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

Superoperator Examples

^{First created in August 2018}

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For example, if $\rho={1\over4}\Ket0\Bra0+{1\over4}\Ket0\Bra1+{1\over2}\Ket1\Bra1 =\begin{bmatrix} {1\over4}&{1\over4}\\ 0&{1\over2}\\ \end{bmatrix}$ and $A=S.$

It is given that $S=\begin{bmatrix}1&0\\0&i\end{bmatrix},~~~ S\otimes I =\begin{bmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&i&0\\ 0&0&0&i\\ \end{bmatrix} ,~~~ I\otimes S =\begin{bmatrix} 1&0&0&0\\ 0&i&0&0\\ 0&0&1&0\\ 0&0&0&i\\ \end{bmatrix} .$

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$\mathcal{L}(S)[\rho] ={1\over4}(S\Ket0)\Bra0+{1\over4}(S\Ket0)\Bra1+{1\over2}(S\Ket1)\Bra1 ={1\over4}\Ket0\Bra0+{1\over4}\Ket0\Bra1+{i\over2}\Ket1\Bra1 =\begin{bmatrix} {1\over4}&{1\over4}\\ 0&{i\over2} \end{bmatrix} =S\rho \mapsto\begin{bmatrix} {1\over4}\\{1\over4}\\0\\{i\over2}\\ \end{bmatrix} .$

$\left(S\otimes I\right)\Ket{\rho\rangle} =\left(S\otimes I\right)\left({1\over4}\Ket{00}+{1\over4}\Ket{01}+{1\over2}\Ket{11}\right) =\begin{bmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&i&0\\ 0&0&0&i\\ \end{bmatrix} \begin{bmatrix} {1\over4}\\{1\over4}\\0\\{1\over2}\\ \end{bmatrix} =\begin{bmatrix} {1\over4}\\{1\over4}\\0\\{i\over2}\\ \end{bmatrix} .$

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$\mathcal{R}(S)[\rho] ={1\over4}\Ket0(\Bra0 S)+{1\over4}\Ket0(\Bra1 S)+{1\over2}\Ket1(\Bra1 S)) ={1\over4}\Ket0\Bra0+{i\over4}\Ket0\Bra1+{i\over2}\Ket1\Bra1 =\begin{bmatrix} {1\over4}&{i\over4}\\ 0&{i\over2} \end{bmatrix} =\rho S \mapsto\begin{bmatrix} {1\over4}\\{i\over4}\\0\\{i\over2}\\ \end{bmatrix} .$

$\left((I\otimes S^T)\Ket{\rho\rangle}\right)^T =\left({1\over4}\Bra{00}+{1\over4}\Bra{01}+{1\over2}\Bra{11}\right)\left(I\otimes S\right) =\begin{bmatrix} {1\over4}&{1\over4}&0&{1\over2} \end{bmatrix} \begin{bmatrix} 1&0&0&0\\ 0&i&0&0\\ 0&0&1&0\\ 0&0&0&i\\ \end{bmatrix} =\begin{bmatrix} {1\over4}\\{i\over4}\\0\\{i\over2}\\ \end{bmatrix} .$

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A 2x2 example with $S$:

$\mathrm{LHS} =\Ket{SP\rangle} =\Ket{ \begin{bmatrix}1&0\\0&i\end{bmatrix} \begin{bmatrix}\psi_1\phi_1&\psi_1\phi_2\\\psi_2\phi_1&\psi_2\phi_2\end{bmatrix} {\large\rangle}} =\Ket{ \begin{bmatrix}\psi_1\phi_1&\psi_1\phi_2\\i~\psi_2\phi_1&i~\psi_2\phi_2\end{bmatrix} {\large\rangle}} =\begin{bmatrix}\psi_1\phi_1\\\psi_1\phi_2\\i~\psi_2\phi_1\\i~\psi_2\phi_2\end{bmatrix} .$

$\mathrm{RHS} =(S\otimes I)\Ket{P\rangle} =\begin{bmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&i&0\\ 0&0&0&i\\ \end{bmatrix} \begin{bmatrix}\psi_1\phi_1\\\psi_1\phi_2\\~\psi_2\phi_1\\~\psi_2\phi_2\end{bmatrix} =\begin{bmatrix}\psi_1\phi_1\\\psi_1\phi_2\\i~\psi_2\phi_1\\i~\psi_2\phi_2\end{bmatrix} .$