# The Quantum Search Algorithm

2018-07-01

The algorithm for finding solutions to the search problem using a quantum computer finds a solution with a much smaller number of steps than a classical computer.

In the case of a quantum computer, when there is only one solution to the problem and $$N$$ elements to search (for example, we want to find a password from $$N$$ possible passwords), the number of operations required is $$O(\sqrt{N})$$, while for classical computers it is $$O(N)$$.
For example, if we have a password consisting of $$6$$ characters, the number of which is equal to $$50$$, then on a classic computer we would need to check $$50^{6}=15625000000$$ possible passwords, whereas on a quantum computer only approximately $$50^{3}=125000$$ iterations are needed to find correct password.

### Definitions

Suppose we have a space of $$N$$ elements that we will search in order to find exactly $$M$$ solutions for a given problem, we assume that $$M \in [1 ,N/2]$$. Instead of looking for elements we will look for the indexes of these elements in the range from $$0$$ do $$N−1$$. In addition, we will simplify the situation assuming that $$N =2^{n}$$, that is, index is stored in $$n$$ bits. The solution of the problem can be described by the function $$f(x)$$, which takes the index $$x$$ as an argument and returns the value $$1$$ if the index matches the solution or $$0$$, if the given index is not a solution to the problem.

Let’s start with defining the operation $$O$$ as follows:

$$$$\left \vert x\right \rangle \left \vert q\right \rangle \overset{O}{ \longrightarrow }\left \vert x\right \rangle \left \vert q \oplus f(x)\right \rangle$$$$
where $$\oplus$$ denotes addition modulo $$2$$. That is, a single qubit $$\left \vert q\right \rangle$$ which is equal to $$\left \vert 0\right \rangle$$ or$$\left \vert 1\right \rangle$$ will be changed only if $$f(x) =1$$, which means that $$x$$ is the solution.

Below we will use qubit:

$$$$\left \vert q\right \rangle =\frac{1}{\sqrt{2}}(\left \vert 0\right \rangle -\left \vert 1\right \rangle )$$$$

which we get from qubit $$\left \vert 0\right \rangle$$ after operation $$HX$$, where:

\begin{align}X =\left (\begin{array}{cc}0 & 1 \\ 1 & 0\end{array}\right ) \\ \end{align}

and

\begin{align} H =\frac{1}{\sqrt{2}}\left (\begin{array}{cc}1 & 1 \\ 1 & -1\end{array}\right )\end{align}

where $$H$$ is called the Hadamard operation.

Therefore, $$O$$ operation can also be defined as follows:

$$$$\left \vert x\right \rangle \left \vert q\right \rangle \overset{O}{ \longrightarrow }( -1)^{f(x)}\left \vert x\right \rangle \left \vert q\right \rangle$$$$

### Algorithm

The goal of the algorithm is to find a solution at the smallest possible amount of the operation of $$O$$.

We start from the computer in state $$\left \vert 0\right \rangle ^{ \otimes n}\left \vert 0\right \rangle$$ ($$\left \vert 0\right \rangle ^{ \otimes n}$$ corresponds to the index $$x=0$$), applying the Hadamard transform $$H^{ \otimes n}$$ to state $$\left \vert 0\right \rangle ^{ \otimes n}$$ and $$HX$$ to last qubit. We get equal superposition state:

$$$$\left \vert \psi \right \rangle =\frac{1}{N^{1/2}}\sum \limits _{x =0}^{N -1}\left \vert x\right \rangle \left \vert q\right \rangle$$$$
The quantum algorithm consists of the repeated application of the $$G$$ operator, whose single application can be described as follows:

1. Apply $$O$$
2. Apply the Hadamard transform $$H^{ \otimes n}$$
3. Phase change to $$-1$$ in each state $$\left \vert x\right \rangle \left \vert q\right \rangle$$, except $$\left \vert 0\right \rangle \left \vert q\right \rangle$$:
$$\left \vert x\right \rangle \left \vert q\right \rangle \longrightarrow -( -1)^{\delta _{x0}}\left \vert x\right \rangle \left \vert q\right \rangle$$
4. Apply the Hadamard transform $$H^{ \otimes n}$$

