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I remember reading somewhere that Witten argued that if the Poincaré symmetry of spacetime were nontrivially combined with internal symmetries, then the S-matrix would be so constrained that the scattering amplitude would be non-vanishing at discrete angles only.

I would like to know more about this:

  1. At which angles would the scattering amplitude be non-vanishing
  2. By crossing symmetry, wouldn't this mean that in the crossed channel, the amplitude be non-vanishing at discrete energies?
  3. Is there a reference that contains more information on this?
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There is a brief mention of something like it here, (although no talk of the discrete angles) and it looks like the original reference would be "E. Witten, in: Proc. Intern. School of Subnuclear Physics, Erice 1981, ed. A. Zichichi" (I have no access to this however). – twistor59 Dec 17 '12 at 8:12
@twistor59 Please put your very helpful comment as an answer so that I can accept it. – QuantumDot Feb 16 '13 at 20:49
OK I fleshed out my understanding of Argyres argument. – twistor59 Feb 17 '13 at 12:24
up vote 5 down vote accepted

Something along the lines of what you want is explained in these lecture notes by Argyres.

As I understand it, the essence of the idea is that it explores the consequence of considering interacting theories in which you have Lorentz symmetries plus "internal" symmetries which are not separated in the usual sense, i.e. the internal symmetry generators don't commute with the Lorentz generators as they do in conventional theories (for example the generator of U(1) gauge transformations conventionally commutes with the generators of Lorentz transformations). The consequence is that the interactions become trivial in the sense that the allowed scattering angles become discrete (and hence vanishing by analyticity of the S matrix).

In a scattering process, the conservation laws resulting from translation invariance and Lorentz invariance leave only the angle of the scattering undetermined. If now, the theory is invariant under a symmetry group that mixes up tranlations, Lorentz transformations and some internal symmetries, then the extra conservation laws place restrictions on the scattering angle, leaving only a discrete allowed set.

Argyres gives the following example:

Take a theory with one of these "mixed up" symmetries - we have a conserved charge $Q_{\mu\nu}$. This is an internal symmetry generator which also has Lorentz indices. It's assumed to be symmetric and traceless (presumably for irreducibility). Its matrix element in a one-particle momentum eigenstate is of the form $$\langle p|Q_{\mu\nu}|p\rangle \propto p_{\mu}p_{\nu}-\frac{1}{4}\eta_{\mu\nu}p^2$$

It is reasonable to assume that in a two particle state, its matrix element is of the form $$\langle p^1p^2|Q_{\mu\nu}|p^1p^2\rangle = \langle p^1|Q_{\mu\nu}|p^1\rangle+\langle p^2|Q_{\mu\nu}|p^2\rangle $$ Considering now a process where a pair of momenta $p^1, p^2$ elastically scatter into $q^1, q^2$ then if $Q_{\mu\nu}$ is conserved $$\langle p^1p^2|Q_{\mu\nu}|p^1p^2\rangle=\langle q^1q^2|Q_{\mu\nu}|q^1q^2\rangle $$ $$\langle p^1|Q_{\mu\nu}|p^1\rangle+\langle p^2|Q_{\mu\nu}|p^2\rangle = \langle q^1|Q_{\mu\nu}|q^1\rangle+\langle q^2|Q_{\mu\nu}|q^2\rangle$$ $$p^1_{\mu}p^1_{\nu}-\frac{1}{4}\eta_{\mu\nu}(p^1)^2+p^2_{\mu}p^2_{\nu}-\frac{1}{4}\eta_{\mu\nu}(p^2)^2 = q^1_{\mu}q^1_{\nu}-\frac{1}{4}\eta_{\mu\nu}(q^1)^2+q^2_{\mu}q^2_{\nu}-\frac{1}{4}\eta_{\mu\nu}(q^2)^2 $$ Hence $$ p^1_{\mu}p^1_{\nu}+p^2_{\mu}p^2_{\nu} = q^1_{\mu}q^1_{\nu}+q^2_{\mu}q^2_{\nu} \ \ (1)$$ Four momentum is also conserved $$p^1_{\mu}+p^2_{\mu} = q^1_{\mu}+q^2_{\mu} $$ Now suppose scattering changes the momenta by $$q^1_{\mu} = p^1_{\mu}+a_{\mu}; \ \ \ q^2_{\mu} = p^2_{\mu}+b_{\mu}$$ Four momentum conservation tells us that $b_{\mu} = -a_{\mu}$. Substituting for the $q$'s in (1) we eventually get $$a_{\mu}(p^1_{\nu}-p^2_{\nu})+ a_{\nu}(p^1_{\mu}-p^2_{\mu})+2a_{\mu}a_{\nu} = 0$$ Since this holds for all values of $p^1_{\mu}, p^2_{\mu}$ we deduce that $a_{\mu}$ must vanish, i.e the interaction is trivial in the sense that $q^1_{\mu} = p^1_{\mu}$ and $q^2_{\mu} = p^2_{\mu}$ i.e. the scattering angle is zero.

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Thanks for this nice and so well explained answer twistor :-) – Dilaton Feb 17 '13 at 15:19
This is wonderful! I went through your logic, I've got a few questions: (1) Why is it reasonable to assume for the two particle state that matrix elements can be split? (2) I went through your logic for the case that Q is just a vector $Q^\mu$. What I found was that your eqn (1) was just 4-momentum conservation, which is not inconsistent. Is it safe to say that Q can not have more than one Lorentz index? – QuantumDot Feb 18 '13 at 3:14
(Not my logic - just parrotting an example due to Argyres!). (1) The mixed up Lorentz/internal symmetry group to which the CM theorem applies is meant to be a symmetry of the S matrix. In this case it treats two particle states like they were tensor products of one particle states - there is no weird non-locality operating. – twistor59 Feb 18 '13 at 8:21
(2) Yes, the example degenerates in the case of a single Lorentz index, but looking at the discussion in the reference preceding this example, I think the point was that if you had an interacting theory with the appropriate symmetry, then you could build a conserved charge $Q_{\mu\nu}$ and hence arrive at the contradiction. – twistor59 Feb 18 '13 at 8:22
@twistor59: What about (local) operators which cause interactions and mix momenta? Since our in and out states are plane waves and not distinct position eigenstates, I don't think Q needs to be nonlocal to mix them, eg: $\phi^4$ interaction. I wonder if it's possible to have examples like eg: Q operating on the particles swaps their momentum. – Siva Feb 20 '13 at 1:37

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