Grover Iteration Homework: Realizing O_psi with Hadamard Operators

[neu]

Homework Statement

One of the operators used in the Grover iteration is:

$$\hat{O_{\psi} }= 2 \mid \psi \rangle \langle \psi \mid - I$$

where $$\mid \psi \rangle = \frac{1}{\sqrt{N}} \Sigma^{N-1}_{x=0} \mid x \rangle$$

Show that the operator:

$$\hat{O_{0} }= 2 \mid 000...0 \rangle \langle 000...0 \mid - I$$

acting on a register of $$log_{2} N$$ qubits, the operator $$\hat{O_{\psi} }$$ can be realized with the use of hadamard operators

The Attempt at a Solution

Now I know that the answer is :

$$\hat{O_{\psi} }=\hat{H}^{\otimes N}\hat{O_{0} }\hat{H}^{\otimes N}$$

and i tried to evaluate this explicity. for example, in the case N =1

$$\hat{H}\hat{O_{0} }\hat{H}= \left( \begin{array}{c c} 1 & 1 \\ 1 & 1 \end{array} \right) - I$$

And the array is the sum

$$\mid 0 \rangle \langle 0\mid + \mid 0 \rangle \langle 1\mid + \mid 1 \rangle \langle 0\mid + \mid 1 \rangle \langle 1 \mid$$

but This is reverse engineering which I'm not happy with. Does anybody know how to derive the relationship $$\hat{O_{\psi} }=\hat{H}^{\otimes N}\hat{O_{0} }\hat{H}^{\otimes N}$$??

 

[Jhero]

I would like to first clarify the notations used in the problem. The operator \hat{O_{\psi}} is defined as the Grover operator, which is used in the Grover iteration algorithm to amplify the amplitude of the desired state in a quantum register. The state \mid \psi \rangle is the equal superposition state, which is a uniform superposition of all the basis states in the register. This state is used as the initial state in the Grover iteration algorithm.

Now, to show the relationship \hat{O_{\psi} }=\hat{H}^{\otimes N}\hat{O_{0} }\hat{H}^{\otimes N}, we can start by evaluating the operator \hat{O_{\psi}} explicitly. We can rewrite \hat{O_{\psi}} as:

\hat{O_{\psi}} = 2 \frac{1}{N} \sum^{N-1}_{x=0} \mid x \rangle \langle x \mid -I

Substituting the value of \mid \psi \rangle in this expression, we get:

\hat{O_{\psi}} = 2 \frac{1}{N} \sum^{N-1}_{x=0} \frac{1}{\sqrt{N}} \mid x \rangle \langle x \mid -I

This can be further simplified to:

\hat{O_{\psi}} = \frac{2}{N} \sum^{N-1}_{x=0} \sum^{N-1}_{y=0} \mid x \rangle \langle y \mid -I

Using the fact that \mid x \rangle \langle y \mid = \hat{H}\mid 0 \rangle \langle 0 \mid \hat{H}^{\dagger} where \hat{H} is the Hadamard operator, we can rewrite the above expression as:

\hat{O_{\psi}} = \frac{2}{N} \sum^{N-1}_{x=0} \sum^{N-1}_{y=0} \hat{H}\mid 0 \rangle \langle 0 \mid \hat{H}^{\dagger} -I

Now, using the property of the Hadamard operator \hat{H}^{\dagger}\hat