# How to map the eigenvectors of a superoperator into the corresponding operators?

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.)

• An eigenvector of a superoperator is a linear operator (a $d\times d$ matrix) Jun 14, 2021 at 6:32
• 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? Jun 14, 2021 at 6:45

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.