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I am a bit confused about resonances in QFT. I am reading Schwarz's QFT book and as far as I understand, if in a reaction the mass of the particle acting as a propagator is bigger than the sum of the masses of the particles interacting, the propagator particle can be on shell, thus the propagator has an imaginary part and doing some math you get the Breit-Winger distribution (this is chapter 24.1.4). However, the particles that were discovered as resonances are not propagators, as far as I understand. For example, in muon-muon scattering to obtain the J/psi particle, the 2 muons interact by exchanging a photon, not a J/psi particle. So the propagator is that corresponding to a photon. Can someone explain to me how does this work? How do resonances from and how does this on-shell propagator comes into play? Thank you!

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  • $\begingroup$ You may want to have a look at Weinberg's QFT, Vol.I, §10.2. $\endgroup$ – AccidentalFourierTransform Oct 26 '18 at 15:03
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This is partly about clarifying the (nonperturbative) connection between field operators and particles, and partly about relating this connection to how perturbative calculations are done.

I'll start with the nonperturbative part. In general, the connection between field operators and particles is not one-to-one in QFT. Fields are what we use to define the model and to construct local observables. Particles are phenomena that the model predicts. This is evident in QCD, where the model and its observables are constructed in terms of quark and gluon fields, but the particles are mesons and baryons. The connection is closer to being one-to-one in QED, but even in QED there are subtleties: electrons always have associated electric fields, and we can have short-lived bound states like positronium. The example you cited ($\mu^+\mu^-\rightarrow J/\psi$) requires a model more analogous to QCD, where the connection is not one-to-one at all.

Continuing with the nonperturbative part: Let $|0\rangle$ be the vacuum state, and let $A(x)$ be some operator constructed from the field operators. To be specific, let's take $A(x)$ to be something like $A(x)=\mu_+(y)\mu_-(x-y)$, where $\mu_\pm(x)$ are muon field operators. The important thing here is that when applied to $|0\rangle$, the operators $\mu_\pm(x)$ do not simply create muon particles. The state $A(x)|0\rangle$ is a superposition of many terms, one of which contains just a pair of oppositely charged muon particles, but other terms contain other stuff, including photons, the $J/\psi$, other mesons, and just about anything else that has the same quantum numbers as a pair of oppositely charged muons. The relative amplitudes of these various contributions change in time (I'm working in the Heisenberg picture here, so the $x$ in $A(x)$ includes a time coordinate), but there is no time at which a simple construct like $A(x)|0\rangle$ contains only a pair of muon particles. To construct such a state would require acting on the vacuum with an operator much more complicated than $A(x)$, and I'm pretty sure that nobody knows exactly which operator that would be.

The key question here is: How can we isolate the contribution from just the particle (or resonance) of interest — say, the $J/\psi$ meson? We don't know how to do this directly in the state-vector $A(x)|0\rangle$, but we can do it indirectly. Here's the idea: write $$ A(x)|0\rangle = \sum_p f(p)|p\rangle + \text{other stuff}, $$ where $|p\rangle$ represents a state with nothing but a single $J/\psi$ with momentum $p$, and $f(p)$ is a complex coefficient, and "other stuff" includes a term with nothing but photons and any other terms that have the same quantum numbers as a pair of oppositely charged muons. We don't know how to express the states $|p\rangle$ expilcitly in terms of field operators, and that's okay. We'll use a trick. To use the trick, we need to remember that the $J/\psi$ is not a stable particle, so the vectors $|p\rangle$ cannot be eigenstates of the Hamiltonian. This means $$ e^{-iHt}|p\rangle = z|p\rangle + \text{other stuff}, $$ where $z$ is some time-dependent complex number with magnitude that must decrease in time, because $|p\rangle$ is not an eigenstate. Empirically, we expect the magnitude to decay exponentially, so $$ e^{-iHt}|p\rangle \approx e^{-\Gamma t}e^{-iE(p)t}|p\rangle + \text{other stuff} $$ for some $\Gamma>0$. There is no violation of unitarity here. As the norm of the $|p\rangle$ term decreases, the norm of the "other stuff" increases, because that's where the decay products are going. I wrote this relationship as an approximation only because the time-dependence of the norm of the $|p\rangle$ term might not be exactly described by the simplistic $\exp(-\Gamma t)$ factor.

