rotation about $x$ and $y$ axis on the Bloch sphere

Question

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

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

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

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.

Answer 1

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

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

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

Answer 2

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.