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Particle in lattice QCD

  1. Sep 16, 2010 #1

    Demystifier

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    Can someone explain to me how the concept of particle is defined in lattice QCD?

    Here are the reasons why it seems problematic to me:
    1) Lattice QCD is based on functional-integral formulation of QFT, which does not contain any operators in the Hilbert space. In particular, it does not contain the particle creation and destruction operators.
    2) It is a non-perturbative theory with confinement, which means that one cannot define particles though the LSZ reduction based on assumption that asymptotic states are free particle states (quarks and gluons).

    References in which these things are explained would be highly desirable.

    Also, if one knows a simpler toy model of a non-perturbative definition of particles in interacting QFT, that might be even more interesting.
     
    Last edited: Sep 16, 2010
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  3. Sep 16, 2010 #2

    tom.stoer

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    I can only talk about the quenched approximation. That means that the fermionic determinant is set to one = all fermion loops are suppressed and fermions behave classically. The quarks are "static", only the gluon degrees of freedom are dynamical. So the quarks are defined as static particles.

    I do not know how lattice QCD (as of today) goes beyond this quenched approximation.
     
  4. Sep 17, 2010 #3

    Demystifier

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    Thanks, but it still does not answer my question, not even approximately. What I want is a representation of a 1-particle state (be it quark, gluon, hadron, glueball, ... whatever) as a state in a Hilbert space. How to find such a representation when
    1) one deals with path integrals that do not even use the concept of Hilbert spaces, and
    2) free particle states are not even a good approximation

    Actually, QCD serves here only as an example. What I want to know is how, in general, one can derive the concept of PARTICLE from a theory of quantum FIELDS. It is well known how to do it with PERTURBATIVE CANONICAL quantization of fields, but the example of lattice QCD is interesting because it is neither perturbative nor canonical.
     
  5. Sep 17, 2010 #4

    tom.stoer

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    OK, understood.

    The canonical quantization is definitly not limited to the perturbative treatment, but can be defined rigorously even non-perturbatively. It is e.g. possible to write down the full, gauge-fixed QCD Hamiltonian w/o unphysical degrees of freedom (zero norm states, ghosts, ...).

    Regarding particle concept in the PI formalism: it's rather simple: every localized and stable (physical) field configuration is a particle. A localized gauge field configuration can e.g. be considered as a glueball.

    The problem is how to prepare and/or identify such configurations on the lattice, especially if fermions are taken into account.
     
  6. Sep 17, 2010 #5

    Demystifier

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    But such a definition of a particle has nothing to do with the usual Fock-space definition of particle in free QFT:
    First, a Fock 1-particle state does NOT need to be localized. (For example, a momentum eigenstate does not have a localized wave function).
    Second, a localized configuration does not need to be a Fock 1-particle state. (For example, two bosons with the same localized wave-packet wave function.)

    What I want is a generalization of the Fock-space definition of particle in non-perturbative QFT. I don't want a "classical" concept of particle based on local configurations.
     
  7. Sep 17, 2010 #6

    tom.stoer

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    I think you can't have a "non-perturbative Fock-space particle". It's self-contradictory.

    A Fock state is something like a state [tex]|ABC\ldots\rangle = A^\dagger B^\dagger C^\dagger|0\rangle [/tex] with A, B, C being field operators creating 1-particle states [tex]|A00\dots}, |0B0\ldots\rangle, |00C\ldots\rangle[/tex] which are solutions of the free theory.

    If you take a non-perturbative solution of the full theory, a soliton for example, you can try to create it via [tex]|soliton\rangle = S^\dagger |0\rangle[/tex]. Of course you can now define [tex]|soliton 1, soliton 2\rangle = S_1^\dagger S_2^\dagger|0\rangle[/tex] but unfortunately it's no longer a solution of the theory. It may not even be close to a solution!

    So even if you are able to define this soliton creation operator, I doubt that you will be able to use it to create reasonable states. I don't even know whether you can span the whole Hilbert space.

    Why do you want to throw away the benefit of the PI formalism to be able to deal with real, localized particles instead of plane waves? Look at lattice gauge theory: they are able to visualize localized gauge field configurations; they can calculate their properties; they can calculate hadron masses w/o using Fock states. In the PI formalism the classical concept of a localized field configuration remains valid even after quantization.
     
    Last edited: Sep 17, 2010
  8. Sep 18, 2010 #7

    strangerep

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    Are you only interested in "particles" at asymptotic times, or for all times?

    If the latter, then I'm guessing you want an operator A s.t.
    [tex]
    [H,A] ~\propto~ A
    [/tex]
    ?
     
  9. Sep 18, 2010 #8
    Particles only make rigorous sense if the particle number operator commutes with the Hamiltonian. However, if the commutator is small, one can still make reasonably sensible statements, ala LSZ and perturbation theory. However, if the commutator is not small, then I would say that is a sign to let go of the picture of particles, because that view will simply make conceptual understanding difficult (even simple states in the system will be a complicated superposition of particles).

