Quantum Computing Asked on February 23, 2021
I was reading about the Grover Search algorithm on https://qiskit.org/textbook/ch-algorithms/grover.html#example. I understood the method but I have a few questions. My question regards the two-qubit case.
Does the diffusion operator $D=2|sranglelangle s|-1$, depend upon the initial state i.e $|+rangle|+rangle$ and the marked state?
Actually I was reading an article https://journals.aps.org/pra/pdf/10.1103/PhysRevA.68.022306, which had an equation
begin{equation}
-U_{S_j}|S_jrangle_{w}=|wrangle
end{equation}
with $U_x=1-2|xranglelangle x|$, $S_1=left(dfrac{0+1}{sqrt{2}}right)^{otimes 2}$, and $w$ is the marked state. The other $S_{j’s}$ can be the states for instance $|+rangle|-rangle$, $|-rangle|-rangle$, $|-rangle|+rangle$ etc. with total such $S_j$ being $16$. My question is how does one make the diffusion operator for a state $|+rangle|-rangle$. As an example from the table in the article it states for instance if $j=2$, $S_2=|+rangle|-rangle$
$$-U_{S_2}|S_1rangle_{10}=-|00rangle,$$
where $10=w$ is the marked state.
Can somebody explain how this equation came? can somebody atleast hint at some references?
it should be clear that the core of the Grover algorithm includes 3 steps
a) prepare initial state $|srangle$
b) apply $U=1-2|omegaranglelangle omega|$
c) apply $D=2|sranglelangle s|-1$
then repeat step b and c
In the original Grover algorithm, the diffusion operator is fixed as $D=2|sranglelangle s|-1$, which you can say it depends upon the initial state. Actually in the image of qiskit textbook, you can see the initial state is $|srangle=H^{otimes n}|0rangle^{n}$
In the paper your reference, it extends the Grover algorithm, especially extends the diffusion operator from $|srangle$ (named as $left|S_{1}rightrangle$) to another 15 states $left|S_{j}rightrangle$
For the equation you mentioned in the case of $j = 2$, the detail derivation is as follows: the tensor product formula can learn from here
Correct answer by kita on February 23, 2021
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