What makes quantum decoherence different from dissipation?

Question

From my understanding quantum decoherence and dissipation are completely different ways of modelling information loss to the environment. Dissipation can be modeled using the Caldeira-Leggett model which uses an effective Hamiltonian and Zurek's decoherence is something else entirely that bypasses the usual unitary evolution of the Schrodinger equation. When are each of these models used? Are they conflicting?

Answer 1

Dissipation and decoherence are general processes that are not limited to specific models proposed by Caldeira-Leggett or Zurek. The terminology usually relates to whether or not energy is lost into the environment. The general set up comprises a small "open system" $A$ placed in contact with a larger environment $B$ via some interaction Hamiltonian $H_{AB}$, so that the total Hamiltonian is $$ H = H_A + H_B + H_{AB}.$$

Decoherence refers specifically to the decay of coherences in the density matrix of the open system $A$. To be precise, we write this density matrix as $$ \rho_A(t) = \sum_{i,j} \rho_{ij}(t) \lvert i\rangle \langle j \rvert,$$ and then decoherence corresponds to decay of the off-diagonal elements $\rho_{ij}(t)$ for $i\neq j$. "Pure decoherence" means that only the off-diagonal elements decay, while the diagonal elements $\rho_{ii}(t)$ are invariant.

Of course, this definition is dependent on the choice of basis vectors $\lvert i\rangle$, so that the concept of decoherence is basis-dependent in general. However, it is frequently convenient to consider the energy eigenbasis of the open system, so that $\lvert i\rangle$ are the eigenstates of $H_A$. Then we have pure decoherence (also called pure dephasing in this basis) if $[H_A,H_{AB}]=0$, which means that the interaction does not change the energy of the open system.

Dissipation corresponds to $[H_A,H_{AB}] \neq 0$, so that the interaction changes the energy of the open system. Then the diagonal elements of $\rho_A$ in the energy eigenbasis, i.e. $\rho_{ii}(t)$, change over time such that energy is lost irreversibly. Generally this process also implies decoherence, since the $\rho_{ij}(t)$ must decay in order to preserve positivity of the density matrix.

Answer 2

Dissipation implies that the energy of a quantum system $S$ under investigation, is spread over the many degrees of freedom of the bath. The dissipation is usually accompanied by fluctuation, i.e. the energy gets redistributed again and again in different ways between $S$ and all the degrees of freedom in the bath. Eventually, it is possible that after some time, all the energy be found, if measured, re-concentrated on $S$.

Now, dissipation doesn't mean decoherence, if we have an enough powerful computer to keep track of the evolution in time of the total system ($S$ + bath) under all the possible configurations, the total system would still be described by a wave-function - no decoherence. If we are interested only in the description of $S$, then no, we can't describe the total system as a product of an isolated wave-function of $S$, and some wave-function of the bath, but as a superposition of products of different states of the system with the corresponding states of the bath.

In absence of such a computing possibility, for obtaining the evolution of $S$ we trace the density matrix of the total system over the different states of the bath, and obtain a density matrix for our system only. But this density matrix is of a mixture of states of the system, not of a wave-function representing a single state.

Now, the decoherence occurs very similarly, i.e. the system $S$ isn't isolated, it comes in contact with an environment. Though, this environment has an infinite number of degrees of freedom, or even infinite and also undefined number (as for instance a macroscopic measurement apparatus which is by itself an open system, exchanging all the time particles with the surroundings). Whether the total system, $S$ + environment continues to have a wave-function is a disputed issue. At this point different interpretations of the quantum theory adopt different positions. I won't get into them, only mention that the Standard Quantum Mechanics regards the state of the system $S$ as decohered.