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

Square Root of Quantum Gates

^{First created in August 2018}

Pauli Gates

To find the square root of a Pauli gate ($\pi/2$ gates), we can use a combination of $S$ ($\pi/4$ gate) and $T$ ($\pi/8$ gate).

To recap:

$R_x(\theta) =e^{-i{\theta\over 2}X} =\begin{bmatrix} \cos{\theta\over 2} & -i\sin{\theta\over 2}\\ -i\sin{\theta\over 2} & \cos{\theta\over 2} \end{bmatrix}.~~ \therefore~ X \approx e^{-i{\pi\over 2}X} =\begin{bmatrix}0&-i\\-i&0 \end{bmatrix} \approx \begin{bmatrix}0&1\\1&0 \end{bmatrix} .$

$R_y(\theta) \approx e^{-i{\theta\over 2}Y} =\begin{bmatrix} \cos{\theta\over 2} & -\sin{\theta\over 2}\\ \sin{\theta\over 2} & \cos{\theta\over 2} \end{bmatrix}.~~ \therefore~ Y \approx e^{-i{\pi\over 2}Y} =\begin{bmatrix}0&-1\\1&0 \end{bmatrix} \approx \begin{bmatrix}0&-i\\i&0 \end{bmatrix} .$

$R_z(\theta) =e^{-i{\theta\over 2}Z} =\begin{bmatrix} e^{-i\theta/2} & 0\\ 0 & e^{i\theta/2} \end{bmatrix}.~~ \therefore~ Z \approx e^{-i{\pi\over 2}Z} =e^{-i{\pi\over2}}\begin{bmatrix}-1&0\\0&i\end{bmatrix} \approx \begin{bmatrix}1&0\\0&-1\end{bmatrix} .$

$S=R_z(\pi/2) =e^{-i{\pi\over4}Z} =\begin{bmatrix} e^{-i\pi/4} & 0\\ 0 & e^{i\pi/4} \end{bmatrix} \approx\begin{bmatrix} 1 & 0\\ 0 & e^{i\pi/2} \end{bmatrix} ,~~$ hence the name $\pi/4$ gate.

$T=R_z(\pi/4) =e^{-i{\pi\over8}Z} =\begin{bmatrix} e^{-i\pi/8} & 0\\ 0 & e^{i\pi/8} \end{bmatrix} \approx\begin{bmatrix} 1 & 0\\ 0 & e^{i\pi/4} \end{bmatrix},~~$ hence the name $\pi/8$ gate.

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$\sqrt{Z}=S.~~ \sqrt{S}=T .$

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$HXHX\approx Y.~~ HX =\Rsr2\begin{bmatrix}1&1\\1&-1\end{bmatrix} \begin{bmatrix}0&1\\1&0\end{bmatrix} =\Rsr2\begin{bmatrix}1&1\\-1&1\end{bmatrix}.~~ (HX)^2 ={1\over2}\begin{bmatrix}0&2\\-2&0\end{bmatrix} \approx Y.~~ \sqrt{Y}\approx HX .$

$HYHY\approx I.~~ HY =\Rsr2\begin{bmatrix}1&1\\1&-1\end{bmatrix} \begin{bmatrix}0&-i\\i&0\end{bmatrix} =\Rsr2\begin{bmatrix}i&-i\\-i&-i\end{bmatrix}.~~ (HY)^2 =-{1\over2}\begin{bmatrix}2&0\\0&2\end{bmatrix} \approx I.~~ \sqrt{I}\approx HY .$

$HZHZ\approx Y.~~ HZ =\Rsr2\begin{bmatrix}1&1\\1&-1\end{bmatrix} \begin{bmatrix}1&0\\0&-1\end{bmatrix} =\Rsr2\begin{bmatrix}1&-1\\1&1\end{bmatrix}.~~ (HZ)^2 ={1\over2}\begin{bmatrix}0&-2\\2&0\end{bmatrix} \approx Y.~~ \sqrt{Y}\approx HZ .$

