collapse into one. So, it’s sort of item
put in that way. And I found it useful here to give people sort of physical intuition about
what we’re doing, what bit is or a qbit is. If you’ve heard of polarization, you’ve probably bought
fancy sunglasses, says that they’re polarized. So, polarization is actually
sort of quantum phenomenon. You can be polarized like
this way or this way, and your glasses will only
be polarized a certain way so only let light polarized in this way
through basically. It turns out that qbits, you can think of this, yeah, a photon being polarized
this way is a zero, and photon being polarized
this way is a one. Superposition means it’s
actually like both, right? So, it could actually go
through both gradients. And then if you measure it, and a way to measure
it is just to fire it through grating and
see whether it goes through, then you can collapse
it to zero or one. So, that’s our intuition pinning this to
a physical concept. Of course, modern, we found that photons
actually make very, very poor qbits, they tend
to collapse to easily. So, we don’t use those, but
you can maybe think of that. Does anyone have
any confusion here, really, about how this measurement process collapse
probability works? Okay, great. We will learn about
multiple qbits. The same is multiple cbits. We represent them by
the tensor product. Why is it doing that? Okay, hold on, laser pointer. Okay, there we go.
Nope, that’s not. Okay. So, if we have two qbits,
and we multiply them, the sum of squares identity
actually is maintained. And so, if we consider our perfectly balanced 50
percent coin-flip qbit. And we tensor with
another coin-flip qbit, we get this four vector with one half in every single column, and one half squared is
of course, one-quarter. So, one-quarter plus one-quarter plus one-quarter plus
one-quarter is one. This maintains the property. And the way to read this is when you measure
these two qbits, it has a one-quarter chance
of collapsing to (0, 0), (0, 1), (1, 0), and (1,1). This is sort of the probability
of when you measure it, finding a one there, and zero everywhere else. So, we could find, it’s a one-quarter probability
of finding one in this position and
all zeroes everywhere else. Does that make sense
for multiple qbits? Okay great. You guys are just trucking
along and no problem. Should be super easy. Yes?>>I have a question.>>Absolutely.>>So, there’s only
going to be one, one left in the vector?>>Yes.>>There couldn’t be two ones?>>There could not two ones because that would
actually violate our a squared plus b squared
equals one constraint.>>Okay.>>But yeah, after measurement
there’ll only be one, one here in this in four vector, which we can then factor
into two bits actually. Okay. Yes?>>You also find collapsing
for that longer vector?>>Yeah, yeah. So, if you measure
these two qbits at once, you can think of this product
state vector is collapsing. And then that’s
just the probability of finding a one here and
a zero everywhere else. It’ll become useful in
the future when we can’t actually factor out the product
state, but that’s anyway. Okay. How about
operations on qbits? We have these qbits, how do
we actually manipulate them? The same way we do with
the classical bits. And all the matrix
operators we’ve learned also work with qbits. So, for example, negation. If we have our qbit that has a 25 percent chance
of collapsing to zero, 75 percent chance
collapse into one, if we run it through
the bit flip operator, now it has a 75 percent chance
of collapse into zero, and 25 percent chance
of collapse into one. So, we can manipulate
these probabilities actually called amplitudes with gates. And you can think of gates, they correspond to some
device scientists have created which is like manipulate
the qbits with a gentle touch. Not enough to collapse them, but enough to
change their state. So, pretty neat. And there are also a couple
important matrix operators, which we haven’t learned
yet because they only really make sense in
the quantum context. Okay, all right. We’re good. Then the most important one
is the Hadamard gate. And what it does is, it
takes a zero or one qbit, and it takes it to
the coin-flip state where it’s in exactly
equal superposition. The matrix looks like this. It as one over root two in every single entry except for a negative in
the bottom right corner. So, if you multiply
that by this, you get one over root
two, one over root two. If you multiply this by one, we get one over root two,
negative one over two. Pretty neat. So, this is how we actually get into superposition. We use the Hadamard device, which scientists have developed, and we can make much use of. So, quiz time. Can anyone tell me why
we need a negative one in the bottom right corner? Why do we need
a negative sign here?>>It’s like a matrix
on bigger provision. I don’t know exactly why, but it is too big.>>Well, the answer basically
has to be reversible. So, if we didn’t have
this negative number here, then both zero and one
would map to the same value, and it would not be reversible. So, we need this negative sign.
Yes, correct. Okay.>>Question.>>Yes?>>Review of superposition, what does that mean exactly?>>Superposition, yeah. So, superposition
means that a qbit is in both states zero and
one at the same time.>>Okay.>>And I want to be really
clear about something here. This doesn’t mean
the qbit is secretly zero or secretly one,
and we don’t know. It means it’s actually
in both at the same time, this is just quantum weirdness. This is just like what
happens at the very, very small level of our world, which is that things don’t have definite values until
we measure them.>>Here you have (0, 1) in superposition
is one over root two, one over root two. And one and superposition
one over root two minus one.>>Yes.>>So, we seem to have different values but they’re representing conceptually
the same thing.>>Right. So, this comes down to the distinction between the quantum
and classical world,. Which is that from
the classical perspective, these two qbits are the same. Both have 50 percent probability of collapsing to zero or one. But, if we stay within
the quantum realm, this sign information is preserved as long as
we don’t measure it, and it can affect
our computations.>>So, it’s got a 50-50 chance, but it has a sense that
it used to be a one?>>Has a sense that
it used to be one? Not really, it just
ends up in this state, which is different within the quantum realm
than this state. These are two different states.>>Okay.>>Yes.>>Both for the 50-50 chance.>>Yes, exactly. Both when you measure it and convert it into the classical world come
down with a 50-50 chance. Yes. But they are different, and they have different
computational properties. Okay. Now, I’m going
to show you something really cool about
the Hadamard gate, which actually takes us out of superposition into
the classical bits. This should be unsurprising. Everything is
its own inverse, right? If we just apply the Hadamard
gate to our value here, our coin-flip
value, we get zero. And for this one, we get one. So, this suggests to us
a quantum computation structure. We can start with
our classical bit values. We put them into superposition, do a bunch of
quantum stuff to them, and at the end if we’re clever, we can transition
them to zero or one so that our whole computation
was deterministic. Pretty neat, right? Okay. Now, I should say that not all
algorithms work this way. There are algorithms
like Shor’s algorithm, which only gives
the right answer of 50 percent of the time. You can’t be clever and always get the right answers,
just not doable. But the one, the problem
we will look at does give this
deterministic property. So, does anyone have
trouble with this concept? We transition out
of superposition without measurement,
pretty interesting.>>Can you go over
this transition from a classical computation to quantum and then
converting it back?>>Yes.>>That’s correct.
Let’s say two plus two, how would that look like in?>>How-?>>Just the simplest
type of math.>>The simplest type of math. What we’re going to go
over that basically, we’re going to go
over a problem. There are certain problems
where computers are better than classical.
Addition is not one of them. It works the exact
same way in both. It should also be said that
all classical computation can be done on
a quantum computer, all you do is just restrict all the values to zero and one, and it works, but it’s kind of a waste because
you can also use all the extra values
that we found. Okay, and we’ve actually sort of define
a state machine here, which I am sure many
of you will appreciate if you’re software engineers
or computer scientists. This is the unit circle which many of you will remember from high school trigonometry, right? It’s a circle of radius
one around the origin. Here’s the X axis, this is the Y axis. You can think of
the top value is the value of the x-value and
the bottom one is the y-value so our zero value
is at position one, zero which is along the X axis. Our one value is position zero, one which is along the Y axis. And this is
the bit flip operator, it defines transitions
around this unit circle. So zero goes to one,
one goes to zero, zero negative one
goes to negative one, zero which when
you collapse them, this is zero, this is
one, and vice versa. And if we look at the superposition values, here’s
something interesting. The bit flip operator really kind of has
no effect on them. For this one, if you
flip it upside down it’s stays itself same
with this over here. This one’s kind of interesting
and that it transitions you all the way
across the unit circle and we’ll make use of that. But yeah, this is pretty much the full action of the bit flip operator on
the unit circle. I should mention here that if we were using complex numbers, this would actually be a sphere. Much more difficult to sort
of intuitively graphically represent so we’re sticking with real numbers, they
suit us just fine. But it’s nice to know that
additional dimension is always available to us if we want to make our computation
even more powerful. And then here’s the action
of the Hadamard gate, which we saw in
the past couple slides. It takes us from
zero to one over two, one over two, one, two, one over two negative
one over two, and kind of symmetrically, it also works in
this way over here. So, we have a nice state machine
that we’ve developed and now we can start running
things on the state machine. Very nice intuitive
sort of representation, you don’t need to do matrix
