### You don’t know how Quantum Computers work!

Ever since I first heard about quantum computing, there was something that bugged me about it. The story goes like this: All operational

computers today are classical computers. They store and operate on information represented

as 1s and 0s (bits) to solve intensive, mathematical problems. Quantum computers, which are being

developed in labs right now, are fundamentally different. They take advantage of quantum

mechanics to do multiple calculations at once. This is done by putting quantum bits into

superpositions so they can be 1 and 0 at the same time. But you have to be careful with

superpositions, because as soon as you read them, they collapse to either a 1 or 0, with

a known probability. If all that was a little too fast, feel free to check out Building the Bits and Qubits, as it explains all of this more clearly. Don’t worry, I’ll wait here. Done? Good! Now we’re on the same page.

Here’s my problem: if you put a qubit in a superposition to do many computations at the same time, wouldn’t you only obtain one of those many calculations when you measure your answer? You can’t get many answers at once, since reading the answer to a quantum computation would destroy most of the information your qubits are storing! How can quantum computers

even be practical if this happens? The problem is that this story doesn’t completely explain what quantum computers actually do. Why don’t we just take a look? To understand what’s happening here, let’s

actually build a quantum algorithm. Qubits can be visually represented as Bloch spheres.

1, 0, and everything in-between. Today, we won’t be needing all of these qubit states, so let’s simplify this representation to a slice of this sphere. Let’s call it the Bloch circle. And just a little aside for those of you following the math, the Bloch circle illustrates all possible qubit states where the probability amplitudes have no imaginary

component. And don’t worry if none of that made sense. this is just so those who already

know the math can follow along. Now let’s begin. Say that I hand over to

you this quantum circuit. It takes in two qubits, and spits out two qubits. Let’s call these x and y. Qubit x always passes right through this circuit unaffected, so if it was a 1, it comes out as a 1, and if it was a 0, it comes out as a 0. Now, here’s where you come in. I’m not gonna tell you what this circuit does to qubit y. I will only tell you that it will do one of 4 possible things. Case A: it does absolutely nothing to it,

just like qubit x. Case B: it flips it, so if it was a 1, it comes out a 0, and if it was a 0, it comes out a 1. Case C: it flips it like before, but only if qubit x is a 1. If instead qubit x is a 0, it does nothing to it. Case D: it flips it only if qubit x is a 0.

If instead it’s a 1, it does nothing to it. So to review: In case A, never flip y. In

case B, always flip y. In case C, only flip y if x is a 1.

In case D, only flip y if x is a 0. In cases C and D, whether qubit y flips or

not depends on qubit x’s value, so let’s call these the needy cases. “Umm, qubit x, sorry to bother you… What should I do to qubit y?” This is not true for cases This is not true for cases A and B, so let’s call them the laid-back cases. They’re the cases where the circuit goes: “Yeah, I don’ really care what you say qubit x, imma jus’ go do my own thing.” By the way, remember this exact terminology.

It’s important! …naw I’m kidding, these names are completely arbitrary! So, with that in mind, here is the circuit

I’m giving to you. By toying around with it, can you figure out if it’s a laid-back

case, or a needy case? Oh and by the way, I rigged the circuit to self-destruct if you

try to open it. So all you can do is pass qubits through it to see what happens. And,

it also self destructs after you use it once. So you only have one chance. Good luck! Okay, that was a little unfair. Let me show

you how to do it. What would happen if we pass in both qubits x and y as 0s? In case A, we would get two unaffected 0s

as expected. In case B, nothing would happen to qubit x,

and qubit y would flip vertically to a 1. In case C, qubit y would only flip if qubit

x is a 1. Qubit x is not a 1, so it’s unaffected. In case D, qubit y would only flip if qubit

x is a 0. Qubit x IS a 0, so it does flip. So, the result will only ever be 00, or 01.

