| Summary: | We propose the following definition of topological quantum phases valid for mixed states: two states are in the same phase if there exists a time independent, fast and local Lindbladian evolution driving one state into the other. The underlying idea, motivated by \cite{Konig2014}, is that it takes time to create new topological correlations, even with the use of dissipation.
We show that it is a good definition in the following sense: (1) It divides the set of states into equivalent classes and it establishes a partial order between those according to their level of ``topological complexity''. (2) It provides a path between any two states belonging to the same phase where observables behave smoothly.
We then focus on pure states to relate the new definition in this particular case with the usual definition for quantum phases of closed systems in terms of the existence of a gapped path of Hamiltonians connecting both states in the corresponding ground state path. We show first that if two pure states are in the same phase in the Hamiltonian sense, they are also in the same phase in the Lindbladian sense considered here.
We then turn to analyse the reverse implication, where we point out a very different behaviour in the case of symmetry protected topological (SPT) phases in 1D. Whereas at the Hamiltonian level, phases are known to be classified with the second cohomology group of the symmetry group, we show that symmetry cannot give any protection in 1D in the Lindbladian sense: there is only one SPT phase in 1D independently of the symmetry group.
We finish analysing the case of 2D topological quantum systems. There we expect that different topological phases in the Hamiltonian sense remain different in the Lindbladian sense. We show this formally only for the $\mathbb{Z}_n$ quantum double models $D(\mathbb{Z}_n)$. Concretely, we prove that, if $m$ is a divisor of $n$, there cannot exist any fast local Lindbladian connecting a ground state of $D(\mathbb{Z}_m)$ with one of $D(\mathbb{Z}_n)$, making rigorous the initial intuition that it takes long time to create those correlations present in the $\mathbb{Z}_n$ case that do not exist in the $\mathbb{Z}_m$ case and that, hence, the $\mathbb{Z}_n$ phase is strictly more complex in the Lindbladian case than the $\mathbb{Z}_m$ phase. We conjecture that such Lindbladian does exist in the opposite direction since Lindbladians can destroy correlations.
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