5
$\begingroup$

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?

$\endgroup$

2 Answers 2

4
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $m$.

  3. See also the Källén–Lehmann spectral representation.

  4. For unstable/quasi-particles, see also the relativistic Breit–Wigner distribution.

$\endgroup$

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.