Does the Hilbert space of the universe have to be infinite dimensional to make sense of quantum mechanics? Otherwise, decoherence can never become exact. Does interpreting quantum mechanics require exact decoherence and perfect observers of the sort which can only arise from exact superposition sectors in the asymptotic future limit?
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2 Answers
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With a finite-dimensional Hilbert space, the whole apparatus of practical QM is lost. Very little is left - no continuous spectra, no scattering theory, no S-matrix, no cross sections. No Dirac equation, no relativity theory, no relation between symmetry and conservation laws, no quantum fields. Almost the whole of the achievements of modern physics would be ruined.
Already the Hilbert space of a single oscillating mode is infinite-dimensional, and the universe contains zillions of them. Fortunately, zillions times infinite is still infinite, but...
Due to superselection sectors, the Hilbert space of QED is already nonseparaple (i.e., has an uncountable basis). The physical Hilbert space of a quantum field theory is the direct integral of the Hilbert spaces corresponding to the different superselection sectors. The direct integral is mathematically well defined, http://en.wikipedia.org/wiki/Direct_integral and gives a nonseparable space once the integral is over a continuum.
In QED it is (at least) the continuum of directions in 3-space. One needs this nonseparable space to define Lorentz transformations of charges states, as charged states moving in different directions are in different superselection sectors. Thus the dimension of the Hilbert space of the universe should be at least the cardinality of the continuum.
Now QED describes the universe with gravitation, weak and strong forces ignored. Unfortunately, very little is known about the Hilbert space of nonabelian gauge theories and quantum gravity, so it is not that clear which cardinality the Hilbert space of the universe will have once we know whether the universe is described by one.
On the other hand, the interpretation of quantum mechanics cannot depend on exact models, as our models of the real world are never exact replicas of the latter.
answered Jun 8, 2012 at 14:25
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1$\begingroup$ The oscillating mode is infinite only if the oscillator extends over all space and to microscopic distances. If you put a cosmological cutoff on space and a Planckian cutoff, then you get a finite Hilbert space dimension. There are no nonseparable Hilbert spaces in quantum mechanics,ever ever, even including superselection sectors. This is not well defined--- the Hilbert space is defined superselection sector by superselection sector. $\endgroup$ Commented Jun 9, 2012 at 2:20
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$\begingroup$ An oscillator is by definition associated with a Hamiltonian defined on a representation of the CCR, which implies infinite dimensions. - The Hilbert space is separable in each sector, but gauge symmetries, field operators, and perturbation theory are defined only in the direct integral, which is nonseparable. It is not even clear whether the Hamiltonian dynamics respects the sectors; lack of this may well be the reason why spacetime dimension 4 is so difficult to analyze. $\endgroup$ Commented Jun 10, 2012 at 10:12
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$\begingroup$ I don't agree with this idea--- the question was about physics, and of course the CCR fail if you imagine a discrete space. The "direct integral" is something which is a little vaguely defined--- you are talking about the infrared photon issues--- this is not clearly sensible, although it is also not clearly un-sensible. But if it is sensible, you have to define it by an explicit limit. $\endgroup$ Commented Jun 11, 2012 at 4:48
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1$\begingroup$ @RonMaimon: Yes, the question was about physics, and if you take away the oscillator representations from physics, very little is left - no continuous spectra, no scattering theory, no S-matrix, no cross sections. The whole apparatus of practical QM is lost. Finite-dimensional Hilbert spaces just cover quantum information theory (not yet really applicable) and speculations in some theoretical quarters about the possible structure of a quantum gravity theory. $\endgroup$ Commented Jun 11, 2012 at 8:20
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1$\begingroup$ @RonMaimon: Fir the same reason, you'd say that in place of Hilbert spaces one should consider finite-diemsnional vector spaces over the rationals, and indeed, only finite sets as even a rational vector space is an idealization. One cannot avoid idealizations in modeling. The best idealization is always the one that gives most easy conceptual and computational access to the situation at hand. All progress in physics was based on such conceptual advances. Going back to finite-d spaces or even finite sets amounts to undoing all that. $\endgroup$ Commented Jun 12, 2012 at 10:39
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The universe probably has an infinite dimensional Hilbert space. However, finite dimensions suffice to "make sense of" quantum mechanics. Or at least finite dimensions are sufficient to gain intuition and to expose the philosophical difficulties of the various interpretations of quantum mechanics. A cat lives in a very high dimensional (or infinite) Hilbert space, but the essence of the Schrodinger's cat paradox can be grasped by just considering it to be a two-state system.
And finite dimensions are sufficient for perfect decoherence. An example is the $|+> \otimes |0>$ state fed into a controlled-not gate in the context of quantum computing. In fact, that simple circuit is actually a pretty good model for understanding the role of observers (the second qubit "observes" or copies the state of the first qubit, thereby causing decoherence). Finite dimensions are a lot easier to think about. You may of course not want to think of you as an observer as being a qubit entangled with the object you were measuring, but that is a can of worms that philosophers will probably still be working out for a very long time.
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$\begingroup$ In a finite-diemsnional Hilbert space there is no decoherence. (All motions are quasiperiodic and nothing can dissipate.) $\endgroup$ Commented Jun 11, 2012 at 8:31
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$\begingroup$ @ArnoldNeumaier: perhaps we have differing ideas of what decoherence means. The example I gave with the CNOT gate is finite dimensional, is not quasiperiodic (unless you consider the constant state $|00>+|11>$ to be periodic), and leads to disappearance of the off-diagonal elements of the density matrix. Tracing out the second qubit leaves the first qubit in a classical state. I have heard people in the quantum information community refer to this as decoherence. Am I wrong? $\endgroup$ Commented Jun 11, 2012 at 12:20
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$\begingroup$ There is an extended discussion of this problem at physicsforums.com/showpost.php?p=3157393&postcount=103 $\endgroup$ Commented Jun 11, 2012 at 12:45
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$\begingroup$ A constant state is of course periodic for every period. In quantum computing, one needs the environment (with its infinite-D Hilbert space) for preparing definite input states. Without that you cannot use a quantum device for computation. - ''Decoherence occurs when a system interacts with its environment in a thermodynamically irreversible way.'' from en.wikipedia.org/wiki/Decoherence $\endgroup$ Commented Jun 11, 2012 at 12:52