I have an issue, perhaps with normalization with the following state. For $\alpha^2 + \beta^2 =1 $, the probabilities in this state does not sum up to 1.
$$|\psi\rangle := \frac{1}{2}\left[\alpha\left(|0\rangle(|x\rangle+|x'\rangle\right) + \beta \left(|1\rangle(|x\rangle-|x'\rangle \right)\right] $$
as $p(0)=\frac{2\alpha^2+2\alpha^2\langle x|x'\rangle}{4}=\alpha^2(\frac{1+\langle x|x'\rangle}{2})$ and $p(1)=\frac{2\beta^2 - 2\beta^2\langle x | x'\rangle}{4} = \beta^2\frac{1-\langle x | x'\rangle}{2}$.
And $p(0)+p(1) = \frac{\alpha^2 + \beta^2 + \alpha^2\langle x| x'\rangle- \beta^2\langle x | x'\rangle }{2} = \frac{1 + (\alpha^2 - \beta^2) \langle x | x'\rangle}{2}$
Long version:
I created the state as follows:
- $|0\rangle |0\rangle$
- Hadamard on first qubit $\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)|0\rangle$
Controlled operations: if 0 on first register I create state $|x\rangle$ otherwise I create state $|x'\rangle$ on second qubit. $\frac{1}{\sqrt{2}}(|0\rangle|x\rangle+|1\rangle|x'\rangle)$
Now a perform a second Hadamard gate and re-group the terms:
$$ \frac{1}{2}\left[\left(|0\rangle(|x\rangle+|x'\rangle\right) + \left(|1\rangle(|x\rangle-|x'\rangle \right)\right] $$
- Now I am ready to do a rotation on the $Y$-axis of $\alpha$ on the first qubit. This leads to: $$ \frac{1}{2}\left[\alpha\left(|0\rangle(|x\rangle+|x'\rangle\right) + \beta \left(|1\rangle(|x\rangle-|x'\rangle \right)\right] $$
Perhaps I am doing something wrong with the normalization? But I don't need any normalization factor as $\alpha^2+\beta^2=1$
Thank you.
