# What makes quantum decoherence different from dissipation?

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?

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.

• @Sofia 1) I don't understand your claim about the environment not having a Hamiltonian. Are you saying that the energy of the environment cannot be defined? That is a very controversial claim. 2) It is not possible to state how the off-diagonal elements decay in general. In the simplest phenomenological models it is exponential, but in almost all physically realistic models the time dependence is more complicated. 3) Dissipation is the irreversible loss of energy by definition. However this is referring to average quantities. There are of course still fluctuations. – Mark Mitchison Mar 12 '15 at 17:02
• Pretty much everything I have written here is a completely uncontroversial description of the terminology as it is used by people working in the field of open quantum systems. I think you and I may be discussing different things if you really object so strongly. In particular, you seem to have interpreted this as a question about quantum foundations/interpretations. I am not talking about philosophy, but rather about the general problem of a quantum system interacting with an environment, as posed in the OP. – Mark Mitchison Mar 12 '15 at 17:03
• @Sofia It is always possible to include the environment in your description by extending your definition of $H_B$. For example, the environment includes both the measuring apparatus and the electromagnetic radiation field. I didn't make any claims about the nature of the environment. I just stated that a description of any open quantum system exists according to $H_A + H_B + H_{AB}$. Are you seriously claiming that this is not correct? If so, please give a counter-example. – Mark Mitchison Mar 12 '15 at 20:41
• I work with the $\alpha$ decay which is a typical case of open system. People don't work with a Hamiltonian of type $H_A + H_B + H_{AB}$, but with a non-Hermitic effective Hamiltonian. And about unitarity there are endless disputes. But bottom line, as I say, they work with effective Hamiltonians. – Sofia Mar 12 '15 at 21:03
• you say "Nevertheless, it is always possible in principle to describe the full system by a Hamiltonian of the type I described." How it is possible? If it were possible physicists would have use it. Did you ever see an $\alpha$ particle returning from the limit of the universe and re-entering the nucleus? And if we would try to produce the $\alpha$-unstable nucleus by inelastic scattering, this process is not the opposite of the decay. As to decoherence, we know how it begins, but not how it ends. – Sofia Mar 12 '15 at 23:34

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.

• Who placed this minus, would explain why? – Sofia Mar 12 '15 at 15:17