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I have been working with FRQI and there is this equation in a paper is given. Can anyone explain how they get that just by multiplying by $\mathcal{H}$ ? $$\mathcal{H}\left(|0\rangle^{\otimes2n+1}\right)=\frac{1}{2^n}|0\rangle\otimes\sum_{i=0}^{2^{2n}-1}|i\rangle$$

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    $\begingroup$ Please try to use Mathjax to display equations and not images. Also please include all information necessary to answer your question. What is FRQI? What is $\mathcal{H}$? Perhaps you can also provide a reference to the paper if you think it is useful. $\endgroup$
    – Rammus
    Commented Jan 3, 2023 at 0:19

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First, note that $\mathcal{H}$ is equivalent to $$ (I\otimes H^{\otimes2n})|0\rangle^{\otimes2n+1} $$

Second, the multi-qubit hadarmard's definition: $$H^{\otimes n}|0\rangle=\frac{1}{\sqrt{2^n}} \sum_{i=0}^{2n-1}|i\rangle $$

For $H^{\otimes2n}$: $$ H^{\otimes 2n}|0\rangle=\frac{1}{2^n} \sum_{i=0}^{2^{2n}-1}|i\rangle $$

Now it easier to see: $$ \mathcal{H}(|0\rangle^{\otimes2n+1}) = (I\otimes H^{\otimes2n})|0\rangle^{\otimes2n+1} = \frac{1}{2^n}|0\rangle\otimes\sum_{i=0}^{2^{2n}-1}|i\rangle$$ where $|0\rangle^{\otimes 2n+1}$ is the initial state

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