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$\require{cancel} \newcommand{\Ket}[1]{\left|{#1}\right\rangle} \newcommand{\Bra}[1]{\left\langle{#1}\right|} \newcommand{\Braket}[1]{\left\langle{#1}\right\rangle} \newcommand{\Rsr}[1]{\frac{1}{\sqrt{#1}}} \newcommand{\RSR}[1]{1/\sqrt{#1}} \newcommand{\Verti}{\rvert} \newcommand{\HAT}[1]{\hat{\,#1~}} \DeclareMathOperator{\Tr}{Tr}$
First created in June 2018
Ref: https://arxiv.org/pdf/1707.03429.pdf
Open Quantum Assembly Language
Andrew W. Cross, Lev S. Biship, John A. Smolin, Jay M. Gambetta
January 10th, 2017
This page is an addendum to "QASM $U$ Gates", aiming at details of how the formulae of the three operators $u_1, u_2$ and $u_3$ come about.
The expression to prove is boxed, followed by the proof, which is deliberately verbose.
All rotations here are around axes on the Bloch Sphere.
Most formulae used here are from the Matrix Exponent page.
When $\displaystyle A^2=I,~~e^{i\theta A} =I\cos\theta+iA\sin\theta,~~$ where $\theta\in\mathbb{R}.$
$R_x(\theta) =\begin{bmatrix} \cos\theta/2 & -i\sin\theta/2\\ -i\sin\theta/2 & \cos\theta/2 \end{bmatrix} .$
$R_y(\theta) =\begin{bmatrix} \cos\theta/2 & -\sin\theta/2\\ \sin\theta/2 & \cos\theta/2 \end{bmatrix} .$
$R_z(\theta) =\begin{bmatrix} e^{-i\theta/2} & 0\\ 0 & e^{i\theta/2} \end{bmatrix} .$
$\displaystyle \boxed{ U(\theta,\phi,\lambda):=R_z(\phi)~R_y(\theta)~R_z(\lambda) =\begin{bmatrix} e^{-i(\phi+\lambda)/2}\cos(\theta/2) & -e^{-i(\phi-\lambda)/2}\sin(\theta/2)\\ e^{i(\phi-\lambda)/2}\sin(\theta/2) & e^{i(\phi+\lambda)/2}\cos(\theta/2) \end{bmatrix} },$
where $R_y(\theta)=e^{-i\frac{\theta}{2}Y}$ and $R_z(\phi)=e^{-i\frac{\phi}{2}Z}.$
Proof:
$R_z(\phi)~R_y(\theta)~R_z(\lambda)\\ =\left(\cos(\phi/2)I-i\sin(\phi/2)Z\right)\cdot\left(\cos(\theta/2)I-i\sin(\theta/2)Y\right)\cdot\left(\cos(\lambda/2)I-i\sin(\lambda/2)Z\right)$
$=\cos(\phi/2)\cos(\theta/2)\cos(\lambda/2)I\\ -i\cos(\phi/2)\cos(\theta/2)\sin(\lambda/2)Z\\ -i\cos(\phi/2)\sin(\theta/2)\cos(\lambda/2)Y\\ -\cos(\phi/2)\sin(\theta/2)\sin(\lambda/2)(iX)\\ -i\sin(\phi/2)\cos(\theta/2)\cos(\lambda/2)Z\\ -\sin(\phi/2)\cos(\theta/2)\sin(\lambda/2)I\\ -\sin(\phi/2)\sin(\theta/2)\cos(\lambda/2)(-i)X\\ +i\sin(\phi/2)\sin(\theta/2)\sin(\lambda/2)(-Y).$