multiplication all the time. Everyone kind of gets this? Okay, great. So here’s an example of something we could run through
the state machine. Here, I’m introducing
something called quantum circuit notation. You can think of
the cubit traveling along this line has these
operations applied to it, is it goes through the boxes. So X is bit flip, H is Haddamard, and then
so we go a bit flip, Haddamard, and
bit flip, Haddamard, bit flip and if
we start out here, we go bit flip, Haddamard, bit flip, Haddamard, bit flip. So we start off at one, zero and get to zero, one, we got to cross the circle. And note, this is
reversible so we can also go this way and get back. So this is kind of how we’ll be doing all of our computations
like this basically, just hopping around
the unit circle. It’s pretty simple,
pretty intuitive, right? It’s just a state machine. We use this in classical
computing all the time. Just has a bit of
weird rules with the claps and whatever
but don’t worry. Okay, now we have everything we need to learn
the simplest problem where a quantum computer outperforms
a classical computer. We learn that Cbits are
just a special case of Qbits, and Qbits are two vectors of complex numbers such that a squared plus b
squared equals one. Within the Qbits is going
to be in superposition, they’re probabilistically
collapsed to Cbits by measurement. Multi-Qbits systems we
learned are tensor products of single Qbit systems
same as with Cbits. Matrices, again, represent
operations on Qbits, and we learn about
the Hadamard gate, very important which takes our zero and one bits to
superposition and back. And finally, we learned that we can think of Qbits
in their operations is forming the state machine
on the unit circle or unit sphere if we’re
using complex numbers. It’s actually called
the bloch sphere, if you want to look that up
for the full complex numbers, but unit circle,
just fine for us. Okay, so we’re gonna learn about something
called the Deutsch oracle. This is originally proposed
by David Deutsch in, was it like 83 or
something like that? And the problem is this. Imagine I show up at your doorstep and I
give you a package, the package is
just this black box, it’s just horribly present. It’s a black box which
has a function on one bit. There’s one of
those four functions on one bit, do you
remember what those are? I said I would ask you
about them. Set zero, set one, identity, indication. Exactly. So one of
those four functions, I don’t tell you which one
is inside this black box. There’s a wire going in and
a wire going out so you can send in values see
where it goes out, but you can actually look
inside the black box, okay? So how many queries on a classical computer
would it take to figure out which function is inside
the black box? Two, exactly. You send in zero,
see what comes out, you send in one,
see what comes out. So that will uniquely
identify which function it is. Now, how many queries
do you think it takes on a quantum computer. You’re wrong, it’s two. I say this to illustrate
an important point, which is quantum computers, they don’t really compute
with all value simultaneously. At the end of
the day, you collapse your Qbit to a single bit
of information, and a single bit of
information is not enough to uniquely identify one of
four functions, right? You need two bits at least So this actually takes two queries
on a quantum computer. Well I’m not too
exciting so far. Now, what if instead of not wanting to know
which function it is, we just want to know
whether the function is constant or variable. The constant functions are, constant zero, constant one, they’re always mapped to
0, always mapped to one. The variable functions
our identity and negation. So how many queries would this take on a classical computer?>>Two.>>Two. You put in zero, if you get zero you
don’t know whether it’s identity or constant zero. If you get one, you don’t
know whether it’s bit flip or constant one, symmetrically with
one, so it always takes two queries on
a classical computer. Even though there
are two categories, so really we only
need a single bit of information to tell us
which category it’s in. But how many queries do you think it takes
on a quantum computer? I kind of burned you last time
so maybe you’re a bit shy.>>I’m going for a one. One, you are correct. We can do this in a single query
on a quantum computer. We can tell whether
its constants or variable. This like undeniably outperforms a classical computer and
we’re going to learn how. Okay, we do with the magic
of superposition. But first we have to define which each of
those four functions look like on a quantum computer. We have an instant
obvious problem with the constant functions. Can anyone tell me what it is?>>Not reversible.>>They’re not reversible, how are we going to write these non-reversible functions
on a reversible computer? This is actually a really common problem in quantum computation. You really often
have to deal with non-reversible functions if you have to write them in
an irreversible way, and the hack we do is this. So originally we
had this, right. We add an input and an output, a single wire going through. And we just imagine
that you put in the value and you
get the function applied to that value, right? But the hack that you always
do on quantum computers is you add an
additional output Qbit. So you actually have two Qbits. We have to rewire our black box. So you have your input
Qbit, which is unchanged, and then the value of
the function on the input Qbit is written to
the output Qbit. Does everyone kind of get this. I think this the thing
that people have the most trouble with,
this is necessity. The reason we have to do this, is so we can write
non-reversible functions in a reversible way. And the way you do that is to have a separate input
and output qubit. So when you see
larger quantum computations, instead of having all the bits go in and have transformations applied to them and
then being measured. You usually have
a separate set of output bits, so the input bits go in, they have their value sort of
written to the output bits, and you measure the output bits. This is a very standard way
of doing quantum computation. So does anyone have
any trouble with this? This is a very difficult
concept. So, yes?>>I didn’t get it.>>Okay. So, do you understand
this model basically, this is what we have now, okay? So we need to rewire it so there are two wires
going through the box now. And one of those wires
we call the input wire, one of those wires we
call the output wire. We assume the output wires just initialize to zero,right?
It’s always zero. Now our input wire is what we
actually give our input in. Like it’d be like zero or
one or whatever value you want to give
input to the function. So, it goes to the box,
it’s just unchanged. We don’t touch the value
of the input wire, but we calculate
the function in here, on this input and then we write its value to
the output wire, okay?>>Black box?>>Yes, we are
rewiring the black box. We have to rewire
the black box in order for the black box
to really work. So the black box I gave
you has to be rewired. Yes.>>Did you use the first input, which is the output of zero?>>Did I use the first input?>>Yeah, because one of them
was kind of identity, right? You put the X out, and
then it was the F or X.>>One of them was
X and the other was>>You need the X out.>>Say we only have this, right? This is
the old way of doing it. We can’t have a non-reversible
function in here, right? For it to be on
the quantum field.>>No, that I get, right?>>Okay, yes.>>Because when that thing
is on the right side, you have two going
in, two going out.>>Yes, two going in,
two going out, yes.>>I get why two going out.>>Yes.>>I don’t get why we have
that zero is also going in, It doesn’t seem that
you’re actually using it to produce any of the others.>>You’re correct,
it’s just we need.>>Symmetry?>>Not really symmetry, it’s just like how we would
write the quantum circuit, we would need
two qubits in general. So, you’re correct that this algorithm value
isn’t really used.>>Is it like we
don’t care a pen off?>>No, we do want it to
be zero because the way that we write
the circuit assumes that this value coming in
is always zero. Yes.>>Why is it ingoing
output wire called output.>>What? Yeah. So, this is
kind of weird that the output of the input
is called input. But, we basically only really
look at this value here, is the basic idea. This is like the real output. The input will
always be unchanged.>>Question. So if I have multis two black
boxes in series, the upper input which
you have labeled Alvap.>>That’s smart,
that’s a good one.>>And the second stage
would be the F of X output of the first stage?>>Okay, so you’re
saying if you had these in series
rather than sort of in parallel to what we’re doing
is that what you’resaying?>>Yeah I want to
do two operations one and then another one. And the black box is
doing some operation.>>The black box is
doing some operation. You can have as many
operations as you want in here inside
this black box basically.>>True, but I’m trying to conceptually
understand whether I would take the F of X is
the output upper right. And that will be the input.>>The input, the next one.>>Absolutely and in fact I said that this black-box
assumes this value is zero, this isn’t maybe throw
in a bit much of you, but actually we can break
that assumption and send it some other value which maybe will get us more information which is in fact
what we’ll be doing. So, the working of black box assumes that the
output input is zero, but we can break
that assumption and be tricky.>>My way of
thinking was I didn’t have any other value
for the first stage, but for subsequent inputs
I do have one. So I could use it if I want?>>Yeah you could, yeah and you might get some interesting
values which actually well. Okay I swear, this is the most difficult part conceptually of
this entire presentation. So if anyone has any trouble
with it please let me know. Ask any questions you can spend as longer time on
this as we want. Yes>>As we go ahead, is there some additional clarifications
because I didn’t get it.>>Okay. Well we’ll
go over all each of the four functions in this way,
so maybe that will help. So, for example, constant
zero, here’s how it would look. X goes in, X goes
out it’s unchanged. Now, the output is always
going to be zero, right? That’s just what
the constant zero does. So we always want this value in the upper right to be zero. If we assume zero goes in, this is our circuit diagram, neither of the two wires make
