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

Rotation by Phase Shift

^{First created in July 2018}

Phase in quantum mechanics refers to the positioning of waves in space and time. Phase difference is between the peaks of various waves and is therefore relative.

For example, for a polarised photon qubit, if $\Ket0$ is represented by horizontal polarisation and $\Ket1$ vertical, then $\Rsr2(\Ket0+\Ket1)$ is polarised diagonally and is "in phase" (peak to peak), and $\Rsr2(\Ket0-\Ket1)$ is "out of phase" (peak to trough). Anywhere in between will be shown as circular or elliptical polarisation.

A global phase shift is like shifting all waves by the same amount, and has no physical effect on the system.

The choice of basis may change the "look" of the state (measurement), like looking at the wave from a different origin, but it has no physical effect on the system as the measurement result is interchangeable. In the $\Rsr2(\Ket0+\Ket1)$ example, a diagonal polariser will allow 100% pass through, which can be converted into a 50-50 chance if measuring, say, by a horizontal polariser on $\Ket0$.

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On a Bloch Sphere, a state vector is expressed as $\Ket\psi=\cos\frac{\theta}{2}\Ket0+e^{i\phi}\sin\frac{\theta}{2}\Ket1.$ Phase shift means the change of the imaginery part of the probability amplitude, as represented by $e^{i\phi}$.

In practice, phase shift is useful to manipulate rotation on the Bloch Sphere. This is done by applying operator $U$ to map $\Ket0$ to the rotation axis, apply a phase shift, followed by $U^{-1}$ to remap to the original frame.

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For example, rotation on the $X$-$Y$ plane is about the $Z$-axis (in the direction from $\Rsr2(\Ket0+\Ket1)$ towards $\Rsr2(\Ket0+i\Ket1)$).

This rotation is implemented as the Phase Shift Gate $R(\phi)= \begin{bmatrix} 1 & 0\\ 0 & e^{i\phi} \end{bmatrix} .$ It is about the $Z$-axis unless explicitly specified (e.g. $R_x(\phi)$ for rotations about the $X$-axis).

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Rotation about other axes can be implemented by rotating $+Z$ to the rotation axis, perform the rotation by the phase shift operator, then rotating it back by the same rotation inverse.

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For example, rotation about the $X$-axis can be done by the Hadamard gates. (Please note that $\because H^2=I,~\therefore H^{-1}=H$.)

$HR_z(\phi)H\\ =\Rsr2 \begin{bmatrix} 1 & 1\\ 1 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & e^{i\phi} \end{bmatrix} \Rsr2 \begin{bmatrix} 1 & 1\\ 1 & -1 \end{bmatrix} \sim {1\over2} \begin{bmatrix} 1 & 1\\ 1 & -1 \end{bmatrix} \begin{bmatrix} e^{-i\phi/2} & 0\\ 0 & e^{i\phi/2} \end{bmatrix} \begin{bmatrix} 1 & 1\\ 1 & -1 \end{bmatrix}\\ ={1\over2} \begin{bmatrix} 1 & 1\\ 1 & -1 \end{bmatrix} \begin{bmatrix} e^{-i\phi/2} & e^{-i\phi/2}\\ e^{i\phi/2} & -e^{i\phi/2} \end{bmatrix} ={1\over2} \begin{bmatrix} e^{-i\phi/2}+e^{i\phi/2} & e^{-i\phi/2}-e^{i\phi/2}\\ e^{-i\phi/2}-e^{i\phi/2} & e^{-i\phi/2}+e^{i\phi/2}\\ \end{bmatrix}\\ ={1\over2} \begin{bmatrix} 2\cos(\phi/2) & -2i\sin(\phi/2)\\ -2i\sin(\phi/2) & 2\cos(\phi/2)\\ \end{bmatrix} =\begin{bmatrix} \cos(\phi/2) & -i\sin(\phi/2)\\ -i\sin(\phi/2) & \cos(\phi/2)\\ \end{bmatrix} .$

To verify, $HR_z(\pi)H= \begin{bmatrix} 0 & -i\\ -i & 0 \end{bmatrix} \sim\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} =X.$

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The "preparation rotation" is not unique. For example, you may achieve the above using $R_y(\pi/2)$ instead of $H$. The inverse is $R_y(-\pi/2)$.

$R_y(-\pi/2)R_z(\phi)R_y(\pi/2) =\begin{bmatrix} \cos{-\theta\over 2} & -\sin{-\theta\over 2}\\ \sin{-\theta\over 2} & \cos{-\theta\over 2} \end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & e^{i\phi} \end{bmatrix} \begin{bmatrix} \cos{\theta\over 2} & -\sin{\theta\over 2}\\ \sin{\theta\over 2} & \cos{\theta\over 2} \end{bmatrix}\\ \sim \Rsr2 \begin{bmatrix} 1 & 1\\ -1 & 1 \end{bmatrix} \begin{bmatrix} e^{-i\phi/2} & 0\\ 0 & e^{i\phi/2} \end{bmatrix} \Rsr2 \begin{bmatrix} 1 & -1\\ 1 & 1 \end{bmatrix}\\ ={1\over2} \begin{bmatrix} 1 & 1\\ -1 & 1 \end{bmatrix} \begin{bmatrix} e^{-i\phi/2} & -e^{-i\phi/2}\\ e^{i\phi/2} & e^{i\phi/2} \end{bmatrix} ={1\over2} \begin{bmatrix} e^{-i\phi/2}+e^{i\phi/2} & -e^{-i\phi/2}+e^{i\phi/2}\\ -e^{-i\phi/2}+e^{i\phi/2} & e^{-i\phi/2}+e^{i\phi/2} \end{bmatrix}\\ ={1\over2} \begin{bmatrix} 2\cos(\phi/2) & 2i\sin(\phi/2)\\ 2i\sin(\phi/2) & 2\cos(\phi/2)\\ \end{bmatrix} =\begin{bmatrix} \cos(\phi/2) & i\sin(\phi/2)\\ i\sin(\phi/2) & \cos(\phi/2)\\ \end{bmatrix} .$

To verify, $R_y(-\pi/2)R_z(\pi)R_y(\pi/2)= \begin{bmatrix} 0 & i\\ i & 0 \end{bmatrix} \sim\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} =X.$