# Why doesn't the Weinberg-Witten theorem forbid collinear photons?

The Weinberg-Witten theorem tells us that any theory that has an effective graviton, i.e. a massless helicity-2 particle as a state in the free-particle Fock space, cannot have a gauge-invariant and Lorentz-covariant stress-energy tensor that gives the graviton nonzero energy. This is intended as a no-go theorem ruling out composite gravitons, because if the theory can be expressed using only particles of spin $$\le 1$$ then it presumably will have such a tensor.

A composite graviton would be a bound state of lower-spin particles such as gauge bosons, with that sum of their spins in the direction of propagation equal to 2. My question is: why does the state need to be bound? Why are we only interested in states that can be called "particles"? QED, for instance, includes massless states of helicity 2: states with 2 photons that just happen to have the same direction and spin. They are not single particles, but they are part of the Hilbert space, and matrix elements exist for them. The argument of Weinberg-Witten would seem to apply. Yet QED does have a covariant stress-energy tensor, and collinear two-photon states do have nonzero energy. Why isn't this forbidden?

I think I have a partial answer: the proof of the WW theorem derives a contradiction by writing down the tensor (at the origin) as an operator on the gauge-fixed Fock space, taking its matrix elements between single-graviton states of different momenta, and taking a limit as the momenta approach coincidence while we shift the Lorentz frame to make the momenta equal and opposite. Thus it's not enough for the graviton to have nonzero expected energy; it must have nonzero matrix elements even between states of unequal momenta. Since we are talking about an operator at a point, this seems like a reasonable assumption. Yet it seems to me that this is where my "collinear photon" case falls out: We can (I think) write the electromagnetic stress energy tensor as a sum of term of the form $$a^\dagger_k a_{k'}$$, meaning that we only get a nonzero matrix element when at most one photon has different momenta between the two states. Since we want states with different directions for the momenta of the collinear particle pairs, we get zero and cannot derive a contradiction.

Is this correct? Or am I perhaps confusing myself by thinking about pure momentum states rather than normalizable wavepackets? This was done in the original proof, but perhaps it introduces problems with more than one particle?

Of course, the really important question is: what changes when the state is bound; i.e. an actual composite graviton? Is it possible that we have a new loophole for the WW theorem, where we can have composite gravitons as long as we somehow force the stress-energy tensor to de diagonal in the momentum?

• A photon is uncharged (under $U(1)$) so it doesn't source its spin-1 current. This is consistent with WW theorem. On the other hand, a gluon is charged (under color transformations) and it does source its corresponding spin-1 current. WW theorem then tells us that its current is then not Lorentz covariant which is true (the current is not gauge-invariant). Similarly, the graviton is charged (under diffeos) and its current (stress tensor) is not Lorentz covariant (the stress tensor is not gauge (diffeo)-invariant.). Commented Apr 20, 2022 at 7:49
• @Prahar Yes, all of that is correct. But you don't seem to be addressing my question, so what are you trying to say? Commented Apr 20, 2022 at 7:54
• Apologies, I think I misread a comment you made in your post and I was trying to correct it. Anyway, QED may have a Lorentz covariant stress tensor but that stress tensor is NOT gauge invariant under diffeos. It is of course gauge-invariant under $U(1)$ symmetry, but once you have a massless spin-2 particle (bound or not) you immediately have diffeomorphisms as a gauge symmetry and all gauge-invariant operators must be invariant under (small) diffeomorphisms. The stress-tensor is not invariant under diffeomorphisms. Commented Apr 20, 2022 at 7:58
• @Qmechanic, would you mind explaining why you added the tag "photons"? I would have thought that's much too broad a topic for this question. Am I missing something about the tag system? Commented Apr 20, 2022 at 8:00
• Hi @Ahron Maline. Welcome to Phys.SE. Yeah, it is not ideal. Feel free to improve the tags. Commented Apr 20, 2022 at 8:04

Okay, I think I am satisfied that the "partial answer" I included in the question is the correct answer. The proof of the WW theorem involves matrix elements of the form $$\langle p|T^{\mu\nu}(x)|p'\rangle,$$ where $$|p\rangle$$ is a momentum eigenstate of the spin-2 particle and $$p,p'$$ are two nearly equal null momenta. The proof relies on this matrix element being nonzero, while it does equal zero for my case of two collinear photons. The operator $$T^{\mu\nu}$$ cannot change the momenta of two different photons, because it is only quadratic in the photon field $$A^\mu$$. Therefore the proof does not apply, and such states can of course exist.
The reason I was dissatisfied with the answer, and called it only "partial", was that I didn't understand why we should be so confident that the above matrix element does not vanish when $$|p\rangle$$ is a single-particle state such as an "emergent graviton" that is a bound state of other fields. If it can vanish when the underlying particles are not bound, then how do we know that creating a bound state will change things?
Now I think I get it: the condition $$\langle p|T^{\mu\nu}(x)|p'\rangle\ne0$$ is basically an intuitive physical requirement. It means that if (hypothetically) there was a an interaction of the form $$T^{\mu\nu}h_{\mu\nu}$$ (and of course there is such an interaction, namely gravity, but we do not require this), then a background field $$h_{\mu\nu}$$ with a small gradient could cause slight changes in the momentum of our particle, making it move along a curved trajectory. This can be taken as a reasonable definition of what is meant by the particle being "charged under $$T^{\mu\nu}$$", and this is what is required for the WW theorem to apply.