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The dimensionality of the state space of a quantum system

  1. Jun 23, 2012 #1

    jcsd

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    Hi, I haven't posted for a while. I've seen this topic come up a few times, but it always seems to me that a few points aren't made clear. Can I just check the following is true?


    1) The state space of a quantum system is always an infinite-dimensional seperable Hilbert space i.e. a Hilbert space with a countably infinite Hilbert dimension.

    2) An infinite-dimensional seperable Hilbert space is an uncountably infinite-dimensional vector space (i.e. has an uncountable Hamel dimension).

    3) Sometimes though subspaces of finite dimension may contain most of the relevant information (e.g. when talking about 'qubits'), so the conceit that the state space is a finite dimensional Hilbert space is used.

    4) Sometimes the opposite conceit is presented, i.e. that the state space is a non-seperable Hilbert space (e.g. when describing the eigenstates of an obseravble with a continuous spectrum). In actuality the state space is still seperable, but an implicit reference is made to a non-seperable Hilbert space which 'looks' a bit like it should be the state space.

    A related question, what *exactly* is the problem with non-seperable Hilbert spaces?

    Thanks
     
  2. jcsd
  3. Jun 24, 2012 #2

    tom.stoer

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    I think this is not true. A simple example are plane waves in L²[-∞,+∞] which makes L² look like a non-separable Hilbert space - which is not the case. L² always has a countable Schauder basis which are e.g. the Hermite functions. In that sense the plane waves are not a basis - and the use of plane waves does not make L² non-separable; there is no 'reference to a non-separable Hilbert space'. The Fourier transformation using plane waves is not an expansion in terms of an 'uncountable basis' but a more general linear functional.

    The Hamel basis allowing every vector to be expanded using a finite subset of the Hamel basis is usually irrelevant for infinite-dimensional spaces; in many (all?) cases it is not even known, b/c it's existence is proved using the axiom of choice which makes the proof non-constructive (at least for infinite-dimensional spaces).

    An example for a non-seprarable Hilbert space is l; I do not know an example in physics where this space is used.

    I guess that even the Fock space constructed from a countable set of separable Hilbert spaces is itself separable, but I am not sure.

    I think this is the most interesting question you are asking. - I don't know the answer! (of course separability makes the math easier, but I am afraid that this is not the answer you are expecting ;-)
     
    Last edited: Jun 24, 2012
  4. Jun 24, 2012 #3

    jcsd

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    Thanks Tom!
    Where do the linear functionals live I suppose my question can be boiled down to. Is it a Hilbert space? is it seperable? (the answer is no?)

    Thanks again, thinking about it, it seems quite obvious that the Hamel basis is fairly useless in such a setting and the orthonormal basis
    I think the tensor product of a finite number of seperable spaces is seperable, but the tensor product of an infinite number of seperable spaces is not.


    My guess is that perhaps you encounter problems in trying to explicitly construct an uncountable orthonormal basis?
     
  5. Jun 24, 2012 #4

    tom.stoer

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    The linear functionals live in the dual H* of the original space H which is usually a separable Hilbert space, too; in case of L² it is identical (the position and momentum space are both L² spaces).
     
  6. Jun 25, 2012 #5

    A. Neumaier

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    Fock spaces are separable.

    But the Hilbert space of interacting quantum field theories containing massless particles (such as QED) is nonseparable, with uncountably many superselection sectors (at least one each for each direction in space, for charges moving in this direction), between which superpositions are not possible.

    Nonseparable Hilbert spaces are technically more difficult to use (but only when one cares about mathematical rigor; limits must be taken via nets rather than sequences).
     
  7. Jun 25, 2012 #6

    tom.stoer

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    Can you comment more on these Hilbert spaces? They cannot be (unitary) equivalent to Fock space, so the question is how they look like?
     
  8. Jun 25, 2012 #7

    jcsd

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    Going back to my original post, one related question: does anything like decoherence occur if we only consider state (sub-)spaces of finite (but very large) dimension?
     
  9. Jun 25, 2012 #8

    jcsd

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    Sorry if my questions seem a bit silly, I'm trying to get a handle on the mathematical settings of quantum mechanics, but don't the postion eigenvectors live in the dual of a subspace of the Hilbert space? Is that always a Hilbert space?
     
  10. Jun 25, 2012 #9
    Now, the level of this discussion is way above my head but I just thought I'd interrupt for a moment. Correct me if I'm wrong, but the above would not be a true statement as far as I understand it. The dimensionality is equal to the number of possible outcomes of a measurement. For position, yes it is infinite-dimensional since position is continuous, but for a spin 1/2 system the Hilbert space is two dimensional since there are only two possible outcomes of a spin measurement.
     
  11. Jun 25, 2012 #10

    jcsd

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    The fact that finite dimensional subspaces of the state space are often useful to study isn't controversial, but I wanted to check, is there anything wrong in principle with finite dimensional state spaces. I know you need infinite dimensions for the uncertainity principle and also I *think* (see above) you need it for decoherence.
     
