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I am trying to understand to which realistic objects one can apply the definition of single-particle states as irreducible representations of the Poincare group. Let me start with a couple of motivational questions.

  1. In $\phi^4$ theory, would we call the states of definite momentum and energy single-particle states?

  2. What are the single-particle states in a theory like QCD, with multiple fields and various charges? Are those what we intuitively think of as "dressed quarks"/"dressed gluons" or do "bound states" count as well?

My intuitive understanding is that one should call a single-particle state the state with the maximum number of charges being defined. For example, in QED, I would say that the single-particle states are the states of definite energy, momentum, and electric charge. How does the presence of multiple fields and charges affect the definition?

CLARIFICATION

I am not asking "what do we call a particle in QCD?", but rather "what is the physical meaning of states defined in Weinberg (2.5.1)?" and "how does this definition work for theories with multiple fields"?

mavzolej
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1 Answers1

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An invariant way to understand single particle states and bound states are to look for poles and resonances in S-matrix elements. In fact you can get this information from the propagator (two-point function), using the Kallen-Lehmann spectral representation, which is defined in an interacting theory. Single particle states appear as poles in the propagator, and unstable bound states appear as resonances.

In $\phi^4$ theory, you can associate the energy and momentum eigenstates with poles in the propagator. This is less scary than it might sound; the propagator is $(k^2+m^2)^{-1}$, which is zero when the field obeys $k^2+m^2=0$, which is the Klein-Gordon equation in momentum space. (Note that in the interacting theory, this $m$ is not the same as the bare mass $m_0$ multiplying $\phi^2$ in the Lagrangian, and additionally there is a wavefunction renormalization condition that should make the residue of this pole equal to one.)

In QCD, the stable particles at low energies will be color-neutral combinations like baryons and hadrons, and "glue balls." The "Yang-Mills mass gap" Millennium prize problem is precisely to show that the spectrum of Yang-Mills theory does not include free massless gluons, even though you would expect it to perturbatively. Quarks and gluons are always off shell. But if we are interested in processes high energies above the $\Lambda_{\rm QCD}$, where QCD becomes weakly coupled due to asymptotic freedom, we can meaningfully talk about free quarks and gluons scattering.

Andrew
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  • When you say "stable particles", is this the same as your definition of "single-particle states" or "bound states"? – mavzolej Aug 12 '21 at 14:54
  • Please see the update to my question. – mavzolej Aug 12 '21 at 15:41
  • @mavzolej Stable particles are states that don't decay. If you think of a proton as a bound state of quarks and gluons, then the proton would appear as a pole but is still a bound state. – Andrew Aug 12 '21 at 16:10
  • The physical meaning of the free states is that they are particle states that propagate at asymptotically large distances and correspond to particles we see in a detector in a scattering experiment. You can map them to poles in the propagator/two point function. The latter is invariant under redefinitions of fields and avoids all the mess about what fields are fundamental or not when talking about QCD. – Andrew Aug 12 '21 at 16:11
  • Around (2.5.1) Weinberg emphasizes that this definition is valid not only for free theories. My question is how to interpret "single-particle state" according to this definition in the interacting theories. – mavzolej Aug 12 '21 at 16:12
  • I guess I don't understand your question. I agree with Weinberg's statement and as he says he doesn't assume he's working with a free theory. What is the issue with it? – Andrew Aug 12 '21 at 16:15
  • I guess when I said "the physical meaning of free states" I should have said "the physical meaning of asymptotic 1 particle states" – Andrew Aug 12 '21 at 16:15
  • There is no issue with Weinberg's definition, but I don't think you answered my question, which is "how to interpret "single-particle state" according to (2.5.1) in the interacting theories?" Are those identical to "stable particles" which you introduced? – mavzolej Aug 12 '21 at 16:22
  • @mavzolej Yes, the single particle states are stable particles, which appear as poles in the propagator. – Andrew Aug 12 '21 at 16:42
  • The advantage to thinking about poles in the propagator, instead of fields, is that it gives you a clean way to think about what's going on even when the particle states are bound states of the fundamental degrees of freedom, which is what happens in QCD. – Andrew Aug 12 '21 at 16:43
  • These definitions are not very useful for me because I am planning to neither solve my theory perturbatively, nor to use the $S$-matrix. I am looking for a definition of single-particle states in terms of field operators and their eigenvalues. – mavzolej Aug 12 '21 at 17:36
  • @mavzolej Well... the definition I'm giving is a non-perturbative definition. It's related to the fields since the two point function $\langle T \phi(x) \phi(y)\rangle$ is exactly what the Kallen-Lehman prescription gives an exact, non-perturbative expression for. The kind of looser more intuitive definition is that you can expand the field in terms of creation and annhilation operators, and when acting on the vacuum at asymptotically early or late times these creation/annhilation operators act like they do in the free theory. But I'm trying to avoid that language. – Andrew Aug 12 '21 at 18:16
  • Thanks for explanations! Can we intuitively say that "poles in the full propagator" correspond to the eigenvalues of the momentum operator? – mavzolej Aug 12 '21 at 23:20
  • @mavzolej I would say "poles in the full propagator" tell you the invariant mass $m^2$ of states in the theory. The invariant mass is one of the Casimir invariants for a representation of the Poincaire group (the other is, essentially, the spin). So given the mass and spin, you can use the Poincaire algebra to construct the 1-particle states, which is what Weinberg Vol 1 Ch 2 is all about. – Andrew Aug 13 '21 at 00:01
  • Awesome! How do we use this definition to tell anything about the values of other charges present in the theory, such as electric charge or color? – mavzolej Aug 13 '21 at 00:13
  • @mavzolej You typically would use other $n$-point functions. The $3$ point function $\langle T A_\mu \bar\psi \psi \rangle$, where $A_\mu$ is the photon and $\psi$ is the electron, measures the electric charge of the electron. Similarly you can use a correlation function like $\langle A^4 \rangle$ (gluon to fourth power) in Yang-Mills theory to measure the gauge coupling. In fact a clever trick to calculate the coupling in YM theory is to look at the two-point function in a non-trivial background field, which probes the 4 point function – Andrew Aug 13 '21 at 00:19
  • To get a non-perturbative numerical answer, one approach would be to write these correlation functions as a path integral and then evaluate the path integral using a lattice (like in lattice QCD). – Andrew Aug 13 '21 at 00:41