8.Shor's Algorithm I: Understanding Quantum Fourier Transform, Quantum Phase Estimation - Part 2

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Qiskit

Qiskit

3 жыл 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: 50
@gerardomoscatelli9035
@gerardomoscatelli9035 3 жыл бұрын
This simple explanation of the 1-qubit QFT as Hadamard gate deserves the prize of the best short QFT explanation ever ! First time I can finally intuitively understand what is going on ! Excellent video.
@Tyokok
@Tyokok Жыл бұрын
oh god! the best QFT explanation ever on KZfaq that I can find. Thank you so much professor!
@dieterrothbayern
@dieterrothbayern 3 жыл бұрын
Explained perfectly. - limited to the essentials - only the math that is necessary - Explicit reference to the important parts of the content - very clear, as far as this is possible with quantum computing This also applies to the following videos in the series.
@horaciodottori3924
@horaciodottori3924 3 жыл бұрын
Abraham, I'm a 78 years old student. I Fully agree with Gerardo Moscatelly, Your explanation of the notation used is just brilliant.
@Taidetunis
@Taidetunis Жыл бұрын
I watched 100 different videos to understand QFT and this is exactly what I was looking for. Thank you so much !
@jialaiz9101
@jialaiz9101 2 жыл бұрын
I LOVEEEE the guy explaining Shor's algo, 1000000 times better than my prof:)
@johannasommer6822
@johannasommer6822 3 жыл бұрын
This has saved my pre-QC exam evening, thank you so so much for these videos!
@japnitahuja
@japnitahuja 8 ай бұрын
The lecturer does an amazing job! I can follow everything easily. The way he explains and breaks down things is simply great!
@mathematicalninja2756
@mathematicalninja2756 Жыл бұрын
Understood the entire quantum Fourier transform and it’s amazing how a qubit encodes amplitude as well as relative phase at the same time
@oscarnadjar7516
@oscarnadjar7516 2 ай бұрын
I have seen, read and study the QFT a lot, finally I understand it, Abraham is amazing. I'll keep with this series, thanks a lot.
@MrDroenix
@MrDroenix 3 жыл бұрын
Appreciate all the work and knowledge you're putting out there! While I may not fully comprehend the knowledge given, it's surely being absorbed over time as exposure to the material happens. Hope to keep motivated in this exciting upcoming field, thank you again
@MrZeeew
@MrZeeew 2 жыл бұрын
Abraham Asfaw is a qisikit gift
@GaugeMcArora
@GaugeMcArora Жыл бұрын
While watching the lecture. When I was 15 minutes in, I had the urge to listen to Mr Tambourine man by Bob Dylan. I’ll be back in 5 minutes.
@sr33dhar
@sr33dhar 2 жыл бұрын
At 43:15, after step 1, I believe Abraham meant we apply a Hadamard to get the exp(2.pi.i.x1/2) phase instead of having applied a URot. Is this a minor slip of the tongue? One could argue that the Hadamard is a special case of the URot though, where k = 1.
@gourabbanerjee9531
@gourabbanerjee9531 10 ай бұрын
now i actually understood QFT including mathematics and physical realization ❤ thank u so much sir you are amazing
@lengooi6125
@lengooi6125 2 жыл бұрын
Very well done. Superb explaination simply awesome stuff and thank you for saving me lots of time trying to figure QFT on my own
@BrendanBarry-tv7uj
@BrendanBarry-tv7uj 4 ай бұрын
@MudahnyaFizik Asked "At 45:40, you claimed that the phase change for |1> of the first qubit in the circuit is the same but in reversed order of the |1> in the QFT. I don't get it. In the QFT, we have only a single phase shift due to the nth qubit. However, in the circuit, we have phase change due to all other qubits. How is the two state the same? " I had the same question until I realized the x in the expanded version shown at 45:40 is the full x, not x_n. Using what we saw for the expanded y: x = sum(k=1 to k=n) x_k * 2^(n-k) = 2^(n-1) * x_1 + 2^(n-2) * x_2 + ... + 2^(n-n) * x_n Therefore expanding out the term shown at 45:40 (|0> + exp(i2pi x/2^n) |1>) = (|0> + exp(i2pi * [2^(n-1) * x_1 + 2^(n-2) * x_2 + ... + 2^(n-n) * x_n] /(2^n)) |1>) Canceling out the numerators using the (1/2^n) we find: (|0> + exp(i2pi x/2^n) |1>) = (|0> + exp(i2pi * [2^(-1) * x_1 + 2^(-2) * x_2 + ... + 2^(-n) * x_n]) |1>) Which is the exact expression shown at 45:37
@ai3t86
@ai3t86 2 жыл бұрын
simplest explanation on this topic. Cheers
@nilakanthameher2406
@nilakanthameher2406 3 жыл бұрын
Thank you so much for such a simple explanation......
