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Hilbert space of QFT

  1. Mar 3, 2012 #1
    Hi everyone. Many texts when describing QFT start immediately discussing about free field theories, Fock spaces etc.. I want to understand general properties of the Hilbert space, and how to find a basis of it, and how to find a particle interpretation. I know there are very mathematical formulations with test functions etc.. but i'm not interested in that.

    I'll tell you my guess. In a QFT we must have Poincare' invariance. So, from it we can have its noether charges, and use them to label the Hilbert space with their eigensates. How many commuting objects can we create? [itex]P_{\mu}[/itex], [itex]P^2[/itex],[itex]W^2[/itex], [itex]W_3[/itex], and plus some commuting [itex]Q_a[/itex] of an internal symmetry. [itex]W_{\mu}[/itex] is [itex]W_{\mu}=\frac{1}{2}\epsilon_{\mu\nu\rho\sigma}M^{\nu\rho}P^{\sigma}[/itex] So, I could create


    where [itex]m^2[/itex] is real non negative (if it is null i should consider helicity), [itex]p_{\mu}[/itex] is any, [itex]s=0,1,2..[/itex] and [itex]\xi_{s}=-s,..s[/itex].

    So I can express a generic element of the Hilbert space as linear combination of [itex]|m^{2},p_{\mu},q_{a},s,\xi_{s}>[/itex] right?
    [itex]|m^{2},p_{\mu},q_{a},s,\xi_{s}>[/itex] could be interpreted as a particle state if it is onshell.
    Is there a way to build a commuting number operator whose eigenvalues identify which particle we are considering? The vacuum would be when the number operator is zero.

    Then, when we impose a certain Lagrangian, its equation of motion will give some contraints on the states, right?

    Any suggestions or text references explicitly dealing with what said are welcome.
    Thank you very much.
  2. jcsd
  3. Mar 3, 2012 #2


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    Weinberg vol-1 is the obvious suggestion for a reference.
  4. Mar 3, 2012 #3
    Yeah, thanks for the reply, but that's a big book. Where does it talk about the general properties of the Hilbert space?

    By the way, looking around I found written that the Hilbert space of QFT is a unitary representation of the Poincarè group. The one particle states form a subspace of the whole Hilbert space and corresponds to an irreducible unitary rep of Poincarè, and it's made of states [itex]\phi(x)|0>[/itex] (or distributions if you like) where [itex]|0>[/itex] is the vacuum. So I know how to create a basis for this space via Wigner's method. But QFT have arbitrary number of particles. So how would that reducible Poincarè rep look like? A direct sum of [itex]n[/itex] particle spaces (and is this an [itex]n[/itex] times tensor product of a 1-particle space?) Even with interaction?

  5. Mar 4, 2012 #4
    I thought Lie groups are represented by operators on the Hilbert space, not the Hilbert space itself.
  6. Mar 4, 2012 #5

    A. Neumaier

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    All separable Hilbert spaces are equivalent; so there is very little to be said generally. The differences are in the representations. Chapter 5 constructs the relevant reps.
    Without interactions, you get the Fock space constructed from the representation space of an irreducible rep. With interactions, you get something nobody knows so far how to describe it. There is only Haag's theorem, which says that it cannot be a Fock space. But all known positively is that, restricted to bounded regions, it is a Fock space. The limit to the unbounded case is mathematically ill-understood and constitutes the infrared problem.
  7. Mar 4, 2012 #6


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    A bit more one understands about the Hilbert space of QFT, at least in QED (and in the perturbative sense of course). The infrared problem shows indeed that the Fock space of free particles is not the correct space to represent asymptotically free states of QED in the (perturbative) S-matrix theory.

    The traditional way to cure this problem is pretty old and goes back to

    F. Bloch and A. Nordsieck. Note on the radiation field of the electron. Phys. Rev., 52:54–59, 1937.

    The idea to cure th IR problem is that in any reaction of charged particles, soft photons are easily created due to the masslessness of photons. The energy resolution of any detector used to measure the corresponding cross section of such a reaction is always finite, and thus one cannot distinguish whether you really detect only the exclusive collision process in question or whether there is one or more additional soft photon(s) emitted with a (total) energy smaller than the detector's energy resolution.

    This is nicely described in modern terms, using Feynman diagrams, e.g., in Weinberg, Quantum Theory of Fields, Vol. 1

    Another more modern approach is to more carefully determine the asymptotic states of such scattering events. That's already apparent in non-relativistic scattering theory of charged particles. Since the photon is massless, the electrostatic force (Coulomb force) is long ranged. The potential only falls with the inverse distance, and thus standard asymptotic theory, assuming freely streaming particles, represented by plane waves (momentum eigen states) as asymptotic states, is not applicable to that case. The way out here is very simple since one can solve the two-body Coulomb problem in non-relativistic quantum theory exactly (for both the bound-state problem as well as Coulomb scattering).

    Unfortunately one cannot solve more complicated problems exactly, but there's an alternative approach, using modified asymptotic states, taking into account the relevant long-range part of the Coulomb interaction in the asymptotic region. This also has the advantage that one can extend this technique to the full QED case. As it turns out the correct asymptotic states for the photons is not the Fock space of the non-interacting theory but coherent states of an indefinite photon number, which is not a Fock space. For the charged particles it's the modified asymptotic states, taking into account the long-range Coulomb interaction.

    This is nicely described in

    P. P. Kulish and L. D. Faddeev, Theoretical and Mathematical Physics
    Volume 4, Number 2, 745-757, DOI: 10.1007/BF01066485


    Mark S. Swanson, Phys. Rev. D 25, 2086–2102 (1982),
  8. Mar 6, 2012 #7

    A. Neumaier

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    I didn't know the second paper (thanks!); it gives in the introduction a nice summary of what has been accomplished.
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