Locality in the scattering amplitude

Question

Early in this talk by Nima Arkani-Hamed, he describes what locality means for scattering amplitudes. "Locality tells you that the only poles in the scattering amplitude occur when the sum of a subset of the momenta of the particles goes on shell."

The talk then goes on to describe his recent research, but right now I'm not worrying about understanding that. I'm just trying to understand the statement quoted above. I know what locality looks like in the Lagrangian density (the terms of the Lagrangian density contain field operators which are all functions of the same spacetime coordinate), but it's not at all obvious to me how that is equivalent to the above statement about scattering amplitudes.

Can you explain it (preferably at about the level of a first graduate course in QFT)?

Answer 1

It is not an answer, but just some hints.

Consider the simplest free QFT with a massless bosonic scalar, the terms in the Lagrangian are local : $\phi(x) \square \phi(x)$. Considering an interacting theory ($\phi^3, \phi^4$). You are interested in calculate scattering amplitudes with incoming particles and outcoming particles. You will use the propagator in $\frac{1}{k^2}$ which form is direcly linked to the above Lagrangian term, and you may note that this propagator has a pole when $k$ is on-shell.

If you consider only a tree-level diagram, the transition amplitude is simply the product of propagators, each propagator could be written $\frac{1}{l^2}$, where $l$ is the sum of some external momenta (at each vertex, you have momentum conservation). So a pole of the scattering amplitude corresponds to the pole of the propagators, and this corresponds by putting on-shell some particular sum of the external momenta.

Now, consider a loop-diagram, with dimensional regularization, like $I(q) \sim g^2 (\mu^2)^\epsilon\int d^{4-\epsilon}p \frac{1}{p^2}\frac{1}{(p-q)^2}$,where $\epsilon$ is $>0$, $q$ is an external momentum. By using the Feynmann formula $\frac{1}{ab} = \int_0^1 \frac{dz}{[az+b(1-z)]^2}$, you will get : $I(q) \sim g^2 (\mu^2)^\epsilon\int_0^1 dz \int (d^{4-\epsilon}p) \dfrac{1}{[p^2 - 2p.q(1-z)+q^2(1-z)]^2}$, and finally :

$I(q) \sim g^2 (\mu^2)^\epsilon ~\Gamma(\frac{\epsilon}{2})\int_0^1 dz \dfrac{1}{[q^2 z(1-z) ]^{\large \frac{\epsilon}{2}}}$

Here, $\epsilon$ is $>0$, so we see that if $q^2=0$, the integral is not defined, so $q^2=0$ should represent a pole for the scattering amplitude.

The relation with the locality could be seen as looking at the Fourier transform (taking $\epsilon=0$) of the scattering amplitude which could be written $I(x) \sim [D(x)]^2$, where $D(x)$ is the propagator in space-time coordinates.

Now, we should hope that any scattering amplitude, with loops, should have poles, which corresponds to some particular sum of the external momenta being on-shell.

Answer 2

Here's the sketch of an attempted explanation, and a work in progress.

Unitarity implies the optical theorem (Ref: Peskin & Schroeder Section 7.3), which says

$$\Im[\mathcal{M}_{i \rightarrow f}] = \sum_{\textrm{m=middle}} \int d\Pi_{\textrm{m}} \mathcal{M}(i\rightarrow m) \mathcal{M}^* (m \rightarrow f)$$ Essentially, you're re-expressing the scattering amplitude $\mathcal{M}(i \rightarrow f)$ as a sum over possible channels, and adding up the residues (of poles) whenever those intermediate states go on-shell. (NB: Since the matrix element is an analytic function of the variables, one should be able to obtain the real part given the imaginary part)

To my understanding, in section 10.2 (Polology) of his QFT-Book1 Weinberg shows that this kind of interpretation (poles when there's some on-shell intermediate state aka a "resonance") holds even when the intermediate states $m$ are bound states (non-perturbative in general) and not necessarily just degrees of freedom in the free theory.

It is the same idea behind the spectral representation of the propagator (a la Kallen-Lehmann. Refs: Peskin & Schroeder Section 7.1, Weinberg QFT1 Section 10.7)

The exact step where/why locality comes in is not yet clear to me.

Ps: I would appreciate help in fleshing out the details of this answer -- maybe making it a community wiki.