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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?

Qmechanic
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Octovian
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2 Answers2

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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.

Arnold Neumaier
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    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. – Ron Maimon Jun 09 '12 at 02:20
  • 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. – Arnold Neumaier Jun 10 '12 at 10:12
  • 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. – Ron Maimon Jun 11 '12 at 04:48
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    @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. – Arnold Neumaier Jun 11 '12 at 08:20
  • @RonMaimon: A direct integral of Hilbert spaces 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 the continuum of directions in 3-space. One needs this nonseparable space already to define Lorentz transformations on charges states. – Arnold Neumaier Jun 11 '12 at 08:29
  • @ArnoldNeumaier: I have to agree with Ron here. Physically there is always a cutoff --- as yourself alludes to, quantum gravity may provide one in both the UV and the IR. The CCR is a nice model of a real oscillator, but clearly no physical oscillator can sustain unbounded energy without some other physics coming into it. As for the lack of decoherence --- analogy: in statistical physics symmetry breaking is a nice fiction since no observed system yet is actually thermodynamic; however deviations are exponentially suppressed. Similarly, the periodic orbits can be exponentially large. – genneth Jun 11 '12 at 10:31
  • @genneth: The point is that one needs the fictions to be able to apply powerful mathematics, without which nothing is computable. A finite-dimensional Hilbert space would also provide only a model, but one that lacks all the structure that makes things computable, and hence the world predictable. – Arnold Neumaier Jun 11 '12 at 11:09
  • @ArnoldNeumaier: It could be approximating an infinite dimensional Hilbert space, and it must be to match observation. The only reason I am not sure is because finite dimensional Hilbert spaces approximate and define the infinite dimensional limit for large systems, and we can't measure the asymptotic infinity. It's very useful for good mathematics results of course. – Ron Maimon Jun 12 '12 at 02:30
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    @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. – Arnold Neumaier Jun 12 '12 at 10:39
  • @ArnoldNeumaier: I agree, and I am not advocating this. But the physics that demands finite Hilbert space is holgraphy with finite area cosmological horizons, and this must be taken seriously, even though it is completely paradoxical. For modelling QED (or massless theories in general) in R^4 you might be right that the right approach is nonseparable analogs of Hilbert space, I don't know for sure. The issue is that if you allow things like a finite density of particles you can get uncomputable computations in the infinite time limit, and so the S-matrix would become uncomputable. – Ron Maimon Jun 12 '12 at 20:41
  • @RonMaimon: A finite density is only needed for thermal systems, and then the S-matrix ceases to exist in any field theory. Instead one wants to compute thermal correlation functions, which are perfectly well-defined after renormalization. – Arnold Neumaier Jun 13 '12 at 11:35
  • @ArnoldNeumaier: That's if you set up a thermal system, and then no S-matrix. But if you make an infinite number of particles arranged in special ways coming in from infinity, you can make a Turing machine in intermediate stages, and you can make it blow itself up if the computation halts, and the S-matrix encode the halting problem solution. This is the type of thing that I am talking about which makes it tough to talk about infinite numbers of particles. – Ron Maimon Jun 13 '12 at 17:44
  • Of course one can play with these things theoretically. But they have no applications yet. And it is unrelated to the question asked - for how can you have infinitely many particles in a finite-dimensional Hilbert space? – Arnold Neumaier Jun 13 '12 at 20:21
  • to "The direct integral is mathematically well defined, and gives a nonseparable space once the integral is over a continuum." : This is wrong, for instance $L^{2}(\mathbb{R})$ is separable. – jjcale May 30 '13 at 16:33
  • @jjcale: But $L^2(R)$ is not a direct integral of nontrivial Hilbert spaces. – Arnold Neumaier Sep 12 '13 at 10:12
  • @Arnold Neumaier : I think, the direct integral of separable Hilbert spaces is again separable. For instance, if you have a unitary representation of a locally compact separable group on a separable Hilbert space, then this Hilbert space can be written as a direct integral over the irreducible representations of the group. – jjcale Sep 12 '13 at 18:54
  • @jjcale: But because of the assumed separability, the measure of this direct integral will be provably discrete. Thus this ''integral'' is a countable direct sum, and not a direct integral in the sense I used the term. – Arnold Neumaier Sep 13 '13 at 09:22
  • Whreas a true direct integral of principal series representations of SL(2) gives a unitary representation on a nonseparable Hilbert space. – Arnold Neumaier Sep 13 '13 at 09:23
  • @Arnold Neumaier : Take the poincare group and a Hilbert space of 2 particles : This is a separable Hilbert space and a continuous direct integral over the one-particle spaces corresponding to the center of mass motion. Or look at $L^{2}(\mathbb{R}^{2})$ which is a continuous direct integral of copies of $L^{2}(\mathbb{R})$. – jjcale Sep 13 '13 at 19:05
  • @jjcale: No. The inner products are very different. – Arnold Neumaier Sep 14 '13 at 13:49
  • To understand what happens, take the direct integral over [0,12] of continuously many copies of a 1-dimensional space, and try to find a countable basis! – Arnold Neumaier Sep 14 '13 at 13:50
  • @Arnold Neumaier : You are on the wrong track, see http://planetmath.org/directintegralofhilbertspaces – jjcale Sep 28 '13 at 17:33
  • @jjcale: Thanks for providing the explicit link. I noticed that I had mixed up the notions of direct sum and direct integral. One should read in my above comments direct sum whenever I wrote direct integral. Then the comments make sense. – Arnold Neumaier Oct 02 '13 at 12:31
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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.

Dan Stahlke
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  • In a finite-diemsnional Hilbert space there is no decoherence. (All motions are quasiperiodic and nothing can dissipate.) – Arnold Neumaier Jun 11 '12 at 08:31
  • @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? – Dan Stahlke Jun 11 '12 at 12:20
  • There is an extended discussion of this problem at http://www.physicsforums.com/showpost.php?p=3157393&postcount=103 – Arnold Neumaier Jun 11 '12 at 12:45
  • 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 http://en.wikipedia.org/wiki/Decoherence – Arnold Neumaier Jun 11 '12 at 12:52