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Interacting theory lives in a different Hilbert space [ ]

  1. Oct 25, 2009 #1


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    In article #34 of a recent thread about Haag's theorem, i.e.,
    a point of view was mentioned which I'd like to discuss further.
    Here's the context:
    I think I see a flaw in the argument above.

    Suppose I want to know the accelerations of the two particles (or
    maybe just their relative acceleration wrt each other). In the
    case of 2 free particles I can certainly ask the question, but
    the answer is always 0. But for two interacting particles, the
    answer is nonzero in general.

    Expressing this in the language of quantum logic and yes-no
    experiments, the question "is the acceleration 0" always yields "yes"
    in the free case, but can yield "no" in the interacting case.
    Similarly, the question "is the acceleration nonzero" always yields
    "no" in the free case but might yield "yes" in the interacting case.

    Denote the 2-free-particle Hilbert space as [itex]H_0[/itex] and the
    2-interacting-particle Hilbert space as [itex]H[/itex].

    Proceeding by contradiction, let's assume that [itex]H_0[/itex] and [itex]H[/itex] are unitarily
    equivalent, i.e., that any basis of [itex]H_0[/itex] also spans [itex]H[/itex]. Assume as well that the
    dynamical variable known as "acceleration" corresponds to (densely
    defined) self-adjoint operators in [itex]H_0[/itex] and [itex]H[/itex], and that the two
    operators are equivalent to each other up to a unitary transformation.

    Every state in [itex]H_0[/itex] is a trivial eigenstate of the acceleration
    operator, with eigenvalue 0. Expressed differently, the acceleration
    operator annihilates every state in [itex]H_0[/itex]. However, we expect to find
    states in [itex]H[/itex] corresponding to nonzero accelerations, i.e., states which
    the acceleration operator does not annihilate.

    This implies that the basis states of [itex]H_0[/itex] cannot span H, contradicting
    the initial assumption. Conclusion: [itex]H_0[/itex] and [itex]H[/itex] are not unitarily

    So it's not enough that the same logical propositions (questions)
    can be asked in both spaces. The spectrum of the corresponding
    operator must also be considered, and whether both spaces
    accommodate the full spectrum.

    Or am I missing something?
  2. jcsd
  3. Oct 26, 2009 #2
    Hi strangerep,

    I must admit that my argument was not correct. Indeed, operators of observables (such as acceleration) have different properties and different commutation relations in the interacting and non-interacting cases. However, still I don't see the reason why these operators cannot coexist in the same Hilbert space. In ordinary non-relativistic quantum mechanics we use the same Hilbert space for both interacting and non-interacting system. What is fundamentally different in the relativistic case and in QFT?
  4. Oct 26, 2009 #3
    That is a good point. Indeed, let us consider two Hamiltonians: one for a couple of free particles and another - with a repulsive potential. Are these Hamiltonians related with a "unitary" transformation? No. (Or, maybe, yes? Consider one-particle case). Yet, there is no problem with the perturbation theory (scattering) as well as with expansion of the exact solution of interacting case in series of free solutions. We have the same Hilbert space but two different basises to span this space.

    In many cases it is so. I can also refer to the simplest 1D Sturm-Liouville problem considered in my publications where the perturbation theory may even give divergent matrix elements whereas the exact solutions are finite, physical, and they co-exist in the same Hilbert space.
    Last edited: Oct 26, 2009
  5. Oct 26, 2009 #4
    In QFT there is an additional subtraction prescription. There is a principal difference before and after subtraction. Subtraction (discarding some corrections) changes the solution. The latter corresponds now to another Hamiltonian - a Hamiltonian without self-action. After that the rest is similar to a non-relativistic case.
    Last edited: Oct 27, 2009
  6. Oct 26, 2009 #5


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    Hi meopemuk.

    Thanks for your answer.

    As I mentioned in my original post, it's not so much about whether such an operator can
    be represented in the Hilbert space(s), but whether the Hilbert spaces in question
    account for the same (subset of) the full spectrum of the operator.

    For finite degrees of freedom, one is saved by well-known theorems about isomorphisms
    between Hilbert spaces, the Stone-von Neumann theorem, and all that. For infinite
    degrees of freedom, continuous spectra, etc, these theorems fail.

    Afaict, it's all about how infinite degrees of freedom leads to unbounded operators.
  7. Oct 26, 2009 #6


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    That sounds like a scenario with finite degrees of freedom. (?)