any modifications, right? It’s a straight wire. Does that sort of makes sense? We will see the next slide, what about for constant one? We want one to go
out here, right? This is our function
output value. And we assume zero
is the input here. All we do is we do
a bit flip right here, and that gets us
the semantics we want. Is anyone having any trouble
with this so far? I swear this is
the most difficult part. So if you get this
you’re golden. Yeah.>>So these labels
output, input, would you call them something like computations and its spare?>>Sure yeah, you could. You can call it the
compute and the spare fit. Yeah, that also
works.They might reduce some confusion about there being an open output and input output.>>Why is it called output?>>Why is it called output? We call this the upper
qubit because it’s just the place that we write
the output of the function. But we could call
it like compute and spare or something like
that, like he suggested. Like this would be
the spare sort of bit that we write the function number 2. Yes>>We need this to do
a reversible computation, right why is this?>>Reversible?>>Can we compare
the reversibility of this likewise the right
one reversible one?>>Absolutely, okay. So let’s consider this one
right here, right? We have our black box with a single wire and say we want to do
the bit flip operator. So, that would just be. Sorry not with
bit flip operator, a bit flip operator
is reversible. So we want to do
the constant zero operator. We always want to
map this to zero. So you need some operation right here, that would
map this to zero. But it’ll be the matrix 1100 but this is not a valid operator
in a quantum computer, because it’s not reversible. So, does that kinda makes sense? It’s just we can’t construct
a device which does this. It’s just not permitted
by physical reality. All this math and
stuff I’m showing you, it’s not like special it’s just, we found that it corresponds to the semantics and workings
of a quantum system, and so it’s a nice way to
reason about it basically. Okay. Of course if we want to write constant
zero here with their two inputs are the two wire model it
would just be blank, right? Or if we want to do this is constant one always maps to one we saw that this
has an X gate here. And this is reversible, right? Okay. So we managed to write a non-reversible function
in a reversible way. If you’ve got
this congratulations, this is the trickiest part. It took me awhile to
figure out why we needed to do this staring at the textbook. Does anyone have
any confusion here? Okay. So remember constants
zero is just blank. Constant one has
a single X gate. Yes.>>There’s a point
you always end up with a four by four reversible matrix by doing this in black box?>>Yes. Black box
is always a four by four reversible
matrix applied to a product state, you’re correct. Although we will be not going into the math in that way
we’ll just be using the state machine to look at it, but the math is very easy
for you to do it yourself. Okay. So let’s look at
identity this is a bit more complicated this is
the CNOT gate, okay? Where this is the control,
and this is the target. So we’ll end up with, so input again is unchanged. We want output we
assume it zero we want to end up with
a value X, right? So if zero is sent in here, then this is the control bit, so the target bit is
just left unchanged, right? So zero comes out. If one is here, one is sent in, and the control bits
one so we flip thright? Is bit, and we end up with one. Does this make sense?
This is kind of tricky it is a big step up
in complexity of our circuit, but everyone’s fine with this? Wow okay. Can anyone guess
how will do negation?>>You can hardset
the control bit to one.>>How can we hardset
control bit to one?>>Two controlled NOTS. They’re.>>Not two controlled NOTS, we just have an X gate there. So, up to here it’s
equivalent to identity, right? And then we just flip the output bit
and then we got NOT X. So, those are
the four circuit values. Okay. Everyone’s good so far? Everyone knows how we
write our four functions? Are you ready to
see how we solve it on a quantum
computer? We use this. So, we’re going to
break the assumption that this value goes into
the black-box as zero. We put, first, we initialize both in putting up
a qubits to zero. We bit flip them so
now they’re both one. We put them through the
Hadamard gate to put them into equal superposition then we
send them into the black-box. One of those four
circuits is applied. Then we do the post-processing, we send them through
the Hadamard gate again and we measure them.
This is measurement. And I claim, and I will
show it to you that if the black-box function
is constant the system will be in state
one-one after measurement. Like this will have value
one, this will have value one. We’re using input as the most
significant bit by the way. And if the black-box function is variable and the system will be in state zero-one
after measurements. So this will be zero,
this will be one. And we will know whether
the function is constant or variable in a single query. Pretty nifty, right? I’m going
to show you how it’s done. First the preprocessing. Remember, we start
with zero then we’d been flipped both in
the Hadamard gate both. So we start here, bit flip
takes us up to here, Hadamard gate down here. So this is the value that
is sent into the black-box. Just remember it’s in the bottom right
on the unit circle. This is the value
right before we send it through one of
those four circuits. Okay? Now, let’s look
at how we do constant-0. So constant-0 remember,
it’s just blank on both and the post-processing
is a single Hadamard gate. So since no gates are
applied they stay in this bottom right value, and since the Hadamard gate is applied as part of
the post-processing, we flipped back up here. When we measure this,
this is one-one. Remember I said
the value would be one-one if it were
a constant function. So we’re one for one, okay? Seem okay so far? All right. Let’s add a bit flip because we’re going to
go over constant-one now. So this is the other
constant function. Now remember, if
we do a bit flip in this state we flip across
the unit circle over here, and the Hadamard gate then takes us down to here,
zero and negative one. When we measure it
we again get one-one. Okay? So we’re two for two. Not bad, right? Okay. Things
get a bit more complicated. Does anyone have any trouble
with this so far? And then, yes?>>I just didn’t get like
why it turns out to one-one?>>Why it works out to one-one?>>Yes.>>That’s a good question.
Okay. So we’re up here, right? If we measure this
then what value do we get? This is just one. Remember, this is
the value of the one bit. So if you measure
one, we get one. Yes. Now down here, if we measure this we also get one because negative one
squared is one. So is 100% probability
of collapsing to one. And so, we can say
this is one-one. This is like
the most significant bit and this is the least
significant bit. So we end up with one-one. Does that answer your question?>>Yeah.>>Okay. Great. All right. Let’s go on to something more
complicated, which is CNOT. This is going to get a bit weird but this is how we
do identity remember? The input bit is
the control bit, the output bit is
the target bit. And don’t panic, I’m going to show you why
this is on the next slide. But the action of
CNOT is to take our input qubit up to here. That’s the green line. This
is weird for two reasons. One is that the CNOT is supposed to leave
the control bit unchanged. Right? But here this is
the only thing that control bit, like the qubit it’s
the only thing that’s changed. And, this sort of
action is a bit odd, but don’t worry I’m going
to show you the math, the reason why this
works on the next slide. But just for now, so we flip from here up to here, and then when we do
the Hadamard gate to post-process we
end up in state zero. So zero-one, pretty neat. Okay, here’s how the math works. We’re applying the CNOT gate to the state one over root two, negative one over root two, and one over a two, negative one
over root two, right? When we tensor these together, we get this ridiculous matrix
which I’m going to factor out one-half,
just to make it readable. So, we have the matrix one, or a spectra is at 1-1-1,1. Right? Now, we remember this familiar matrix
from the CNOT slide. When we apply this matrix
to this vector we get this. Right? We flip
the bottom two rows. Now, when we factor this, remember the products
that we can factor it into
the individual state, we get this tensor product. If you ignore everything
in the middle and just compare our start values
to our end values, this is the output or
least significant bit. This is unchanged. Right? This is
still one 1/2 -1/2. The input bit which
started out as 1/2 -1/2 ends up as
1/ 2 1 over root two. This is just the action
of the gate. So this is kind
of a limitation of our visualizer that
we can’t really show the state transition
but I hope I’ve convinced you that this arrow
is in fact what happens. Does anyone have
any questions about this? Okay so we’re three
for three right. I said the measure value
would be zero-one, if it were a
variable function and indeed it is as I said. So for three-three.Three
for three. Let’s go over the very last one
which is negation. It’s a CNOT and then a bit flip. Right? Let’s see how that works. Again we have this arrow. Taking us up to here and the bit flip takes us
across the unit circle and we do the Hadamard post-processing to
get us down to here. When we measure these,
we get zero-one. Four for four, huh? Single query, outperformed classical computation,
pretty nuts, right? Pretty nuts. Okay, so we did it. We now, we have outperformed
classical computing. Classical computing
is in the dust. It’s done, it’s dead. So we might think like, yeah I see how the math works
and what’s the intuition? I’m going to tell
you the intuition. The intuition is the difference within the categories was
a single negation gate. So if we go back to here, so we have constant
zero and constant one. The only difference
between them is a single negation gate. When you apply negation gate
in a superposed state, it doesn’t really
have any effect. Right? So we kind of neutralize the effect
of the negation gate. So we neutralize
the difference within the categories and
then we magnify the difference between
the categories which is that the variable functions
have a CNOT and the constant
functions do not. So we magnify the effect of the CNOT and minimize
the effect of the negation. And that’s a
powerful thing about quantum computing which