  12. Jun 25, 2012 #11
    Ah, sorry, I see now that no. 3 in your first post deals with what I said. I mostly reacted to the word "always" in "state space of a quantum system is always an infinite-dimensional...Hilbert space" and didn't read your post accurately enough. I'm afraid I can't answer your questions, but I follow the discussion with interest!
     
  13. Jun 25, 2012 #12

    strangerep

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    They're not silly at all.

    Since you're an SA, I wasn't sure which reference material you've studied, but the early chapter(s) of Ballentine gives a useful introduction to dual space, rigged Hilbert space, and its use for unbounded operators like position/momentum.

    Indeed. That's the Hellinger-Toeplitz theorem: an unbounded operator cannot be well-defined on the entire Hilbert space ##H,## but only on a subspace thereof. Let's call it the "small" space ##\Omega##. In this case we have ##\Omega \subset H##.

    No, ##\Omega## is not norm-complete, hence not a Hilbert space. In general we have
    $$
    \Omega \subset H \subset \Omega^*
    $$where the asterisk denotes topological dual.

    In the case of a Hilbert space, one can prove ##H = H^*## (which is the essence of the Riesz representation theorem, iirc).

    The (generalized) eigenvectors of position/momentum span ##\Omega^*## (Gel'fand-Maurin nuclear spectral theorem), hence can also be used to decompose arbitrary elements of ##\Omega## and ##H##.

    I think that's not necessarily the case. In general, uncertainty principles arise from noncommutation of observables, and this can occur in the finite-dim case also. But I guess you meant the position/momentum UP, which involves unbounded operators, hence inf-dim spaces.
     
    Last edited: Jun 25, 2012
  14. Jun 26, 2012 #13

    jcsd

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    Thanks Strangerep, that is incredibly helpful! :)
     
  15. Jul 9, 2012 #14

    A. Neumaier

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    Lacking a mathematically consistent description of QED, an answer must be somewhat hand-waving. The analysis of infrared problems by Kibble, Kulish and Faddeev, etc. reveals that the correct mathematical description of infrared clouds is by means of coherent states. Coherent states whose classical fields do not differ asymptotically at infinite spatial distances can be mapped onto each other (ignoring topological details) by a unitary Bogolubov transformation, hence belong to the same superselection sector. The classes of unitarily inequivalent representations are therefore labelled by the asymptotic characteristics of states, namely charges and the equivalence classes of coherent fields (photon clouds) they generate. While the charges are countable (and also present in massive theories; this is the DHR theory), the asymptotic coherent fields form a manifold (at least an [\tex]S^2[/tex]), which makes the Hilbert space nonseparable. Probably it is a direct integral of separable Hilbert spaces, with the integral labelled by charges and their direction of motion; but there may be more to it.
     
  16. Jul 9, 2012 #15
    I don't know anything about Hamel dimension, but I think 2) is false! A Hilbert space is separable if and only if it has a countable basis. That claim (and some discussion about uncountable-dimensional Hilbert spaces) is here.

    For many practical purposes, e.g. my thesis, the problem with non-separable Hilbert spaces is that they are a pain to work with. For finite-dimensional spaces, the spectral theorem is essentially just cool matrix tricks. For countably-infinite-dimensional spaces, some of the matrix-flavored ideas are not too hard to generalize. For uncountably-infinite-dimensional spaces, the spectral theorem gets quite a bit more technical.
     
  17. Jul 9, 2012 #16

    tom.stoer

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    Your statement that "a Hilbert space is separable if and only if it has a countable basis" is correct for the Schauder basis; but an infinite dimensional Banach space always has an uncountable Hamel basis - which is different from the Schauder basis, of course.
     
  18. Jul 9, 2012 #17
    I see. Is one (or neither) of these a standard choice for Hilbert spaces? I noticed that Wikipedia's Hamel basis article mentions orthogonal bases for Hilbert spaces as an "alternative" to Hamel bases.

    I had previously assumed that the Hilbert-space definition of "basis" was the only one used in practice. Thanks for pointing out the distinction.
     
  19. Jul 10, 2012 #18

    tom.stoer

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    A Hamel basis B is a set of vectors which provides a representation of arbitrary vectors v of the vector space V in terms of a linear combination of a finite subset of this basis, i.e.

    [tex]\forall v \in V:\; v=\sum_{n=1}^N v_n\,b_n;\;b_n\in\mathcal{B}[/tex]

    In QM we never use the Hamel basis but always a Schauder basis for which N is replaced ∞. Afaik the Hamel basis is not even known for most spaces b/c its existence is proved using the axiom of choice and is therefore nonconstructive. So for Hilbert spaces and in QM the word "basis" means "Schauder basis" - in most cases ;-)
     
  20. Jul 11, 2012 #19

    A. Neumaier

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    Hamel bases are virtually useless in applications, whereas orthonormal bases are their bread and butter.
     
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