    If the answer was "yes", both Hamiltonians would have the same spectrum, so that
    seems not correct.

    I'm not familiar with the word "detraction" in this context.
    Did you mean "renormalization"?
  8. Oct 27, 2009 #7
    I don't see any big problem with unbounded operators. Physically, most relevant operators of observables are bound to be unbounded (pardon the pun). For example, the operator of position is unbounded, because the universe is infinite. The operator of momentum is unbounded, because there is no limit on the value of momentum.

    If some mathematical formalism (functional analysis, theory of Hilbert spaces, etc.) has a problem with unbounded operators, then I consider this a problem of math rather than physics. Perhaps our math is build on unrealistic axioms (out of convenience). Perhaps, we need to be more creative and consider more general structures such as non-separable Hilbert spaces or non-standard Hilbert spaces, or something like that. In this particular case (unbounded operators) I don't think that math troubles indicate some new physics. I believe that physical considerations must take the lead, and math must follow.

    Another point is that in the "dressed particle" approach to QFT (which is formulated in terms of particles rather than fields) the number of relevant degrees of freedom is always finite, because any realistic physical system is made of a finite number of particles.
  9. Oct 27, 2009 #8
    Yes, they call it "renormalization" although it is subtraction or discarding, to tell the truth .
    I thought you had read my argumentations in "RiR".

    P.S. I meant "subtraction", of course. My poor English...
    Last edited: Oct 27, 2009
  10. Oct 27, 2009 #9
    strangerep & meopemuk-> One crucial flaw of that argument is a purely logical one. While it is true that we can ask the same questions in both an interacting and a free theory, this does not imply at all that the underlying mathematical structure must be the same. Indeed, I can ask the same questions , say about particle scattering, in a classical, non quantum theory. Of course, physically the answers will be incorrect, but I can still ask those questions and obtain answers. But also, to an interacting theory I can ask questions a free theory simply cannot answer. For example, considering an out of equilibrium gas of particles, I could ask "How long will it take for the system to reach thermal equilibrium?". Clearly, if the particles cannot interact, an out of equilibrium system will never thermalize.

    meopemuk -> strangerep nailed the answer: it's about the number of degrees of freedom. In ordinary QM you always have a finite nr of degrees of freedom and the Stone - Von Neumann theorem tells you, more or less, that all the representations of the canonical commutation relations are unitarily equivalent. That's why no one bothers talking about a different Hilbert space when dealing with QM other than the usual [tex]L^2(\mathbb{R}^n) [/tex]. It's not so easy in QFT. If the nr of degrees of freedom is infinite, as it is when you deal with fields, the theorem no longer applies and it is in fact a physically relevant question as to what the appropriate Hilbert space is. And this holds for "simple" free theories. So the fundamental difference between QM and QFT is the number of degrees of freedom, not "additional subtraction procedures".

    The problem with unbounded operators is precisely that they can have infinite expectation value in certain states. Sure, a purely mathematical problem. But if I'm to extract physics from some maths, then my maths better yield unique and finite answers or I'm in trouble. And it's not difficult to see how manipulating infinities can give you "paradoxes". That's the problem with unbounded operators: their maths is not clear yet we need them for physics.

    And to say "consider a Hamiltonian with two particles" you must really specify if you're talking about QM or QFT. Actually, you don't need to: QFT is a multiparticle theory from the very beginning. So you're talking about QM, where things are "mathematically easy", so to say.

    Finally, will now give you two more physical argument as to why the free field theory and the interacting one live in different Hilbert spaces. (Am still kinda struggling with the mathematical arguments myself :) )

    Let's first clear what the Hilbert space of a (vacuum) free theory is. It's the usual Fock space, and the vectors of this space are multiparticle states. It is important however to observe now that these particles do not interact. So in the usual Fock space there is no space for interaction! Think about a self interacting particle propagating through space. The self-interaction will sure modify the one particle state of free theory and this state is in no sense a linear combination of multiparticle states of the same Fock state as all the "virtual particles" are also interacting whereas the Fock space ones are not.

    Second, consider a system in thermal equilibrium at some finite temperature T. Such a state cannot be obtained in any way from the multiparticle states of the vacuum Fock space. Thermal states are not excitations of the vacuum state. Think about the quark-gluon plasma in the early Universe. The particle density in such states is non zero whereas the vacuum multiparticle states are zero density states. So even for free theories, the vacuum Hilbert space and a finite temperature Hilbert space will be different.