By combining steps 2, 3 and 4 we get:

$$$$H^{ \otimes n}(2\vert 0\rangle \left \vert q\right \rangle \left \langle q\right \vert \left \langle 0\right \vert -I)H^{ \otimes n} =2\left \vert \psi \right \rangle \left \langle \psi \right \vert -I$$$$
Therefore, the $$G$$ operator can be expressed by means of a formula $$G =(2\left \vert \psi \right \rangle \left \langle \psi \right \vert -I)O$$.
Let’s define two orthonormal states:

\begin{align}\left \vert \alpha \right \rangle =\frac{1}{\sqrt{N -M}}\sum _{x}\left \vert x\right \rangle \left \vert q\right \rangle \\\left \vert \beta \right \rangle =\frac{1}{\sqrt{M}}\sum _{y}\left \vert y\right \rangle \left \vert q\right \rangle \end{align}

where sum over $$y$$ is sum over all indexes which are solutions to the search problem, and sum over $$x$$ is sum over all indexes which are not solutions to the search problem. It is easy to show that the following equality occurs:

$$\left \vert \psi \right \rangle =\sqrt{\frac{N -M}{N}}\left \vert \alpha \right \rangle +\sqrt{\frac{M}{N}}\left \vert \beta \right \rangle$$

By calculating the matrix elements $$G$$ in the basis states $$\left \vert \alpha \right \rangle$$ and $$\left \vert \beta \right \rangle$$ we obtain, that $$G$$ is the operator of rotation in the two-dimensional space spanned by these states:

$$G =\left (\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right )$$

where $$\cos \theta =(N -2M)/N$$. State $$\left \vert \psi \right \rangle$$ can be expressed by $$\theta$$ as:

$$\left \vert \psi \right \rangle =\cos \frac{\theta }{2}\left \vert \alpha \right \rangle +\sin \frac{\theta }{2}\left \vert \beta \right \rangle$$

Applying of $$G$$ $$k$$ times we obtain:

$$G^{k}\left \vert \psi \right \rangle =\cos \left (\frac{2k +1}{2}\theta \right )\left \vert \alpha \right \rangle +\sin \left (\frac{2k +1}{2}\theta \right )\left \vert \beta \right \rangle$$

Therefore, by applying the operator $$G$$ many times, we turn the state $$\left \vert \psi \right \rangle$$ close to the state $$\left \vert \beta \right \rangle$$, so observation yields the correct answer with high probability.
Because the initial state has the form $$\left \vert \psi \right \rangle =\sqrt{\frac{N -M}{N}}\left \vert \alpha \right \rangle +\sqrt{\frac{M}{N}}\left \vert \beta \right \rangle$$, so in the space of two basis states $$\left \vert \alpha \right \rangle$$ and $$\left \vert \beta \right \rangle$$ rotation of the initial state by an angle $$\arccos \sqrt{M/N}$$ transform it into the state $$\left \vert \beta \right \rangle$$. Let $$[z]$$ be the nearest integer for the real number z. Thus, the number of rotations made by $$G$$:

$$$$R =\left [\genfrac{}{}{}{}{\arccos \sqrt{M/N}}{\theta }\right ]$$$$

transform the state $$\left \vert \psi \right \rangle$$ close to the solution $$\left \vert \beta \right \rangle$$. Because $$R\leq [\pi /2\theta ]$$, so an upper bound on $$R$$ is taken from a lower bound on $$\theta$$. Because $$\theta /2 \geq \sin \theta /2 =\sqrt{\frac{M}{N}}$$, hence we get the upper bound of the required amount of iteration:

$$R \leq \left [\frac{\pi }{4}\sqrt{\frac{N}{M}} \, \right ]$$

Finally we get, that $$R =O(\sqrt{N/M})$$. Therefore only for one solution, $$M=1$$, $$R =O(\sqrt{N})$$, in the classic case we would get $$R=O(N)$$.

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