If you actually did the experiment, and got 01, that means that qubit y would have flipped,

and you would know that this circuit is either case B or case D. But which is it? Would y

have flipped because it always flips, or would the circuit have been listening for instructions

from qubit x, and it happened to say to flip qubit y? Would something else have happened

if x was different? We could then then try passing in x as a 1 instead to see what happ… Whoops, the circuit gets destroyed after a single use, remember? And, we still don’t know if the circuit was a laid-back or a needy case. In fact, if you just pass 1s and 0s

into this circuit, you would always need to use it at least twice to get an answer. But

we can’t do that, because I’m a mean person and made it explode after 1 use. Let’s try thinking of something else. Now,

the problem we had before is that we didn’t know if qubit y flipped, because it always

flips, or it only did that because qubit x was a 0. We didn’t know if it would flip

if x was a 1. So, let’s see what would happen if we put qubit x into a superposition, so

that it’s a 1 and a 0 at the same time. That makes sense. With that, we can see what

would happen when qubit x is a 1 AND when qubit x is a 0 at the same time. To do that

let’s use the Hadamard gate. The Hadamard gate is useful because it can take qubits

into and out of super positions. Let’s use it to take a qubit 0, and turn it into a qubit

right. Qubit right, is half-way 0 and half-way 1. Or as many like to say, it’s 0 and 1

at the same time. If you’re doing the math, qubit right represents this quantum state. That should do it. We’re testing both scenarios

at once. Let’s see what would happen. In case A, nothing ever happens to either

qubit, so they are unaffected. In case B, qubit y flips no matter what qubit

x is. So it flips. In case C, qubit y flips only if qubit x is

a 1. But it’s both a 1 AND a 0. So does it flip? Does it both flip and not flip at

the same time? Something interesting happens here. Qubit y goes into a superposition, but

it isn’t tied back to its own opposing state. It’s tied back to the state of qubit x.

In other words, qubits x and y come out in an entangled state. They can no longer be

considered separate from each other. Now, measuring either of these qubits will cause

both to collapse to 1 or 0. In this particular case, they will always collapse to the same

value. Here’s the reasoning: If we found that qubit y flipped, it would because qubit

x was a 1, since that is the only scenario in which qubit y WILL flip in case C. Similarily,

if we found that qubit y didn’t flip, it would be because qubit x was a 0, since that

is the only case in which qubit y will NOT flip. In case D, qubit y flips only if qubit x is a 0. Again, qubit x is kind of both 1 and

0 at the same time. In this case, the qubits come out entangled again, but they will always

collapse to opposing values. If qubit y flips, it was only because qubit x was a 0. If qubit

y doesn’t flip, it was only because qubit x was a 1. At first, it looks like we may have solved

the problem. The qubits come out entangled in only the needy cases, and not the laid-back

cases. So all we need to do now is check whether or not the qubits are entangled, and we have

our answer! So what would happen if we tried it? What would happen if we put in qubit right

for x, and qubit 0 for y. Say we find that qubit y came out flipped. We would know that

we don’t have case A, because it never flips in that case. And let’s say qubit x comes

out a 1. Wait, was that because it was entangled and became a 1 when we measured qubit y (as

in case C), or was it because it randomly collapsed to a 1 from an entirely separate

superposition (as a possible case B). Like last time, we don’t know if we have a laid-back

case or a needy case, because there is simply no way to check if two individual qubits

are entangled or not. This is the problem I was talking about earlier. Sure, you can

do multiple calculations at the same time by putting qubits into superpositions. There

is information about two separate calculations in these qubits, but you lose that information

when you read them. Well it turns out, there is a very clever way of getting around this… Okay, enough stalling! This is the solution.

Set x to qubit right, and set y to qubit left, both in superpositions. If you’re doing

the math, qubit left represents this quantum state. In case A, nothing happens.

In case B, y always flips. But flipping qubit left top to bottom just results in qubit left

again, so oddly, nothing happens. In case C, the strangest thing occurs. The

classical description of the circuit says that qubit x never changes. But, somehow,

the circuit actually flips qubit x left to right, even though nothing is really supposed

to happen to it! The same thing happens in case D. As for how this happens, hold that thought

for just a sec. We’re really close to solving the problem. Now, one last obstacle remains. As a reminder,

we are trying to figure out whether the circuit is laid-back or needy. In the laid-back cases,

qubit x faces right. In the needy cases qubit x faces left. But we can’t measure this

qubit yet, because superpositions collapse randomly to a 1 or 0. All we need to do, is

pass it through the Hadamard gate once more. Now, qubit x only comes out as a 0 when the circuit is laid-back, and only comes out as a 1 when the circuit is needy. Problem solved. So let’s do that. Qubit x comes out

as a 1. Therefore, we know that this circuit is either case C or case D. But no matter

what, it was a needy case. This is called Deutsch’s algorithm. It depends on the fact that qubit y is specifically

facing left. This mechanism results in the ability for qubit x to flip left-to-right.

