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The set of linear operators acting on a $d$ dimensional Hilbert space, $H$ form a vector space, called operator space $\mathcal{L}(H)$. Elements of operator space are $d \times d$ matrices. Now the set of linear operators on the operator space, $\mathcal{L}(\mathcal{L}(H))$ form another vector space of dimension $d^4$. Elements of this space are called superoperators. Superoperators are $d^2 \times d^2$ matrices. Their eigenvectors form independent directions in the operator space and should form a complete basis if full rank.

My question is, how do I map these eigenvectors which are $d^2 \times 1$ column vectors to $d \times d$ matrices? (Since Operator space is a space of $d\times d$ matrices. Intuitively, there is a correspondence.)

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  • $\begingroup$ An eigenvector of a superoperator is a linear operator (a $d\times d$ matrix) $\endgroup$ Jun 14 '21 at 6:32
  • $\begingroup$ Superoperator is a $d^2 \times d^2$ matrix. When you find eigenvectors, you get $d^2 \times 1$ column vectors. How to transform them to $d \times d$ matrices? $\endgroup$
    – Raman
    Jun 14 '21 at 6:45
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A super-operator is a linear operator on $\mathcal L(\mathcal H)$. Its eigenvectors are elements of $\mathcal L(H)$. If you want to use matrix algebra, you have to pick a basis for $\mathcal L(H)$, compute the matrix of the superoperator in this basis, then diagonalise it. The $d^2\times 1$ eigenvectors you find actually represent elements of $\mathcal L(H)$ using the basis you chose earlier.

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