9. Shor's Algorithm I: Understanding Quantum Fourier Transform, Quantum Phase Estimation - Part 3
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- Опубликовано: 15 сен 2024
- Lecturer: Abraham Asfaw
Lecture Notes and Labs: qiskit.org/lea...
#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.
Name and authors of books introduced in the video
0:22 "Quantum Computation and Quantum Information" by Nielsen and Chuang
0:45 "Quantum Computer Science" by Mermin
2:39 "Quantum Computing for Computer Scientists" by Yanofsky and Mannucci
3:10 "Picturing Quantum Processes" By Coecke and Kissinger
Thank you for documenting these!
Hello Abe, you are a true S.T.A.R, thank you for all suggestions, best from Amsterdam Jordaan
outstanding video sir thanks❤ such a complicated topic you explained so nicely smoothly ❤
I think at 33:00 it should be x=(2^n)*Θ/(2π). The counting register then delivers 0≤(2^n)*Θ/(2π)≤(2^n)-1 and thus 0≤Θ≤2π*(1-2^(-n)).
BTW, at 23:00 the case is n=1 because then QFT=H and H=H(dagger) and the 1bit counting register only delivers the values 0 or π.
At 35:50 this is nonsense :-) because the probs have to be ≤ 1.
Yeah and the number in the ket being 1000 times bigger has no connection to the probability of the state being 1000 bigger. I dont see how this improves the precision compared to the one qubit QPE. I think we can tell the phae psi by directly measuring the state after apply inverse QFT and that state is equal to 2^n*theta/2pi, from which we get theta
Hi, I carefully read your first comment. And that's exactly where I got lost. My understanding is like yours, since Abe use theta_psi to replace 2pi*x/ 2^n. And also would you please also help explain at 34:11, how is inverse QFT applied here and got final state | 2^n*theta_psi >. I don't have clue. Is inverse QFT like cancel out all it 2pi/2^n at 31:21? so that only leave theta_psi and 2^n? compare to 21:34, where is our |0> and |1> states? Thank you so much if you could advice!
@erwinmarschall8879 Yes correct. At 33:00, it should be x, instead Abe writes it as Theta_psi
the number in the ket being 1000 times bigger has no connection to the probability of the state being 1000 bigger. I dont see how this improves the precision compared to the one qubit QPE. I think we can tell the phae psi by directly measuring the state after apply inverse QFT and that state is equal to 2^n*theta/2pi, from which we get theta
At 21:42, In the Quantum Phase Estimation (QPE) circuit, I understood the maths, but I have a conceptual problem... The circuit U is not in the path of the first qubit |0>. In fact, the qubit |0> controls the output of U (its the controlling input of the U gate). But when you measure the qubit |0>, the phase created by U affects the measurement of the first qubit. This seems quite counter-intuitive. What's the explanation for this? Is it due to entanglement created by the H and controlled U gates?
Very good question! Usually, as you said, the target qubit's state gets changed by the application of a controlled unitary operation on it, while the control qubit state remains unchanged.
But there are some instances where the control qubit's phase will change, while the target qubit will remain unchanged. This is a bit counter-intuitive, until you do the math yourself. The best example of such a phenomenon is called 'phase kickback'. Look it up and try to follow the math to understand how it works!
Tum toh bohot mast kaam karta hai Abraham bhai! #IndianMeme for appreciation!! Abe....you rock!
Thank you for sharing knowledge
Hello Abe, first of all thank you for this very inspiring course. But I have a question: what if the scaled eigevalue we find after the QFT^-1 transformation, exceeds the size of the classical bits of the circuit? I mean, with n bits we can represent up to 2^n-1, what happens if theta_phi * 2^n > 2^n-1? Am I missing something? Thank you again
The phase has a period of 2pi. What I mean by that is that e^(2pi + theta) = e^(theta) because of complex numbers and how phase works. This should mean that anything with theta_phi*2^n > 2^n -1 would have an equivalent within the interval; some phases with an integer 2pi difference would look the same.
An important note is that in section 3.8 of the qiskit online textbook there is a factor of 2pi outside of theta; There seems to be a /2pi missing for the state expression |2^n * theta_psi>
The /2pi factor is important because it clarifies/tells us that the period is 2^n. If n = 4 qubits this means |17> = |1> for the final result corresponding to phases theta_psi = 17pi/16 and pi/16 respectively. I think that the final measured output will be |1> for both theta_psis as like you said 17 cannot be expressed as only 4 qubits.
I think at 33:00 it should be x=(2^n)*Θ/(2π). The counting register then delivers 0≤(2^n)*Θ/(2π)≤(2^n)-1 and thus 0≤Θ≤2π*(1-2^(-n)).
BTW, at 23:00 the case is n=1 because then QFT=H and H=H(dagger) and the 1bit counting register only delivers the values 0 or π.
2:26 important reason of choosing a book
What does it really mean to apply U^(2^n - i) ? Are we exponentiating U ? Why ?
No, We are applying U to that many times in series.
Jitni tareef karo, utni kam h.
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