If qubit y was facing right instead of left, nothing ever happens to qubit x. It’s a

lot like how multiplying a number by 1 keeps you where you are, but multiplying by -1 flips your sign. Likewise, qubit y facing right keeps qubit x where it is, but qubit

y facing left flips qubit x left-to-right. But why is this? Left and right are both half-way

0 and half-way 1. Why should something so drastically different happen under each case?

Well, left and right do each collapse to 1 or 0 with the same probability. But mathematically,

they are as different from each other as 0 is different from 1. Qubit right is half-0-half-1

in-phase, and qubit left is half-0-half-1 out-of-phase. They are, in a sense, “opposites”.

For those following the math, [0,1] and [left,right] each form an orthonormal basis. So, while the in-phase state has no effect

on qubit x, the out-of-phase state has the ability to flip it. And if this still doesn’t make much sense, which it probably shouldn’t, just remember, as Richard Feynman once said:

“If you think you understand quantum mechanics, you don’t understand quantum mechanics.”

After all, the mathematics we are dealing with here involves transformations through

4-dimensional vector spaces, so it’s a little tricky to come up with good analogies. This

terminology zoo and this flipping piece of card are honestly all I got. And if any of you can, please research quantum

computing, go through the math, and find a better way to explain this. It really bugs

me when I get the math but I can’t explain the essence of it. What kind of channel is

this? Frame of Matrix Algebra? Quantum Mechanics… you have defeated me. Moving on… Since we’ve solved our problem,

here’s what the individual circuits actually look like on the inside, made out of elementary

quantum gates. And here are their classical analogs. So, what have we learned from all of this?

That there is no hope in understanding quantum weirdness? I don’t know the answer to that

question, but that isn’t the point of this video. The point is that the answers extracted

from quantum computations contain no more information than the answers extracted from

classical computations. With classical bits, running the circuit once, only ever narrowed

our answer down to 2 possible cases. Using a quantum algorithm did not change this, but

it WAS able to use quantum mechanics to cleverly twist the question we were asking to result

in information that we found more useful. Now, you may be wondering one more important

question. What was the point of this? If I already built this quantum circuit, I obviously

knew that I gave you a needy case. Why did I keep that fact secret and make you figure

it out? When would finding out something that was already knowable in the first place be

practical? Well, this is precisely what a quantum computer does best. Take this example. I mentioned in the last

video how quantum computers can break current computational security techniques. Today’s

internet security depends on the fact that is very hard to factor a large number which

is a product of two primes. It’s not so bad with a small 2-factor number like 15,

but it’s practically impossible for a classical computer to factor large ones. If someone

can factor this giant number, the security fails. How does a quantum computer break this?

It turns out that it’s possible to create a quantum circuit which implements some modular

arithmetic with this giant number. It’s a lot like the circuit I gave you. If we pass

in superpositions to it, we are able to obtain a result, which we can then use in some classical

arithmetic to quickly find the giant number’s prime factors. Classical security has been

broken. This is known as Shor’s algorithm. It’s been done as a proof of concept, but

it shouldn’t actually be a real threat anytime soon, because the physical implementation

of quantum computers is very difficult, and is currently being researched. But it does

put pressure on humanity to develop more robust security techniques. Maybe quantum security

techniques? In each example, we had a problem with an

answer that was computable, and was already a property of the very quantum circuit that

was already built. We do not need quantum computers to solve these problems in principle.

They just enable us to solve some problems, such as these, in less time. There’s another, more optimistic one called

Grover’s algorithm, which finds the location of an item in an unordered list. Classically,

because this list is randomly organized, you would need to check each item, one by one,

until you found the answer. Grover’s algorithm allows us to find it much faster. If we had,

say around 10 000 items, instead of performing up to 10 000 queries, you would only need

a measly 100. This opens up numerous possibilities in database theory. There are more quantum algorithms, but I think

we’ve already learned the essentials today. Quantum computers cannot do anything that

classical computers can’t already do in principle. Quantum computers cannot give you

multiple answers to the multiple computations it may be doing at once. What they can do,

is twist the questions we can ask it, to aim the answer down to something more relevant. Quantum computing is hard, which may be part

of the reason it’s still being researched. But be on the look out for computationally intensive software being partly running on quantum processors. They may be just around the corner. KABOOM! đ

Finally something that i could understand, that paper flipping trick still amazes me even though i had enough amusement with my own example of it.

