I am in the process of understanding a proof. First, the following is said there:
$$H\begin{pmatrix}1\\0\\\vdots\\0\end{pmatrix}=\frac{1}{\sqrt{N}}\begin{pmatrix}1\\1\\1\\1\end{pmatrix}$$
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 $N=4$
$$H\begin{pmatrix}1\\1\\1\\1\end{pmatrix}=H\begin{pmatrix}1\\0\\0\\0\end{pmatrix}+H\begin{pmatrix}0\\1\\0\\0\end{pmatrix}+H\begin{pmatrix}0\\0\\1\\0\end{pmatrix}+H\begin{pmatrix}0\\0\\0\\1\end{pmatrix}$$
$$=\frac{1}{\sqrt{N}}\begin{pmatrix}1\\1\\1\\1\end{pmatrix}+\frac{1}{\sqrt{N}}\begin{pmatrix}1\\-1\\1\\-1\end{pmatrix}+\frac{1}{\sqrt{N}}\begin{pmatrix}1\\1\\-1\\-1\end{pmatrix}+\frac{1}{\sqrt{N}}\begin{pmatrix}1\\-1\\-1\\1\end{pmatrix}$$
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:
$$H\begin{pmatrix}1&1&…&1\\1&1&…&1\\\vdots&\ddots&\ddots&\vdots\\1&1&…&1\end{pmatrix}H=\sqrt{N}\begin{pmatrix}1&1&…&1\\0&0&…&0\\\vdots&\ddots&\ddots&\vdots\\0&0&…&0\end{pmatrix}H=N\begin{pmatrix}1&0&…&0\\0&0&…&0\\\vdots&\ddots&\ddots&\vdots\\0&0&…&0\end{pmatrix}$$
Maybe someone of you can bring light into the darkness and help me a bit …
Best Answer
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.