$\require{cancel} \newcommand{\Ket}[1]{\left|{#1}\right\rangle} \newcommand{\Bra}[1]{\left\langle{#1}\right|} \newcommand{\Braket}[1]{\left\langle{#1}\right\rangle} \newcommand{\Rsr}[1]{\frac{1}{\sqrt{#1}}} \newcommand{\RSR}[1]{1/\sqrt{#1}} \newcommand{\Verti}{\rvert} \newcommand{\HAT}[1]{\hat{\,#1~}} \DeclareMathOperator{\Tr}{Tr}$

Rotation - Bloch Sphere

^{First created in July 2018}

On this page, we will explain into more details on rotations about various axes, eventually arriving at a general formula of rotation about an arbitrary state vector.

Ref: http://www.vcpc.univie.ac.at/~ian/hotlist/qc/talks/bloch-sphere-rotations.pdf

The Density Operator

Let $\Ket{\psi}=\cos\frac{\theta}{2}\Ket0+e^{i\phi}\sin\frac{\theta}{2}\Ket1~$. $\quad\rho=\Ket{\psi}\Bra{\psi}$ projects a pure state onto $\Ket{\psi}$.

$\rho =\Ket{\psi}\Bra{\psi} =\Ket{\psi}\otimes\Ket{\psi}^\dagger =\left(\cos\frac{\theta}{2}\Ket0+e^{i\phi}\sin\frac{\theta}{2}\Ket1\right)\left(\cos\frac{\theta}{2}\Bra0+e^{-i\phi}\sin\frac{\theta}{2}\Bra1\right)$

$= \begin{bmatrix} \cos^2\frac{\theta}{2} & e^{-i\phi}\sin\frac{\theta}{2}\cos\frac{\theta}{2}\\ e^{i\phi}\sin\frac{\theta}{2}\cos\frac{\theta}{2} & \sin^2\frac{\theta}{2} \end{bmatrix}$

$= \begin{bmatrix} \frac{1+\cos\theta}{2} & e^{-i\phi}\frac{\sin\theta}{2}\\ e^{i\phi}\frac{\sin\theta}{2} & \frac{1-\cos\theta}{2} \end{bmatrix}$

$= \frac{1}{2} \begin{bmatrix} 1+\cos\theta & (\cos\phi-i\sin\phi)\sin\theta\\ (\cos\phi+i\sin\phi)\sin\theta & 1-\cos\theta \end{bmatrix}$

$= \frac{1}{2} \begin{bmatrix} 1+\cos\theta & \cos\phi\sin\theta-i\sin\phi\sin\theta\\ \cos\phi\sin\theta+i\sin\phi\sin\theta & 1-\cos\theta \end{bmatrix}$

$= \frac{1}{2} \left( \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & \cos\phi\sin\theta\\ \cos\phi\sin\theta & 0 \end{bmatrix} + \begin{bmatrix} 0 & -i\sin\phi\sin\theta\\ i\sin\phi\sin\theta & 0 \end{bmatrix} + \begin{bmatrix} \cos\theta & 0\\ 0 & -\cos\theta \end{bmatrix} \right)$

$= \frac{1}{2} \left( I + X\cos\phi\sin\theta + Y\sin\phi\sin\theta + Z\cos\theta \right) .$

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Let $\mathbf{r}_\rho=(r_x,r_y,r_z)=(\cos\phi\sin\theta,\sin\phi\sin\theta,\cos\theta)$ and $\sigma=(X,Y,Z)$.

$\boxed{\rho=\frac{1}{2}\left(I+\mathbf{r}_\rho\cdot\sigma\right).}$

To visualise how $\rho$ projects an arbitrary state vector $\Ket\alpha$ onto $\Ket\psi$, let's rotate $\Ket\alpha$ by $\pi$ about $\Ket\psi$ through the operator $\left(\mathbf{r}_\rho\cdot\sigma\right)$ onto $\Ket{\alpha'}$. $I$ represents the original $\Ket\alpha$. The "average" of the two is the projection $\rho\Ket\alpha$.

An interesting fact here is that the three Pauli Operators ($X$, $Y$ and $Z$) each rotates about its corresponding axis by $\pi$. The inner product $\left(\mathbf{r}_\rho\cdot\sigma\right)$ performs the same operation (rotating by $\pi$) but about $\mathbf{r}_\rho$.

The $\rho$ projector's "Bloch vector" $\mathbf{r}_\rho$ is the Cartesian vector in the above context.

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?? In general, given a unitary operation $U$ and an arbitrary state $\Ket\psi$, $U_\psi=\mathbf{r}\cdot(U_x,U_y,U_z)$, where $\mathbf{r}$ is the Cartesian co-ordinate of $\Ket\psi$ on the Bloch Sphere, and $U_x, U_y$ and $U_z$ are to apply $U$ about $X$-axis, $Y$-axis and $Z$-axis respectively.

Rotation Operators

The definition of a unitary operator $U$ is $U^\dagger U=UU^\dagger=I$. When it operates on a projector $P=\Ket\psi\Bra\psi$, the result is $\big(U\Ket\psi\big)\big(\Bra\psi U^\dagger\big)=UPU^\dagger$.

