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?


1 Answer 1


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.

  • $\begingroup$ 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 $\endgroup$ Jun 3, 2021 at 22:22
  • 1
    $\begingroup$ @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. $\endgroup$ Jun 3, 2021 at 22:31
  • 1
    $\begingroup$ okay, great, thanks a lot :) $\endgroup$ Jun 3, 2021 at 22:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.