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Let H be the Hadamard operator. $$ H = (\left| 0 \right> \left< 0 \right| + \left| 0 \right> \left< 1 \right| + \left| 1 \right> \left< 0 \right| -\left| 1 \right> \left< 1 \right| )$$ prove that. $$H^{\otimes n} \left| 0 \right>^{\otimes n} = \sum_{i=0}^{2^n -1} \left| i \right>$$ Now, it is evident that this works for $n=1$ and $n=2$, because we know that. $$ \left| 0 \right>^{\otimes 2} = \left| 0 \right> \otimes \left| 0 \right> = \left| 00 \right> $$ Then. $$ H^{\otimes2}\left| 00 \right> = (\left| 0 \right> \left< 0 \right| + \left| 0 \right> \left< 1 \right| + \left| 1 \right> \left< 0 \right| -\left| 1 \right> \left< 1 \right| )(\left| 0 \right> \left< 0 \right| + \left| 0 \right> \left< 1 \right| + \left| 1 \right> \left< 0 \right| -\left| 1 \right> \left< 1 \right| )\left| 00 \right>$$

$$H^{\otimes2}\left| 00 \right> = \frac{1}{2}(\left< 00 \right|\left| 00 \right>\left| 00 \right> + \left< 00 \right|\left| 00 \right>\left| 01 \right> + \left< 00 \right|\left| 00 \right>\left| 10 \right> + \left< 00 \right|\left| 00 \right>\left| 11 \right>)$$

$$H^{\otimes2}\left| 00 \right> = \frac{1}{2}(\left| 00 \right> + \left| 01 \right> + \left| 10 \right> + \left| 11 \right>) = \frac{1}{2}\sum^{3}_{i=0}\left| i \right> = \frac{1}{\sqrt{2^2}}\sum^{2^2 -1}_{i=0}\left| i \right> $$ but I do not know how to proof that it works for $n = k + 1$.

Any clue is very welcome, thank you in advance for your time and advices.

For more information about bra-ket notation, or the Hadamard operator you can consult those links.

Mario Vega
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    Could you possibly define the objects in your question? Also it is typically expected that the question itself be not only in the title, but also formally stated in the body section of the post. If these objects are absolutely standardized and there exists absolutely no confusion as to the interpretation of these symbols, please link the appropriate Wikipedia page, for example, for those not familiar with the precise definitions. – Merkh Sep 16 '19 at 07:29
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    I have to follow up with another comment then. First, $|i\rangle$ can mean $|00\rangle,|01\rangle,...,|11\rangle$ when $i = 0,1,2,3$. I understand this in principle, but nonetheless the ket states are somewhat ambiguous under this notation, perhaps making the proof more difficult than if a better notation was employed. Second, is the problem even stated correctly? Should the RHS be $\frac{1}{\sqrt{2^n}} \sum_{i=0}^{2^n-1} |i\rangle$. Or is the last equation for $n=2$ incorrect? – Merkh Sep 16 '19 at 08:05
  • Hi @Merkh, first at all, thank you so much for your guidance, I tried to improve the question following your advices. Now, I think I could add a few examples if you think it is a good idea for the Keth notation, but in the literature is quite common to see it expressed like that (as far I understand). And regarding the RHS, as far as I can see is stated correctly, for the example of $n=2$ you end up with 4 kets with the same values as the LHS, both sides multiplied by 1/2, but I don't know if I could improve it somehow to make it more clear, what do you think? thanks again! – Mario Vega Sep 16 '19 at 08:19

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Perhaps it suffices to show that when represented as a matrix in the standard basis, the first row of $H_n := H^{\otimes n}$ is just a row containing all $1$'s, or when properly normalized, containing all $\frac{+1}{2^{n/2}}$. This follows directly as a consequence of $H_n$ being the tensor product $H_n = H_{1} \otimes H_{n-1}$. See https://en.wikipedia.org/wiki/Hadamard_transform and follow the steps to arrive at the formula for the entries of $H_n$ as \begin{equation} (H_n)_{i,j} = \frac{(-1)^{i\cdot j}}{2^{n/2}} \end{equation} where $i = 0$ for the upper row, and $i \cdot j$ is the bitwise dot product.

Relating back to the question at hand, the notation $H^{\otimes n} | 0 \rangle^{\otimes n}$ means ``take the coefficients of the first row of $H_n$, and weight the states $|j\rangle$ by their respective entry, for $j=0,1,\dots,2^n-1$." Since each entry of this row is $\frac{1}{2^{n/2}}$, this will be precisely the RHS that you want to show. Therefore, the crux of the proof is really showing that $H_n = H_1 \otimes H_{n-1}$, from which the formula for $(H_n)_{i,j}$ follows.

Merkh
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