The definition of a square matrix exponential is $e^X=\sum_{k=0}^\infty\frac{1}{k!}X^k,$ where $X^0\equiv I.$

For a self-inverse operator $A$ (including all three Pauli operators and the Hadamard), $A^2=I$ and therefore

$e^{i\theta A} =\sum_{k=0}^\infty\frac{1}{k!}(i\theta A)^k =\sum_{k=0}^\infty\frac{1}{(2k)!}(i\theta)^{2k} I+\sum_{k=0}^\infty\frac{1}{(2k+1)!}(i\theta)^{(2k+1)} A =\cos(\theta)~I+i\sin(\theta)~A .$

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The Pauli $X$, $Y$ and $Z$ operators rotate the state by $\pi$ on the Bloch Sphere. Their general form rotating about each axis by $\theta$ is given here:

$\boxed{ R_x(\theta)=e^{-i\frac{\theta}{2}~X}\\ R_y(\theta)=e^{-i\frac{\theta}{2}~Y}\\ R_z(\theta)=e^{-i\frac{\theta}{2}~Z} }$

You can show the same for rotation about the Hardmada "axis" $\left(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}\right)$:
$R_h(\theta)=e^{-i\frac{\theta}{2}~H}$

Note: The negative sign is an intrinsic part of the exponent, not an indication of direction.

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Example 1: To rotate $\Ket0$ by $\pi/2$ anti-clockwise about the $X$-axis would come to
$R_x(\pi/2)\Ket0 =\cos(-\pi/4)~I+i\sin(-\pi/4)~X =\left(\frac{1}{\sqrt{2}}~I-i\frac{1}{\sqrt{2}}~X\right)\Ket0 =\frac{1}{\sqrt{2}} \begin{bmatrix} 1 & -i\\ -i & 1 \end{bmatrix} \Ket0 =\frac{1}{\sqrt{2}}(\Ket0-i\Ket1) .$

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Example 2: To rotate $\Ket0$ by $\pi/2$ clockwise about the Hardmada "axis"
$R_h(-\pi/2)\Ket0 =\cos(\pi/4)~I+i\sin(\pi/4)~H =\left(\frac{1}{\sqrt 2}~I+i\frac{1}{\sqrt 2}~H\right)\Ket0 =\left( \frac{1}{\sqrt 2} \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} + \frac{i}{2} \begin{bmatrix} 1 & 1\\ 1 & -1 \end{bmatrix} \right) \Ket0 =\left(\frac{1}{\sqrt 2}+\frac{i}{2}\right)\Ket0+\frac{i}{2}\Ket1\\ \sim\frac{2}{\sqrt 3}\left(\frac{1}{\sqrt 2}-\frac{i}{2}\right)\left[\left(\frac{1}{\sqrt 2}+\frac{i}{2}\right)\Ket0+\frac{i}{2}\Ket1\right] =\frac{\sqrt 3}{2}\Ket0+\left(\frac{1}{2\sqrt 3}+\frac{i}{\sqrt 6}\right)\Ket1\\ =\cos(\pi/6)\Ket0+\left(\frac{1}{\sqrt 3}+i\frac{\sqrt 2}{\sqrt 3}\right)\sin(\pi/6)\Ket1 \approx\cos(\pi/6)\Ket0+e^{0.3 i}\sin(\pi/6)\Ket1 .$

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It can be shown that rotation about Bloch vector $\HAT n$ is $\large\boxed{R_{\HAT n}(\theta)=e^{-i\frac{\theta}{2}~(\HAT n\cdot\sigma)}}~$, where $\HAT n=(n_x,n_y,n_z)$ and $\sigma=(X,Y,Z)$.

To repeat example 2 where the Hadamard $\HAT h=\left(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}\right)$,

$R_{\HAT h}(-\pi/2)\Ket0 =e^{i\pi/4~(\HAT h\cdot\sigma)} =\cos(\pi/4)~I+i\sin(\pi/4)\left(h_xX+h_yY+h_zZ\right) =\cos(\pi/4)~I+i\sin(\pi/4)\left(\frac{1}{\sqrt{2}}X+\frac{1}{\sqrt{2}}Z\right)\\ =\cos(\pi/4)~I+i\sin(\pi/4)\frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1\\ 1 & -1 \end{bmatrix} =\left( \frac{1}{\sqrt 2} \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} + \frac{i}{2} \begin{bmatrix} 1 & 1\\ 1 & -1 \end{bmatrix} \right) .$

It will produce identical result.

Universal Operator

If we include a global phase shift factor into the rotation operator, we can produce any Bloch vector.

$\large\boxed{U(\gamma,\theta,\HAT n)=e^{i\gamma}R_{\HAT n}(\theta).}$

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For example, the Hadamard operator is a rotation of $\theta=\pi$ about $\left(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}\right)$.

$R_{\HAT h}(\pi) =\cos(-\pi/2)~I+i\sin(-\pi/2)H =\cos(-\pi/2) \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} +i\sin(-\pi/2)\frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1\\ 1 & -1 \end{bmatrix} =-i\frac{1}{\sqrt 2} \begin{bmatrix} 1 & 1\\ 1 & -1 \end{bmatrix} .$

By observation, a phase shift of $~e^{i~\pi/2}=i~$ is required. $H=e^{i~\pi/2} \frac{-i}{\sqrt 2} \begin{bmatrix} 1 & 1\\ 1 & -1 \end{bmatrix} =\frac{1}{\sqrt 2} \begin{bmatrix} 1 & 1\\ 1 & -1 \end{bmatrix} .$

So $H=U\left(\pi/2,\pi,\left(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}\right)\right).$