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We have the 3 following gates :

$$ H = \dfrac{1}{\sqrt{2}}\begin{bmatrix}1 & 1 \\ 1 & -1 \end{bmatrix} $$ $$ R(\varphi) = \begin{bmatrix}1 & 0 \\ 0 & e^{-i\varphi} \end{bmatrix} $$ $$ R(\psi) = \begin{bmatrix}1 & 0 \\ 0 & e^{i\psi} \end{bmatrix} $$

and we want to construct a one-bit circuit that produces the final state $$ |\Xi\rangle = \cos {\varphi\over{2}} |0\rangle + e^{i\psi}\sin {\varphi\over{2}} |1\rangle $$

I do not understand how a factor $\cos {\varphi\over{2}} $ can appear in front of $|0\rangle$, can someone help me ?

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Starting from $|0\rangle$ if you 'mix' the amplitudes with $H$ then rotate by $R(\phi)$ and then 'unmix' using $H$ again you'll have transfered the phase $\phi$ to the amplitude of $|0\rangle$. i.e. starting with $H|0\rangle$:

$$|0\rangle\overset{H}{\to}\frac1{\sqrt{2}}\Bigl(|0\rangle + |1\rangle\Bigr)$$ then you rotate in $Z$ by $\phi$: $$\frac1{\sqrt{2}}\Bigl(|0\rangle + |1\rangle\Bigr)\overset{R(\phi)}{\to}\frac1{\sqrt{2}}\Bigl(|0\rangle + e^{-i\phi}|1\rangle\Bigr)$$ then H again: $$\frac1{\sqrt{2}}\Bigl(|0\rangle + e^{-i\phi}|1\rangle\Bigr)\overset{H}{\to}\frac12\Bigl((1+e^{-i\phi})|0\rangle + (1-e^{-i\phi})|1\rangle\Bigr)$$

Now the amplitude of $|0\rangle$ is \begin{align} \frac14\|1 + e^{-i\phi}\|^2 &= \frac14\Bigl(1+\cos(\phi)-i\sin(\phi)\Bigr) \Bigl(1+\cos(\phi)+i\sin(\phi)\Bigr)\\ &= \frac12\Bigl(\cos(\phi) + 1\Bigr)\\ &= \cos^2(\frac{\phi}2) \end{align} and similarly the amplitude of $|1\rangle$ is $\sin^2(\frac{\phi}2)$ so you can write this state as $$\cos\bigl(\frac{\phi}2\bigr)|0\rangle + e^{i\theta}\sin\bigl(\frac{\phi}2\bigr)|1\rangle$$ where $e^{i\theta}$ is their relative phase which ends up being $i = e^{i\frac{\pi}2}$.

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  • $\begingroup$ Thank you very much ! Great explanation ! $\endgroup$
    – Bozu
    Apr 26, 2022 at 12:56

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