Lindblad Equation - Diagonalization

Diagonalization

Since the matrix is positive, it can be diagonalized with a unitary transformation u:

u^\dagger h u = \begin{bmatrix} \gamma_1 & 0 & \cdots & 0 \\ 0 & \gamma_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \gamma_{N^2-1} \end{bmatrix}

where the eigenvalues are non-negative. If we define another orthonormal operator basis

we can rewrite the Lindblad equation in diagonal form

This equation is invariant under a unitary transformation of the Lindblad operators and constants,

and also under the inhomogenous transformation

However, the first transformation destroys the orthonormality of the operators (unless all the are equal) and the second transformation destroys the tracelessness. Therefore, up to degeneracies among the, the of the diagonal form of the Lindblad equation are uniquely determined by the dynamics so long as we require them to be orthonormal and traceless.

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