1
$\begingroup$

$$R_x(\theta)= \begin{bmatrix} \cos \left( \frac{\theta}{2} \right) & -i\sin \left( \frac{\theta}{2} \right) \\ -i\sin \left( \frac{\theta}{2} \right) & \cos \left( \frac{\theta}{2} \right)\end{bmatrix}$$

$$R_y(\theta)= \begin{bmatrix} \cos \left( \frac{\theta}{2} \right) & \sin \left( \frac{\theta}{2} \right) \\ \sin \left( \frac{\theta}{2} \right) & \cos \left( \frac{\theta}{2} \right)\end{bmatrix}$$

$$R_z(\theta)=\begin{bmatrix} \exp\left(-i\frac{\theta}{2}\right) & 0 \\ 0 & \exp\left(-i\frac{\theta}{2}\right)\end{bmatrix}$$

This is the rotation gates as matrices on the Bloch sphere. It was easy to show that $R_z$ to be the rotation about $z$ axis on the Bloch sphere. However, I can't find a way to show that $R_x$ and $R_y$ are rotations about $x$ and $y$ axis respectively and I can't find any solutions on the google.... Could anyone show me why $R_x$ and $R_y$ are rotations about $x$ and $y$ axis.

$\endgroup$

2 Answers 2

1
$\begingroup$

I will leave $R_y(\theta)$ as an exercise for you.

Below is a verification of $R_x(\theta)$ is a rotation about the $x$-axis.

Denote the Pauli matrices as $$X=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, Y=\begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}, Z=\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} $$

We have the following expression for $R_x(\theta)$ $$R_x(\theta) = \cos \left( \frac{\theta}{2} \right)I-i\sin\left( \frac{\theta}{2}\right)X$$

Consider $$\rho=\frac12(I+r_xX+r_yY+r_zZ)$$

$\begin{align}\rho' &= R_x(\theta)\rho R_x(\theta)^\dagger \\ &=\frac12(I+r_xR_x(\theta)XR_x(\theta)^\dagger+r_yR_x(\theta)YR_x(\theta)^\dagger+r_zR_x(\theta)ZR_x(\theta)^\dagger)\end{align}\tag{1}$

Now, let's analyze each separate term:

\begin{align}R_x(\theta)XR_x(\theta)^\dagger &= \left(\cos \left( \frac{\theta}{2} \right)I-i\sin\left( \frac{\theta}{2}\right)X\right)X\left(\cos \left( \frac{\theta}{2} \right)I+i\sin\left( \frac{\theta}{2}\right)X\right)\\ &=\left(\cos \left( \frac{\theta}{2} \right)X-i\sin\left( \frac{\theta}{2}\right)I\right)\left(\cos \left( \frac{\theta}{2} \right)I+i\sin\left( \frac{\theta}{2}\right)X\right)\\ &=\cos^2 \left( \frac{\theta}{2}\right)X-i\sin \left(\frac{\theta}{2} \right)\cos \left(\frac{\theta}{2} \right)I+i\sin \left(\frac{\theta}{2} \right)\cos \left(\frac{\theta}{2} \right)I+\sin^2 \left( \frac{\theta}{2}\right)X\\ &= X\end{align}

\begin{align}R_x(\theta)YR_x(\theta)^\dagger &= \left(\cos \left( \frac{\theta}{2} \right)I-i\sin\left( \frac{\theta}{2}\right)X\right)Y\left(\cos \left( \frac{\theta}{2} \right)I+i\sin\left( \frac{\theta}{2}\right)X\right)\\ &=\left(\cos \left( \frac{\theta}{2} \right)Y+\sin\left( \frac{\theta}{2}\right)Z\right)\left(\cos \left( \frac{\theta}{2} \right)I+i\sin\left( \frac{\theta}{2}\right)X\right)\\ &=\cos^2 \left( \frac{\theta}{2}\right)Y+\sin \left(\frac{\theta}{2} \right)\cos \left(\frac{\theta}{2} \right)Z+\sin \left(\frac{\theta}{2} \right)\cos \left(\frac{\theta}{2} \right)Z-\sin^2 \left( \frac{\theta}{2}\right)Y\\ &= \cos (\theta) Y + \sin(\theta)Z\end{align}

\begin{align}R_x(\theta)ZR_x(\theta)^\dagger &= \left(\cos \left( \frac{\theta}{2} \right)I-i\sin\left( \frac{\theta}{2}\right)X\right)Z\left(\cos \left( \frac{\theta}{2} \right)I+i\sin\left( \frac{\theta}{2}\right)X\right)\\ &=\left(\cos \left( \frac{\theta}{2} \right)Z-\sin\left( \frac{\theta}{2}\right)Y\right)\left(\cos \left( \frac{\theta}{2} \right)I+i\sin\left( \frac{\theta}{2}\right)X\right)\\ &=\cos^2 \left( \frac{\theta}{2}\right)Z-\sin \left(\frac{\theta}{2} \right)\cos \left(\frac{\theta}{2} \right)Y-\sin \left(\frac{\theta}{2} \right)\cos \left(\frac{\theta}{2} \right)Y-\sin^2 \left( \frac{\theta}{2}\right)Z\\ &= -\sin (\theta) Y + \cos(\theta)Z\end{align}

\begin{align}\rho'&=\frac12 \left(I+r_xX+r_y(\cos (\theta) Y + \sin(\theta)Z)+r_z(-\sin (\theta) Y + \cos(\theta)Z)\right)\\ &= \frac12 \left(I+r_xX+(r_y(\cos (\theta) - r_z \sin (\theta)) Y + (r_y\sin(\theta) + r_z\cos(\theta))Z\right)\\\end{align} Hence if we write

$$\rho'=\frac12 \left(I+r_x'X+r_y'Y+r_z'Z\right)$$

$$\begin{bmatrix}r_x' \\ r_y' \\ r_z'\end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\ 0 & \cos \theta & - \sin \theta\\ 0 & \sin \theta & \cos \theta\end{bmatrix}\begin{bmatrix}r_x \\ r_y \\ r_z\end{bmatrix}$$

$\endgroup$
0
0
$\begingroup$

I'm a bit late to the party, but here's another way to prove that $R_x(\alpha)$ is a rotation about the $x$ axis on the Bloch sphere.