is that you can change the action of
various logic gates by putting yourself in superposition in different sorts
of states and stuff. So that’s pretty
neat. And this problem you may say like,
“Okay well yeah. Great. Whatever.”
It’s really contrived. When do I ever do
this in real life? That’s a good
question and it was contrived back when
it was announced, I think was 1993, but
then a generalized version of this function was found which it takes n bits as
input, not a single bit. And the n bits either all
map to the same value, sorry, like that’s two to
the n possible values, they all map to the same value or they equally
map to zero or one. They’re constant or
balanced it’s called. And so it’s kind of just
generalized version of this. It turns out we can still
solve that on a single query. Exponential speedup, huh? Exponential speedup. A single query versus on a classical computer you
have to try all two to the n values before you know whether it is constant
or variable. Pretty nuts. Okay? Then there’s a variant of that problem building everyone’s kinda like
building on top of each other, it’s called Simon’s
periodicity problem. It’s again you have
like a black box and you’re trying to figure
out some properties of the function and
then once that came around Shor he looked at
this and he’s like, “Oh! We can use this to
factor large integers.” And so he came up with
Shores algorithm and that’s when the whole field
really caught fire. So I hope I’ve
convinced you that well this individual problem is kind of contrived
and not that useful, it’s the base, like
the foundation of some very interesting
computational properties. Did that make sense? Anyone have any questions? You’re all like
convinced that we just crushed a classical
computer right here? Yes?>>On the Microsoft site there’s an article where
the girl was saying, with quantum computers
you can take programs that take billions of years to run and run them in
a week. Is this… [Croostalk].>>This could be
that. I mean yeah. We have an exponential speedup. If you have like, I don’t know. What it would take something
like a billion years to run? Like if you have
an 80 bit function, like to the power of 80 versus
like a single query that’s probably a billion years
versus a week something like that. So.>>A question. So I
understand speedup but this is the speedup in answering
this question from the output coming
to get the input. Right? Whether the function
was a constant function or.>>Yes.>>Right? But in what context is
that question unimportant?>>Great question. Why do we care about
this sort of thing? So for example in
Shor’s algorithm you can construct a question like does this integer have a certain factor in
structure basically? And you can tell a property
like whether it has that property or not basically
which will allow you to, and if it does have
that property you can factor it pretty much. It’s very very hand-waverly explaining how
Shor’s algorithm works. So you can construct
a lot of problems in terms of does this function have
a certain property or not, which you can find
efficiently on a quantum computer. I hope, Yes?>>Do you mean
this decision helps?>>Mainly decision problems. Although, as we know from
theoretical computer science, decision problems can be
usually be turned into like generally
finding the answer to problems and vice versa. So, yeah. Okay, pretty good so far. To recap the whole presentation, we learned how to model
classical computation with basic linear algebra,
we learned about qbits, superposition, and
the Hadamard gate, and finally we learned about
the Deutsch Oracle problem, where quantum outperforms
classical, pretty nifty? Okay, now the bonus topics. With all the tools we have these will be a breeze for you. You’ll never have to read
some pop science articles on quantum entanglement
and like try and muddle through there
a horrible metaphor. You’ll actually know
the math, all right? You’ll know the math,
that is incredible. And we’ll go on quantum
teleportation, which is super, super nifty like
it is mind-blowing. Okay, quantum entanglement. You asked a great question, a very impressing question
earlier which is, “Can we always factor the product state
into the qbit state?” When we cannot do that, that means that
the qbits are entangled. For example this state, if we have qbits in this state
which is one over, two, 001 over root two, if we try and
factor this, we can. So let’s try and factor this. If we can factor it, we should be able to write it as a tensor product of
two matrices, right? Where A times C is
one over root two, A times D is zero, B times C is zero, and B times D is
one over root two. So we have this set
of equations right? Now, A times D equals zero that means A or D has to equal zero. If A equals zero then AC will be equal
zero and it doesn’t, so A cannot equal zero. If D equals zero then BD would
equal zero but it doesn’t, so D cannot equal zero. So the only solution is there is no solution. We
cannot factor this. This is an percent
unfactorable state. And what this means is we
cannot separate these qbits. They have no individual value, their value only make sense together, they’re
called entangled. And the way we might
think of this is that if we measure this state, it has a 50 percent chance
of collapsing to zero, zero and 50 percent chance
of collapsing to one,one, so the qbits are
coordinating, right? Does anyone have
any confusion with this sort of mathematical
representation? Okay, let’s talk
about what it means. Well, let’s not talk
about what it means yet, you might be like, “Well, Andrew I can sling
any old value into the vector how do you
actually entangled qbits?” It’s really simple. If we have two qbits, we just put one qbit through the Hadamard gate to
put in superposition, and then that qbit is
the control bit in C naught. Two gates, that’s all we need. Here’s how it works
mathematically. We started with zero zero, we put the most significant bit
in superposition, right? And then here’s our familiar
matrix, seen our matrix. If we apply it to
the product state, we get this. This is the state from
the previous slide. So really simple
to entangle qbits. Okay? What does this mean? I mean they seem
to be coordinating, or like communicating,
or something in some way, these two qbits. And if you measure one qbit, it collapses the other
in coordinated state. Now the weird thing, here’s how we’re going do
really weird stuff here, okay? This happens across
huge distances. China just this year, managed to do this between Earth and a satellite in space. They entangled qbits, put
one half on the satellite, it was launched into space. When they measure them,
they’re the same value. It happens across
huge distances. It happens faster than
light, it is instantaneous. A 2013 experiment
entangled qbits, moved them very far apart, synchronizing with
an atomic clock, they measured them
very close together. So close together and time
that it would have taken like 10,000 times longer
to travel between them. So they aren’t communicating
with any sort of radiation anything
like that, okay? They are instantaneous,
it’s faster than light. And you might be like “Well, it’s pretty obvious
what’s going on here. When the qbits were entangled, they just decided ahead of time what they’re
gonna do, right?” Well I got to
explain everything, but that’s called
hidden variable theory, this is hidden variable theory. It’s the same reason that
when a qbits in superposition, it isn’t secretly zero, or secretly one, it’s
both, it’s the same thing. Hidden variable
theory, we’re not going to over the proof
why it doesn’t work, but basically here’s
this guy named John Bell, and he showed that if hidden
variable theory were true, it had some really
weird implications, and therefore it’s unattainable. This is just the state, this
is how the universe works. They coordinate
faster than light, they just do and that’s nice. And you might be like, “Well, I’ve been told
my entire life there’s a universal speed limit,
the speed of light, right? Like what’s up with that? We can’t just chuck it all out.” And this is in fact a big source of consternation to Einstein, the idea that
locality wasn’t true, and then it was discovered that there’s a really
important qualifier. You can have faster-than-light
coordination, but you cannot have
faster-than-light communication, so information cannot be
communicated. So think about it. If I entangle two qbits,
I give you one, we go to opposite ends of the universe, if
I collapse mine, all I know is that
you collapse yours, and when you collapse yours,
you will see the same thing. That there’s no information that can be communicated
there really. So, faster than
light coordination is okay, faster than like light
communication, not okay. Yes?>>Coordination does not
involve communication?>>It does in a way, but communication here is a technical term
which means like, I have a bit of information
that I can transfer to you, pretty much. So I use the term
coordination to try and give a name to something
which is not that, but still involves
like some kind of coordination I guess. Like I said, there is
a limitation of our language, all we can do is look at
the math and see what it does, and our language it was
not developed for this. Anything I try and tell you is fundamentally a lie,
except for the math.>>I have a question.>>Yes.>>So let’s say I have
two entangled qbits.>>Yes.>>And I measure one
and I know its value. Does that mean
that the other one is could potentially
still not be measured, but when it is measured it
will have the same value?>>Right, when you measure one it instantly
collapses the other.>>I’m sorry, that was
my question earlier.>>Yeah, good question. I said that’s
a simplification again of language but you can
think of it that way. Yes, sorry you had a question.>>So there’s no way to tell that something has been
collapsed or not then.>>Excellent question. If you could tell whether something’s
been collapsed or not, you could measure one and
use that as synchronization, faster than light
synchronization, right? Yeah you can’t. The process of telling whether something is in superposition or not
would just collapse it.>>It seems like you will
be able to tell through some of the sort of
quantum computation, like you do some computation
on that, they was collapsed, they would only be
this state and if it wasn’t it would sometimes
do this, sometimes do that.>>You would think so, but
in fact it is not the case.>>I believe you I was
just creating imaginations, is there any intuational ways
