# How does einselection fit into quantum decoherence?

My understanding of decoherence is the following:

There is a system state vector $$|\Psi\rangle_s$$ and the environment state vector $$|\Psi\rangle_\epsilon$$ which are combined to form $$|\Psi\rangle_s \otimes |\Psi\rangle_\epsilon = |\Psi \rangle_{s \epsilon}$$. If the interaction Hamiltonian $$\hat{H}_{s \epsilon}$$is time-independent the combined state evolves such that $$|\Psi(t) \rangle_{s \epsilon} = e^{- i \hat{H}_{s \epsilon} t / \hbar} |\Psi \rangle_{s \epsilon} \, .$$ Although the system state vector doesn't act unusually, its density matrix $$\text{Tr}_{\epsilon}(\rho_{s \epsilon})$$ has an unusual property, its off diagonals approach zero extremely quickly. As a byproduct, the entropy of this matrix increases.

I'm trying to understand how einselection/superselection fits into this picture. Is it merely that only in a particular basis the off diagonals get reduced? Somehow this fits with observables, but I'm not sure.

By the way, decoherence is sometimes criticized because it refers to a particular factorization of the Hilbert space, which is artifical. That's a valid criticism of that way of formulating decoherence, but it can also be formulated in a more general way that doesn't require any factorization of the Hilbert space at all. (I'm bringing this up because it may help answer the question more fully.) Instead of choosing a factorization of the Hilbert space, we only need to recognize (1) which observables are associated with the system of interest — which we have to do anyway in order to use quantum theory at all — and (2) which observables could feasibly be measured using the limited resources that are actually available in the real universe. If $$A$$ is some observable associated with the system of interest, then any state-vector can always be written as a linear combination of eigenstates of $$A$$. We say that decoherence (specifically "einselection" in $$A$$'s eigenbasis) has occurred whenever no future feasibly-measurable observable can mix those eigenspaces with each other. In that case, if the non-mixing property is close enough to being exact, we can safely apply Born's rule with respect to $$A$$.