Binary codes

Fun fact: The machine is actually Case C, because we know that it's Needy, and that it's either Case B or Case C.

I think you could make the Bloch circle an actual term if you submitted it to the right channels. What those channels are, I don't know, but it's definitely an option.

Subscribed right away.

My only question is how do you find out if the qbits are entangled?

DIGITAL WITCHCRAFT!!!

And people think accounting is hard to understand…

ok

7:51 Putin

So everything we know….we actually do not know…. ouch. My heart's broken.

If x is true, then y is flipped. Or it could be that if x is false, then y is flipped. Sometimes the value of y is just flipped between true and false independent of x. Your job is to figure out what is going on. The solution is to make x be right and y be left. Even if a quantum computer is built, how many people will be able to write software for it?

Don't give up haha, jk jk. Try explaining the movement (spin) of a qubit through time. I am by faaaaar not an expert but i have a ton of questions and have no clue who to ask. Could the 'messy' mechanics of quantum particle be simply explaind as particle movent trough time, but considering that 'time' is just another special dimension?

I dont even understand regular computers

MAYBE "you don't know how quantum computers work!" BECAUSE YOU WON'T EXPLAIN IN COMPARISON, HOW NONE QUANTUM COMPUTERS WORK. WAIT. IT WAS EXPLAINED (VIDEO POSITION; 0:41 / 15:48). THERE IS NO FLIPPING-FLIPPING. IT'S JUST THAT YOU WOULD HAVE A NEW TYPE OF SWITCHING COMPONENT FOR COMPUTERS, DEVOID OF USING TRANSISTORS. THE SWITCHING SYSTEM IS SO NEW, THAT IT'S BEEN THEORIZED THAT 'IT WILL BE CAPABLE OF ACCESSING 1 AND 0 AT THE SAME TIME'. IMAGINE IF YOU WILL WHAT IS SAID OF âPARTICLES BEING AT DIFFERENT PLACES AT THE SAME TIMEâ. HOW DO YOU SUPPOSE THAT'S POSSIBLE? IF IT'S TRUE AT ALL. SIMPLE, FREQUENCY. SERIOUSLY, I IGNORED THAT TRUTH FROM THE DAY I LEARNED OF IT.

I SPENT YEARS WORKING ON IDEAS OF MY OWN AND HAVE SEEN FREQUENCIES AS HIGH AS 43GHZ @ 3dBâCONTINUOUSâ. AND THE ONLY THING I EVER REALIZED THAT MAKES MORE SENSE THAN HIGH FREQUENCIES, IS PERMEABILITY. â43GHZ @ 3dBâCONTINUOUSââ, MEANS, A PARTICULAR SIMULATED âPERMEABILITYâ, DROPS AS FREQUENCY GOES UP. TO SOLVE THAT PROBLEM, HIGHER AND HIGHER AMPLITUDES OF CURRENT OR VOLTAGES MAY BE USED TO MANIPULATE PARTICLESâ IN THAT CASE, BY USAGE OF CURRENT, YOU END UP WITH HEAT. NO ONE KNOWS WHAT HAPPENS WITH THE USAGE OF HIGH VOLTAGE LOW CURRENT, THEY HAVE THEIR THEORIES AND HYPOTHESISâ. HENCE, HOW CAN SOMETHING BE MORE THAN ONE PLACES AT THE SAME TIME SO TO SPEAK, IF THEY CANâT EVEN GET TO A NEW LOCATION?

IN THE FIRST PLACE, NO ONE IN THOSE PROJECTS ARE TAKING THOSE TINY STEPS PEOPLE SPEAK OF. THEY WANT IT ALL FAST.

TRY TO FIND OUT WHATâS THE HIGHEST Hz MANAGED ALREADY? WHAT'S THE âPERMEABILITYâ âGRAPHENE TECHNOLOGYââ? AND HOW MANY QUBITS WERE GENERATED AS WORKABLE âENTANGLEDâ CHARACTERS?

