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Heuristics for definitions of open and closed quantum dynamics
I've been reading some of the literature on "open quantum systems" and it looks like the following physical interpretations are made:
Reversible dynamics of a closed quantum system are represented by a group of automorphisms (of a $C^∗$-algebra, von Neumann algebra, or $B(H)$, depending on whom you read)
Irreversible (but still deterministic) dynamics of a closed system are represented by a semigroup of endomorphisms
The dynamics of an open system (which are non-deterministic, hence also irreversible) are represented by a semigroup of unital completely positive maps.
I'm trying to motivate these interpretations based on my limited familiarity with quantum mechanics. If you use the formalism of QM wherein states are represented by density operators, then you get something like this:
In the Schrodinger picture, a unitary time evolution is represented by $\rho \mapsto U(t) \rho U(-t)$, where $U(t) = e^{-itH/\hbar}$, and a measurement corresponding to an operator $X = \sum_\lambda \lambda P_\lambda$ is represented by $\rho \mapsto \sum_\lambda P_\lambda \rho P_\lambda$.
In the Heisenberg picture, a unitary time evolution is represented by $Y \mapsto U(-t) Y U(t)$, and a measurement by $Y \mapsto \sum_\lambda P_\lambda Y P_\lambda$.
The unitary time evolution in the Heisenberg picture is an automorphism group on $B(H)$, and the measurement process is a unital completely positive map. So I can intuit that, in an open quantum system (i.e. one which is constantly "measured" by interacting/entangling with the environment) the dynamics should be represented by a semigroup of unital completely positive maps. And certainly the idea that reversible dynamics for a closed system corresponds to an automorphism group make sense. What I still don't quite see is why you would represent irreversible dynamics of a closed system by a semigroup of endomorphisms. How does the difference between closed and open systems correspond to maps on $B(H)$ which are multiplicative or not? Also, is there an easily understandable example (for someone with a lot more math than physics) of where a closed system with irreversible dynamics might come from?
quantum-optics
asked Mar 29, 2012 in Theoretical Physics by Dave Gaebler (20 points) [ no revision ]
"a measurement corresponding to an operator X=∑λλPλ is represented by ρ↦∑λPλρPλ" doesn't seem to be correct, i.e. post-measurement result is given by one of $P_{\lambda}\rho P_{\lambda}$ (plus normalization), for $\lambda$ chosen nondeterministically.
commented Mar 29, 2012 by Marcin Kotowski (405 points) [ no revision ]
Where did you read that the completely positive map representing the dynamics of an open quantum system has to be unital? I didn't think that was necessary.
commented Mar 30, 2012 by Peter Shor (780 points) [ no revision ]
From a geometric point of view, the definition of reversible quantum dynamics is precisely the usual one for any dynamical system: $\mathcal{L}_X(\omega) = 0$, where $X$ is the dynamical flow, $\omega$ is the symplectic structure, and $\mathcal{L}$ is the Lie derivative, that is, the flow is a symplectomorphism (and an isometry). Whereas in contrast, the stochastic flow associated to Lindblad (open) dynamics is in general *not* a symplectomorphism (nor is it an isometry); thus open-system quantum dynamics also is similar to any other open-system dynamics.
commented Apr 2, 2012 by John Sidles (485 points) [ no revision ]
@MarcinKotowski I thought the density-operator formulation allowed you to keep track of all the possible outcomes at once, weighted by probability...the formula I had appears on http://en.wikipedia.org/wiki/Density_operator#Measurement (but of course a more authoritative source would be better!)
commented Apr 9, 2012 by Dave Gaebler (20 points) [ no revision ]
An open system in the sense you describe is not a system coupled to the environment, but the idealized version of this system where the environment has been eliminated using a Markov approximation, so that there is a closed dynamics on the system itself, without reference to the environment (and hence to measurement), and without memory (which is enforced by the Markov approximation). [To see how the environment is eliminated, see, e.g., cond-mat/0011204.]
Closed dynamics means that you can express the time derivative of the state of the system in terms of the current state and its past, and lack of memory means that you in fact have a differential equation for the states rather than an integrodifferental equation. (The above characterizxation would not work in the Heisenberg picture.)
If there are time-dependent external fields, the characterization you inquire about doesn't hold. But if there are no time-dependent external fields then the differential equation is autononous, and hence describes a semigroup.
For finitely many degrees of freedom, it is in fact a group, but running the dynamics backwards is highly unstable. For infinitely many degrees of freedom, running the dynamics backwards is ill-posed.
Thus the semigroup property has nothing to do with quantum mechanics (which just contributes complete positivity). Think of the heat equation as an example of a semigroup describing a dissipative (i.e., open) classical system.
But the form of the resulting dynamics is specifically quantum, typically a Lindblad equation; see, e.g., quant-ph/0302047.
answered Mar 30, 2012 by Arnold Neumaier (14,437 points) [ no revision ]
This doesn't really answer the question: how can a closed system with irreversible dynamics arise naturally in a physical system?
@Peter Shor: I don't think that was the original question. But your question is answered in the two papers I quoted, which show how the unitary dynamics including the environment gives rise to a reduced dissipative dynamics if the Markov approximation is assumed.
commented Mar 30, 2012 by Arnold Neumaier (14,437 points) [ no revision ]
In quantum mechanical formalisms in which discrete operations are unraveled as continuous differential flows on a state-space manifold the distinction is geometrically natural:
Closed quantum systems are Hamiltonian flows
Open quantum systems are Lindbladian flows
The former (smooth) flows are induced by the symplectic structure of the state-space, and the latter (stochastic) flows are induced by the metric structure of the state-space. Carlton Caves' on-line notes Completely positive maps, positive maps, and the Lindblad form works through the details. In particular, the former flows are exact symplectomorphisms and isometries, while the latter flows in general are neither.
The two flows interact as follows: a metric "lowering" followed by a symplectic "raising" defines a complex structure that (in conventional Hilbert coordinates) is readily shown to be the familiar multiplicative $i$ of quantum mechanics. Ashtekar and Schilling's review Geometrical formulation of quantum mechanics gives further details.
A concrete engineering advantage of this post-Hilbert dynamical framework is that it captures most of the "quantum goodness" of Hilbert space, and yet is well-posed on manifolds that have lower dimensionality than Hilbert space (upon which trajectory integration is numerically tractable).
answered Apr 2, 2012 by John Sidles (485 points) [ no revision ]
p$\hbar$ysicsO$\varnothing$erflow