Now, consider the time-ordered correlation function $$ \langle 0|T\,B(y)A(x)|0\rangle, $$ where $A$ and $B$ are operators chosen so that $A|0\rangle$ and $B|0\rangle$ include terms describing the initial and final (respectively) configurations of interest, such as an incoming pair of muons and an outgoing spray of whatever. We can use the LSZ trick to select the right incoming/outgoing particle configurations. I'll be very schematic here, because this is already worked out in several places. (I don't have access to Schwartz's QFT book right now, but it's in Weinberg's book that AccidentalFourierTransform cited, at least for the special case $\Gamma=0$, and allowing $\Gamma\neq 0$ is an easy generalization.) Write the $T$-ordering in terms of $\theta$-functions and insert a complete set of $J/\psi$ momentum eigenstates to get \begin{align*} \langle 0|T\,B(y)A(x)|0\rangle &\sim \theta(y_0-x_0)\sum_p\langle 0|B(y)|p\rangle\,\langle p|A(x)|0\rangle \\ &+ \theta(x_0-y_0)\sum_p\langle 0|A(x)|p\rangle\,\langle p|B(y)|0\rangle + \text{other stuff}. \end{align*} We're inserting $J/\psi$ eigenstates because we want to isolate the contribution from the $J/\psi$ resonance. Everything else is lumped into "other stuff". Now write $x=(t,\mathbf{x})$ to separate the temporal and spatial components of $x$, and use $$ \langle 0|A(x)|p\rangle = e^{-\Gamma t}e^{-iE(p)t}\langle 0|A(0,\mathbf{x})|p\rangle, $$ and likewise for $\langle 0|B(y)|p\rangle$. Now take the Fourier transform to go from $x,y$ to the energy/momentum domain. As shown elsewhere (like Weinberg's book), this reveals the $J/\psi$ pole, with an imaginary part due to $\Gamma$. This pole off the real axis shows up as a resonance in scattering cross-sections. (At least, it looks like a pole when we ignore the "other stuff". With a little hand-waving and empirical input, we could probably almost justify this.)

Everything I've said so far is nonperturbative, except maybe the LSZ part. Now I'll connect it to perturbation theory. Working to some low order in a small coupling expansion, the photon "pole" (or whatever the massless version of a pole is) will show up in a propagator because the photon corresponds closely to one of the fields that was used to construct the model. But we won't see the $J/\psi$ pole/resonance in a small coupling expansion (at least not without some other fancy trickery, like what is mentioned on page 153 in Peskin and Schroeder's Introduction to Quantum Field Theory), because it doesn't correspond closely to any individual field that was used to construct the model.

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  • $\begingroup$ I am still a little bit confused about the nature of the $J/\psi$-states. You say they are not eigenstates, and i understand that their usefulness lies in them being 'approximate eigenstates' in some sense. Later on, however, you say they are not poles, but i thought one may show in some contexts, for example non-relativistic scattering theory, exactly the existence of such poles, which then give rise to resonances. $\endgroup$ – Lorenz Mayer Oct 29 '18 at 9:09
  • $\begingroup$ @LorenzMayer The model I had in mind (but didn't specify clearly) is one whose Lagrangian has quark, lepton, and gauge fields (e.g., the SM) but no one field specifically for to the $J/\psi$. Nonperturbatively, there is a pole/resonance (by the 1st argument I sketched). But at low orders in a simple small-coupling expansion, we won't see it. This model's Feynman diagrams have quark/gluon lines, but no $J/\psi$ line. We could presumably recover the $J/\psi$ pole/resonance by summing some class of diagrams to all orders (fig 10.3 in Weinberg v1), or we could use a different model. $\endgroup$ – Chiral Anomaly Oct 29 '18 at 12:58
  • $\begingroup$ ok, i guess the source of my confusion might have been a misunderstanding. I thought you were saying that resonances are not poles. $\endgroup$ – Lorenz Mayer Oct 29 '18 at 13:52
  • $\begingroup$ @LorenzMayer Good comment. I didn't mean to imply that. I was subconsciously thinking of "pole" as the mathematical term and "resonance" as the physical manifestation of it, but the wording of my answer obscured that connection. I edited the answer a little bit to fix this. Thanks! $\endgroup$ – Chiral Anomaly Oct 30 '18 at 0:07

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