@nastyavicodin6229
@nastyavicodin6229 8 ай бұрын
Thanks! You are a good teacher
@Radical9535
@Radical9535 Жыл бұрын
something i completely understand whow! ty!
@Itachi__Uchiha-ck4mk
@Itachi__Uchiha-ck4mk 2 жыл бұрын
All I can say is Best presentation 👍
@MudahnyaFizik
@MudahnyaFizik 3 жыл бұрын
At 45:40, you claimed that the phase change for |1> of the first qubit in the circuit is the same but in reversed order of the |1> in the QFT. I don't get it. In the QFT, we have only a single phase shift due to the nth qubit. However, in the circuit, we have phase change due to all other qubits. How is the two state the same?
@alessandroc.4543
@alessandroc.4543 8 ай бұрын
I'm having the same problem
@martingrillo6956
@martingrillo6956 2 жыл бұрын
In the example (green) till 26:24 - 3 Qubits are used; N is 2^3 = 8 but QFT|x> takes it's decimal value x="five" into account (in the exponent), while building the 3 parts contianing scalar product and ignores the binary representation of the number - which says: my middle term is zero... That looks like an error atm.
@granteberle4879
@granteberle4879 3 жыл бұрын
These videos are amazing! I really appreciate you guys posting these lectures as I could not participate in the class this summer. I have a quick question I ran into during the notes on the QFT. At 16:25, Abe writes a y^{k} (superscript k). Should this term be a y_{k} (subscript k)? I think he might have been distracted by explaining the cancelation of the 2^{n} because he writes it as y_{k} later on. I don't blame him... not an easy topic to cover so quickly and he did an amazing job👍
@ethiopianqubit
@ethiopianqubit 3 жыл бұрын
You got it Grant, that's a typo. Thanks for catching it! I was looking at the comments to see of people are following along while also writing this down, must have been distracted a little.
@noelsajenmathew5472
@noelsajenmathew5472 Жыл бұрын
@@ethiopianqubit ABE there should be a "x" with it also pls correct if i am wrong
@tiyashaghosh9933
@tiyashaghosh9933 3 жыл бұрын
Hi, this course has been amazing so far! but it would be of great help for us who couldn't attend the live lectures and doing the course now, if we could see those live chats. it would be helpful because we can also look for some great questions apart from those which were being answered on the lectures. Also, I don't know if it is technically possible or not but if it is, I will be highly obliged if the team looks after this matter. Thank you.
@harshitgupta9211
@harshitgupta9211 3 жыл бұрын
At 43.16, you have applied the rotation matrix conditioned on x2 but if that rotation matrix rotates the state by e^((2(pi) i)/ 4), why does the coefficient of the power of e becomes multiplied by x2 if we do that? Shouldn't it be just x1 ?
@trayangospodinov5973
@trayangospodinov5973 10 ай бұрын
When we change the basis, what we essentially get is smaller and smaller rotation. I assume this has physics interpretation. My concern is the follows: If we try to make fourier basis of let's say 1000 bits, wouldn't we required phase that (if it has physical meaning, I suppose it does) is smaller than the planck's constant, making it practically unimplementable? Therefore not much more useful, than classical computer (cuz we will be capped at basis with cardinality let's say 200-300). I want to clarify, that my physics background is nonexistent. I'm sure I didn't state my question as one competent in the field do, but I hope it's understandable.
@abduhu8437
@abduhu8437 2 жыл бұрын
Thank you! Please! Which software do you use for those lectures?
@paymanmahmoudi5015
@paymanmahmoudi5015 23 күн бұрын
Can anyone explain what the swap gate was for in the end?
@vishalkumarrajeshbhaigohel654
@vishalkumarrajeshbhaigohel654 6 ай бұрын
at time stamp 45:38 how the first expression is exactly equal to the derivation. It has more terms in the form of exp multiplied to the original term that we are looking for.
