Is it possible to write down a KL-like formula for massless particles (in particular, the photon)?

The usual proof of the theorem assumes (see http://www.thphys.uni-heidelberg.de/~weigand/QFT1-13-14/SkriptQFT1.pdf, page 44) that we can boost to a frame where the particle is at rest, and I couldn't find any proof that won't do this.


Write the spectrum of the Hamiltonian as $$ H|p,\lambda\rangle=E_p(\lambda)|p,\lambda\rangle $$ $$ \boldsymbol P|p,\lambda\rangle=\boldsymbol p|p,\lambda\rangle $$ where $\lambda$ is any set of parameters that is needed to distinguish different states with the same energy-momentum.

Write the completeness relation of the eigenstates as $$ |\Omega\rangle\langle\Omega|+\sum_\lambda\int \widetilde{\mathrm d k}\ |p,\lambda\rangle\langle p,\lambda|=\mathbb I $$ and the exact propagator as $$ \boldsymbol\Delta(x,y)=\langle \Omega|T\ \phi(x)\phi(y)|\Omega\rangle $$

Insert the identity between the fields: $$ \boldsymbol\Delta(x,y)=\sum_\lambda\int \widetilde{\mathrm d k}\ \langle \Omega|\phi(x)|p,\lambda\rangle\langle p,\lambda|\phi(y)|\Omega\rangle $$ where I take the fields to have null vev.

Next, write $\phi(x)=e^{iPx}\phi(0)e^{-iPx}$, and boost to a frame where $p=(m,0,0,0)$, to get $\langle \Omega|\phi(x)|p,\lambda\rangle=\langle \Omega|\phi(0)|0,\lambda\rangle e^{-ipx}$. From this, the KL theorem follows.

Is it possible to circumvent this boost to get an alternative proof of the theorem? Or maybe the theorem is false for massless particles?

EDIT: Partial solution

See Wikipedia: https://en.wikipedia.org/wiki/Mass_gap

It seems that in general KL is false for gauge theories. In the case of the photon, the theorem is true, but it has to be checked explicitly. I don't know if this remains true in QCD.

Can anyone confirm this? Is the mass-gap problem related to my question? Is this (partial) solution correct?

  • 1
    $\begingroup$ Why do you actually need to boost at all? As far as I understand it's not actually need it and I think the proof goes through with $\langle\Omega|\phi(0)|p,\lambda\rangle$ as well. See e.g. chapter 10.7 in Weinberg volume I of QFT. $\endgroup$
    – TwoBs
    Commented Oct 1, 2015 at 19:15

1 Answer 1


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:


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


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

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


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

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.

  • 1
    $\begingroup$ I think this pretty much solves it. Im uncertain about a specific point, though: We take $P^\mu$ to be the exact momentum operator, and write 1-particle states as $P^\mu|p\rangle=p^\mu|p>$. Now, from first principles, how can we know that the eigenvalues satisfy $p^2=\text{const}$? In an interacting theory, we could expect to get corrections to the energy-momentum relation, in addition to mass renormalisation, right? If we want an axiomatic approach, we should prove somehow that 1-p states satisfy $(H^2-\boldsymbol P^2)|p\rangle\propto|p\rangle$. Is this possible?Is this important at all? $\endgroup$ Commented Oct 3, 2015 at 11:13
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    $\begingroup$ Also, in regards to the photon, I believe the Lorentz transformation properties of the fields are a bit more complicated that in the scalar field. If I recall, (see Weinberg I), massless vector fields transform as $A\to \Lambda A+\partial \Omega$, for $\Omega$ a scalar operator. Anyway, I'll try to tackle the general case (any sort of field), and if I get sensible results, Ill post them here for future reference. Bis dann, thank you very much, cheers :) $\endgroup$ Commented Oct 3, 2015 at 11:21
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    $\begingroup$ i've edited the answer to address some of these questions, hope it helps! $\endgroup$
    – Fra
    Commented Oct 3, 2015 at 13:44
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    $\begingroup$ The pole in $p^2=0$ might be isolated, the branch cut starts at $M^2_{th}$ at the threshold for the creation of multiparticle states. For istance the photon which is not self interacting will have the threshold at the mass of the electron thus in this case you can isolate the one particle contribution from the continuum. (pag 220 Itzkinson-Zuber) In general this is not true and you might not be able to isolate the one particle contribution. In the case of the photon the fact that it remains massless depends upon the fact that the $U(1)$ symmetry reamins unbroken. (i.e. gauge invariance) $\endgroup$
    – Fra
    Commented Oct 6, 2015 at 10:15
  • 2
    $\begingroup$ you can say the symmetry keeps the pole where it is. Again in the case of $M^2_{th}=0$ there is a branch-point singularity and you can't separate the one particle state from the continuum. $\endgroup$
    – Fra
    Commented Oct 6, 2015 at 10:18

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