First, consider the following way to map every vector $\mathbf v$ of $\mathbb C^2$ to $\mathbb{R}^3$: Normalize $\mathbf v$ to $\mathbf v_0$ and compute the point $\mathbf p_0$ on the Bloch sphere to which $\mathbf v_0$ would be mapped. Then map $\mathbf v$ to $\|\mathbf v\|\cdot \mathbf p_0$. Using this mapping, every function from $\mathbb C^2$ to $\mathbb C^2$ induces a function from $\mathbb{R}^3$ to $\mathbb{R}^3$.

Now, the function described by the Pauli matrix $X$ maps $\bigl(\cos(\theta/2),\mathrm{e}^{\mathrm{i}\phi}\sin(\theta/2)\bigr)^T$ to $\bigl(\mathrm{e}^{\mathrm{i}\phi}\sin(\theta/2),\cos(\theta/2)\bigr)^T$ which is physically indistinguishable from $\bigl(\cos(\pi/2-\theta/2),\mathrm{e}^{-\mathrm{i}\phi}\sin(\pi/2-\theta/2)\bigr)^T$. Therefore, the induced map on $\mathbb{R}^3$ maps the point with spherical coordinates $(r,\theta,\phi)$ to the one with $(r,\pi-\theta,-\phi)$. This is obviously a linear map.

Because $R_x(\alpha)$ is $\cos(\alpha/2)I-\mathrm{i}\sin(\alpha/2)X$, the map induced by $R_x(\alpha)$ is also linear and to see what it does we only need to check what it does to the three base vectors of $\mathbb{R}^3$.

The base vector in $z$ direction has $\phi=0$ and $\theta=0$ and corresponds to $(1,0)^T$ in $\mathbb{C}^2$. Its image under $R_x(\alpha)$ is $\bigl(\cos(\alpha/2),\mathrm{e}^{-\mathrm{i}\pi/2}\sin(\alpha/2)\bigr)^T$ and the image of the base vector under the induced map is thus $\phi'=-\pi/2$ and $\theta'=\alpha$.

The base vector in $y$ direction has $\phi=\pi/2$ and $\theta=\pi/2$ and corresponds to $\bigl(\cos(\pi/4),\mathrm{i}\sin(\pi/4)\bigr)^T$ in $\mathbb{C}^2$. Its image under $R_x(\alpha)$ is \begin{align*} &\begin{pmatrix} \cos(\alpha/2)\cos(\pi/4)+\sin(\alpha/2)\sin(\pi/4) \\ -\mathrm{i}\sin(\alpha/2)\cos(\pi/4)+\mathrm{i}\cos(\alpha/2)\sin(\pi/4) \end{pmatrix} \\ = &\begin{pmatrix} \cos(-\alpha/2)\cos(\pi/4)-\sin(-\alpha/2)\sin(\pi/4) \\ \mathrm{i}\sin(-\alpha/2)\cos(\pi/4)+\mathrm{i}\cos(-\alpha/2)\sin(\pi/4) \end{pmatrix} = \begin{pmatrix} \cos(\pi/4-\alpha/2) \\ \mathrm{e}^{\pi/2} \sin(\pi/4-\alpha/2) \end{pmatrix} \end{align*} and the image of the base vector under the induced map is thus $\phi'=\pi/2$ and $\theta'=\pi/2-\alpha$.

Finally, the base vector in $x$ direction has $\phi=0$ and $\theta=\pi/2$ and corresponds to $\bigl(\cos(\pi/4),\sin(\pi/4)\bigr)^T$ in $\mathbb{C}^2$. Its image under $R_x(\alpha)$ is \begin{align*} &\begin{pmatrix} \cos(\alpha/2)\cos(\pi/4)-\mathrm{i}\sin(\alpha/2)\sin(\pi/4) \\ -\mathrm{i}\sin(\alpha/2)\cos(\pi/4)+\cos(\alpha/2)\sin(\pi/4) \end{pmatrix} \\ = &\frac1{\sqrt2}\begin{pmatrix} \cos(-\alpha/2)+\mathrm{i}\sin(-\alpha/2) \\ \cos(-\alpha/2)+\mathrm{i}\sin(-\alpha/2) \end{pmatrix} = \frac1{\sqrt2}\begin{pmatrix} \mathrm{e}^{-\mathrm{i}\alpha/2} \\ \mathrm{e}^{-\mathrm{i}\alpha/2} \end{pmatrix} = \mathrm{e}^{-\mathrm{i}\alpha/2} \begin{pmatrix} \cos(\pi/4) \\ \sin(\pi/4) \end{pmatrix} \end{align*} which is physically indistinguishable from $\bigl(\cos(\pi/4),\sin(\pi/4)\bigr)^T$. Thus, the image of this base vector under the induced map is $\phi'=0$ and $\theta'=\pi/2$, i.e., the vector doesn't move.

All three base vector are mapped to where they should be mapped by a rotation by $\alpha$ about the $x$ axis.

$\endgroup$

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.