DAMN, THIS VIDEO WAS SINCE 2015. I DON'T THINK IN THIS CASE IT MATTERS THOUGH.

9:50 I didn't understand why that happened, when you pass the qubit x through a hadamard gate, and it goes to 0 in laid-back, and 1 in needy cases. Shouldn't it be random and collapse to either 0 or 1, independent of the circuit ?

Yes I do.

I came here three years later cause supposedly Google managed to make a Quantum Computer.

does this means that future ourselves doesn't want us to learn quantum computing, because in future its becomes more weird.

If you think weirdness is like entanglement.

I am a CE/CS major and your videos are really putting some light on topics that i never knew about. Thank you!

I try to understand what âs happened here but no luck. Basically, why quantum computer so fast is because one qbit represents four states(up down left right) while in a normal computer ,one bit can only represent two states(zero or one)right?? However there is no way to figure out left and right because it will change to up or down randomly if you read it? In this video, we can only figure out laid (case a and b )or needy(case c and d), but we still canât figure out which case. so how quantum computer work? The thing is , it just fall back to âone bitâ data , normal one bit represent zero or one, while qbit represent laidback or needy……

well everything's alright but…

when am i going to use it?

and

for what?

can calculations correct emotions?

Hmm, Just around the corner was almost exactly 4 years it seems (if Google's implementation pans out).

I got lost at superposition because i pause it to sleep for a second

well, very nice desciption, probably one maaningfull of 1000 around… so its also that qubit inputs are in fact 2D complex numbers (very hard to encode there, btw), so in fact analog signal where the outputs of quantum gates are always by-design collapsed digital results only? so in fact, it can be simulated also as analog computer ALU made of op-amps? (well, 1D real numbers? very aproximated explanation??)

Honestly bro, you explained qbit the best, with first video ofcourse. Other video seems just repeating another explanation which lead to another confusion. At first I thought it just doesnt make sense to process a random rotating qbit and how qbit become 1 and 0 at the same time. Now i can sleep.

Was all that applied in the Google machine? And if so, does all that implies that the particles used are entangled (you need to create them)? In order to predict the states, it sounds like they should be.

when I feel like someone is watching my deep understanding of quantum concepts it vanishes.

Ohhhh, so quantum computers are magic? Cool, thanks.

ny friend i,m from indonesia and you from america .if i go to america what am i…a citizen or a toerist….so my friend i will always be a 0 and you will be the one or vice versa……..i think i understand Quantum Computers

I understand quantum.

Couldn't you use normal bits to emulate a qubit. You could have a huge array of qubits, with each storing either a 1 or a 0 with additional probability data tacked on. If you want to entangle 2 qubits just add more information saying adopt the state of an entangled qubit only if its state is 1 or 0 with no probability data. All these "quGates" could be programmed in, using if statements. If someone know all the behaviors of these quantum particles you could build a "fake" quantum computer out of a regular one, without providing the conditions needed with a "real" quantum computer.

The only real question with this video is, how do you get an answer and what kind of data would you provide?

Wait a minute ,if i dont understand quantum mechanics does that mean i know it or i dont ,but deep down i know i dont know a heck about it ,but if i say i dont know quantum mechanics am i in superposition state of 4 possibilities where i know quantum mechanics,i know a little about quantum mechanics,i dont know a little about quantum mechanics ,and i dont know anything about quantum mechanics .

But what if i say i know quantum mechanics ,does that mean i dont know quanyum mechanics or i know quantum mechanics or in between .

But i already know the answer to that and i left it to you the reader to find out ,tricky ah .

This is short written form about the shitty video above .no one is going to get anything out of it.

Thanks for your effort making this video, it helped me understand quantum better. But I don't buy into all this mysteriousness and elitism of quantum. If concepts are explained in the correct sequence, there's no reason why anyone shouldn't understand them. That learning sequence should start with explaining every pertinent property of a qubit. Its not clear why you chose the label "right" for a qubit in superposition until we reached solution time, when you introduced the possibility of a qubit being "left". This is counter-intuitive for anyone who has worked with sign-less bits and should be addressed up front.

About floating point representation of irrational numbers overlapping using qbits. Is it possible to represent, for example, sqrt (2) without loss of precision on a quantum computer before reading and collapsing the overlay?