Thanks for the comment. 2*(5/8)* 2 pi = (20/8) pi = (5/2) pi. Since this represents a rotation of (5/2) pi around a circle it is equivalent to rotating (1/2) pi around the circle (since 2 pi is the entire circle we can remove that from the rotation and still land on the same position).
For the notation at 4:11 it'd be clearer to have the thing that is to the power of j_1 in brackets because it was momentarily confusing to see it as the power of the power, cause that'd actually lead to a multiplier of e
Why is the intial rotation 5/8 times 2*pi? Is it because of the fact that 5 is the 5th out of 8 possibilites the number has? And why are we considering like this? I mean, partitioning 360 into total number of possibilites the number can have?
Thanks for the comment! The QFT transforms the state from being in the computational basis (0's and 1's with no superposiiton) to being in superposition with the initial 0's and 1's being encoded in the phase of the qubit. The first qubit will always have phase x/2^n * 2pi, where x is the initial number of the state and n is the number of qubits. We do this because since this is a quantum circuit, it is reversible; this is used in reverse, where we encode the answer to a problem into the phase of qubits and then use the QFT to read the phase/answer. The next video of the course goes through the QPE which uses the QFT.
Hi, big thanks for this video. What does the QFT of some set of qubits actually mean? As in what does that set of qubits represented on screen actually represent? with a DFT you input a sequence of amplitudes, and the DFT returns a sequence where each complex element represents the amplitude of the sine and cosine waves associated with it's index right? So how does inputting the number 7 into a QFT mean anything? Sorry I'm just a little confused
I have a confusion at 5:56 . There is this |k> where k ranges from 0 to 2^(n)-1. I think the k represents the decimal form of of the binary state of the j string, right? Like |000> = |0>, |001>=|1>, |100>=|4>, |111>=|7>. Am I correct or I am missing something?
I sincerely respect your effort to make videos helpful to make sense!
This is simple yet elegant
Im impressed. New subscriber here
You are the best. Short and precise
thanks a lot for that great explanation and the animation
Problem Sets for this Course: drive.google.com/drive/folders/1A-RHTQFRY_pipVfItQBxMU-xEexRESQj?usp=sharing
Thanks for Watching!
At 1:50, 2*(5/8)* 2 pi = 5 pi /2, not pi/2. Hence the last one is 5 pi, not pi. What am I missing?
Thanks for the comment. 2*(5/8)* 2 pi = (20/8) pi = (5/2) pi. Since this represents a rotation of (5/2) pi around a circle it is equivalent to rotating (1/2) pi around the circle (since 2 pi is the entire circle we can remove that from the rotation and still land on the same position).
5* pi/2 = (2n+1)* pi/2 = pi/2
Trigonometry basics
5 (1/2) pi = 5 quarter turns = 1 quarter turn
Thank you, great videos. By the way I dont see any problem set for 3.6 alluded to in the beginning.
For the notation at 4:11 it'd be clearer to have the thing that is to the power of j_1 in brackets because it was momentarily confusing to see it as the power of the power, cause that'd actually lead to a multiplier of e
Why is the intial rotation 5/8 times 2*pi? Is it because of the fact that 5 is the 5th out of 8 possibilites the number has? And why are we considering like this? I mean, partitioning 360 into total number of possibilites the number can have?
Thanks for the comment! The QFT transforms the state from being in the computational basis (0's and 1's with no superposiiton) to being in superposition with the initial 0's and 1's being encoded in the phase of the qubit. The first qubit will always have phase x/2^n * 2pi, where x is the initial number of the state and n is the number of qubits. We do this because since this is a quantum circuit, it is reversible; this is used in reverse, where we encode the answer to a problem into the phase of qubits and then use the QFT to read the phase/answer. The next video of the course goes through the QPE which uses the QFT.
@@quantum-soar Thanks a lot for the explanation and of course, for the videos too..they helped me a lot for my Quantum exam.
Hi, big thanks for this video.
What does the QFT of some set of qubits actually mean? As in what does that set of qubits represented on screen actually represent?
with a DFT you input a sequence of amplitudes, and the DFT returns a sequence where each complex element represents the amplitude of the sine and cosine waves associated with it's index right?
So how does inputting the number 7 into a QFT mean anything? Sorry I'm just a little confused
In the formula at 5:40, is the j term in the exponent in binary form or decimal form? I don't quite get it.
If you work it out for multiple qubits, you will see that j is in decimal form that satisfies the formula to be true.
I have a confusion at 5:56 . There is this |k> where k ranges from 0 to 2^(n)-1. I think the k represents the decimal form of of the binary state of the j string, right? Like |000> = |0>, |001>=|1>, |100>=|4>, |111>=|7>. Am I correct or I am missing something?
Thanks for the comment! Yes that is correct k is the decimal form.
Love your videos. Just curious, what software(s) do you use to create your videos?
Please we need simon' s algorithm and grover' s search algorithm
Anyone else here after the paper by Yilei Chen?
Me 😅