Källén-Lehmann representation is just a way to expand in the momentum basis the two point correlation function of a local operator $\hat{O}(x)$, it holds true for massive and massless theories alike except for non Abelian gauge theory in which the situation is a bit more complicated. Let's demonstrate the K-L formula with a more general proof: let's start with the one particle state

$$F(p)=\langle \Omega|\phi(0)|p,\lambda\rangle$$

which we'll see doesn't depend at all on $p$ . To be able to prove this result you just need to know how a scalar field transforms. Let $U(\Lambda)$ be an unitary representation of the Lorentz group, then:

$$U^{\dagger}(\Lambda)\phi(0)U(\Lambda)=\phi(0)$$

and the single particle state transforms like:

$$|\Lambda p,\lambda\rangle=U(\Lambda)| p ,\lambda\rangle$$

Thus, you can write

$$F(\Lambda p)=\langle \Omega|\phi(0)|\Lambda p,\lambda\rangle=\langle \Omega|\phi(0)U(\lambda)| p,\lambda\rangle=\langle \Omega|U(\Lambda)U^{\dagger}(\Lambda)\phi(0)U(\lambda)| p,\lambda\rangle$$

but $U^{\dagger}(\Lambda)\phi(0)U(\lambda)=\phi(0)$ and by definition $| \Omega\rangle=U(\Lambda)| \Omega\rangle$ then you have

$$F(\Lambda p)=F(p)$$

which means that $F=F(p^2)$ is a function of a Lorentz scalar. Now, you can write $p^2=m^2$ ($m^2$ might be zero; the important thing is $p^2=\text{const.}$) so you have that

$$F=F(m^2)= \sqrt{Z}$$

Note that all steps so far have been carried out without assuming the spectrum of the theory has any mass gap, indeed the states we boosted are completely arbitrary and not in some rest frame.

For multi-particle states you can just use

$$\langle O(x)O(0)\rangle=\int_{0}^{\infty} d\mu^2 \rho(\mu^2) \ i\Delta_F(x,\mu^2) \tag 1$$

where the propagator has a fictitious mass $\mu^2$, in fact in this case the continuum of states starts right in $p=0$ and there is no forbidden gap between the one particle state and the multi-particle states.

To demonstrate the general formula one starts from: (from now on, I will forget about time ordering and stuff and just consider $\Delta_F^+=\Delta_F$ just to skip unnecessary passages, also I'll assume from the start that the theory is massless!)

$$\langle O(x)O(0)\rangle=\langle 0|O(x)\sum_n |n,\lambda\rangle\langle
n,\lambda|O(0)|0\rangle=\sum_n |\langle 0|O(0) |n,\lambda\rangle|^2 e^{-iP_n \cdot x}$$

where we define the identity as

$$\mathbb{I}=|0\rangle\langle 0|+\int \frac{d^3p}{(2\pi)^3}\frac{1}{2|p|}\sum_{\lambda}|p,\lambda\rangle\langle p,\lambda|+...$$

and have chosen the covariant normalization of states (i.e. the measue $d\sigma=\frac{d^3p}{(2\pi)^3}\frac{1}{2|p|}$ is lorentz invariant and states transform as above) and $P_n$ is the momentum associated with the n-th particle state.
Now let's write

$$\langle O(x)O(0)\rangle=\int d^4q \ e^{-iqx}\ \sum_n |\langle 0|O(0) |n,\lambda\rangle|^2 \delta^4(q-P_n)$$

And introduce

$$\rho(q)= \sum_n |\langle 0|O(0) |n,\lambda\rangle|^2 \delta^4(q-P_n)$$

It is easy to prove that

$$\rho(\Lambda q)=\rho(q) \rightarrow \rho=\rho(q^2)$$

by using the fact that the $\delta$ is invariant and that the identity $\mathbb I$ doesn't care if the states are boosted or not, i.e. $\mathbb I= \sum_n |n,\lambda\rangle\langle
n,\lambda|=U(\Lambda)\mathbb I U^{\dagger}(\Lambda)=\sum_n |\Lambda n,\lambda\rangle\langle
\Lambda n,\lambda|=\mathbb I$ where we have used the unitarity of the Lorentz transformation representation $U \ U^{\dagger}=\mathbb I$) moreover its support is contained in the spectrum of the theory so

$$\rho(q)=0 \qquad \text{ if } \qquad q^0<0\\
\rho(q)=0 \qquad \text{ if } \qquad q^2<0$$

Let's reparametrize $\rho$ as follows: $\rho(q^2)\rightarrow \frac{\theta(q^0)}{(2\pi)^3}\rho(q^2)$ in order to make obvious the second condition on $\rho$. Now we can write

$$\langle O(x)O(0)\rangle=\int \frac{d^4q}{(2\pi)^3} \theta(q^0)\rho(q^2) \ e^{-iqx}=\int d\mu^2\int \frac{d^4q}{(2\pi)^3} \theta(q^0)\rho(q^2) \ e^{-iqx} \delta(q^2-\mu^2)$$

where $\mu$ is really just an arbitrary parameter. Now we can see that

$$i\Delta_F(x,\mu^2)=\int \frac{d^4q}{(2\pi)^3} \theta(q^0)\ e^{-iqx} \delta(q^2-\mu^2)=\int \frac{d^3q}{(2\pi)^3}\frac{1}{2\sqrt{|q|^2+\mu^2}} e^{-ixq}$$

It's clear now that we can write the two point function as in $(1)$, to make manifest the presence of the one particle state the decomposition of $\rho$ will be

$$\rho=\frac{Z}{(2\pi)^3}\delta(q^2)+\rho_{M}$$

(where $M$ stands for multi-particle states) then we will have

$$\langle O(x)O(0)\rangle=Z \ i\Delta_F(x,0)+\int_{0}^{\infty} d\mu^2 \rho_{M}(\mu^2) \ i\Delta_F(x,\mu^2)$$

where

$$i\Delta_F(x,0)=\int \frac{d^3p}{(2\pi)^3}\frac{1}{2|p|} e^{-ixp}$$

As far as i know the same argument and proof (with some tweaks to take in account the polarization degrees of freedom) goes for the photon two point function. For non-Abelian gauge theories it's not possible to have a well-defined $\rho$ by itself because it is necessary to take into account the ghost fields which have the role of canceling out unphysical degrees of freedom of the gluon propagator. That's why i think it's not very practical to use this formalism in the context of non-Abelian gauge fields where it's easier to operate from a path integral approach.

EDIT (answer to comment):

The K-L decomposition is done using a basis of the momentum operator representing the intermediate states between the initial and final position of the particle thus when you consider the one particle contribution it means it is simply the free particle propagating without any intermediate states (no loops if you look at it perturbatively). Then the only changes from the free theory are the physical mass being influenced by the interaction and the renormalization constant Z which means that the probability of having a one particle contribution is no longer 1 as it was in the free theory where the field $\phi$ could just produce one particle states. This also means that being a particle propagating by itself with no force acting upon it, when you consider the one particle contribution it will have a well defined energy $p^0=E(\textbf{p})=\sqrt{\textbf{p}^2+m^2_{phys}}$.

If you want a very clear derivation for the fermionic version (spin 1/2) just look at Itzykson and Zuber's book on QFT. They also have a derivation for the photon propagator which I find really clunky yet is the only one I've found, having looked for it in about ten books on the subject.

This post imported from StackExchange Physics at 2016-05-31 07:23 (UTC), posted by SE-user Fra