$=\cos(\phi/2)\cos(\theta/2)\cos(\lambda/2)I\\ -i\cos(\phi/2)\cos(\theta/2)\sin(\lambda/2)Z\\ -i\cos(\phi/2)\sin(\theta/2)\cos(\lambda/2)Y\\ -i\cos(\phi/2)\sin(\theta/2)\sin(\lambda/2)X\\ -i\sin(\phi/2)\cos(\theta/2)\cos(\lambda/2)Z\\ -\sin(\phi/2)\cos(\theta/2)\sin(\lambda/2)I\\ +i\sin(\phi/2)\sin(\theta/2)\cos(\lambda/2)X\\ -i\sin(\phi/2)\sin(\theta/2)\sin(\lambda/2)Y.$
$=\cos(\theta/2)(\cos(\phi/2)\cos(\lambda/2)-\sin(\phi/2)\sin(\lambda/2))I\\ -i\sin(\theta/2)(\cos(\phi/2)\sin(\lambda/2)-\sin(\phi/2)\cos(\lambda/2))X\\ -i\sin(\theta/2)(\cos(\phi/2)\cos(\lambda/2)+\sin(\phi/2)\sin(\lambda/2))Y\\ -i\cos(\theta/2)(\cos(\phi/2)\sin(\lambda/2)+\sin(\phi/2)\cos(\lambda/2))Z.$
$=\cos(\theta/2)\cos((\phi+\lambda)/2)I\\ +i\sin(\theta/2)\sin((\phi-\lambda)/2)X\\ -i\sin(\theta/2)\cos((\phi-\lambda)/2)Y\\ -i\cos(\theta/2)\sin((\phi+\lambda)/2)Z$
$= \begin{bmatrix} \cos(\theta/2)\cos((\phi+\lambda)/2) -i\cos(\theta/2)\sin((\phi+\lambda)/2) & i\sin(\theta/2)\sin((\phi-\lambda)/2) -\sin(\theta/2)\cos((\phi-\lambda)/2)\\ i\sin(\theta/2)\sin((\phi-\lambda)/2) +\sin(\theta/2)\cos((\phi-\lambda)/2) & \cos(\theta/2)\cos((\phi+\lambda)/2) +i\cos(\theta/2)\sin((\phi+\lambda)/2) \end{bmatrix}$
$=\begin{bmatrix} e^{-i(\phi+\lambda)/2}\cos(\theta/2) & -e^{-i(\phi-\lambda)/2}\sin(\theta/2)\\ e^{i(\phi-\lambda)/2}\sin(\theta/2) & e^{i(\phi+\lambda)/2}\cos(\theta/2) \end{bmatrix} .$
To visualise the effect of a $U$ gate, on the Bloch Sphere rotate $\lambda$ (arg3) about axis $Z$, $\theta$ (arg1) about $Y$, then again $\phi$ (arg2) about $Z$.
As a result, $R_x(2n\pi)=R_y(2n\pi)=R_z(2n\pi)=I.$
$\boxed{ u_1(\lambda):= \begin{bmatrix} 1 & 0\\ 0 & e^{i\lambda} \end{bmatrix} \sim U(0,0,\lambda)=R_z(\lambda).}$
Proof: Definition. No proof required.
$\boxed{u_2(\phi,\lambda) :=U(\pi/2,\phi,\lambda) \scriptsize =\begin{bmatrix} e^{-i(\phi+\lambda)/2}\cos(\pi/4) & -e^{-i(\phi-\lambda)/2}\sin(\pi/4)\\ e^{i(\phi-\lambda)/2}\sin(\pi/4) & e^{i(\phi+\lambda)/2}\cos(\pi/4) \end{bmatrix} =\Rsr2\begin{bmatrix} e^{-i(\phi+\lambda)/2} & -e^{-i(\phi-\lambda)/2}\\ e^{i(\phi-\lambda)/2} & e^{i(\phi+\lambda)/2} \end{bmatrix} \sim\Rsr2\begin{bmatrix} 1 & -e^{i\lambda}\\ e^{i\phi} & e^{i(\phi+\lambda)} \end{bmatrix} }.$
Proof:
$u_2(\phi,\lambda) =R_z(\phi+\pi/2)~R_x(\pi/2)~R_z(\lambda-\pi/2)\\ =\begin{bmatrix} e^{-i(\phi/2+\pi/4)} & 0\\ 0 & e^{i(\phi/2+\pi/4)} \end{bmatrix} \begin{bmatrix} \cos\pi/4 & -i\sin\pi/4\\ -i\sin\pi/4 & \cos\pi/4 \end{bmatrix} \begin{bmatrix} e^{-i(\lambda/2-\pi/4)} & 0\\ 0 & e^{i(\lambda/2-\pi/4)} \end{bmatrix}$
