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I'm trying some example with the rotation gates and stuck here:

$$\langle \psi| \sigma_z |\psi \rangle = \langle 0 | R_x(\phi_1)^\dagger R_y(\phi_2)^\dagger \sigma_z R_y(\phi_2) R_x(\phi_1) | 0 \rangle = \cos(\phi_1)\cos(\phi_2). $$

How did they get $\cos(\phi_1)\cos(\phi_2)$?

I pleased to know some clear steps to have an intuition about it. Thanks!

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  • $\begingroup$ Use Euler's formula: $e^{i\theta \hat{A}} = cos(\theta)I+isin(\theta)\hat{A}$, where $\hat{A}$ is the matrix satisfy $\hat{A}*\hat{A}=I$. $\endgroup$
    – narip
    Jun 24 at 12:44
  • $\begingroup$ Thanks for your answer, I have just solved it your way but it's become complex quickly. Is there any short solution? $\endgroup$
    – Monad
    Jun 24 at 15:35
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Maybe it is best to just work out the calculation step-by-step.

First, let $U = R_y(\phi_2) R_x(\phi_1) $, and $|\psi \rangle = U|0\rangle$.

The goal is to calculate $ \langle \psi| \sigma_z |\psi \rangle$ where $\sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$.

  1. Computing $|\psi \rangle$ explicitly.

\begin{align} U|0 \rangle &= R_y(\phi_1)R_x(\phi_2) |0\rangle =\begin{pmatrix} \cos{\dfrac{\phi_1}{2}} & -\sin\dfrac{\phi_1}{2} \\ \sin\dfrac{\phi_1}{2} & \cos\dfrac{\phi_1}{2} \end{pmatrix} \begin{pmatrix} \cos\dfrac{\phi_2}{2} & -i\sin\dfrac{\phi_2}{2} \\ -i\sin\dfrac{\phi_2}{2} & \cos\dfrac{\phi_2}{2} \end{pmatrix} \begin{pmatrix}1 \\ 0 \end{pmatrix}\\ &= \begin{pmatrix} \cos{\dfrac{\phi_1}{2}} & -\sin\dfrac{\phi_1}{2} \\ \sin\dfrac{\phi_1}{2} & \cos\dfrac{\phi_1}{2} \end{pmatrix} \begin{pmatrix}\cos\dfrac{\phi_2}{2} \\ -i\sin\dfrac{\phi_2}{2} \end{pmatrix}\\ &= \begin{pmatrix} \cos\dfrac{\phi_1}{2}\cos\dfrac{\phi_2}{2} + i\sin\dfrac{\phi_1}{2}\sin\dfrac{\phi_2}{2} \\ \sin\dfrac{\phi_1}{2}\cos\dfrac{\phi_2}{2} - i\cos\dfrac{\phi_1}{2}\sin\dfrac{\phi_2}{2} \end{pmatrix} = |\psi\rangle\\ \end{align}

  1. Compute $\sigma_z|\psi\rangle$ $$\sigma_z|\psi\rangle = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} \cos\dfrac{\phi_1}{2}\cos\dfrac{\phi_2}{2} + i\sin\dfrac{\phi_1}{2}\sin\dfrac{\phi_2}{2} \\ \sin\dfrac{\phi_1}{2}\cos\dfrac{\phi_2}{2} - i\cos\dfrac{\phi_1}{2}\sin\dfrac{\phi_2}{2} \end{pmatrix} = \begin{pmatrix} \cos\dfrac{\phi_1}{2}\cos\dfrac{\phi_2}{2} + i\sin\dfrac{\phi_1}{2}\sin\dfrac{\phi_2}{2} \\ -\sin\dfrac{\phi_1}{2}\cos\dfrac{\phi_2}{2} + i\cos\dfrac{\phi_1}{2}\sin\dfrac{\phi_2}{2} \end{pmatrix} = |\phi \rangle $$

  2. Compute $ \langle \psi| \sigma_z |\psi \rangle = \langle \psi | \phi \rangle $

\begin{align} \langle \psi | \phi \rangle &= \begin{pmatrix} \cos\dfrac{\phi_1}{2}\cos\dfrac{\phi_2}{2} - i\sin\dfrac{\phi_1}{2}\sin\dfrac{\phi_2}{2} & \sin\dfrac{\phi_1}{2}\cos\dfrac{\phi_2}{2} + i\cos\dfrac{\phi_1}{2}\sin\dfrac{\phi_2}{2} \end{pmatrix} \begin{pmatrix} \cos\dfrac{\phi_1}{2}\cos\dfrac{\phi_2}{2} + i\sin\dfrac{\phi_1}{2}\sin\dfrac{\phi_2}{2} \\ -\sin\dfrac{\phi_1}{2}\cos\dfrac{\phi_2}{2} + i\cos\dfrac{\phi_1}{2}\sin\dfrac{\phi_2}{2} \end{pmatrix} \\ &= \cos^2\dfrac{\phi_1}{2}\cos^2\dfrac{\phi_2}{2} + \sin^2\dfrac{\phi_1}{2}\sin^2\dfrac{\phi_2}{2} - \sin^2\dfrac{\phi_1}{2}\cos^2\dfrac{\phi_2}{2} - \cos^2\dfrac{\phi_1}{2}\sin^2\dfrac{\phi_2}{2} \\ &= \cos(\phi_1)\cos(\phi_2) \hspace{1 cm} \textrm{(Trig identities manipulation)} \end{align}

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  • $\begingroup$ If I follow this way, I can reduce the number of calculations as below: $\langle 0|M|0\rangle = a$ with a is the top-left element of $2x2$ $M$ matrix. So we can just focus on it. $\endgroup$
    – Monad
    Jun 25 at 1:34

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