about the case?>>Is there intu can I say intu?>>Shut up and do the math?>>Shut up and calculate, that’s a good like a way out. I’m not sure I can’t
say it intuitively, I can say that if you do put a pencil and paper
it, try it yourself, try and come with some sequence of
operations you can do on the other qbit
which would reveal it as having being collapsed
or not then maybe but, yes.>>Doesn’t this coordination
communication conversation that we’re having
violate causality or it doesn’t?>>Yes, this is
the critical thing, it does not violate causality. If we could communicate then
it would violate causality, but since we coordinated
does not violate causality.>>But instead
what you said was, if you measure
the other one is going to be that as well, right?>>Yes.>>The act of measurement
is in your control, right?>>Yes.>>So how can that happen
on the other side of the qbit if you don’t have
communication because that’s.>>Well let’s define
what communication is. Communication is I have a bit, either zero or one and I’m able to send it
to you in someway.>>Somehow yeah.>>Yes.>>The state that
is not collapsed, be communicated to
the other state, this is what I’m saying.>>Yeah I mean this
is a language problem. Qbits are like communicating but we can’t get information
through them basically. You might be able to say that. I use that term coordinating to try and avoid this problem. Other times in literature you’ll just see
the word correlated. Which tries to avoid
that whole problem.>>Coordination, it state
the word here when you are using it, it has an action
that you have performed.>>Yes, you control when you measure in qubits
it collapses the other. But there’s no way
you can change causality in
that other qubits frame. Basically, there’s no way
you can get information from your frame into that frame which would affect
their causality. All we can know is, we have generated the same random numbers
basically, yeah.>>It’s kind of like
pre-coordination so these are.>>I want to avoid that,
pre-coordination is the qubits decided ahead of
time which is not the case.>>Until you measure.>>Yeah.>>Or until these.>>At the time they measure they decide whether they’re going to zero or one and they somehow communicate that coordinate, correlate, whatever,
to the other qubits and it goes in the same. Yes.>>So, if you cannot communicate what’s the practical
application of quantum?>>Great question.
We’re going to get into quantum teleportation
on the very next slide. First, I want to tell
a funny story about this. So, you may have heard the phrase spooky
action at a distance. This is Einstein coin that
he was referring to this. So, Einstein,
Podolsky and Rosen are three physicists who actually came up with the idea
of quantum entanglement. They call it the EPR paradox, there like the workings of quantum mechanics
result in this. This is obviously
absurd therefore, quantum mechanics is garbage, was the basic argument. But then the experiments came
out and they’re like nope, this is how it works, the universe is in fact absurd, locality is broken
and so it’s kind of funny that a deep
pair of paradox is originally designed to
discredit quantum mechanics, it turned out to be the actual way that the universe works, I think it’s
a great little story. Okay. Any other questions
about entanglement, yes.>>Let’s say the same
problem, the qubits.>>Yes.>>If we say that this has a certain probability collapsing to zero
collapsing to one, and then we observe that
it collapses to one.>>Yes.>>How can we justify that
our presumption that it had a probability of collapsing
to zero was valid?>>Excellent question. The answer is used a whole bunch
of times and you see that you get a 50 percent
distribution of values.>>Exactly they won’t be
occurring at the same time.>>Experimentally, we know that they occur
at the same time. With the 2013 experiment. It entangled qubits took
them far apart it used.>>No, I mean the same qubits.>>Yes.>>It happened at time X.>>Yes.>>And now you take
another pair of qubits and this is another pair of qubits but not
necessarily the same ones.>>Yes.>>So, until it
happened at that time, that space or whatever, you can’t really tell that this was happening
with probability.>>One half.>>Yes.>>You’re questioning, okay so, they could have
decided ahead of time for example maybe or.>>Yeah, or probably
the language of probability itself is not active.>>It works. Probability
works I mean, it’s just the amplitude
squared gives you the classical
probability of it collapsing on
zero or one I guess. I’m not a 100 percent sure
what you’re driving at, so.>>So, The question is once
observed then we rationalize.>>Once observed then
we rationalize, yes. And then if we do the experiment a whole bunch of times we see that it’s about
a 50 percent chance of them both being
zero or both being one. Which in fact, I will
show that to you live on a quantum computer
that we do that, okay. All right, let’s move on teleportation This is
the actual use which you asked about and you’ve probably read about
quantum teleportation again in pop science articles. Needless to say, their
explanation was complete trash, made no sense at all,
but now it will make sense to you, okay? It’s the process by which we take
an arbitrary qubit state, and transfer it across space
and time somewhere else. So, you use entangled
qubits as a sort of bridge to send a qubit
from one place to another. And here’s a very
important thing it’s called the
no-cloning theorem. You can transfer qubit states so you can cut and paste them, but you cannot copy them, you cannot copy and paste a qubit state,
you cannot clone it. This is called the
no-cloning theorem. It’s actually pretty simple
to prove you just show that any off-such operation would not be reversible or
its own inverse, but we’re not going
to go do that. And, here’s an
important thing, the teleportation despite using a faster than light phenomenon
of entanglement, is not itself the whole protocol is not itself faster than light for communication because you actually must exchange
two classical bits. So, if I’m teleporting
my qubit to you in addition, to us having
an entangled qubit pair, I also have to send you
two bits of information and we’ll show why that
is in the next slide. So, everyone kind of
conceptually grasp the purpose of
quantum teleportation? Okay. So, here’s what
the circuit looks like. We’re not going to go
over the math, there are appendices to
this slide deck which you can see
after the class to go over the math
but it gets a bit complicated so we’re
not going to do it. So, we have the qubit
that we want to teleport here we call it T
and its value is phi, is just this generic qubit value
can be any value. And then, we have these two qubits A and B which are both
initialized to zero. Now, you will recall this
is the entanglement circuit. So these two qubits are
entangled, all right? Now after they’re entangled, we’ll say we have Alice and Bob. Alice wants to send her
entangled qubit to Bob. So, Alice has these two qubits, Bob has this qubit. And these two qubits
are entangled. So, the qubits are like these two qubits
are separated right after being
entangled, basically. Now, Alice then entangles her qubit she wants to teleport with
the other two qubits. So it’s a three qubit
entangled system. If you were to write out the product state you
wouldn’t be able to factor it into three qubits. Then she puts this qubit
through the hadamard gate. Finally, Alice measures
both qubits in her possession, and this results in
two classical bits. These are the two bits that
Alice has to send to Bob, and the measurement result of this bit determines whether Bob has to run his qubit through
an X or bit flip gate. And the measurement result of this qubit to controls whether
Bob has to run this qubit through a new type
of gate which we haven’t seen before
is called the phase flip gate or Z and this is
what the matrix looks like. And basically, once Bob has applied those two gates
or neither gate, if both of these end up been zero Bob doesn’t have
to apply anything, he’ll end up somewhat magically with Alice’s qubit value
she wants to teleport. Pretty nifty. And you might
be like while we already exchanging a qubit here
why do we even care like we’re obviously just
giving a qubit to Bob, can’t Alice just give
her a qubit right there? So, what you can do is you can pre-entangle a whole bunch of qubits a few billion
qubits or something. And then you ship them by mail
they’re called EPR halves. You ship them by mail or
something so Alice and Bob have a big repository of
these entangled qubits, and then anytime Alice
wants to send to a qubit to Bob she can just use up
one of the EPR pairs. And at that point, that EPR pair has been collapsed can no longer be used so
it’s non-renewable resource. But Alice can just send as many qubits as she
wants over to Bob. Okay? Now, you might be saying, “Well I don’t really get this.” The math is in the appendix if you really want to go see it. It gets pretty complicated, but I think I did
an okay job of explaining it. Okay, you might be like “Okay.” Oh by the way, I forgot to
say, really important thing. You may have heard
that we can simulate quantum computers on
a classical computer but it takes exponential
slowdown or exponential memory. This is the reason,
if two qubits become entangled you have to keep their full
product state around. So, if N qubit with
become entangled you have a vector of size two to the N you have
to keep in memory. And that is why, it takes exponential memory to simulate a quantum computer on
a classical computer. Pretty nifty, okay. So maybe you’re like, “Okay this is all pretty interesting, further
learning goals.” You can learn the
Deutsch-Jozsa algorithm that’s the one with N bits
that I talked about, and also Simon’s periodicity
problem which is another, I figured out some property
about this function. We’re not going to talk about
the complexity stuff but anyway you can also learn Shor’s algorithm,
Grover’s algorithm, Quantum cryptographic
key exchange, that one’s actually
pretty simple I recommend looking at that. You can learn how
they’re actually implemented in physical terms, if you care about that. An important thing quantum
error correction since these systems are so small
like a spare cosmic gray could just come out of
nowhere and just wreck your entire computation you need very very stringent
error Correction Schemes. So, I think
theoretically it takes five physical qubits for a single logical qubit like
we’ve been operating that. In practice, is looking
more like we’ll need one or two hundred to get a qubit of 100 percent probability like we’ve been
