Reflection operation: quantum operation that is reverse of itself
Rotation operation: quantum operation that is not reverse of itself
Remember Grover's SAT Algorithm: assuming unique string x \in \{0, 1\}^n satisfy f(x) = 1.
Now we aim for a geometric interpotation of Grover's algorithm.
Observations:
Reflection accross mean: The reflection only changes one bit, and therefore it is a reflection accross a hyperplane (1 dimension lower than the space). In high demension, the reflected version is very close in \cos(\theta) to the original vector.
If ... then Minus: It is also a reflection, but accross a line. This is because the uniform superposition \begin{bmatrix}1, 1, 1, 1, ..., 1\end{bmatrix} is the only vector that doesn't get reflected.
Imagine the horizontal axis denotes everything else that does not satisfy the the function f. Vertical axis is the bit we sill keep reflecting. We start from uniform position |\phi\rangle, and do "If ... then Minus" (reflection across |\alpha_{\perp}\rangle), and do "Reflection Accross Mean" (reflection across \simeq |\phi\rangle, although not exact)
If we use un-normalized representation, imagine the circle to be length \sqrt{N}
If you do too many times of V and W, for example, twice of amount, you will go pass our goal. In each combo V \cdot W, the angle \theta of the vector got changed by a constant amount. In numerical representation, we observe we get the the goal with slower and slower speed, this is because of how we measure the successfulness: we measure it by probability. At the start, we increase the probability to go to our goal a lot, but it becomes slower and slower as two angle approaches even though we change it by a constant amount every time.
The total amount of rotation is: 90 \degree - \theta/2 \simeq \pi/2 since, according to the graph above |\alpha_{\perp}\rangle \perp |\alpha\rangle.
Assume |v\rangle = V|\phi\rangle = [1,...,1,-1,1,...,1], |u\rangle = |\phi\rangle = [1,...,1]
We then ask: how much \theta did we rotate when we do one "combo"?
If you want to do k rotations instead of 1, you will rotate by about \frac{2k}{\sqrt{N}}. Therefore, if you want to rotate \frac{\pi}{2}, you need k = \frac{\pi}{4}\sqrt{N}
If you want to do k rotations instead of 1, you will rotate by about 2k\sqrt{\frac{m}{N}}$. Therefore, if you want to rotate \frac{\pi}{2}, you need k = \frac{\pi}{4}\sqrt{\frac{N}{m}} \in O(\frac{1}{\sqrt{p}})
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