# Why is the scalar QED vertex not inconsistent with angular momentum conservation?

In scalar QED, there is a vertex where two spin $$0$$ particles come in and a spin $$1$$ photon comes out. Naively this can't possibly be consistent with angular momentum conservation, because two spin $$0$$ things can't add up to spin $$1$$.

It is claimed here that this is okay because the photon is "off-shell", so it has spin $$0$$. I don't believe this argument. While it is true that the usual formulation of QED has longitudinal photons, the whole point is that they decouple. And it's perfectly possible to formulate QED with only physical states from the start. Being off-shell is weird, but not so weird that it can reach outside the Hilbert space and produce a new scalar state out of nowhere.

A more sensible possibility, in my opinion, is that the two spin zero particles must have orbital angular momentum $$\ell = 1$$. But that seems rather complicated, and I don't know how to show that.

What is the resolution to this puzzle?

• Are you talking about the vertex, or an impossible scalar-scalar fusion? In 2, why don't you actually write the Lorentz invariant vertex and the Feynman rule and make your Lorentz misgivings explicit? – Cosmas Zachos Jan 1 '18 at 21:30
• @CosmasZachos I'm talking about the vertex with two spin $0$ particles and one spin $1$ particle, as shown here. I know that this vertex by itself is forbidden by momentum conservation, but I would imagine that's independent of the angular momentum part I'm asking about. – knzhou Jan 1 '18 at 21:36
• The coupling is Lorentz (and and its rotation subgroup) invariant and you should be able to see that from it... – Cosmas Zachos Jan 1 '18 at 21:47
• @CosmasZachos I can see that, but I don't know why my other argument is wrong -- how can angular momenta of $0$, $0$, and $1$ add up to $0$? – knzhou Jan 1 '18 at 21:49
• You are only adding intrinsic spins, skipping orbital angular momentum?? If you look at the vertex on the bottom of the 2nd transparency of Steve G's notes you adduced, the gradient translates to a momentum, which is dotted to the photon... Thats both a Lorentz and a rotation scalar. If you wish, the intrinsic angular momentum of the photon is matched by the orbital angular momentum of that vertex/combination. – Cosmas Zachos Jan 1 '18 at 21:54

Indeed, the two spin $0$ particles must have angular momentum $\ell = 1$. The crucial feature is that, unlike the case of regular QED, the scalar QED interaction $\mathcal{H}_{\text{int}} \sim A_\mu \phi \partial^\mu \phi$ contains a derivative, so it is proportional to the momentum $p^\mu$.
Then a first-order transition amplitude from scalar and photon states $|i \rangle, |i' \rangle$ to $|f \rangle, |f' \rangle$ is $$\langle f, f' | \mathcal{H}_{\text{int}} | i, i' \rangle \sim \langle f | p^\mu \phi_{\mathbf{p}} \phi_{\mathbf{0}} | i \rangle \, \langle f'| A_\mu | i' \rangle.$$ Focusing on the first term here, by the Wigner-Eckart theorem, the total angular momentum of the final state is the sum of the angular momentum of the initial state and that of the transition operator. Since the $\phi$ fields and $\phi$ particles are scalars, and the vector $p^\mu$ carries spin $1$, that means one unit of angular momentum is transferred to the orbital angular momentum of the $\phi$ particles, for each scalar QED vertex.
• Why does the vector $p^\mu$ carry spin 1? The longitudinal part of a gauge field is precisely the spin zero part, so I'd expect $p^\mu$ to carry spin zero. (Recall that the vector representation of the Lorentz group, when restricted to spatial rotations, coincides with the $0\oplus1$ representation of the Orthogonal group. In other words, a Lorentz vector contains both a spin zero and a spin 1 part). – AccidentalFourierTransform May 29 '18 at 20:36
• @AccidentalFourierTransform By your second sentence, are you saying that $p^\mu A_\mu$ has spin $0$? I would agree with that, and propose that the individual parts $p^\mu$ and $A_\mu$ that matter for this process each carry spin $1$. For example, we can work in temporal gauge, where $A_0 = 0$, so only the spin $1$ part of each survives. Since orbital angular momentum is gauge invariant this reasoning should hold for all gauges. (Of course, tell me if I'm misinterpreting something you're saying!) – knzhou May 29 '18 at 20:51
• No, I'm saying that $A_\mu=A_\mu^T+\partial_\mu\phi$. The first term is the spin 1 part, and the second part is the spin zero part. In terms of polarisation vectors, $\epsilon_\mu=\epsilon_\mu^T+p_\mu$. The momentum $p_\mu$ represents the spin zero part of a vector. More generally, a gradient cannot create spin 1 particles: $\langle 0|\partial_\mu\phi|\boldsymbol p\rangle=p_\mu\langle0|\phi|\boldsymbol p\rangle=0$, with $|\boldsymbol p\rangle$ a spin 1 particle. – AccidentalFourierTransform May 29 '18 at 21:03
• @AccidentalFourierTransform Now I'm confused, I think we're saying the same thing. We agree $A_\mu, p_\mu \in 0 + 1$, and we agree $p_\mu A^\mu \in 0$. What I think your decomposition says is that in the product $p_\mu A^\mu$, which could potentially consist of $(0 + 1) \times (0 +1)$, the only piece that contributes is the $1 \times 1$, from which we get the $0$. And that's what I said in my comment. Note that in my answer, $p_\mu$ is not referring to the longitudinal polarization of $A_\mu$. It means the $p_\mu$ in the product $p_\mu A^\mu$. – knzhou May 29 '18 at 21:20
• I agree that $A_\mu\in 0\oplus 1$, and that $p_\mu A^\mu\in 0$. But I disagree that $p_\mu\in0\oplus1$. Rather, $p_\mu\in0$. The spin 1 rep. is orthogonal to the spin 0 rep., and therefore $p_\mu$ picks up only the spin 0 part of $A_\mu$. If $p_\mu$ had spin 1, it would add to the spin 1 part of $A_\mu$, and we would have $p_\mu A^\mu\in 0\oplus 1\oplus2$, which is clearly false. Thus, my initial comment: $p_\mu$ has spin zero, not spin one. – AccidentalFourierTransform May 29 '18 at 21:31