Lecture 016 - Geometry of Grover

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".

Geometry of Grover and Approximate Sharp-SAT

Reflection operation: quantum operation that is reverse of itself

Rotation operation: quantum operation that is not reverse of itself

Geometry of Grover's SAT Algorithm

Remember Grover's SAT Algorithm: assuming unique string $x \in \{0, 1\}^n$ satisfy $f(x) = 1$.

1. We prepare uniform superposition
2. We can convert any code into $AND/OR/NOT$ circuit
3. Do If $C(x_1, x_2, ..., x_n)$ then Minus
4. Reflect accross the mean
5. Go to 3.

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 $|\alpha_{\perp}\rangle \perp |\alpha\rangle$.

Knowing the Probability of Success is $\frac{1}{N}$

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"?

\begin{align*} \cos \theta =& \langle v' | u' \rangle\\ =& \frac{1}{\sqrt{N}} \frac{1}{\sqrt{N}} \langle v | u \rangle\\ =& \frac{1}{N} ((N - 1)(1) + 1(-1))\\ =& \frac{N - 2}{N}\\ =& 1 - \frac{2}{N}\\ =&1-2p\\ \cos^2 \theta =& (1 - \frac{2}{N})^2\\ =& 1 - \frac{4}{N} + \frac{4}{N^2}\\ \simeq& 1 - \frac{4}{N} \tag{by $\frac{4}{N^2}$ negligible}\\ 1-\sin^2 \theta =& 1 - \frac{4}{N}\\ \sin \theta =& \frac{2}{\sqrt{N}}\\ \theta \simeq& \frac{2}{\sqrt{N}}\tag{by $\theta$ small}\\ \end{align*}

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 $k = \frac{\pi}{4}\sqrt{N}$

Knowing the Probability of Success is $\frac{m}{N}$

\begin{align*} \cos \theta =& \langle v' | u' \rangle\\ =& \frac{1}{\sqrt{N}} \frac{1}{\sqrt{N}} \langle v | u \rangle \\ =& \frac{1}{N} ((N - m)(1) + m(-1))\\ =& \frac{N - 2m}{N}\\ =& 1 - \frac{2m}{N}\\ \theta =& 2\sqrt{\frac{m}{N}}\\ \end{align*}

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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