# How do I get this FRQI equation? [duplicate]

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 $$\mathcal{H}$$ ? $$\mathcal{H}\left(|0\rangle^{\otimes2n+1}\right)=\frac{1}{2^n}|0\rangle\otimes\sum_{i=0}^{2^{2n}-1}|i\rangle$$

• 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. Jan 3 at 0:19

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