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Non-separable Hilbert spaces

  1. Apr 17, 2013 #1
    What goes wrong if you try to do QM/QFT with a non-separable Hilbert space? Why do the Wightman axioms stipulate a separable space?

    And I need something else cleared up: The Hilbert space of non-trivial QFTs are indeed non-separable right?
     
  2. jcsd
  3. Apr 17, 2013 #2
    bumpity bump
     
  4. Apr 17, 2013 #3

    DarMM

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    On a non-seperable Hilbert Space, just because a Group has a unitary representation on that Hilbert Space, the generators might not have representations as Self-Adjoint operators. So for example there would be unitary operator implementing rotations, but no self-adjoint observable for angular momentum.

    No, they are separable.
     
  5. Apr 18, 2013 #4
    How does the Hilbert space of a QFT end up being separable when at every spacetime point you have a harmonic oscillator? That seems like it should result in an uncountable number of basis elements to me.
     
  6. Apr 18, 2013 #5

    rubi

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    That's not related. You can also have strongly continuous representations on a non-separable Hilbert space and non-strongly continuous representations on a separable Hilbert space.

    These distributional states aren't part of the Hilbert space itself for the same reason that delta-distributions aren't L^2 functions in usual quantum mechanics.
     
  7. Apr 18, 2013 #6

    Jano L.

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    Is is hard to work with uncountable basis, so one thing people invented in QFT is to restrict the volume to finite value and expand everything as sum of Fourier components. These are countable. This is, of course, at the expense of the generality of description of the resulting field (some field configurations are thrown out, namely those which do not have Fourier representation, e.g. very wild or divergent functions...).
     
  8. Apr 18, 2013 #7

    dextercioby

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    Due to the continuous spectrum of the 4-momentum operator for fields modeled over an unbounded domain of space-time, the so-called >Hilbert space of a QFT< is rather a distribution space in a rigged Hilbert space (Gel'fand triplet).
     
  9. Apr 18, 2013 #8

    DarMM

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    Sorry, that should read the Weyl algebra. The generators of the Weyl algebra might have no representation on a non-separable Hilbert Space. In other words, there are no strongly continuous representations of the Weyl algebra on a non-separable Hilbert Space.
     
  10. Apr 18, 2013 #9

    stevendaryl

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    That's part of the answer. There's another part, and I couldn't find a definitive article about it from Googling the internet, but here's the way I understand it:

    The Hilbert space that is usually used for QFT considers all states to be in some sense perturbations of the vacuum state. That is, although QFT allows for an indefinite number of particles, the total system state is viewed as a superposition of 1-particle states, 2-particle states, 3-particle states, etc. What you can't represent with such a Hilbert space is a space with an actually infinite number of particles. To describe infinitely many particles would require a non-separable Hilbert space, I believe. (You can still get a countable basis for an infinite number of particles if there is a "vacuum" state in which all particles have the lowest possible energy level. Then the infinite number of particles in their ground state can be eliminated through redefining the creation and annihilation operators.)
     
  11. May 1, 2013 #10

    DarMM

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    I realised that this explanation might be seen as somewhat vague, so I thought I would expand on it. The Weyl Algebra can be understood in many ways, but for quantum mechanics it is essentially any combination of unitary operators representing translations in position or momentum space.

    The generators of the Weyl algebra are the position and momentum operators themselves, or in quantum field theory they are the field and its conjugate momentum field.

    So, on a non-seperable Hilbert space it is possible for the field/position operator to be ill-defined, which would make calculations extremely difficult. Of course we "know" that position is a well-defined observable, so we can ignore non-seperable Hilbert spaces when it comes to quantum mechanics.

    In fact if you do quantum mechanics from the abstract algebraic point of view you can actually prove that most temporal automorphisms (jargon word in algebraic quantum theory for time evolution) have GNS representations* as self-adjoint operators on a seperable Hilbert space.

    In field theory non-seperable representations can occur for thermal states.

