7. Shor's Algorithm I: Understanding Quantum Fourier Transform, Quantum Phase Estimation

7. Shor's Algorithm I: Understanding Quantum Fourier Transform, Quantum Phase Estimation - Part 1

Рет қаралды 69,992

Күн бұрын

Lecturer: Abraham Asfaw
Lecture Notes and Labs: qiskit.org/learn/intro-qc-qh
#Qiskit
This course is an introduction to the world of quantum computing, with an exploration of some of the key quantum algorithms and their implementations using quantum circuits, as well as the quantum hardware that is designed to run these algorithms. The course was first offered during the Qiskit Global Summer School in July 2020 as a two-week intensive summer school.

Пікірлер: 33

@Chronosaur 3 жыл бұрын

I reckon when he first published his paper on this, loads of people were like "are you Shor?"

3 жыл бұрын

You are truly a great teacher! I love the way you make complicated stuffs look so easy :-)

@tomgraupner171 3 жыл бұрын

This series of six videos reg. Shor's and prerequesites is soooo awesome. I'm teaching myself Quantum Mechanics and QC and there's lots of material around. These videos belong to the top of my personal list! THANKS A LOT !!!

@pranshisaxena6063 2 жыл бұрын

This the best explanation, I have found so far. Thank you!

@_oubliette 3 жыл бұрын

This is literally invaluable, thank you so much

@fleurette210 3 жыл бұрын

Awesome! Thank you for the free content!

@saikrishnasunkam4344 3 жыл бұрын

I tried learning QC like 6 years ago and gave up after two weeks because I had no clue what the books were talking about. God I wish I had this legend teaching me back then.

@satishchandramedi9729 3 жыл бұрын

Same for me I have heard about this quantum computing in 2016 and only managed to learn basics at that time.....now learning slowly by digging deeper with the help of these lectures

@ferashamdan4252 3 жыл бұрын

Thank you. An excellent tutor

@yevonnaelandrew9553 3 жыл бұрын

Thank you for the content!

@anthonyvenable3048 Жыл бұрын

Thanks so much for the video, it really helps!

@nastyavicodin6229 8 ай бұрын

Thank you for very good explanation!

@senkum80 2 жыл бұрын

Thanks a lot. Saved my time

@sakuranooka 3 жыл бұрын

22:22 According to your definition, the sine function would not be periodic. I think you should replace the "iff" by a simple "if". Actually the "iff" should be elsewhere, namely: f(x) is periodic with period p "iff" {f(x) = f(y) "if" |x - y| = k p for all k in N"}

@pragneshprajapati774 2 жыл бұрын

Thank you !!!

@xShadzHD 3 жыл бұрын

I'm confused at around 23:30 when you define the strict definition. Surely when x = pi/3 say, and y = 2pi/3 then f(x) = f(y), x does not equal y, but mod(x-y) does not equal an integer multiplied by the period? Should the implication only be one way?

@ethiopianqubit 3 жыл бұрын

The moment you move slightly away from pi/3 in either direction, you'll see that the same thing doesn't hold at sin(2pi/3). In other words, sin(pi/3 + eps) =/= sin(2pi/3 + eps), but instead sin(pi/3 + eps) = sin(pi/3 + 2*pi + eps). That's why the implication goes both ways, but in this case the period p is 2pi and not (2pi/3-pi/3 = pi/3) as you have in your comment.

@vardhan254 Ай бұрын

35:41 FOR QFT

@jxchtajxbt53 2 жыл бұрын

If you consider orthogonal matrices the same as Unitary than OK. But in the space of real numbers Orthogonal matrices preserve norms.

@aaroningram9465 3 жыл бұрын

Super easy..so far😀

@mavihs26 Ай бұрын

how can we see the animation is there a new link?

@pghislain 2 жыл бұрын

4G telecom uses information spreading and FT >>> send ... receive

@user-jq5bz4wp2e 9 ай бұрын

Is the If and only If right? I can see that f(x) = f(y) if |x-y| is a multiple of the period. But I don't see that this is true only if |x-y| is a multiple of the period.

@infinity-and-regards 3 жыл бұрын

23:46 The implication does not go both ways right?

@sagardollin475 3 жыл бұрын

In this case it goes both ways, for every time you start at x and cross a distance of k*p , you end up at the same value of the function given by f(x) irrespective of what x was to begin with.

@danielraymond3045 3 жыл бұрын

I think you're right. sin(0) == sin(pi) but that is not an integer "k" multiples of "p" (2*pi) away from the cycle.

@ethiopianqubit 3 жыл бұрын

@@danielraymond3045 Right. But the moment you move slightly away from 0 in either direction, you'll see that the same thing doesn't hold at sin(pi). In other words, sin(0 + eps) =/= sin(pi + eps), but instead sin(0 + eps) = sin(0 + 2*pi + eps).

@ethiopianqubit 3 жыл бұрын

@@sagardollin475 You got it Sagar

@danielraymond3045 3 жыл бұрын

@@ethiopianqubit I think I'm missing something. f(x) == f(y) ONLY if |x - y| = kp. So, if there was a case where f(x) == f(y) and |x - y| =/= kp, then the double implication breaks, no?

@BigDaddyGuagg 3 жыл бұрын

God bless. You saved my ass.

@SampleroftheMultiverse 5 ай бұрын

U Shape Wave Thanks for your informative and well produced video. Hello,You and your viewers might find my quantum-like analog interesting and or useful. I have been trying to describe the “U” shape wave that is produced in my amateur science mechanical model in the video linked below. I hear if you over-lap wave together using Fournier Transforms, it may make a “U” shape or square wave. Can this be correct representation Feynman Path Integrals? In the model, “U” shape waves are produced as the loading increases and just before the wave-like function shifts to the next higher energy level. Your viewers might be interested in seeing the load verse deflection graph in white paper found elsewhere on my KZfaq channel. Actually replicating it with a sheet of clear folder plastic and tape and seeing it first hand is worth the effort. kzfaq.info/get/bejne/rdhyptSaqpbad2g.htmlsi=waT8lY2iX-wJdjO3

@alextran8906 2 жыл бұрын

hmm handwriting is hard to read for people with disabilities and foreign students. IBM should hire a Educational Technology to not create unnecessary cognitive overload.