In Grover, we want to know how much angle in subspace we rotate by doing each "combo" and how many times we should perform the "combo".
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: accross a hyperplane (1 dimension lower than the space)
If ... then Minus: It a reflection (only change 1 bit) accross a |\phi_\perp\rangle since we know the reflection does nothing to |\phi\rangle, |\phi^\perp\rangle.
Reflection accross mean: It is a reflection because the uniform superposition \begin{bmatrix}1, 1, 1, 1, ..., 1\end{bmatrix} is the only vector that doesn't get reflected since every entry's amplitude is mean amplitude.
Imagine the horizontal axis denotes everything else that does not satisfy the the function f (x-coordinate denotes the amplitude). 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 ^\circ - \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}})
Conclusion: The angle we rotate by doing one "combo" is \theta = 2\sqrt{p}. And we need to rotate \frac{\pi}{4}\sqrt{\frac{1}{p}} many times.
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