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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)?

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  • $\begingroup$ I don't have an answer but I think it is related to LSZ reduction, hope this might help. $\endgroup$ – Jia Yiyang Sep 23 '13 at 6:47
  • $\begingroup$ Just thinking out loud -- I wonder if this is related to a statement he made in his last paper on the subject: physics.stackexchange.com/questions/54354/… $\endgroup$ – Siva Sep 23 '13 at 18:47
  • $\begingroup$ I guess by now I should know that Weinberg's QFT books have everything and more! :-) Check Section 10.2 (Book1), on Polology. $\endgroup$ – Siva Sep 23 '13 at 19:01
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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.

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    $\begingroup$ It looks along the right lines to me. You can get similar conclusions from the standpoint of the old analytic S matrix theory: there the causality (which translate to locality) conditions are encoded in the analytic structure of the S matrix, which includes a simple pole for single particle exchange, two poles linked by a branch cut for two particle exchange etc. The simple pole forces the exchanged particle on shell. The multiple poles force a consistent network of momentum exchanges, a bit like Kirchoff's laws. $\endgroup$ – twistor59 Sep 23 '13 at 16:50
  • $\begingroup$ It's clear that propagators going on-shell will contribute poles to the S-matrix. I feel the non-trivial part is that these are the only possible poles. @Trimok's argument is suggestive, but does not quite seem watertight to me, especially when the theory might be non-perturbative. $\endgroup$ – Siva Sep 23 '13 at 18:52
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    $\begingroup$ @twistor59, Could you suggest a readable introduction to S-matrix methods where I might find an explanation of your statement? The connections to Kirchoff's laws seem particularly intriguing :-) Ps: By "readable" I mean language and notation that is not particularly arcane and reasonably in sync with modern conventions. $\endgroup$ – Siva Sep 23 '13 at 18:56
  • $\begingroup$ @Siva I was thinking along the lines of the discussion in Bjorken and Drell chapter 18. $\endgroup$ – twistor59 Sep 23 '13 at 19:36
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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.

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  • $\begingroup$ Regarding locality and the lagrangian density, I might not want to depend on that formulation (of locality), since such a Lagrangian formulation of the theory (which assumes a preferred basis for the degrees of freedom) might only make sense when you're perturbing around the free theory and using the states of the free theory as your Fock space basis. I think the claim/result is more general than that. $\endgroup$ – Siva Sep 23 '13 at 19:55
  • $\begingroup$ As said in the different Nima Arkani-Hamed talks, locality (or causality) dictates the poles, and unitarity dictates the behaviour at the poles, that is factorization (as, in the optical theorem). $\endgroup$ – Trimok Sep 24 '13 at 8:28

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