# Why propagator pole is associated to the mass?

We say that the pole of the all-orders photon propagator, $$\frac{1}{q^2[1+\Pi(q^2)]}$$ doesn't shift if $$\Pi(q^2=0)$$ is regular. This amounts to say that the photon remains massless to all orders in perturbation theory. Conversely, the fermion propagator, $$\frac{i}{\not p -m+\Sigma(p)+i\varepsilon}$$ has $$\Sigma(p)$$ which shifts the pole hence the mass is corrected and must be renormalized.

Why the propagator pole is associated to the mass?

I believe the simplest way to understand this is via the Kallen-Lehmann spectral representation, which basically teaches you about the spectrum of the theory-in a nonperturbative way-by looking at the analytic structure of the propagator in momentum space.

You begin by considering an interacting theory, with an interacting vacuum $$|\Omega\rangle$$(related to the free vacuum by the usual trick of taking time evolution to infinity). We have an interacting hamiltonian $$H$$, and we need a complete set of eigenvectors for this to describe our theory. Note that these will also be simultaneous eigenvectors of the momentum, since $$[P,H]=0$$(this is a statement about translation invariance of the theory). You can look at Peskin's chapter 7 for a detailed derivation, but the essential result is that the free theory completeness relation gets modified to

$$(1)=|\Omega\rangle\langle\Omega|+\sum_{\lambda_0}\int \frac{d^3p}{(2\pi)^3}\frac{1}{2E_p(\lambda)}|\lambda_p\rangle \langle\lambda_p|$$

Where $$|\lambda_p\rangle$$ is the state obtained by boosting some zero momentum state $$|\lambda_0\rangle$$. There can be a lot of such zero momentum states, each will give rise to a set of its own boosted states, and so we sum over $$\lambda_0$$ in the completeness relation. $$E_p(\lambda)=\sqrt{|p|^2+m_\lambda^2}$$ is the dispersion relation for $$|\lambda_p\rangle$$, and it is now clear what the interpretation of $$m_\lambda$$ is- it is the energy of a ZERO MOMENTUM state $$\lambda_0$$- this is just we call the mass.

We can now massage this equation-look at Peskin for details, to find the 2 point function-

$$\langle\Omega|\phi(x)\phi(y)|\Omega\rangle=\sum_{\lambda}\int\frac{d^4p}{(2\pi)^4}\frac{1}{p^2-m_\lambda^2} e^{-ip(x-y)}Z$$

Where $$Z^2=|\langle\Omega|\phi(0)|\lambda_0\rangle|^2$$ is what we interpret as the wavefunction renormalization. It is now clear what the poles of this expression are- they are precisely the masses $$m_\lambda$$. We can write this in momentum space and the fourier transform-the spectral density-takes the form-

$$\rho(M^2)=Z\rho(M^2-m_{1p}^2)+...$$

where the first pole obviously comes at the 1-partucle states' masses-this is the smallest mass so it comes first. There will be further contributions from bound states, and a cut starting from multiparticle states. The punchline is, the non analyticities are associated with masses in the theory. These non analyticities are just the poles of the propagator.

Here is a heuristic argument.

1. It's a fact that a full connected propagator/2-point correlation function $$\tilde{G}_c$$ of a Lorentz-invariant theory tend to have a simple pole $$\tilde{G}_c \propto \frac{1}{p^2-m^2}$$, where for the sake of the argument $$m$$ is some constant.

2. Given that a correlation function/scattering amplitude is a measure of how likely a process occur, it is natural to associate an infinity/a pole with the production of a particle with physical mass $$m$$.