$=\Rsr2 \begin{bmatrix} e^{-i(\phi/2+\pi/4)} & 0\\ 0 & e^{i(\phi/2+\pi/4)} \end{bmatrix} \begin{bmatrix} 1 & -i\\ -i & 1 \end{bmatrix} \begin{bmatrix} e^{-i(\lambda/2-\pi/4)} & 0\\ 0 & e^{i(\lambda/2-\pi/4)} \end{bmatrix}$
$=\Rsr2 \begin{bmatrix} e^{-i(\phi/2+\pi/4)} & 0\\ 0 & e^{i(\phi/2+\pi/4)} \end{bmatrix} \begin{bmatrix} e^{-i(\lambda/2-\pi/4)} & -ie^{i(\lambda/2-\pi/4)}\\ -ie^{-i(\lambda/2-\pi/4)} & e^{i(\lambda/2-\pi/4)} \end{bmatrix}$
$=\Rsr2\begin{bmatrix} e^{-i(\phi/2+\lambda/2)} & -ie^{-i(\phi/2-\lambda/2+\pi/2)}\\ -ie^{i(\phi/2-\lambda/2+\pi/2)} & e^{i(\phi/2+\lambda/2)} \end{bmatrix} =\Rsr2\begin{bmatrix} e^{-i(\phi/2+\lambda/2)} & -e^{-i(\phi/2-\lambda/2)}\\ e^{i(\phi/2-\lambda/2)} & e^{i(\phi/2+\lambda/2)} \end{bmatrix}$
$\sim\Rsr2\begin{bmatrix} 1 & -e^{i\lambda}\\ e^{i\phi} & e^{i(\phi+\lambda)} \end{bmatrix} \ldots$ QED
$\boxed{ u_3(\theta, \phi,\lambda):=U(\theta, \phi,\lambda)=R_z(\phi+3\pi)~R_x(\pi/2)~R_z(\theta+\pi)~R_x(\pi/2)~R_z(\lambda) \sim\begin{bmatrix} \cos\theta & -e^{i\lambda}\sin\theta\\ e^{i\phi}\sin\theta & e^{i(\phi+\lambda)}\cos\theta\\ \end{bmatrix} .}$
Proof:
$u_3(\theta, \phi,\lambda)=R_z(\phi+3\pi)~R_x(\pi/2)~R_z(\theta+\pi)~R_x(\pi/2)~R_z(\lambda)$
$\scriptsize =\begin{bmatrix} e^{-i(\phi/2+3\pi/2)} & 0\\ 0 & e^{i(\phi/2+3\pi/2)} \end{bmatrix} \begin{bmatrix} \cos\pi/4 & -i\sin\pi/4\\ -i\sin\pi/4 & \cos\pi/4 \end{bmatrix} \begin{bmatrix} e^{-i(\theta+\pi)/2} & 0\\ 0 & e^{i(\theta+\pi)/2} \end{bmatrix} \begin{bmatrix} \cos\pi/4 & -i\sin\pi/4\\ -i\sin\pi/4 & \cos\pi/4 \end{bmatrix} \begin{bmatrix} e^{-i\lambda/2} & 0\\ 0 & e^{i\lambda/2} \end{bmatrix}$
$= \begin{bmatrix} ie^{-i\phi/2} & 0\\ 0 & -ie^{i\phi/2} \end{bmatrix} \Rsr2 \begin{bmatrix} 1 & -i\\ -i & 1 \end{bmatrix} \begin{bmatrix} -ie^{-i\theta/2} & 0\\ 0 & ie^{i\theta/2} \end{bmatrix} \Rsr2 \begin{bmatrix} 1 & -i\\ -i & 1 \end{bmatrix} \begin{bmatrix} e^{-i\lambda/2} & 0\\ 0 & e^{i\lambda/2} \end{bmatrix}$
$={1\over2}\left( ie^{-i\phi/2} \begin{bmatrix} 1 & 0\\ 0 & -e^{i\phi} \end{bmatrix} \begin{bmatrix} 1 & -i\\ -i & 1 \end{bmatrix} \left(-ie^{-i\theta/2}\right) \begin{bmatrix} 1 & 0\\ 0 & -e^{i\theta} \end{bmatrix} \begin{bmatrix} 1 & -i\\ -i & 1 \end{bmatrix} e^{-i\lambda/2} \begin{bmatrix} 1 & 0\\ 0 & e^{i\lambda} \end{bmatrix}\right)$
$\displaystyle ={e^{-i(\phi+\theta+\lambda)/2}\over2}\left( \begin{bmatrix} 1 & 0\\ 0 & -e^{i\phi} \end{bmatrix} \begin{bmatrix} 1 & -i\\ -i & 1 \end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & -e^{i\theta} \end{bmatrix} \begin{bmatrix} 1 & -i\\ -i & 1 \end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & e^{i\lambda} \end{bmatrix}\right)$
