1
$\begingroup$

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 ...

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.