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$\sqrt{X} =HSH ={1\over2} \begin{bmatrix}1&1\\1&-1\end{bmatrix} \begin{bmatrix}1&0\\0&i\end{bmatrix} \begin{bmatrix}1&1\\1&-1\end{bmatrix} ={1\over2} \begin{bmatrix}1&1\\1&-1\end{bmatrix} \begin{bmatrix}1&1\\i&-i\end{bmatrix} ={1\over2} \begin{bmatrix}1+i&1-i\\1-i&1+i\end{bmatrix} .$

$(HSH)^2=HSHHSH=HSSH=HZH ={1\over2}\begin{bmatrix}1&-1\\1&1\end{bmatrix} \begin{bmatrix}1&1\\1&-1\end{bmatrix} ={1\over2}\begin{bmatrix}0&2\\2&0\end{bmatrix} =X .$

T Gate

The challenge from the IBM Q Experience - Basic Circuit Identities and Larger Circuits page is

Arbitrarily good approximations exist, so can you find a better one? How might you use these circuits to construct an approximate controlled-$T$ unitary transformation?

I am trying to answer the two challenges here.

Square Root of T

The problem is to find a unitary transformation to map $\Ket0$ to $\Ket0$ and $\Ket1$ to $e^{i\pi/8}\Ket1$.

For $\sqrt{T}$, IBM Q has an approximation: $Q=HTHTHSTHTHSTHTHTH$.

$\sqrt{i}=\Rsr2(1+i)$ and $\sqrt{-i}=\Rsr2(1-i).~~$ (We take the primary solution with smallest positive argument.)

$S=\begin{bmatrix}1&0\\0&i\end{bmatrix}.~~ T=\sqrt{S}=\begin{bmatrix}1&0\\0&\sqrt{i}\end{bmatrix}.~~ \sqrt{T}=\begin{bmatrix}1&0\\0&\sqrt[4]{i}\end{bmatrix} .$

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KEPT

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$TH =\begin{bmatrix}1&0\\0&\sqrt{i}\end{bmatrix} \Rsr2\begin{bmatrix}1&1\\1&-1\end{bmatrix} =\Rsr2\begin{bmatrix}1&1\\\sqrt{i}&-\sqrt{i}\end{bmatrix} .$

$THTH ={1\over2}\begin{bmatrix}1&1\\\sqrt{i}&-\sqrt{i}\end{bmatrix}^2 ={1\over2}\begin{bmatrix} 1+\sqrt{i}&1-\sqrt{i}\\ \sqrt{i}-i&\sqrt{i}+i\end{bmatrix} .$

$STHTH =\begin{bmatrix}1&0\\0&i\end{bmatrix} {1\over2}\begin{bmatrix} 1+\sqrt{i}&1-\sqrt{i}\\ \sqrt{i}-i&\sqrt{i}+i\end{bmatrix} ={1\over2}\begin{bmatrix} 1+\sqrt{i}&1-\sqrt{i}\\ i\sqrt{i}+1&i\sqrt{i}-1\end{bmatrix} .$

$(STHTH)^2 ={1\over4}\begin{bmatrix} 1+\sqrt{i}&1-\sqrt{i}\\ i\sqrt{i}+1&i\sqrt{i}-1\end{bmatrix}^2 ={1\over4}\begin{bmatrix} 1+2\sqrt{i}+i+i\sqrt{i}+1+1-\sqrt{i}&1-i+i\sqrt{i}-1+1+\sqrt{i}\\ i\sqrt{i}+1-1+\sqrt{i}-i-1&i\sqrt{i}+1+1-\sqrt{i}-i-2i\sqrt{i}+1\\ \end{bmatrix} ={1\over4}\begin{bmatrix} 3+\sqrt{i}+i+i\sqrt{i}&1+\sqrt{i}-i+i\sqrt{i}\\ -1+\sqrt{i}-i+i\sqrt{i}&3-\sqrt{i}-i-i\sqrt{i}\\ \end{bmatrix} .$

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$Q^2 =(HTHTHSTHTHST~HTHTH)~(HTHTH~STHTHSTHTHTH) =(HTHTHSTHTHST~HTH)~S~(HTH~STHTHSTHTHTH) =(HTHTHSTHTHST~HT)~\sqrt{X}~(TH~STHTHSTHTHTH) .$