using pretty much. So that’s something
to keep in mind. When we see Google’s
coming out with a 30 qubit computer that maybe doesn’t mean
as much as we’d hope. We’ve made only guys like two or three really
high-quality qubits. And also quantum programming
language design which I will go over
because I’ll show you a Q sharp example but anyway,
recommended textbooks. This one is
my absolute favorite, has the same title, has the name of the talk, Quantum Computing for
Computer Scientists. This algorithm is by
our friend David Mermin, the shut up and calculate guy. It is a meat grinder of a textbook but I started
out with this one. It was horrifying,
it was just like, but it skips a lot
of steps basically. It’s more symbolic manipulation. You won’t see as many
vectors and matrices. I like to write
out the vectors and matrices to see what’s
going on which gets really unwieldy when
you’re writing out like a 64 by 64 vector or matrix. But I like to have
both of these. So, I’ll use this to fill
in the gaps with this. This has more interesting sides,
stuff like that. This is also in the MS library. There’s another one called Quantum Computing
Gentle Interaction, also in the MS library. I’ve heard mixed reviews or semi decent reviews of it. I
haven’t looked at it myself. There is also Mix of
Quantum Development kit. They have the docs. They’re actually
pretty nice thing called the Quantum Computing Simulator
and Q Sharp as we’ll see. Now, I said I would
end the talk on some skepticism which
there’s an article, came out I think was Quantum Magazine pretty
prominent recently, interview with
a mathematician who believes that physically realizable quantum computers cannot exist. His basic argument simplifying
is that the amount of noise in the system grows exponentially with
the number of qubits. So we cannot get a really
large qubit system because our computation will
just keep collapsing in the middle of it and it’s just like
an exponential term. And so, I’ll be very
sad if this happened. I’d feel like the universe
is against me. It’d be really,
it’d be very sad if we could not realize
the benefits of Quantum Computing,
like cosmetically sad. But anyway, you can’t argue with the universe and yeah you can read that article
for some skepticism. I think his name is Gil Kali,
is the mathematician. There’s some appendices but I’m going to do some demos now. Okay, the first demo is just the Doge Oracle
problem in Q sharp. So Q sharp is based on F-Sharp and we have this function
I’ve written, IsBlackBoxConstant. You’ve taken a black box
and they tell you whether it’s constant. Oh, gosh, sorry.
Here, of course. Okay. We have our function
IsBlackBoxConstant where it takes in
a black box and it just returns the Bool. Is
it constant or not? And it just uses
this same protocol we went over. It allocates two qubits, calls one input one
output, clears them, sets them both to zero,
does the preprocessing, flips them, puts them
through the Hadamard gate. Then it sends them
into the black box. Afterward, it again
put some through the Hadamard gate
and then it measures both of them and again if
the input results is one, then it’s constant
otherwise its variable. Okay? We have our black boxes defined here. So, ConstantZero. Again, remember it’s nothing. For ConstantOne we
flip the output bit. For Identity we run
a CNOT gate with input control up it is target and Negation reducing
our next, okay? And we just have these, like IsZeroConstantZero I call. Is BlackBoxConstant
with ConstantZero? And then we have
our classical driver. Any quantum computation
will have a sort of classical computer to
tell it what to do, and so I ask for each of those four,
well, what’s going on? And we can run it and it
runs on a simulator and we see the output,
IsConstantZero constant? True. ConstantOne constant? True. Identity constant? False. Negation constant? False. So there you go. Yeah we wrote the Doge article, ran it on a real
quantum simulator, you saw that it worked out. Okay, the final demo of the day and then I’ll
let you get out of here is here we go. This is called the IBM
quantum experience. There is a real like world quantum computer
that IBM has built. It only has five qubits but
they just let you online just drag gates onto this quantum circuit diagram and run it on a real
quantum computer. So I thought it’d be pretty
nifty if we demonstrated entanglement live for
you on stage. Let’s see. So, if you remember
the first thing we do is we drag a Hadamard gate onto one debt and then we use
the control not, oh no, okay. So, the way that quantum computers are
constructed you can only usually put CNOT between certain debts basically
it’s just kind of, otherwise it’s just
mechanically impossible to see. Can we do that? Nope, okay. How about, success. Okay, we did it. This is the CNOT gate. You can see varying
representations of it. It’s like the opass. And I say we’ll run it on a quantum computer, oh
right, measurement gates. I forgot. We need to add the two measurement
gates obviously. So we’re measuring both these qubits
after entangling them. And what we expect is to
see them predominantly been 0-0 and 1-1 right?
Well, let’s test it. We’ll call it that and we can save the result from cache but I want to do
it live for you guys, so I’m going to run
a new execution. And you can imagine right
now deep at IBM Research Lab nestled within
a dilution refrigerator operating at slightly
above absolute zero, we have real qubits being entangled and
measured thousands of times for our entertainment.
That’s pretty cool.>>It is.>>That’s pretty cool. Okay, let’s see if it’s, oh here we go, we
got the results. And executions
it’s pending, okay. So it might take
a couple minutes. We’re going to see now. Does anyone have
any questions while we’re waiting for this? Yes.>>I don’t know what question. You said this experiment where this sentiment tangled qubit
to space and the satellite?>>Yes.>>Are we good at storing
like qubits for a long time? I’m really surprised
they were able to do that. I thought
they kind of…>>Yeah, yeah, that
was actually a lot. They will manage to send
the entangled qubit up file laser but, yes.>>Okay.>>I don’t think
we’re that good at storing qubits for awhile. I think they’re
fairly short-lived.>>[inaudible]>>Yeah, they actually, this Chinese
satellite experiment, they teleported
a qubit up there. So they sent one EPR half up
by a laser and then they use that EPR half to teleport the qubit up there. Pretty cool.>>That’s awesome.>>Yeah.>>So we can send
entangled qubits by laser?>>You can send entangled
qubits by laser, yes.>>Okay, that’s even cool.>>Yeah. Let’s see,
it’s still not. Okay, that’s fine. Anyway, yes.>>You mentioned that there’s this new article by the mathematician that says
the noise has grown faster.>>Yes.>>What is the whole purpose of a topological quantum
computer resist that.>>Of course. Any corner computer will
have issues with noise. The topological quantum computer is supposed to be better than its IBM counterparts in terms of noise but it’ll
still grow exponentially. Like it can still
be better but it still runs into
exponential barrier. And so the next couple of years are really sort of
make or break for quantum computing because
we’ll now have the ability to create computers which run into this exponential term if
it exists, if it exists. And so we will see with
the next couple of years whether quantum computing is
really possible or not. It’s nervous and it makes
me nervous but yeah, because we’ll run into this limit probably
like this year. If Google does not manage to demonstrate
quantum supremacy it’ll be a bit scary
I guess but yeah. Let’s see, we don’t have, I’m
just going to refresh this.>>Question.>>Yes.>>Do we know what Google uses Quantum theorem
for, like what?>>What they use it for?>>Yeah.>>I don’t believe we do.
I think they just have…>>Is it experimental mostly?>>Actually no. They haven’t announced
what they use it for. My manager is actually
playing around with the D-Wave simulator to…>>The Canadian company.
Is it the Canadian company?>>D-Wave is Canadian company. Yeah, everyone out there is like really scammy
and not really quantum but everyone
kind of believes that they’re quantum but
it’s a different paradigm. It’s like optimization basically but it uses quantum computing. They have a simulator
you can use. My manager has been using it to kind of play around with
finding routes in a network. It has general applications like that. Well, this
is taking a lot.>>The data is publicly
available somewhere?>>Yeah, it’s in the invitation and I think it’ll
probably be posted on their resonant website
or posted online.>>So that application is like a requesters type
of application?>>Yeah, yeah, it’s like… It’s optimized. You’re able to express the network graph
in terms that like an optimization result
would be like the least energy path
in terms of like the least cost path
through the graph sort of. That’s different than
the way the [inaudible].>>Open Shortest Path.>>Come on, why isn’t it going? Pending. I’ll be
really sad if this doesn’t finish in time. Oh well.>>So on D-Wave, how are presentations
centered you could do classically faster than
when they were doing. So, they’ve improved in that or?>>I don’t know actually I
haven’t looked into that. They’re probably about
comparable I would say. Because given that
it’s just like optimization and especially
with the explosion of machine learning there’s been a lot of development in
making optimization better. Optimization is like you
have a surface and you’re trying to find the highest point
or the lowest point. So there’s been a lot of
research that goes into that. We’ll just run it
from cache and okay. Here we go, okay. So here’s
the results from cache. Unfortunately our real execution
didn’t finish yet but. So we see that indeed they entangled some different
qubit than ours but we end up with 0-0 and 1-1 almost exclusively but then
there are some in the middle because of
the error rate, right? So this is on
a real quantum computer. We have some error rate where it doesn’t exactly
work according to our model all the time
but pretty cool, these are entangled and they’re measured and we have that. That’s the
experimental data right there that entanglement
is a real thing. Yeah. Okay, that concludes
the presentation. Thank you very much. I hope you all feel as though you
followed the whole thing, no real confusion and feel that you could
keep on learning. This isn’t like
mad genius science. This is very accessible science. So, thank you very much.