@BrendanBarry-tv7uj
@BrendanBarry-tv7uj 4 ай бұрын
@MudahnyaFizik Asked a similar question: "At 45:40, you claimed that the phase change for |1> of the first qubit in the circuit is the same but in reversed order of the |1> in the QFT. I don't get it. In the QFT, we have only a single phase shift due to the nth qubit. However, in the circuit, we have phase change due to all other qubits. How is the two state the same? " I had the same question until I realized the x in the expanded version shown at 45:40 is the full x, not x_n. Using what we saw for the expanded y: x = sum(k=1 to k=n) x_k * 2^(n-k) = 2^(n-1) * x_1 + 2^(n-2) * x_2 + ... + 2^(n-n) * x_n Therefore expanding out the term shown at 45:40 (|0> + exp(i2pi x/2^n) |1>) = (|0> + exp(i2pi * [2^(n-1) * x_1 + 2^(n-2) * x_2 + ... + 2^(n-n) * x_n] /(2^n)) |1>) Canceling out the numerators using the (1/2^n) we find: (|0> + exp(i2pi x/2^n) |1>) = (|0> + exp(i2pi * [2^(-1) * x_1 + 2^(-2) * x_2 + ... + 2^(-n) * x_n]) |1>) Which is the exact expression shown at 45:38
@jasonandrewismail2029
@jasonandrewismail2029 2 жыл бұрын
Fantastic explanation. May I know the name of the presenter so I can search for other videos he has created.
@user-pp7bl1mx5c
@user-pp7bl1mx5c 11 күн бұрын
QFT에서는 n번째 큐비트로 인해 단일 위상 변이만 발생합니다. 47:00 회로에서는 다른 모든 큐비트로 인해 위상 변화가 있습니다. 두 상태가 어떻게 동일합니까?
@Chronosaur
@Chronosaur 3 жыл бұрын
I heard Peter Shor's actually a really bad lecturer - you're never Shor what he's talking about
@sumaiamannan2635
@sumaiamannan2635 Жыл бұрын
I am trying to draw this circuit but got an attribute error “QuantumCircut has no cu1”, how can I fix it please help asap.
@leonmozambique533
@leonmozambique533 3 жыл бұрын
where can we get the code mentioned in the end that was posted in the discord. I’m not in the discord
@shamanbhattacharyya9285
@shamanbhattacharyya9285 3 жыл бұрын
Wait. Y= |4> is a state in the lab frame
@davidwilkie9551
@davidwilkie9551 3 жыл бұрын
Not to complain (again) about " A Rose by any other name will smell as Sweet" and QM-TIMESPACE is the Engineering of Spacetime, a phase-locked substantion shaping of Math-Phys-Chem and Geometry here-now-forever formatting of /by e-Pi-i sync-duration, ie lines of sight projection-drawing (E=mC²) density-intensity superposition quantization of time-timing Superspin, In-form-ation. The Measurement Problem, or Uncertainty Principle consequence of simultaneous Duality of multiphase-locked holographic photon-phonon universally is probably a jargon for Amateur Observer's compositions of coherence-cohesion objectives, so the dominant Nomenclatures of Scientific Analysis are in a state of confusion, which can only be sorted out by talented Artists who have a natural appreciation of macro-micro Quantum-fields Perspective. Eg the common sensing of phenomena is directly related to Neurological arrangements in Brains, and Computational Information is Quantum Chemistry. So reference Sir Rodger Penrose, and Robert Sapolsky etc, before guessing about the logarithmic probability dominant resonance conglomeration connection..? This same does not apply to Computational Methodologies Sciencing, but thanks for the attempt to make the subject more accessible.
@sakuranooka
@sakuranooka 3 жыл бұрын
Yes
@mariak8480
@mariak8480 2 жыл бұрын
Had a stroke trying to read this
@douglasbeveridge8718
@douglasbeveridge8718 3 жыл бұрын
Where can we get the demo-2.ipynb . The MYQFT is very different from the one used by qiskit in the qiskit book
@qiskit
@qiskit 3 жыл бұрын
qiskit.org/learn/intro-qc-qh
@syfulakash2987
@syfulakash2987 3 жыл бұрын
Can you please help me with the tensor product part from 19:25 timestamp? I didn't understand how the instructor introduced those tensor product...
@MudahnyaFizik
@MudahnyaFizik 3 жыл бұрын
Try to do the summation part by part (using the binary y instead of decimal y).
@sakuranooka
@sakuranooka 3 жыл бұрын
In all the derivation in blue through 19:54, due to the product xy in the exponent of the exponential function, the formulas only make sense for x a scalar. After 19:54 you suddenly claim that "we went from |x> = |x1 x2... xn>". What would be the meaning of the product xy in the exponent if |x> really was |x1 x2 ... xn>? And then you go on and use green arrows to map each one of the factors of the tensor product |x> to the tensor factors of |x_tilde>, even though the whole derivation of |x_tilde> is only meaningful if x is assumed to be a scalar. You lose me here...
@mariak8480
@mariak8480 2 жыл бұрын
x1, x2, …, xn in the exponential are just the scalar values that stand in the statevector in it’s starting position, i.e. |x1,x2,…xn>. If you have the dirac brackets, then yes, we’re talking about vectors. If not, scalars.
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