$\sim {1\over2} \begin{bmatrix} 1 & -i\\ ie^{i\phi} & -e^{i\phi} \end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & -e^{i\theta} \end{bmatrix} \begin{bmatrix} 1 & -ie^{i\lambda}\\ -i & e^{i\lambda} \end{bmatrix} ={1\over2} \begin{bmatrix} 1 & -i\\ ie^{i\phi} & -e^{i\phi} \end{bmatrix} \begin{bmatrix} 1 & -ie^{i\lambda}\\ ie^{i\theta} & -e^{i(\theta+\lambda)} \end{bmatrix}$
$={1\over2} \begin{bmatrix} 1+e^{i\theta} & -ie^{i\lambda}+ie^{i(\theta+\lambda)}\\ ie^{i\phi}-ie^{i(\phi+\theta)} & e^{i(\phi+\lambda)}+e^{i(\phi+\theta+\lambda)} \end{bmatrix} ={1\over2} \begin{bmatrix} 1+e^{i\theta} & -ie^{i\lambda}(1-e^{i\theta})\\ ie^{i\phi}(1-e^{i\theta}) & e^{i(\phi+\lambda)}(1+e^{i\theta}) \end{bmatrix}$
$\displaystyle ={1\over2}\cdot \frac{1+e^{i\theta}}{\cos(\theta/2)} \left( \frac{\cos(\theta/2)}{1+e^{i\theta}} \begin{bmatrix} 1+e^{i\theta} & -ie^{i\lambda}(1-e^{i\theta})\\ ie^{i\phi}(1-e^{i\theta}) & e^{i(\phi+\lambda)}(1+e^{i\theta}) \end{bmatrix} \right) .$
The factor $\frac{1+e^{i\theta}}{2\cos(\theta/2)} =\frac{1+\cos\theta+i\sin\theta}{2\cos(\theta/2)} =\frac{1+2\cos^2(\theta/2)-1+2i\sin(\theta/2)\cos(\theta/2)}{2\cos(\theta/2)}\\ =\frac{\cos^2(\theta/2)+i\sin(\theta/2)\cos(\theta/2)}{\cos(\theta/2)} =\cos(\theta/2)+i\sin(\theta/2)~~\text{is a unit-vector}.$
Also, $\frac{1-e^{i\theta}}{1+e^{i\theta}} =\frac{1-\cos\theta-i\sin\theta}{1+\cos\theta+i\sin\theta} =\frac{(1-\cos\theta-i\sin\theta)(1+\cos\theta-i\sin\theta)}{(1+\cos\theta)^2+\sin^2\theta} =\frac{(1-i\sin\theta)^2-\cos^2\theta)}{1+2\cos\theta+\cos\theta^2+\sin^2\theta}$
$=\frac{1-2i\sin\theta-\sin^2\theta-\cos^2\theta}{2+2\cos\theta} =\frac{-i\sin\theta}{1+\cos\theta} =\frac{-2i\sin(\theta/2)\cos(\theta/2)}{1+2\cos^2(\theta/2)-1} =\frac{-i\sin(\theta/2)}{\cos(\theta/2)}.$
So we have $\frac{\cos(\theta/2)}{1+e^{i\theta}}\cdot i(1-e^{i\theta})=\sin(\theta/2) .$
$\displaystyle \therefore u_3(\theta, \phi,\lambda)\sim \frac{\cos(\theta/2)}{1+e^{i\theta}} \begin{bmatrix} 1+e^{i\theta} & -ie^{i\lambda}(1-e^{i\theta})\\ ie^{i\phi}(1-e^{i\theta}) & e^{i(\phi+\lambda)}(1+e^{i\theta}) \end{bmatrix} =\begin{bmatrix} \cos(\theta/2) & -e^{i\lambda}\sin(\theta/2)\\ e^{i\phi}\sin(\theta/2) & e^{i(\phi+\lambda)}\cos(\theta/2) \end{bmatrix} \ldots$ QED
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