100 Replies to “Quantum Computing for Computer Scientists

  1. Can't we just use that cnot gate in a classical computer and find out if the BB has a constant function or a variable function in a single query?

  2. I'm studying for a Quantum mechanics final, watched this to procrastinate and honestly ended up understanding the Bloch Sphere a lot better because of it

  3. @ 1:11:25 he states that teleporting the value of a qbit requires the regular transmission of two measurements. My question is why don't you just transmit the measurement of the original qbit when you need to know it's value instead of transmitting the measurement of two values and then measuring the teleported qbit. What is the advantage of teleportation?

  4. Thanks for showing the math and programming side. You also dropped the hype and media nonsense.
    The applied engineering the hardware -problems will make or brake Q-computing. Which is what I thought from the beginning. Thanks for stating it clearly but the terminology does need work you need a linguist good at inventing new terms for new concepts and building a new language. The words have to be completely new so they can fit in a all new system of thought.

    I wish someone would lecture and do the same for AI.

  5. On the matter of coordination x communication: think of it as if you know your bit is 0 then my bit is already decided for you but not yet revealed for me so I haven’t gained any information, the cat is still both dead and alive because no measurement has been made

  6. As to the input/output confusion at 36:24
    The words i/o refer to the names of two variables: i is used to store the input data, o is used to retrieve the result. The apostrophe ' indicates the state before or after applying the gate BB.
    This is different from the one qbit gate, where there was no variable name for the one qbit, and i/o described the state like the apostrophe ' in the second example.

  7. Did this get speeded up by You-Tube or after filming. He is speaking at about 190wpm or faster and it is too too fast and almost makes him sound like a motorboat or like he is on a drug. I stopped watching films that are speeded up. It’s so unnatural. Too bad because I think this looks great but it is ridiculously fast.

  8. This presentation is fantastic. This is the first one after which I'm legitimately interested in learning more about quantum computing instead of thinking "Man, that's weird"

  9. This is amazingly understandable, thank you! Now I'll be a little better prepared to read the Nielson/Chuang book.

  10. Thanks a lot!
    All those "popular science" explanations that try to oversimplify are in fact just confusing even more.

  11. doesn;t hadamard gate break the quantum coherence, and quantum computer ability to perform all operations in superposition state? I am mainly thinking of using it in the chain of perrforming multiple operations.

  12. quantum will never beat classical computing on all calculations, there's all kinds of physical systems we could leverage to calculating niche problems. I just dont get the hype. If you had a stack of i9's the size of a quantum computer cooled with liquid nitrogen, it would kick quantum arse. No?

  13. My Millennial colleagues really need to stop their poor habit of asking/saying "Right?" There's some use for this word when you're teaching something, but if you're in the midst of explaining one part of a larger puzzle, asking "Right" doesn't exactly help considering you can't always understand one part of the puzzle until you learn more parts of the puzzle — if that makes sense. Which is ironic, because of all things he's trying to teach Quantum Theory and its applications in Computer Science. It shouldn't be taught classically, using traditional logic at every step to qualify what's true (1) or false (0), when you DON'T KNOW THE ANSWER YET. That's literally the "black box." If you're asking "right" after explaining something that is not readily apparent using traditional logic — you should expect that the answer in the individual's mind will be "yes? no? I don't know." Which is ironic, because you shouldn't be teaching them to always apply logic to a science that has a lot of illogic in it — yet as a system of illogical subsystems, the outcome with those taken as a whole becomes logical… Assbackwards approach to teaching this, my friend. Teach them the outcome and work backwards if you expect them to 1) put together the puzzle pieces and 2) expect to understand how that puzzle was solved and how those puzzle pieces being such as they are, solved the larger puzzle… It's called reverse engineering. Better to teach Quantum Theory using reverse engineering, because scientists are still trying to figure out how the hell it works. Because it's still a THEORY and is actively being explored, with many unsolved paradoxes that seemingly disprove the theory's premises entirely — yet the outcomes say otherwise. So don't expect a layman to comprehend until more concepts are introduced to them. It must be explained to students, constantly, that we still can't explain all of this perfectly because it's not a settled science (though no science is truly settled), and thus they need to roll with it sometimes. That should always be a given. If not, we will not get anywhere in terms of accelerating our scientific discovery..

  14. Wow!!! Microsoft has a Research Division? Who knew? I knew that they had a kick-ass marketing division that is without rival. And their legal department is absolutely stellar. (Or should I say quasi-legal dept?). But honestly, I thought that the only research that they did was how to steal intellectual property. This is news to me.

  15. 1:06:39 If I measure one of the entangled qubits, can someone else know when I measured it, causing their qubit to collapse? Or do they not know WHEN I measured it. If they can know, then FTL communication would be possible. You could measure qubits in short and long intervals which could be interpreted as 0's and 1's, which would enable the transfer of data.

  16. I feel like this whole thing could have been condensed to "a quantum computer is a unitary matrix" and "the input/output is a unit vector"

  17. Finally a real explanation of Quantum computing. Popular stupid analogies of flipping coins don't work. I am not a computer scientists and this video finally helped me understand what QC really are

  18. I don't understand the claim that the quantum computer outperformed the classical computer because it only took 'one query', because you've added another input and output. Isn't that 'equivalent' in some sense as the cost of a second query?

  19. It was hard to follow for me but my interest in this has grown due to this. I did run through this multiple times before it started making sense. I am still not 100% there but getting there. QC is really cool. I love it. Nature is so weird and wonderful. This was a great video. Thanks!

  20. For anyone else stumbling upon this, here's the slides

  21. nice talk, but he should stick to the computing part and leave the physics to the xperts (especially around 1hr in +- a couple of minutes) … What he calls "coordination" and claims to have no term yet due to language limitations is actually called "causality"… And Bell's theorem does not do away with locality but with local realism! Huge difference! I could go on here, but I dont want to bring in negativity to a topic and effort worth applauding.
    I suppose it's a good thing, physicists don't watch 90mins videos with "for computer scientists" in the title ^^ otherwise the comment section probably would look somewhat different^^

  22. 2 Questions:
    1. How do we interpret the probabilities if we allow complex numbers? For example, if the spin of particle is in a superposition represented by (√2/2) + (√2/2)i, what will we expect to see when we make a measurement?
    2. How does this model deal with more than 2 q-bits? Would an n-qubit system be represented by a vector of length n for each q-bit, or of length 2^n for each possible binary string of length n?