    Yep, things are complicated in QFT - it's why you get $1m to prove the mass gap conjecture for Yang-Mills theories.

    bob_for_short -> You insist on saying there are "problems" with renormalization because it involves subtraction of infinities. Read the book by Bogoliubov to see QFT without subtractions of infinities but with renormalization. Or read Scharf's book on "Finite QED" - as the title suggests, there is not a single infinity in that book. Or check out Epstein and Glaser papers about causal perturbation theory, where again renormalization is done without any subtraction of infinities. Renormalization is a (fairly) well understood issue in QFT and it's related to the mathematical nature of the various objects encountered in QFT, i.e. distributions.
  11. Oct 27, 2009 #10
    I read it in 1979.
    And I am a re-distribution master.
  12. Oct 27, 2009 #11
    Hi DrFaustus,

    I agree that in the traditional formulation of QFT particles do self-interact; "physical" particles of interacting theory are different from "bare" particles of free theory. I can also agree that this self-interaction makes definition of the interacting Hilbert (Fock) space problematic.

    However, the more I think about the idea of "self-interacting particles" the less attractive it looks. The idea of "self-interacting QFT vacuum" is even less attractive.

    Fortunately, there exists an alternative formulation of QFT, which I find more acceptable philosophically. In this formulation interactions do not modify 0-particle and 1-particle states of the theory. There is no distinction between "bare" and "physical" states. The interacting theory can happily occupy the Fock space of the free theory. This is called the "dressed particle" approach, which stems from the old work

    O. W. Greenberg, S. S. Schweber, "Clothed particle operators in simple models of quantum field theory", Nuovo Cim., 8 (1958), 378.

    There are only few researchers who take this approach seriously, but I hope this number will grow. The "dressed particle" approach can yield exactly the same renormalized S-matrix as traditional QFT. One side benefit is that calculations of the "dressed particle" S-matrix do not involve divergences.
  13. Oct 27, 2009 #12
    Drunk, you mean?
    I don't remember anything from that time.
  14. Oct 27, 2009 #13


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    DarMM wrote a couple of interesting posts about the Hilbert space of the interacting theory here. See #50 and #54. If I understand him correctly, the usual Fock space of non-interacting states is equivalent to (the completion of?) the vector space of square-integrable distributions, and the only thing that changes about that space when we introduce interactions is the measure we use to define the integrals.

    I don't really understand any of this myself, so I don't have much to contribute here, but I hope to learn this stuff some day.
  15. Oct 27, 2009 #14


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    I think this is a mistake. Any time evolution operator that doesn't leave every n-particle subspace invariant would by definition be an interaction, and you can certainly define such operators.
  16. Oct 27, 2009 #15
    Let me summarize: Interacting theory lives just in a different Hilbert space, so its Hamiltonian is alive and kicking.
  17. Oct 27, 2009 #16


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    IIUC, that's why people use rigged Hilbert space -- which is larger than ordinary
    Hilbert space. In the space of tempered distributions (of which ordinary Hilbert space
    is a subspace), one can define a distribution-valued inner product like
    \langle p | q \rangle ~=~ \delta(p-q)
    and also obtain a usable generalized spectral theorem.

    Other people prefer the C*-algebra approach which can also be regarded
    as "larger" than the usual Hilbert space methods (but it would take ages
    for me to make that statement precise. :-)

    Maybe someone will get that $1M prize one day when they
    figure out which maths is right.
  18. Oct 27, 2009 #17


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    Yes, but it was some time ago and I didn't remember
    the word "detraction". Since you intended "subtraction",
    I now understand what you meant. :-)
  19. Oct 27, 2009 #18


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    I think that point could only be valid if the full (non-perturbative) Hamiltonian of the
    dressed theory turned out to be a finite polynomial in the a/c operators. But if one
    needs ever-longer more complicated terms at each perturbation order, one ends up
    with an inf-dim theory, afaict.
  20. Oct 27, 2009 #19
    An "inf-dim theory" is not a problem per se if the series can be summed up into a certain function, as in my electronium potential. The a/c operators are in the denominator and in such a combination (Q) that is integrable in the perturbation theory: V(r,Q) = 1/|r + εeQ|.
    Last edited: Oct 28, 2009
  21. Oct 27, 2009 #20


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    You've got to make sure that the full Hamiltonian still participates correctly
    in a representation of the Poincare algebra, and that the Hamiltonian is
    densely well-defined on the multiparticle Fock space.
    Last edited: Oct 28, 2009
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