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For a real scalar field $\phi$, the Kallén-Lehmann representation of a two-particle propagator in a translation invariant theory in QFT is:

$G(x_1,x_2) = \sum_{m, \; other \; quantum \; numbers} \int \frac{d^4p}{\left( 2 \pi \right)^4} e^{ip(x_1-x_2)} \frac{i}{p^2-m^2+i \epsilon} \lvert <\Omega| \phi(0) | m,\; \pmb{p}=\pmb{0}, \; other \; quantum \; numbers> \rvert^2$,

where the sum over $m$ is performed over all the invariant masses allowed by the theory.

I get that if the theory allowed bound states then I should sum over the masses of bound states as well, which would then show up as poles of the propagator, but why should't I sum over the masses of unbound states as well? After all a state of two unbound particles would have a mass of $2m_{\phi}$, which should then appear as a pole in the propagator. Is it because if the two particles are unbound $\lvert <\Omega| \phi(0) | m, \; \pmb{p}=\pmb{0}, \; other \; quantum \; numbers> \rvert^2$ is just zero, therefore it cancels the pole?

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The sum over states that you've written is supposed to be literally the sum over all states, with the restriction that the total spatial momentum of each state should be $\vec p=0$. You are in principle summing over bound and unbound states. For example if there was a bound state of two $\phi$ particles which had Lorentz-invariant mass $M^2$, this state would show up in the sum.

I don't understand why you think there is some preference for bound states in that sum, and that we're somehow neglecting unbound states. For example, in that sum, we could have a state with two unbound $\phi$ particles with opposite spatial-momenta so that $\vec p_{\textrm{total}}=0$. The energy of such a state would be $E_{\textrm{total}}=2\sqrt{m^2+|\vec p|^2}$ and the mass-squared would be $m^2$. However the state of a single $\phi$ particle having with $\vec p\neq 0$ would not be present in the sum.

Edit: Take a look at Figure 7.2 on page 214 of Peskin & Schroeder. It shows a typical plot of the spectral density function, which is essentially what you're asking about. It illustrates the idea that I conveyed in this answer.

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  • $\begingroup$ My lecturer said that if we assume that our theory does not have bound states then the only mass left in the sum should be that of a single particle, meaning the dressed mass of the fundamental particle of the theory. He then added that such a thing is true provided the coupling constant of the theory is small enough. Now I think I might have missed something during the lecture because what you just said makes total sense to me $\endgroup$ – OutrageousKangaroo Jun 3 at 22:22
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    $\begingroup$ @OutrageousKangaroo if there are no bound states, then the only isolated term will indeed be the single-particle, having dressed mass $m^2$. The 2-particle, 3-particle, etc. will form continuum contributions to that sum. For example, in the 2-particle case, a given particle can have arbitrary 3-momentum $\vec p$, so long as the other particle has the exact opposite 3-momentum $-\vec p$ to give total 3-momentum $\vec P = \vec p - \vec p = 0$. The "mass" (square of the system's total momentum) will be $M=E=2\sqrt{m^2+|\vec p|^2}$, where $|\vec p|^2$ is continuous and arbitrary. $\endgroup$ – Arturo don Juan Jun 3 at 22:31
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    $\begingroup$ okay, great, thanks a lot :) $\endgroup$ – OutrageousKangaroo Jun 3 at 22:36

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