  23. In reference to the confusion in 43:14 regarding naming confusion. Just wondering, is this confusion typical because traditionally, as you mentioned earlier, in a classical computer, you change the variable. Hence the variable wire changes from "input" to "output". However, in the quantum world, the variable stays in place while you change the program. Hence, the wire, which represents a variable instead of a function, is always called the output wire or the input wire. I wonder if it is this shift in perspective that causes the confusion?

  24. I'd like to see things like human choice shown as a Qbit in a superposition (Do/Do Not) until the physical action occurs which is the collapsed stated. I think this kind of thinking is going to revolutionize our way of thinking let alone our way of computing.

  25. 43:24 I think the output input thing could be more easily understood by C programmers, since we have the concept of returning through parameters.

  26. 1:10:07 The guy is questioning whether or not superposition actually exists. Is the probabilities due to the fact that we have imprecise tools that cannot observe the particles, or are the particles actually in multiple states in the same time? Copenhagen interpretation says that particles truly are in multiple positions at the same time, and alternative theories like pilot wave theory says that there are things that exist, that we cannot precisely measure. (I don't know if the math behind pilot wave theory has been fully worked out yet)

  27. 9:55 "…only use operations that are their own inverses"
    He is wrong there, the phase shift gate S being a counterexample.

  28. Ok… if I have 2 qubits and I entangle the qubits they can be in a state of either entangled or not 1 or 0. If I take the qubits and I separate them I now have 2 qubits and they are apart. I use a robot to push a button if the qubit is not entangled any more. If I wanted the robot to push the button I would measure the one qubit and the other one would collapse making the button get pushed. How is not a transfer of information.

  29. I think it would help in the Black Box problem to draw the CNOT gate as having 2 inputs and 2 outputs so the fact that it changes the input makes more sense.

  30. What if I use two input bits in a classical computer, one is always zero as in QC, other one as 1:
    Const 1 -> 11
    Const 0 -> 00
    Identity -> 10
    Negation -> 01
    Then use XOR gate on the output I will get 0 for constant functions and 1 for variable functions in one query. And since QC was allowed to use 2 qbits in the problem, I think it is fair to give CC that much if we are comparing them.Where is the speed up we are talking about ?

  31. 1:06:45 So does this mean that I physically separate two entangled qubits, run some operation on one (so that its probability of returning 0 or 1 changes) and then if I collapse the one qubit the second qubits' probability will get updated "somehow"?

  32. Actually As I understand quantum computing is just a different concept of laying/extracting bits using schrödingers state function?? With a bit of advancement and time I think we can reflect and implement those Qbits principles onto the current classic computing?

  33. Polarization could also be a classical phenomenon once you say ‘photons’ it becomes a quantum phenomenon:)

  34. Great video but don’t you think opening parallel gates to other worlds is a bit much just to play minesweeper 10,000x slower than my Commodore 64?

  35. This is supposed to be for scientists, but it's the video from which I learned the most of Q.C. out of them all. All other videos are just barely understandable analogies. And I'm not a scientis, nor a programmer.

  36. The whole discussion on the deutsch oracle (36:00) would be much clearer if he finished the arrows and adapted his termology: how must we interpret a system where ‘output’ goes in and ‘input’ comes out?

  37. Your ability to make predictions is increased by exponential levels when you have Quantum Computing Technology.

  38. The observation that all metaphors ring wrong for describing quantum algorithms bothered me for a while as well (3:40) however shutup and calculate is bothering in its own way 🙂

  39. well, but two people can coordinate, like that they start a spaceship to meet in the middle faster than light

  40. 25:27 is world-shattering. So quantum physics is not just, naively, "all possibilities are realized", it's "all collectively reversible possibilities are realized"?

  41. So for clarification, the constraints of the Deutsch oracle problem allow there to be usage of multiple bits? It's not required that there is just a single bit used for a classical single bit operation as per the constraints of the problem?

  42. Hi I have a question. For the 4 operations, the preprocessing was done, with the assumption that both starts with both values as 0. However when we were solving for variable, X, the preprocessing was applied from the starting point of 0. But X is not 0, right? Shouldn't the preprocessing have started from somewhere else?

  43. Hope to find out what he means by neutralising the difference within the categories, and magnifying the difference between them. What are the categories he is referring to? How can I tell it is neutralised, how can I tell it is magnified?

  44. When you have n qbits that are being represented with a single vector through an n-fold tensor product of the product states, there are probabilities for the system to collapse with a 1 at one of the 2^n numbers in the vector. How does this n-qbit system collapse? Can all qbits be measured at once, are they required to be measured all at once? And what does the 1 in the product state represent? With a single qbit, there are probabilities for collapsing to 1 and collapsing to 0. What do the probabilities correspond to with a larger product state of multiple qbits?

  45. Watching this 15 months after upload. Anyone know if Google has reached "quantum supremacy" i.e. do we know if the resulting noise makes the quantum computer unusable?

  46. At 1:00:00, using entanglement for communication is possible in one case that I can think of. If I take a Q-bit, which is entangled to my friend's Q-bit. imagine me traveling to some far place. I tell my friend that if the Q-bit collapses to 1, then perform task A. And if it collapses to 0, then perform task B. And I would know which task he performed because of the Q-bit that I have. This is communication even though I don't send information implicitly…what do u guys think…?

  47. Quantum Supremacy – still as far in the future as ever. What the last year has shown is that the scaling problem is far larger than anyone had anticipated, and by implication, perhaps the foundation expectations of quantum computing may be flawed – if not impossible.

  48. Great intro to quantum programming, but the tangent at the end gives false information. Bell's inequality proved that there can be no hidden variable theory only under the assumption that the universe is local. The later discovery of quantum entanglement showed that Bell's inequality actually was proof that all theories must be non-local. Hidden variable theories are NOT disproved by Bell's inequality, they merely fell out of popularity after the initial misinterpretation of Bell's inequality. Bohmian mechanics is a perfect example of a hidden variable theory that is consistent with all known quantum mechanical experiments thus far.

  49. Bell's inequality proves that there aren't hidden binary variables but it still doesn't prove there is any "teleportation" or "spooky action at a distance", you just collapse the wave and measure the same state both sides 100% of the time. Nothing was teleported, we just can't explain this action with binary variables which is quite normal because we are measuring quantum particles not tennis balls.

  50. There is no way to entangle q-bits if they weren't created from the same particle before. The fact that there aren't binary hidden variables doesn't mean they are being teleported faster than the speed of light. You need to entangle the particles before. No magic happening just quantum mechanics

  51. Honestly the usage of the word "teleportation" should be banned from QM. Let's keep this term for Star Trek. There is NO INFORMATION TELEPORTED, Alice and Bob have random results, they need to communicate via normal channel about the operation they performed on the q-bit to have the same sequence of bits, this is not clear in this presentation. No need of mystification of quantum physics with Teleportation, flying saucers, time travel, flying fairies and unicorns

  52. In the Teleportation slide where the output is psi after the operator X or Z is applied there is a very important piece of information missing. Bob needs to know what operator is valid to have the correct sequence. Thus there is NO TELEPORTATION

  53. 0 & 1, meaning the symbolism used representing female and male principle. How dare you! The outer "glow" surrounding the earth is been as if a lockdown surrounding the earth. Fools. 1619-2019 🙂 Yeah. Live with the knowledge that your mathematical equations have always been about minimizing what is universally available to us all. Good luck in your endeavors. Your time is short! 🙂

  54. If I write a SQL query that does the filter before the join, I can get a tremendous speedup in processing. Isn't the pre-processing you do for Deutsch-Jozsa the same sort of thing? How is this considered a fundemental accelleration of processing?

  55. Holy sh*t.. The audience is not buying his weasel words – they are obviously not afraid to ask the most obvious question. If I have an "entangled coin" and split it, and separate the two halves by great distances.. Why is it so amazing that measuring one half reveals "tails"? Wow! faster than the speed of light the other half is "heads"? Seriously? How much money are they scamming for this snake oil? Lol

  56. Thanks for the vid. A lot of good info here.

    If anyone wants more detail on exactly how the teleportation process works, check out this other vid:

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