    *Algebraic field theory views the operators as the primary objects and states are viewed simply as linear maps from the operators to their expectation values. The GNS theorem is a fundamental result which states that if you take some algebra of operators:
    [tex]\mathcal{A}[/tex]
    and a state on that algebra:
    [tex]\rho[/tex]
    defined as a map from elements of the algebra to their expectation values:
    [tex]\rho(A), \quad A \in \mathcal{A}[/tex]

    Then the action of this state on this algebra can always be replicated by a ket in some Hilbert space and bounded operators on that Hilbert space. Specifically we can always find:
    [tex]\Psi_{\rho}, \quad \mathcal{H}_{\rho}, \qquad \Psi_{\rho} \in \mathcal{H}_{\rho}[/tex]
    that is a Hilbert space and a vector element of that Hilbert space,
    and we can also construct a map:
    [tex]\pi : \mathcal{A} \rightarrow \mathcal{B}(\mathcal{H}_{\rho})[/tex]
    mapping the algebra into the bounded operators over that Hilbert space, such that:
    [tex]\rho(A) = \left(\Psi_{\rho}, \pi(A) \Psi_{\rho}\right)_{\mathcal{H}_{\rho}}[/tex]
    This is the GNS theorem.
    In other words, using Hilbert spaces becomes a theorem in the Algebraic approach. The GNS theorem proves that Hilbert space, their elements and their operators, can be used as tools in computing maps on the algebra of observables.

    Now of course often several different states result in the same [tex]\mathcal{H}_{\rho}[/tex] You say that such states are in the same folium. Time evolution can only move you around inside a given folium, which is why in "normal" quantum mechanics and field theory we only work with Hilbert spaces, time evolution always keeps you inside one folium.

    However for deeper issues in field theory, for instance if you want to look at exactly what is the relationship between a free and interacting theory then you need to look at different folia at the same time, which the standard formalism is very bad at doing.
     
  12. May 1, 2013 #11

    strangerep

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    Thank you. :smile:

    Is there another formalism which is better at doing that and which is explainable to people who are not already experts in algebraic QFT? I tend to be somewhat overwhelmed when I try to read papers on that stuff. :uhh:
     
  13. May 2, 2013 #12

    DarMM

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    Unfortunately not strangerep. It might be easier to understand the algebraic approach if you think of finite dimensional quantum systems. There all observables are complex matrices and all states are matrices of trace 1. If we call a state [tex]\rho[/tex] and an observable [tex]A[/tex] then the expectation value is just:
    [tex]\rho(A) = tr(\rho A)[/tex]
    The space of all complex matrices is a C*-algebra and so a state is just a linear map on a C*-algebra. In this case the map is given by the trace. However the trace requires an underlying space, in this case [tex]\mathbb{C}^{n}[/tex], this is the Hilbert space.

    However you can "detach" yourself from this space by just considering the algebra of complex matrices as an abstract algebra and states as maps on that abstract algebra. By the way, a C*-algebra is really just any algebra that "similar" to the algebra of complex matrices.

    All that's going on in QFT is that in that case, if you want to "return" to the:
    [tex]\rho(A) = tr(\rho A)[/tex]
    picture, with some Hilbert space the trace is performed in, then the Hilbert space you return to depends on the state. Whereas if the algebra has a finite number of generators (such as is the case with the example at the beginning of this post or non-relativistic quantum mechanics) you can prove the Hilbert space used to represent the map as trace is always the same for all states.

    In the Algebraic approach the Hilbert space exists only as concrete way to calculate:
    [tex]\rho(A)[/tex]
    by representing it as a trace. For certain states of course it will turn out that the trace simplifies to:
    [tex]\rho(A) = tr(\rho A) = (\Psi_{\rho}, A \Psi_{\rho})[/tex]
    with [tex]\Psi_{\rho}[/tex] some state in the Hilbert space used to carry the state. So the state can be represented by vectors in the Hilbert space and your back to normal quantum theory.

    By the way, you might have noticed a similarity to the density matrix formalism of quantum mechanics. The Algebraic approach is really just the density matrix formalism made rigorous and more general. If you remember that some papers might be easier to read.
     
    Last edited: May 2, 2013
  14. May 2, 2013 #13

    strangerep

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    Yes, I'm ok with the algebraic approach in the simpler cases. The stuff you said about looking at "different folia at the same time" was what caught my eye.
     
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