I am in the process of understanding a proof. First, the following is said there:


This is clear so far, because the Hadamard transformation applied to the state $|0...0\rangle$ produces an equally distributed superposition. I know that.

The exciting part is actually here: $$H\begin{pmatrix}1\\1\\1\\1\end{pmatrix}=\sqrt{N}\begin{pmatrix}1\\0\\\vdots\\0\end{pmatrix}$$

I've already thought a few things about that, but I'm not quite sure why it really is. Here is my train of thought:

Say $N=4$ $$H\begin{pmatrix}1\\1\\1\\1\end{pmatrix}=H\begin{pmatrix}1\\0\\0\\0\end{pmatrix}+H\begin{pmatrix}0\\1\\0\\0\end{pmatrix}+H\begin{pmatrix}0\\0\\1\\0\end{pmatrix}+H\begin{pmatrix}0\\0\\0\\1\end{pmatrix}$$


If you like, this results in: $N\cdot \frac{1}{\sqrt{N}}|00\rangle=\sqrt{N}|00\rangle$ to stay in the example $2|00\rangle$.

Honestly, that does not quite convince me (That's just the case for N = 4).

Why I do this is because I would like to understand the following:


Maybe someone of you can bring light into the darkness and help me a bit ...


The rows of a Hadamard matrix $H$ are orthogonal to each other, and each has Euclidean length $\sqrt N$.
[As a side note, that means $HH^T=N\cdot I$, in finite dimension it implies $\exists H^{-1}=\frac1N\cdot H^T$, so $H^TH=N\cdot I$, and that also the columns of $H$ are orthogonal to each other.]

In your setting, $H$ seems to be normed, i.e. it's $\frac1{\sqrt N}H$ with the above $H$.

Now, assuming the first row is $(1,1,\dots,1)$ (or its normed multiple), then it's orthogonal to every other row of $H$, which means that the product $H\pmatrix{1\\1\\ \vdots\\1}$ will have all zeroes except for the first row.


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