    Something we often start forgetting as we move up in abstraction: only energy eigenvalues and eigenstates are physical (pedant cutoff: up to usual isomorphisms, etc.); the labels which we attach to them are just labels.
     
  10. Sep 19, 2010 #9
    In practice, we work with quark and gluon fields defined across the entire spacetime lattice, with our observables being various correlation functions of these fields.

    But it sounds like you're interested in more general considerations, rather than the details of how we perform our calculations. Perhaps you would find relevant the classic 1977 paper by Luscher, "Construction of a selfadjoint, strictly positive transfer matrix for Euclidean lattice gauge theories", http://dx.doi.org/10.1007/BF01614090 or http://ccdb4fs.kek.jp/cgi-bin/img_index?7611148 [Broken].
     
    Last edited by a moderator: May 4, 2017
  11. Sep 19, 2010 #10

    tom.stoer

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    I think we should define what "particles" are

    - lumps of energy confined to some small region of space, like solitons
    - states in a representation of the Poincare group, defined via its mass, spin etc.
    - plane wave states in Fock space
    - ...
     
  12. Sep 20, 2010 #11

    Demystifier

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    Thanks, that's helpfull.
     
  13. Sep 20, 2010 #12

    Demystifier

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    I've seen many times such an argument, but I never liked it. After all, what is so special about the Hamiltonian operator? OK, if the number operator does not commute with the Hamiltonian, then the number of particles is not conserved. But so what? If some quantity is not conserved, it does not mean that this quantity does not make sense.
     
  14. Sep 20, 2010 #13

    Demystifier

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    It is a difficult question. But let us, at least for the moment, replace it with a more practical question: How a particle is defined in lattice QCD? It seems that the answer is the following: A hadron particle is the lowest energy state with given values of flavor quantum numbers. Then two such states localized at very different locations (quantum solitons with a negligible overlap) would approximately correspond to a 2-particle state.
     
  15. Sep 20, 2010 #14

    tom.stoer

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    Fine!

    The basic disctinction made is between hadrons and quarks. One can e.g. write down a Hamiltonian for QCD using Fock space quarks and gluons in order to describe localized hadrons.

    So essentially one mixes two different concepts of "particles". But this is fine as long as everybody agrees what is meant by "particle".
     
  16. Sep 20, 2010 #15

    Demystifier

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    One additional question: Is meson (or barion) in lattice QCD allways determined by a 2 (or 3) point function, i.e., a correlation function of 2 (or 3) quark fields?

    If yes, then the claim that meson (or barion) contains 2 (or 3) quarks is justified.
     
  17. Sep 20, 2010 #16

    Demystifier

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    Excellent, I love it! It seems to be exactly what I wanted (but was unable to spell it explicitly). And I don't see a reason why such A wouldn't exist for ANY H.

    Do you know any paper in which such A was constructed explicitly for a nontrivial interacting theory?
     
    Last edited: Sep 20, 2010
  18. Sep 20, 2010 #17

    tom.stoer

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    afaik large-N QCD deals with an approx. like [tex]\bar{q}Xq ~ <\bar{q}Xq> + \hat{Q} + \ldots[/tex] where the first operator is bilinear in the quarks with some Dirac and color matrix X, the vev is some condensate and Q is a meson fluctuation operator. One can show that in an appropriate limit the meson fluctions couple only weakly, so this should be related to the above mentioned case; at least in 1+1 dim. QCD the Hamiltonian can be rewritten in terms of these weakly coupled meson fluctuations.
     
  19. Sep 20, 2010 #18
    By "two-point function", we refer to the propagation of a hadron (either a meson or a baryon) from one point of spacetime to another. A "three-point function" adds another operator to the correlation function, to probe for example the form factors of this hadron.

    But to address what you mean as opposed to what you say, the mesonic creation and annihilation operators always involve two quark fields, while the baryonic operators involve three. There would be no way to obtain the appropriate quantum numbers otherwise.
     
  20. Sep 20, 2010 #19

    Demystifier

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    Thanks for this clarification. Are these two quark fields (defining a mesonic operator) fields at the SAME spacetime point? If yes, does it lead to UV divergences on a lattice? (I guess not, because in a continuum Dirac delta(0) is infinite, but on the lattice Cronecker delta(0) is finite.)
     
    Last edited: Sep 20, 2010
  21. Sep 20, 2010 #20

    Demystifier

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    Let me further develop the idea of strangerep. Assume that all eigenstates of H are known. These eigenstates can be labeled by some discrete (as well as some continuous) labels. One of these discrete labels (or a combination of them) has values n=0,1,2,3 ..., where n=0 corresponds to the ground state. Then we can say that the number of particles in a given eigenstate is simply the number n.

    The problem, of course, with such a definition of particles is that it is highly non-unique, unless some additional criteria is posed on the choice of the appropriate discrete label n. Perhaps it was not such a good idea as it seemed to me at first.
     
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