Poles of the propagator

1. Jul 7, 2012

eoghan

Hi!
From "Le Bellac, Quantum and statistical field theory, 10.5.2-Massive vector field":
"The longitudinal part of the propagator $k_{\mu}D^{\mu\nu}$ has no pole at
$k^2=m^2$, so the longitudinal part doesn't constitute a dynamical degree of freedom."

I have two questions:
1) Why the propagator doesn't represent a dynamical degree of freedom if it hasn't any pole?
How do you demonstrate that physical particles correspond to the pole of the propagator?

2) The propagator $D^{\mu\nu}$ is a rank-2 tensor. The longitudinal part is $k_{\mu}D^{\mu\nu}$ and it is a vector, so, how can it be a propagator?

2. Jul 7, 2012

fzero

The components of the vector field satisfy the Klein-Gordan equation

$$(-\partial_\nu \partial^\nu + m^2) A_\mu =0.$$

By Lorentz-invariance, the mass appearing there must be the same for all components. The momentum-space propagator is the Fourier transform of the 2-point function $\langle A_\mu(x)A_\nu(y)\rangle$. Because of the KG equation above, all physical components (and linear combinations of them) of the propagator must either have a pole at $m^2$ or vanish.

The longitudinal part is a linear combination of propagators, or equivalently, the propagator for a linear combination of components of $A_\mu$. Since there is no pole, the corresponding combination of vector fields, $k^\mu A_\mu$ cannot satisfy a non-trivial KG equation. So it cannot be a physical degree of freedom.

3. Jul 8, 2012

Einj

Also you can study the spectral representation by Kallen-Lehmann (for example in Bjorken-Drell book). It is an exact result (not a perturbative one) and it shows that any Green function have always a pole at the physical mass of the particle.

4. Jul 11, 2012

eoghan

I've read the Kallen-Lehmann representation and I've understood why the propagator has poles. However I don't fully understand fzero's answer. I don't understand this passage